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Balbharti Maharashtra State Board 12th Physics Textbook Solutions Chapter 15 Structure of Atoms and Nuclei Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Physics Solutions Chapter 15 Structure of Atoms and Nuclei

In solving problems, use m_e = 0.00055 u = 0.5110 MeV/c², m_p = 1.00728 u, m_n = 1.00866u, m_H = 1.007825 u, u = 931.5 MeV, e = 1.602 × 10^-19 C, h = 6.626 × 10^-34 Js, ε₀ = 8.854 × 10^-12 SI units and me = 9.109 × 10^-31 kg.

1. Choose the correct option.

i) In which of the following systems will the radius of the first orbit of the electron be the smallest?
(A) hydrogen
(B) singly ionized helium
(C) deuteron
(D) tritium
Answer:
(D) tritium

ii) The radius of the 4th orbit of the electron will be smaller than its 8th orbit by a factor of
(A) 2
(B) 4
(C) 8
(D) 16
Answer:
(B) 4

iii) In the spectrum of hydrogen atom which transition will yield longest wavelength?
(A) n = 2 to n = 1
(B) n = 5 to n = 4
(C) n = 7 to n = 6
(D) n = 8 to n = 7
Answer:
(D) n = 8 to n = 7

iv) Which of the following properties of a nucleus does not depend on its mass number?
(A) radius
(B) mass
(C) volume
(D) density
Answer:
(D) density

v) If the number of nuclei in a radioactive sample at a given time is N, what will be the number at the end of two half-lives?
(A) \(\frac{N}{2}\)
(B) \(\frac{N}{4}\)
(C) \(\frac{3N}{4}\)
(D) \(\frac{N}{8}\)
Answer:
(B) \(\frac{N}{4}\)

2. Answer in brief.

i) State the postulates of Bohr’s atomic model.
Answer:
The postulates of Bohr’s atomic model (for the hydrogen atom) :

1. The electron revolves with a constant speed in acircular orbit around the nucleus. The necessary centripetal force is the Coulomb force of attraction of the positive nuclear charge on the negatively charged electron.
2. The electron can revolve without radiating energy only in certain orbits, called allowed or stable orbits, in which the angular momentum of the electron is equal to an integral multiple of h/2π, where h is Planck’s constant.
3. Energy is radiated by the electron only when it jumps from one of its orbits to another orbit having lower energy. The energy of the quantum of elec-tromagnetic radiation, i.e., the photon, emitted is equal to the energy difference of the two states.

ii) State the difficulties faced by Rutherford’s atomic model.
Answer:
(1) According to Rutherford, the electrons revolve in circular orbits around the atomic nucleus. The circular motion is an accelerated motion. According to the classical electromagnetic theory, an accelerated charge continuously radiates energy. Therefore, an electron during its orbital motion, should go on radiating energy. Due to the loss of energy, the radius of its orbit should go on decreasing. Therefore, the electron should move along a spiral path and finally fall into the nucleus in a very short time, of the order of 10^-16 s in the case of a hydrogen atom. Thus, the atom should be unstable. We exist because atoms are stable.

(2) If the electron moves along such a spiral path, the radius of its orbit would continuously decrease. As a result, the speed and frequency of revolution of the electron would go on increasing. The electron, therefore, would emit radiation of continuously changing frequency, and hence give rise to a con-tinuous spectrum. However, atomic spectrum is a line spectrum.

iii) What are alpha, beta and gamma decays?
Answer:
(a) A radioactive transformation in which an α-particle is emitted is called α-decay.
In an α-decay, the atomic number of the nucleus decreases by 2 and the mass number decreases by 4.
Example : \({ }_{92}^{238} \mathrm{U} \rightarrow{ }_{90}^{234} \mathrm{Th}+{ }_{2}^{4} \alpha\)
Q = [m_u – m_Th – m_α]c²

(b) A radioactive transformation in which a β-particle is emitted is called β-decay.
In a β^–-decay, the atomic number of the nucleus increases by 1 and the mass number remains unchanged.
Example : \({ }_{90}^{234} \mathrm{Th} \rightarrow{ }_{91}^{234} \mathrm{~Pa}+{ }_{-1}^{0} e+\bar{v}_{\mathrm{e}}\)
where \(\bar{v}_{\mathrm{e}}\) is the neutrino emitted to conserve the momentum, energy and spin.
Q = [m_u – m_Th – m_α]c²
In a β⁺-decay, the atomic number of the nucleus decreases by 1 and the mass number remains unchanged.
Example : \(\begin{aligned}
&30 \\
&15
\end{aligned} P \rightarrow{ }_{14}^{30} \mathrm{Si}+{ }_{+1}^{0} e+v_{\mathrm{e}}\)
where v_e is the neutrino emitted to conserve the momentum, energy and spin.
Q = [m_P – m_Si – m_e]c²
[Note : The term fi particle refers to the electron (or positron) emitted by a nucleus.]

A given nucleus does not emit α and β-particles simultaneously. However, on emission of α or β-particles, most nuclei are left in an excited state. A nucleus in an excited state emits a γ-ray photon in a transition to the lower energy state. Hence, α and β-particle emissions are often accompanied by γ-rays.

iv) Define excitation energy, binding energy and ionization energy of an electron in an atom.
Answer:
(1) Excitation energy of an electron in an atom : The energy required to transfer an electron from the ground state to an excited state (a state of higher energy) is called the excitation energy of the electron in that state.

(2) Binding energy of an electron in an atom is defined as the minimum energy that should be provided to an orbital electron to remove it from the atom such that its total energy is zero.

(3) Ionization energy of an electron in an atom is defined as the minimum energy required to remove the least strongly bound electron from a neutral atom such that its total energy is zero.

v) Show that the frequency of the first line in Lyman series is equal to the difference between the limiting frequencies of Lyman and Balmer series.
Answer:
For the first line in the Lyman series,
\(\frac{1}{\lambda_{\mathrm{L} 1}}=R\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)=R\left(1-\frac{1}{4}\right)=\frac{3 R}{4}\)
∴ v_L1 = \(\frac{c}{\lambda_{\mathrm{L} 1}}=\frac{3 R_{c}}{4}\), where v denotes the frequency,
c the speed of light in free space and R the Rydberg constant.
For the limit of the Lyman series,

Hence, the result.

Question 3.
State the postulates of Bohr’s atomic model and derive the expression for the energy of an electron in the atom.
Answer:
The postulates of Bohr’s atomic model (for the hydrogen atom) :
(1) The electron revolves with a constant speed in acircular orbit around the nucleus. The necessary centripetal force is the Coulomb force of attraction of the positive nuclear charge on the negatively charged electron.
(2) The electron can revolve without radiating energy only in certain orbits, called allowed or stable orbits, in which the angular momentum of the electron is equal to an integral multiple of h/2π, where h is Planck’s constant.
(3) Energy is radiated by the electron only when it jumps from one of its orbits to another orbit having lower energy. The energy of the quantum of elec-tromagnetic radiation, i.e., the photon, emitted is equal to the energy difference of the two states.

Consider the electron revolving in the nth orbit around the nucleus of an atom with the atomic number Z. Let m and e be the mass and the charge of the electron, r the radius of the orbit and v the linear speed of the electron.

According to Bohr’s first postulate, centripetal force on the electron = electrostatic force of attraction exerted on the electron by the nucleus
∴ \(\frac{m v^{2}}{r}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Z e^{2}}{r^{2}}\) ……………. (1)
where ε₀ is the permittivity of free space.
∴ Kinetic energy (KE) of the electron
= \(\frac{1}{2} m v^{2}=\frac{Z e^{2}}{8 \pi \varepsilon_{0} r}\) ………….. (2)
The electric potential due to the nucleus of charge +Ze at a point at a distance r from it is
V = \(\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{\mathrm{Ze}}{r}\)
∴ Potential energy (PE) of the electron
= charge on the electron × electric potential
= – e × \(\frac{1}{4 \pi \varepsilon_{0}} \frac{Z e}{r}=-\frac{Z e^{2}}{4 \pi \varepsilon_{0} r}\) …………….. (3)
Hence, the total energy of the electron in the nth orbit is
E = KE + PE = \(\frac{-Z e^{2}}{4 \pi \varepsilon_{0} r}+\frac{Z e^{2}}{8 \pi \varepsilon_{0} r}\)
∴ E = \(-\frac{Z e^{2}}{8 \pi \varepsilon_{0} r}\) ………….. (4)
This shows that the total energy of the electron in the nth orbit of the atom is inversely proportional to the radius of the orbit as Z, ε₀ and e are constants. The radius of the nth orbit of the electron is
r = \(\frac{\varepsilon_{0} h^{2} n^{2}}{\pi m Z e^{2}}\) …………….. (5)
where h is Planck’s constant.
From Eqs. (4) and (5), we get,
E_n = \(-\frac{Z e^{2}}{8 \pi \varepsilon_{0}}\left(\frac{\pi m Z e^{2}}{\varepsilon_{0} h^{2} n^{2}}\right)=-\frac{m Z^{2} e^{4}}{8 \varepsilon_{0}^{2} h^{2} n^{2}}\) ……………… (6)
This gives the expression for the energy of the electron in the nth Bohr orbit. The minus sign in the expression shows that the electron is bound to the nucleus by the electrostatic force of attraction.
As m, Z, e, ε₀ and h are constant, we get
E_n ∝ \(\frac{1}{n^{2}}\)
i.e., the energy of the electron in a stationary energy state is discrete and is inversely proportional to the square of the principal quantum number.
[ Note : Energy levels are most conveniently expressed in electronvolt. Hence, substituting the values of m, e, £0 and h, and dividing by the conversion factor 1.6 × 10^-19 J/eV,
E_n ≅ \(-\frac{13.6 Z^{2}}{n^{2}}\) (in eV)
For hydrogen, Z = 1
∴ E_n ≅ \(-\frac{13.6}{n^{2}}\) (in eV).

Question 4.
Starting from the formula for energy of an electron in the n^th orbit of hydrogen atom, derive the formula for the wavelengths of Lyman and Balmer series spectral lines and determine the shortest wavelengths of lines in both these series.
Answer:
According to Bohr’s third postulate for the model of the hydrogen atom, an atom radiates energy only when an electron jumps from a higher energy state to a lower energy state and the energy of the
quantum of electromagnetic radiation emitted in this process is equal to the energy difference between the two states of the electron. This emission of radiation gives rise to a spectral line.

The energy of the electron in a hydrogen atom,
when it is in an orbit with the principal quantum
number n, is
E_n = \(-\frac{m e^{4}}{8 \varepsilon_{0}^{2} h^{2} n^{2}}\)
where m = mass of electron, e = electronic charge, h = Planck’s constant and = permittivity of free space.

Let E_m be the energy of the electron in a hydrogen atom when it is in an orbit with the principal quantum number m and E, its energy in an orbit with the principal quantum number n, n < m. Then
E_m = \(-\frac{m e^{4}}{8 \varepsilon_{0}^{2} h^{2} m^{2}}\) and E_n = \(-\frac{m e^{4}}{8 \varepsilon_{0}^{2} h^{2} m^{2}}\)
Therefore, the energy radiated when the electron jumps from the higher energy state to the lower energy state is
E_m – E_n = \(\frac{-m e^{4}}{8 \varepsilon_{0}^{2} h^{2} m^{2}}-\left(-\frac{m e^{4}}{8 \varepsilon_{0}^{2} h^{2} n^{2}}\right)\)
= \(\frac{m e^{4}}{8 \varepsilon_{0}^{2} h^{2}}\left(\frac{1}{n^{2}}-\frac{1}{m^{2}}\right)\)
This energy is emitted in the form of a quantum of radiation (photon) with energy hv, where V is the frequency of the radiation.
∴ E_m – E_n = hv
∴ v = \(\frac{E_{m}-E_{n}}{h}=\frac{m e^{4}}{8 \varepsilon_{0}^{2} h^{3}}\left(\frac{1}{n^{2}}-\frac{1}{m^{2}}\right)\)
The wavelength of the radiation is λ = \(\frac{c}{v^{\prime}}\)
where c is the speed of radiation in free space.
The wave number, \(\bar{v}=\frac{1}{\lambda}=\frac{v}{c}\)
\(\bar{v}=\frac{1}{\lambda}=\frac{m e^{4}}{8 \varepsilon_{0}^{2} h^{3} c}\left(\frac{1}{n^{2}}-\frac{1}{m^{2}}\right)=R\left(\frac{1}{n^{2}}-\frac{1}{m^{2}}\right)\)
where \(R\left(=\frac{m e^{4}}{8 \varepsilon_{0}^{2} h^{3} c}\right)\) is a constant called the Ryd berg constant.

This expression gives the wave number of the radiation emitted and hence that of a line in hydrogen spectrum.

For the Lyman series, n = 1,m = 2, 3, 4, ………… ∞
∴ \(\frac{1}{\lambda_{\mathrm{L}}}=R\left(\frac{1}{1^{2}}-\frac{1}{m^{2}}\right)\) and for the shortest wavelength line m this series, \(\frac{1}{\lambda_{\mathrm{Ls}}}=R\left(\frac{1}{1^{2}}\right)\) as m = ∞.
For the Balmer series, n = 2, m = 3, 4, 5, … ∞.
∴ \(\frac{1}{\lambda_{\mathrm{B}}}=R\left(\frac{1}{4}-\frac{1}{m^{2}}\right)\) and for the shortest wavelength line in this series, \(\frac{1}{\lambda_{\mathrm{Bs}}}=R\left(\frac{1}{4}\right)\) as m = ∞
[Note: Johannes Rydberg (1854—1919), Swedish spectroscopist, studied atomic emission spectra and introduced the idea of wave number. The empirical formula \(\bar{v}=\frac{1}{\lambda}=R\left(\frac{1}{n^{2}}-\frac{1}{m^{2}}\right)\) where m and n are simple integers, is due to Rydberg. When we consider the finite mass of the nucleus, we find that R varies slightly from element to element.]

Question 5.
Determine the maximum angular speed of an electron moving in a stable orbit around the nucleus of hydrogen atom.
Answer:
The radius of the ,ith Bohr orbit is
r = \(\frac{\varepsilon_{0} h^{2} n^{2}}{\pi m Z e^{2}}\) ………….. (1)
and the linear speed of an electron in this orbit is
ν = \(\frac{Z e^{2}}{2 \varepsilon_{0} n h}\) …………… (2)
where ε ₀ permittivity of free space, h ≡ Planck’s constant, n ≡ principal quantum number, m ≡ electron mass, e electronic charge and Z ≡ atomic number of the atom.
Since angular speed ω = \(\frac{v}{r}\), then from Eqs. (1) and (2), we get,
ω = \(\frac{v}{r}=\frac{Z e^{2}}{2 \varepsilon_{0} n h} \cdot \frac{\pi m Z e^{2}}{\varepsilon_{0} h^{2} n^{2}}=\frac{\pi m Z^{2} e^{4}}{2 \varepsilon_{0}^{2} h^{3} n^{3}}\) ………………. (3)
which gives the required expression for the angular speed of an electron in the nth Bohr orbit.
From Eq. (3), the frequency of revolution of the electron,
f = \(\frac{\omega}{2 \pi}=\frac{1}{2 \pi} \times \frac{\pi m Z^{2} e^{4}}{2 \varepsilon_{0}^{2} h^{3} n^{3}}=\frac{m Z^{2} e^{4}}{4 \varepsilon_{0}^{2} h^{3} n^{3}}\) …………….. (4)
as required.
[Note : From Eq. (4), the period of revolution of the electron, T = \(\frac{1}{f}=\frac{4 \varepsilon_{0}^{2} h^{3} n^{3}}{m Z e^{4}}\). Hence, f ∝ \(\frac{1}{n^{3}}\) and T ∝ n³].

Obtain the formula for ω and continue as follows :

This is required quantity.

Question 6.
Determine the series limit of Balmer, Paschen and Bracket series, given the limit for Lyman series is 912 Å.
Answer:
Data : λ_L∞ = 912 Å

as n = 5 and m = ∞
From Eqs. (1) and (2), we get,
\(\frac{\lambda_{\mathrm{Pa} \infty}}{\lambda_{\mathrm{L} \infty}}=\frac{R_{\mathrm{H}}}{R_{\mathrm{H}} / 9}\) = 9
∴ λ_Pa∞ = 9λ_L∞ = (9) (912) = 8202 Å
\(\frac{\lambda_{\mathrm{Pf} \infty}}{\lambda_{\mathrm{L} \infty}}=\frac{R_{\mathrm{H}}}{R_{\mathrm{H}} / 25}\) = 25
∴ λ_Pf∞ = 25λ_L∞ = (25) (912) = 22800 Å
This is the series limit of the pfund series.

Question 7.
Describe alpha, beta and gamma decays and write down the formulae for the energies generated in each of these decays.
Answer:
(a) A radioactive transformation in which an α-particle is emitted is called α-decay.
In an α-decay, the atomic number of the nucleus decreases by 2 and the mass number decreases by 4.
Example : \({ }_{92}^{238} \mathrm{U} \rightarrow{ }_{90}^{234} \mathrm{Th}+{ }_{2}^{4} \alpha\)
Q = [m_u – m_Th – m_α]c²

(b) A radioactive transformation in which a β-particle is emitted is called β-decay.
In a β^–-decay, the atomic number of the nucleus increases by 1 and the mass number remains unchanged.
Example : \({ }_{90}^{234} \mathrm{Th} \rightarrow{ }_{91}^{234} \mathrm{~Pa}+{ }_{-1}^{0} e+\bar{v}_{\mathrm{e}}\)
where \(\bar{v}_{\mathrm{e}}\) is the neutrino emitted to conserve the momentum, energy and spin.
Q = [m_u – m_Th – m_α]c²
In a β⁺-decay, the atomic number of the nucleus decreases by 1 and the mass number remains unchanged.
Example : \(\begin{aligned}
&30 \\
&15
\end{aligned} P \rightarrow{ }_{14}^{30} \mathrm{Si}+{ }_{+1}^{0} e+v_{\mathrm{e}}\)
where v_e is the neutrino emitted to conserve the momentum, energy and spin.
Q = [m_P – m_Si – m_e]c²
[Note : The term fi particle refers to the electron (or positron) emitted by a nucleus.]

A given nucleus does not emit α and β-particles simultaneously. However, on emission of α or β-particles, most nuclei are left in an excited state. A nucleus in an excited state emits a γ-ray photon in a transition to the lower energy state. Hence, α and β-particle emissions are often accompanied by γ-rays.

Question 8.
Explain what are nuclear fission and fusion giving an example of each. Write down the formulae for energy generated in each of these processes.
Answer:
Nuclear fission is a nuclear reaction in which a heavy nucleus of an atom, such as that of uranium, splits into two or more fragments of comparable size, either spontaneously or as a result of bombardment of a neutron on the nucleus (induced fission). It is followed by emission of two or three neutrons.
The mass of the original nucleus is more than the sum of the masses of the fragments. This mass difference is released as energy, which can be enormous as in the fission of ²³⁵U.
Nuclear fission was discovered by Lise Meitner, Otto Frisch, Otto Hahn and Fritz Strassmann in 1938.

The products of the fission of ²³⁵U by thermal neutrons are not unique. A variety of fission fragments are produced with mass number A ranging from about 72 to about 138, subject to the conservation of mass-energy, momentum, number of protons (Z) and number of neutrons (N). A few typical fission equations are

A type of nuclear reaction in which lighter atomic nuclei (of low atomic number) fuse to form a heavier nucleus (of higher atomic number) with the’ release of enormous amount of energy is called nuclear fusion.

Very high temperatures, of about 107 K to 108 K, are required to carry out nuclear fusion. Hence, such a reaction is also called a thermonuclear reaction.

Example : The D-T reaction, being used in experimental fusion reactors, fuses a deuteron and a triton nuclei at temperatures of about 10⁸ K.

(2) The value of the energy released in the fusion of two deuteron nuclei and the temperature at which the reaction occurs mentioned in the textbook are probably misprints.]

Question 9.
Describe the principles of a nuclear reactor. What is the difference between a nuclear reactor and a nuclear bomb?
Answer:
In a nuclear reactor fuel rods are used to provide a suitable fissionable material such as \(\begin{gathered}
236 \\
92
\end{gathered}\)U. Control rods are used to start or stop the reactor. Moderators are used to slow down the fast neutrons ejected in a nuclear fission to the appropriate lower speeds. Material used as a coolant removes the energy released in the nuclear reaction by converting it into thermal energy for production of electricity.

In a nuclear reactor, a nuclear fission chain reaction is used in a controlled manner, while in a nuclear bomb, the nuclear fission chain reaction is not controlled, releasing tremendous energy in a very short time interval.
[Note : The first nuclear bomb (atomic bomb) was dropped on Hiroshima in Japan on 06 August 1945. The second bomb was dropped on Nagasaki in Japan on 9 August 1945.]

Question 10.
Calculate the binding energy of an alpha particle given its mass to be 4.00151 u.
Answer:
Data : M = 4.00151 u, = 1.00728 u,
m_n = 1.00866 u, 1 u = 931.5 MeV/c²
The binding energy of an alpha particle
(Zm_p + N_n -M)c²
=(2m_p + 2m_n -M)c²
= [(2)(1.00728u) + 2(1.00866 u) – 4.00151 u]c²
= (2.01456 + 2.01732 – 4.00151)(931.5) MeV
= 28.289655 MeV
= 28.289655 × 10⁶ eV × 1.602 × 10^-19 J
= 4.532002731 × 10^-12 J

Question 11.
An electron in hydrogen atom stays in its second orbit for 10^-8 s. How many revolutions will it make around the nucleus in that time?
Answer:
Data : z = 1, m = 9.1 × 10^-31 kg, e = 1.6 × 10^-19 C, ε₀ = 8.85 × 10^-12 C² / N.m², h = 6.63 × 10 ^-34 J.s, n = 2, t = 10^-8 s
The periodic time of the electron in a hydrogen atom,

Let N be the number of revolutions made by the electron in time t. Then, t = NT.
∴ N = \(\frac{t}{T}=\frac{10^{-8}}{3.898 \times 10^{-16}}\) = 2.565 × ⁷

Question 12.
Determine the binding energy per nucleon of the americium isotope \(_{95}^{244} \mathrm{Am}\) , given the mass of \({ }_{95}^{244} \mathrm{Am}\) to be 244.06428 u.
Answer:
Data : Z = 95, N = 244  – 95 = 149,
m_p = 1.00728 u, m_n = 1.00866 u,
M = 244.06428 u, 1 u = 931.5 MeV/c²
The binding energy per nucleon,

= 7.3209 MeV/nucleon

Question 13.
Calculate the energy released in the nuclear reaction \({ }_{3}^{7} \mathrm{Li}\) + p → 2α given mass of \({ }_{3}^{7} \mathrm{Li}\) atom and of helium atom to be 7.016 u and 4.0026 u respectively.
Answer:
Data: M₁ (\({ }_{3}^{7} \mathrm{Li}\) Li atom)= 7.016 u, M₂ (He atom)
= 4.0026 u, m_p = 1.00728 u, 1 u = 931.5 MeV/c²
∆M = M₁ + m_p – 2M₂
= [7.016 + 1.00728 – 2(4.0026)]u
= 0.01808 u = (0.01808)(931.5) MeV/c²
= 16.84152 MeV/c²
Therefore, the energy released in the nuclear reaction = (∆M) c² = 16.84152 MeV

Question 14.
Complete the following equations describing nuclear decays.

Answer:
(a) \({ }_{88}^{226} \mathrm{Ra} \rightarrow{ }_{2}^{4} \alpha+{ }_{86}^{222} \mathrm{Em}\)
Em (Emanation) ≡ Rn (Radon)
Here, α particle is emitted and radon is formed.

(b) \({ }_{8}^{19} \mathrm{O} \rightarrow e^{-}+{ }_{9}^{19} \mathrm{~F}\)
Here, e^– ≡ \({ }_{-1}^{0} \beta\) is emitted and fluorine is formed.

(c) \(\underset{90}{228} \mathrm{Th} \rightarrow{ }_{2}^{4} \alpha+{ }_{88}^{224} \mathrm{Ra}\)
Here, α particle is emitted and radium is formed.

(d) \({ }_{7}^{12} \mathrm{~N} \rightarrow{ }_{6}^{12} \mathrm{C}+{ }_{1}^{0} \beta\)
\({ }_{1}^{0} \beta\) is e⁺ (positron)
Here, β⁺ is emItted and carbon is formed.

Question 15.
Calculate the energy released in the following reactions, given the masses to be \({ }_{88}^{223} \mathrm{Ra}\) : 223.0185 u, \({ }_{82}^{209} \mathrm{~Pb}\) : 208.9811, \({ }_{6}^{14} C\) : 14.00324, \({ }_{92}^{236} \mathrm{U}\) : 236.0456, \({ }_{56}^{140} \mathrm{Ba}\) : 139.9106, \({ }_{36}^{94} \mathrm{Kr}\) : 93.9341, \({ }_{6}^{11} \mathrm{C}\) : 11.01143, \({ }_{5}^{11} \mathrm{~B}\) : 11.0093. Ignore neutrino energy.

Answer:

(a) \({ }_{88}^{223} \mathrm{Ra} \rightarrow{ }_{82}^{209} \mathrm{~Pb}+{ }_{6}^{14} \mathrm{C}\)
The energy released in this reaction = (∆M) c²
= [223.0185 – (208.9811 + 14.00324)j(931 .5) MeV
= 31.820004 MeV

(b) \({ }_{92}^{236} \mathrm{U} \rightarrow{ }_{56}^{140} \mathrm{Ba}+{ }_{36}^{94} \mathrm{Kr}+2 \mathrm{n}\)
The energy released in this reaction =
(∆M) c² = [236.0456 – (139.9106 + 93.9341 + (2)(1 .00866)1(93 1 .5)MeV
= 171.00477 MeV

(c) \({ }_{6}^{11} \cdot \mathrm{C} \rightarrow{ }_{5}^{11} \mathrm{~B}+e^{+}\) + neutrino
The energy released in this reaction = (∆M) c²
= [11.01143 – (11.0093 + O.00055)](931.5) MeV
= 1.47177 MeV

Question 16.
Sample of carbon obtained from any living organism has a decay rate of 15.3 decays per gram per minute. A sample of carbon obtained from very old charcoal shows a disintegration rate of 12.3 disintegrations per gram per minute. Determine the age of the old sample given the decay constant of carbon to be 3.839 × 10^-12 per second.
Answer:
Data: 15.3 decays per gram per minute (living organism), 12.3 disintegrations per gram per minute (very old charcoal). Hence, we have,
\(\frac{A(t)}{A_{0}}=\frac{12.3}{15.3}\), λ = 3.839 × 10^-12 per second

Question 17.
The half-life of \({ }_{38}^{90} \mathrm{Sr}\) is 28 years. Determine the disintegration rate of its 5 mg sample.
Answer:
Data: T_1/2 = 28 years = 28 × 3.156 × 10⁷ s
=8.837 × 10⁸s, M = 5 mg =5 × 10^-3g
90 grams of \({ }_{38}^{90} \mathrm{Sr}\) contain 6.02 × 10²³ atoms
Hence, here, N = \(\frac{\left(6.02 \times 10^{23}\right)\left(5 \times 10^{-3}\right)}{90}\)
= 3.344 × 10¹⁹ atoms
∴ The disintegration rate = Nλ = N\(\frac{0.693}{T_{1 / 2}}\)
= \(\frac{\left(3.344 \times 10^{19}\right)(0.693)}{8.837 \times 10^{8}}\)
= 2.622 × 10¹⁰ disintegrations per second

Question 18.
What is the amount of \({ }_{27}^{60} \mathrm{Co}\) necessary to provide a radioactive source of strength 10.0 mCi, its half-life being 5.3 years?
Answer:
Data : Activity = 10.0 mCi = 10.0 × 10^-3 Ci
= (10.0 × 10^-3)(3.7 × 10¹⁰) dis/s = 3.7 × 10⁸ dis/s
T_1/2 = 5.3 years = (5.3)(3.156 × 10⁷) s
= 1.673 × 10⁸ s
Decay constant, λ = \(\frac{0.693}{T_{1 / 2}}=\frac{0.693}{1.673 \times 10^{8}} \mathrm{~s}^{-1}\)
=4.142 × 10^-9 s^-1
Activity = Nλ
∴ N = \(\frac{\text { activity }}{\lambda}=\frac{3.7 \times 10^{8}}{4.142 \times 10^{-9}} \text { atoms }\)
= 8.933 × 10¹⁶ atoms
=60 grams of \({ }_{27}^{60} \mathrm{Co}\) contain 6.02 × 10²³ atoms
Mass of 8.933 × 10¹⁶ atoms of \({ }_{27}^{60} \mathrm{Co}\)
= \(\frac{8.933 \times 10^{16}}{6.02 \times 10^{23}} \times 60 \mathrm{~g}\)
= 8.903 × 10^-6 g = 8.903 µg
This is the required amount.

Question 19.
Disintegration rate of a sample is 10¹⁰ per hour at 20 hrs from the start. It reduces to 6.3 × 10⁹ per hour after 30 hours. Calculate its half life and the initial number of radioactive atoms in the sample.
Answer:
Data : A (t₁) = 10¹⁰ per hour, where t₁ = 20 h,
A (t₂) = 6.3 × 10⁹ per hour, where t₂ = 30 h
A(t) = A₀e^-λt ∴ A(t₁) = A₀e^-λt₁ and A(t₂) = A_oe^-λt₂

∴ 1.587 e^10λ ∴ 10λ =2.3031og₁₀(1.587)
∴ λ = (0.2303)(0.2007) = 0.04622 per hour
The half life of the material, T_1/2 = \(\frac{0.693}{\lambda}=\frac{0.693}{0.04622}\)
= 14.99 hours
Now, A₀ = A (t₁)e^λt₁ = 10¹⁰e^{(0.04622)(20)}
= 10¹⁰ e^0.9244
Let x = e^0.9244 ∴ 2.3031og₁₀x = 0.9244
∴ 1og₁₀x = \(\frac{0.9244}{2.303}\) = 0.4014
∴ x = antilog 0.4014 = 2.52
∴ A₀ = 2.52 × 10¹⁰ per hour
Now A₀ = N₀λ ∴ N₀ = \(\frac{A_{0}}{\lambda}=\frac{2.52 \times 10^{10}}{0.04622}\)
= 5.452 × 10¹¹
This is the initial number of radioactive atoms in the sample.

Question 20.
The isotope ⁵⁷Co decays by electron capture to ⁵⁷Fe with a half-life of 272 d. The ⁵⁷Fe nucleus is produced in an excited state, and it almost instantaneously emits gamma rays.
(a) Find the mean lifetime and decay constant for ⁵⁷Co.
(b) If the activity of a radiation source ⁵⁷Co is 2.0 µCi now, how many ⁵⁷Co nuclei does the source contain?
(c) What will be the activity after one year?
Answer:
Data: T_1/2 = 272d = 272 × 24 × 60 × 60s = 2.35 × 10⁷ s,
A₀ = 2.0uCi = 2.0 × 10^-6 × 3.7 × 10¹⁰
= 7.4 × 10⁴ dis/s
t = 1 year = 3.156 × 10⁷ s
(a) T_1/2 = \(\frac{0.693}{\lambda}\) = 0.693 τ ∴ The mean lifetime for
⁵⁷Co = τ = \(\frac{T_{1 / 2}}{0.693}=\frac{2.35 \times 10^{7}}{0.693}\) = 3391 × 10⁷ s
The decay constant for ⁵⁷Co = λ = \(\frac{1}{\tau}\)
= \(\frac{1}{3.391 \times 10^{7} \mathrm{~s}}\)
= 2949 × 10^-8 s^-1

(b)A₀ = N₀A ∴ N₀ = \(\frac{A_{0}}{\lambda}\) = A₀τ
= (7.4 × 10⁴)(3.391 × 10⁷)
= 2.509 × 10¹² nuclei
This is the required number.

(c) A(t) = A₀e^-λt = 2e^{-(2.949 × 10-8)(3.156 × 107)}
= 2e^-0.9307 = 2 / e^0.9307
Let x = e^0.9307 ∴ Iog_ex = 0.9307
∴ 2.303log₁₀x = 0.9307
∴ log₁₀x = \(\frac{0.9307}{2.303}\) = 0.4041
∴ x = antilog 0.4041 = 2.536
∴ A (t) = \(\frac{2}{2.536}\) μCi = 0.7886 μCi

Question 21.
A source contains two species of phosphorous nuclei, \({ }_{15}^{32} \mathrm{P}\) (T_1/2 = 14.3 d) and \({ }_{15}^{32} \mathrm{P}\) (T_1/2 = 25.3 d). At time t = 0, 90% of the decays are from \({ }_{15}^{32} \mathrm{P}\) . How much time has to elapse for only 15% of the decays to be from \({ }_{15}^{32} \mathrm{P}\) ?
Answer:

∴ (0.04846 – 0.02739) t = 2.303 (2.1847 – 0.4771)
∴ t = \(\frac{(2.303)(1.7076)}{0.02107}\) = 186.6 days
This is the required time.

Question 22.
Before the year 1900 the activity per unit mass of atmospheric carbon due to the presence of ¹⁴C averaged about 0.255 Bq per gram of carbon. (a) What fraction of carbon atoms were ¹⁴C? (b) An archaeological specimen containing 500 mg of carbon, shows 174 decays in one hour. What is the age of the specimen, assuming that its activity per unit mass of carbon when the specimen died was equal to the average value of the air? Half-life of 14C is 5730 years?
Answer:
0.693
Data: T_1/2 = 5730y ∴ λ = \(\frac{0.693}{5730 \times 3.156 \times 10^{7}} \mathrm{~s}^{-1}\)
= 3.832 × 10^-12 s^-1, A = 0.255 Bq per gram of carbon in part (a); M = 500 mg = 500 × 10^-3 g,
174 decays in one hour \(\frac{174}{3600}\) dis/s = 0.04833 dis/s in part (b) (per 500 mg].

(a) A = Nλ ∴ N = \(\frac{A}{\lambda}=\frac{0.255}{3.832 \times 10^{-12}}\)
= 6.654 × 10¹⁰
Number of atoms in 1 g of carbon = \(\frac{6.02 \times 10^{23}}{12}\)
=5.017 × 10²²
\(\frac{5.017 \times 10^{22}}{6.654 \times 10^{10}}\) = 0.7539 × 10¹²
∴ 1 ¹⁴C atom per 0.7539 × 10¹² atoms of carbon
∴ 4 ¹⁴C atoms per 3 × 10¹² atoms of carbon

(b) Present activity per gram = \(\)
= 0.09666 dis/s per gram
A₀ = 0.255 dis/s per gram
Now, A(t) = A₀e^-λt

This is the required quantity.

Question 23.
How much mass of ²³⁵U is required to undergo fission each day to provide 3000 MW of thermal power? Average energy per fission is 202.79 MeV
Answer:
Data: Power = 3000 MW = 3 × 10⁹ J/s
∴ Energy to be produced each day
=3 × 10⁹ × 86400 J each day
= 2.592 × 10¹⁴ J each day
Energy per fission = 202.79 MeV
= 202.79 × 10⁶ × 1.6 × 10^-19 J = 3,245 × 10^-11 J
∴ Number of fissions each day
= \(\frac{2.592 \times 10^{14}}{3.245 \times 10^{-11}}\) × 10²⁴ each day
0.235 kg of ²³⁵U contains 6.02 × 10²³ atoms
7988 x 1024
∴ M = \(\left(\frac{7.988 \times 10^{24}}{6.02 \times 10^{23}}\right)\) (o.235) = 3.118 kg
This is the required quantity.

Question 24.
In a periodic table the average atomic mass of magnesium is given as 24.312 u. The average value is based on their relative natural abundance on earth. The three isotopes and their masses are \({ }_{12}^{24} \mathrm{Mg}\) (23.98504 u), \({ }_{12}^{25} \mathrm{Mg}\) (24.98584 u) and \({ }_{12}^{26} \mathrm{Mg}\) (25.98259 u). The natural abundance of \({ }_{12}^{24} \mathrm{Mg}\) is 78.99% by mass. Calculate the abundances of other two isotopes.
[Answer: 9.3% and 11.7%]
Answer:
Data : Average atomic mass of magnesium =

12th Physics Digest Chapter 15 Structure of Atoms and Nuclei Intext Questions and Answers

Use your brain power (Textbook Page No. 336)

Question 1.
Why don’t heavy nuclei decay by emitting a single proton or a single neutron?
Answer:
According to quantum mechanics, the probability for these emissions is extremely low.

Balbharti Maharashtra State Board 12th Physics Textbook Solutions Chapter 1 Rotational Dynamics Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Physics Solutions Chapter 1 Rotational Dynamics

1. Choose the correct option.

i) When seen from below, the blades of a ceiling fan are seen to be revolving anticlockwise and their speed is decreasing. Select correct statement about the directions of its angular velocity and angular acceleration.
(A) Angular velocity upwards, angular acceleration downwards.
(B) Angular velocity downwards, angular acceleration upwards.
(C) Both, angular velocity and angular acceleration, upwards.
(D) Both, angular velocity and angular acceleration, downwards.
Answer:
(A) Angular velocity upwards, angular acceleration downwards.

ii) A particle of mass 1 kg, tied to a 1.2 m long string is whirled to perform vertical circular motion, under gravity. Minimum speed of a particle is 5 m/s. Consider following statements.
P) Maximum speed must be 5 5 m/s.
Q) Difference between maximum and minimum tensions along the string is 60 N. Select correct option.
(A) Only the statement P is correct.
(B) Only the statement Q is correct.
(C) Both the statements are correct.
(D) Both the statements are incorrect.
Answer:
(C) Both the statements are correct.

iii) Select correct statement about the formula (expression) of moment of inertia (M.I.) in terms of mass M of the object and some of its distance parameter/s, such as R, L, etc.
(A) Different objects must have different expressions for their M.I.
(B) When rotating about their central axis, a hollow right circular cone and a disc have the same expression for the M.I.
(C) Expression for the M.I. for a parallelepiped rotating about the transverse axis passing through its centre includes its depth.
(D) Expression for M.I. of a rod and that of a plane sheet is the same about a transverse axis.
Answer:
(B) When rotating about their central axis, a hollow right circular cone and a disc have the same expression for the M.I.

iv) In a certain unit, the radius of gyration of a uniform disc about its central and transverse axis is \(\sqrt{2.5}\). Its radius of gyration about a tangent in its plane (in the same unit) must be
(A) \(\sqrt{5}\)
(B) 2.5
(C) 2\(\sqrt{2.5}\)
(D) \(\sqrt{12.5}\)
Answer:
(B) 2.5

v) Consider following cases:
(P) A planet revolving in an elliptical orbit.
(Q) A planet revolving in a circular orbit.
Principle of conservation of angular momentum comes in force in which of these?
(A) Only for (P)
(B) Only for (Q)
(C) For both, (P) and (Q)
(D) Neither for (P), nor for (Q)
Answer:
(C) For both, (P) and (Q)

X) A thin walled hollow cylinder is rolling down an incline, without slipping. At any instant, the ratio ”Rotational K.E.:
Translational K.E.: Total K.E.” is
(A) 1:1:2
(B) 1:2:3
(C) 1:1:1
(D) 2:1:3
Answer:
(D) 2:1:3

2. Answer in brief.

i) Why are curved roads banked?
Answer:
A car while taking a turn performs circular motion. If the road is level (or horizontal road), the necessary centripetal force is the force of static friction between the car tyres and the road surface. The friction depends upon the nature of the surfaces in contact and the presence of oil and water on the road. If the friction is inadequate, a speeding car may skid off the road. Since the friction changes with circumstances, it cannot be relied upon to provide the necessary centripetal force. Moreover, friction results in fast wear and tear of the tyres.

To avoid the risk of skidding as well as to reduce the wear and tear of the car tyres, the road surface at a bend is tilted inward, i.e., the outer side of the road is raised above its inner side. This is called banking of road. On a banked road, the resultant of the normal reaction and the gravitational force can act as the necessary centripetal force. Thus, every car can be safely driven on such a banked curve at certain optimum speed, without depending on friction. Hence, a road should be properly banked at a bend.

The angle of banking is the angle of inclination of the surface of a banked road at a bend with the horizontal.

ii) Do we need a banked road for a two wheeler? Explain.
Answer:
When a two-wheeler takes a turn along an unbanked road, the force of friction provides the centripetal force. The two-wheeler leans inward to counteract a torque that tends to topple it outward. Firstly, friction cannot be relied upon to provide the necessary centripetal force on all road conditions. Secondly, the friction results in wear and tear of the tyres. On a banked road at a turn, any vehicle can negotiate the turn without depending on friction and without straining the tyres.

iii) On what factors does the frequency of a conical pendulum depend? Is it independent of some factors?
Answer:
The frequency of a conical pendulum, of string length L and semivertical angle θ, is
n = \(\frac{1}{2 \pi} \sqrt{\frac{g}{L \cos \theta}}\)
where g is the acceleration due to gravity at the place.
From the above expression, we can see that

1. n ∝ \(\sqrt{g}\)
2. n ∝ \(\frac{1}{\sqrt{L}}\)
3. n ∝ \(\frac{1}{\sqrt{\cos \theta}}\)
   (if θ increases, cos θ decreases and n increases)
4. The frequency is independent of the mass of the bob.

iv) Why is it useful to define radius of gyration?
Answer:
Definition : The radius of gyration of a body rotating about an axis is defined as the distance between the axis of rotation and the point at which the entire mass of the body can be supposed to be concentrated so as to give the same moment of inertia as that of the body about the given axis.

The moment of inertia (MI) of a body about a given rotation axis depends upon

1. the mass of the body and
2. the distribution of mass about the axis of rotation. These two factors can be separated by expressing the MI as the product of the mass (M) and the square of a particular distance (k) from the axis of rotation. This distance is called the radius of gyration and is defined as given above. Thus,
   

Physical significance : The radius of gyration is less if I is less, i.e., if the mass is distributed close to the axis; and it is more if I is more, i.e., if the mass is distributed away from the axis. Thus, it gives the idea about the distribution of mass about the axis of rotation.

v) A uniform disc and a hollow right circular cone have the same formula for their M.I., when rotating about their central axes. Why is it so?
Answer:
The radius of gyration of a thin ring of radius Rr about its transverse symmetry axis is
K_r = \(\sqrt{I_{\mathrm{CM}} / M_{\mathrm{r}}}\) = \(\sqrt{R_{\mathrm{r}}^{2}}\) = R_r
The radius of gyration of a thin disc of radius R_d about its transverse symmetry axis is

Question 3.
While driving along an unbanked circular road, a two-wheeler rider has to lean with the vertical. Why is it so? With what angle the rider has to lean? Derive the relevant expression. Why such a leaning is not necessary for a four wheeler?
Answer:
When a bicyclist takes a turn along an unbanked road, the force of friction \(\vec{f}_{\mathrm{s}}\) provides the centripetal force; the normal reaction of the road \(\vec{N}\) is vertically up. If the bicyclist does not lean inward, there will be an unbalanced outward torque about the centre of gravity, f_s.h, due to the friction force that will topple the bicyclist outward. The bicyclist must lean inward to counteract this torque (and not to generate a centripetal force) such that the opposite inward torque of the couple formed by \(\vec{N}\) and the weight \(\vec{g}\), mg.a = f_s.h₁

Since the force of friction provides the centripetal force,
f_s = \(\frac{m v^{2}}{r}\)
If the cyclist leans from the vertical by an angle 9, the angle between \(\vec{N}\) and \(\vec{F}\) in above figure.

Hence, the cyclist must lean by an angle
θ = tan^-1\(\left(\frac{v^{2}}{g r}\right)\)

When a car takes a turn along a level road, apart from the risk of skidding off outward, it also has a tendency to roll outward due to an outward torque about the centre of gravity due to the friction force. But a car is an extended object with four wheels. So, when the inner wheels just get lifted above the ground, it can be counterbalanced by a restoring torque of the couple formed by the normal reaction (on the outer wheels) and the weight.

Question 4.
Using the energy conservation, derive the expressions for the minimum speeds at different locations along a vertical circular motion controlled by gravity. Is zero speed possible at the uppermost point? Under what condition/s? Also prove that the difference between the extreme tensions (or normal forces) depends only upon the weight of the object.
Answer:
In a non uniform vertical circular motion, e.g., those of a small body attached to a string or the loop-the-loop manoeuvers of an aircraft or motorcycle or skateboard, the body must have some minimum speed to reach the top and complete the circle. In this case, the motion is controlled only by gravity and zero speed at the top is not possible.

However, in a controlled vertical circular motion, e.g., those of a small body attached to a rod or the giant wheel (Ferris wheel) ride, the body or the passenger seat can have zero speed at the top, i.e., the motion can be brought to a stop.

Question 5.
Discuss the necessity of radius of gyration. Define it. On what factors does it depend and it does not depend? Can you locate some similarity between the centre of mass and radius of gyration? What can you infer if a uniform ring and a uniform disc have the same radius of gyration?
Answer:
Definition : The radius of gyration of a body rotating about an axis is defined as the distance between the axis of rotation and the point at which the entire mass of the body can be supposed to be concentrated so as to give the same moment of inertia as that of the body about the given axis.

The moment of inertia (MI) of a body about a given rotation axis depends upon

1. the mass of the body and
2. the distribution of mass about the axis of rotation. These two factors can be separated by expressing the MI as the product of the mass (M) and the square of a particular distance (k) from the axis of rotation. This distance is called the radius of gyration and is defined as given above. Thus,

Physical significance : The radius of gyration is less if I is less, i.e., if the mass is distributed close to the axis; and it is more if I is more, i.e., if the mass is distributed away from the axis. Thus, it gives the idea about the distribution of mass about the axis of rotation.

The centre of mass (CM) coordinates locates a point where if the entire mass M of a system of particles or that of a rigid body can be thought to be concentrated such that the acceleration of this point mass obeys Newton’s second law of motion, viz.,
\(\vec{F}_{\mathrm{net}}\) = M\(\overrightarrow{\mathrm{a}}_{\mathrm{CM}}\), where \(\vec{F}_{\mathrm{net}}\) is the sum of all the external forces acting on the body or on the individual particles of the system of particles.

Similarly, radius of gyration locates a point from the axis of rotation where the entire mass M can be thought to be concentrated such that the angular acceleration of that point mass about the axis of rotation obeys the relation, \(\vec{\tau}_{\mathrm{net}}\) = M\(\vec{\alpha}\), where \(\vec{\tau}_{\text {net }}\) is the sum of all the external torques acting on the body or on the individual particles of the system of particles.

Question 6.
State the conditions under which the theorems of parallel axes and perpendicular axes are applicable. State the respective mathematical expressions.
Answer:
The theorem of parallel axis is applicable to any body of arbitrary shape. The moment of inertia (MI) of the body about an axis through the centre mass should be known, say, I_CM. Then, the theorem can be used to find the MI, I, of the body about an axis parallel to the above axis. If the distance between the two axes is h,
I = I_CM + Mh² …(1)
The theorem of perpendicular axes is applicable to a plane lamina only. The moment of inertia I_z of a plane lamina about an axis-the z axis- perpendicular to its plane is equal to the sum of its moments of inertia I_x and I_y about two mutually perpendicular axes x and y in its plane and through the point of intersection of the perpendicular axis and the lamina.
I_z = I_x + I_y …. (2)

Question 7.
Derive an expression that relates angular momentum with the angular velocity of a rigid body.
Answer:
Consider a rigid body rotating with a constant angular velocity \(\vec{\omega}\) about an axis through the point O and perpendicular to the plane of the figure. All the particles of the body perform uniform circular motion about the axis of rotation with the same angular velocity \(\vec{\omega}\). Suppose that the body consists of N particles of masses m₁, m₂, …, m_n, situated at perpendicular distances r₁, r₂, …, r_N, respectively from the axis of rotation.

The particle of mass m₁ revolves along a circle of radius r₁, with a linear velocity of magnitude v₁ = r₁ω. The magnitude of the linear momentum of the particle is
p₁ = m₁v₁ = m₁r₁ω
The angular momentum of the particle about the axis of rotation is by definition,
\(\vec{L}_{1}\) = \(\vec{r}_{1}\) × \(\vec{p}_{1}\)
∴ L₁ = r₁p₁ sin θ
where θ is the smaller of the two angles between \(\vec{r}_{1}\) and \(\vec{p}_{1} \text { . }\)
In this case, θ = 90° ∴ sin θ = 1
∴ L₁ = r₁p₁ = r₁m₁r₁ω = m₁r₁²ω
Similarly L₂ = m₂r₂²ω, L₃ = m₃r₃²ω, etc.
The angular momentum of the body about the given axis is
L = L₁ + L₂ + … + L_N

where I = \(\sum_{i=1}^{N} m_{i} r_{i}^{2}\) = moment of inertia of the body about the given axis.
In vector form, \(\vec{L}\) = \(I \vec{\omega}\)
Thus, angular momentum = moment of inertia × angular velocity.
[Note : Angular momentum is a vector quantity. It has the same direction as \(\vec{\omega}\).]

Question 8.
Obtain an expression relating the torque with angular acceleration for a rigid body.
Answer:
A torque acting on a body produces angular acceleration. Consider a rigid body rotating about an axis passing through the point O and perpendicular to the plane of the figure. Suppose that a torque \(\vec{\tau}\) on

the body produces uniform angular acceleration \(\vec{\alpha}\) along the axis of rotation.
The body can be considered as made up of N particles with masses m₁, m₂, …, m_N situated at perpendicular distances r₁, r₂, …, r_N respectively from the axis of rotation, \(\vec{\alpha}\) is the same for all the particles as the body is rigid. Let \(\vec{F}_{1}\), \(\vec{F}_{2}\), …, \(\vec{F}_{N}\) be the external forces on the particles.
The torque \(\vec{\tau}_{1}\), on the particle of mass m₁, is
\(\vec{\tau}_{1}\) = \(\vec{r}_{1}\) × \(\vec{F}_{1}\)
∴ τ₁ = r₁F₁ sin θ
where θ is the smaller of the two angles between \(\vec{r}_{1}\) and \(\vec{F}_{1} .\)

where

is the moment of inertia of the body about the axis of rotation.
In vector form, \(\vec{\tau}\) = I\(\vec{\alpha}\)
This gives the required relation.
Angular acceleration \(\vec{\alpha}\) has the same direction as the torque \(\vec{\tau}\) and both of them are axial vectors along the rotation axis.

Question 9.
State and explain the principle of conservation of angular momentum. Use a suitable illustration. Do we use it in our daily life? When?
Answer:
Law (or principle) of conservation of angular momentum : The angular momentum of a body is conserved if the resultant external torque on the body is zero.
Explanation : This law (or principle) is used by a figure skater or a ballerina to increase their speed of rotation for a spin by reducing the body’s moment of inertia. A diver too uses it during a somersault for the same reason.

(1) Ice dance :
Twizzle and spin are elements of the sport of figure skating. In a twizzle a skater turns several revolutions while travelling on the ice. In a dance spin, the skater rotates on the ice skate and centred on a single point on the ice. The torque due to friction between the ice skate and the ice is small. Consequently, the angular momentum of a figure skater remains nearly constant.

For a twizzle of smaller radius, a figure skater draws her limbs close to her body to reduce moment of inertia and increase frequency of rotation. For larger rounds, she stretches out her limbs to increase moment of inertia which reduces the angular and linear speeds.

A figure skater usually starts a dance spin in a crouch, rotating on one skate with the other leg and both arms extended. She rotates relatively slowly because her moment of inertia is large. She then slowly stands up, pulling the extended leg and arms to her body. As she does so, her moment of inertia about the axis of rotation decreases considerably,and thereby her angular velocity substantially increases to conserve angular momentum.

(2) Diving :
Take-off from a springboard or diving platform determines the diver’s trajectory and the magnitude of angular momentum. A diver must generate angular momentum at take-off by moving the position of the arms and by a slight hollowing of the back. This allows the diver to change angular speeds for twists and somersaults in flight by controlling her/his moment of inertia. A compact tucked shape of the body lowers the moment of inertia for rotation of smaller radius and increased angular speed. The opening of the body for the vertical entry into water does not stop the rotation, but merely slows it down. The angular momentum remains constant throughout the flight.

Question 10.
Discuss the interlink between translational, rotational and total kinetic energies of a rigid object that rolls without slipping.
Answer:
Consider a symmetric rigid body, like a sphere or a wheel or a disc, rolling on a plane surface with friction along a straight path. Its centre of mass (CM) moves in a straight line and, if the frictional force on the body is large enough, the body rolls without slipping. Thus, the rolling motion of the body can be treated as translation of the CM and rotation about an axis through the CM. Hence, the kinetic energy of a rolling body is
E = E_tran + E_rot ……. (1)

where E_tran and E_rot are the kinetic energies associated with translation of the CM and rotation about an axis through the CM, respectively.

Let M and R be the mass and radius of the body. Let ω, k and i be the angular speed, radius of gyration and moment of inertia for rotation about an axis through its centre, and v be the translational speed of the centre of mass.

Question 11.
A rigid object is rolling down an inclined plane. Derive expressions for the acceleration along the track and the speed after falling through a certain vertical distance.
Answer:
Consider a circularly symmetric rigid body, like a sphere or a wheel or a disc, rolling with friction down a plane inclined at an angle 9 to the horizontal. If the frictional force on the body is large enough, the body rolls without slipping.
Let M and R be the mass and radius of the body. Let I be the moment of inertia of the body for rotation about an axis through its centre. Let the body start from rest at the top of the incline at a

height h. Let v be the translational speed of the centre of mass at the bottom of the incline. Then, its kinetic energy at the bottom of the incline is

Let a be the acceleration of the centre of mass of the body along the inclined plane. Since the body starts from rest,

Starting from rest, if t is the time taken to travel the distance L,

[Note : For rolling without slipping, the contact point of the rigid body is instantaneously at rest relative to the surface of the inclined plane. Hence, the force of friction is static rather than kinetic, and does no work on the body. Thus, the force of static friction causes no decrease in the mechanical energy of the body and we can use the principle of conservation of energy.]

Question 12.
Somehow, an ant is stuck to the rim of a bicycle wheel of diameter 1 m. While the bicycle is on a central stand, the wheel
is set into rotation and it attains the frequency of 2 rev/s in 10 seconds, with uniform angular acceleration. Calculate
(i) Number of revolutions completed by the ant in these 10 seconds.
(ii) Time taken by it for first complete revolution and the last complete revolution.
[Ans:10 rev., t_first = \(\sqrt{10}\)s, t_last = 0.5132s]
Answer:
Data : r = 0.5 m, ω₀ = 0, ω = 2 rps, t = 10 s

(i) Angular acceleration (α) being constant, the average angular speed,
ω_av = \(\frac{\omega_{\mathrm{o}}+\omega}{2}\) = \(\frac{0+2}{2}\) = 1 rps
∴ The angular displacement of the wheel in time t,
θ = ω_av ∙ t = 1 × 10 = 10 revolutions

(ii)

The time for the last, i.e., the 10th, revolution is t₁ – t₂ = 10 – 9.486 = 0.514 s

Question 13.
Coefficient of static friction between a coin and a gramophone disc is 0.5. Radius of the disc is 8 cm. Initially the
centre of the coin is 2 cm away from the centre of the disc. At what minimum frequency will it start slipping from there? By what factor will the answer change if the coin is almost at the rim?
(use g = π² m/s²)
[Ans: 2.5 rev/s, n₂ = \(\frac{1}{2}\)n₁]
Answer:
Data : µ_s = 0.5, r₁ = π cm = π × 10^-2 m, r₂ = 8 cm = 8 × 10^-2 m, g = π² m/s²
To revolve with the disc without slipping, the necessary centripetal force must be less than or equal to the limiting force of static friction.

The coin will start slipping when the frequency is

The minimum frequency in the second case will be \(\sqrt{\frac{\pi}{8}}\) times that in the first case.
[ Note The answers given in the textbook are for r₁ = 2 cm.]

Question 14.
Part of a racing track is to be designed for a radius of curvature of 72 m. We are not recommending the vehicles to drive faster than 216 kmph. With what angle should the road be tilted? At what height will its outer edge be, with respect to the inner edge if the track is 10 m wide?
[Ans: θ = tan^-1 (5) = 78.69°, h = 9.8 m]
Answer:
Data : r = 72 m, v₀ = 216 km/h, = 216 × \(\frac{5}{18}\)
= 60 m/s, w = 10 m, g = 10 m/s²
tan θ = \(\frac{v_{\mathrm{o}}^{2}}{r g}\) = \(\frac{(60)^{2}}{72 \times 10}\) = \(\frac{3600}{720}\) = 5
∴ θ = tan^-1 5 = 78°4′
This is the required angle of banking.
sin θ = \(\frac{h}{w}\)
∴ h = w sin θ = (10) sin 78°4′ = 10 × 0.9805
= 9.805 m
This gives the height of the outer edge of the track relative to the inner edge.

Question 15.
The road in the example 14 above is constructed as per the requirements. The coefficient of static friction between the tyres of a vehicle on this road is 0.8, will there be any lower speed limit? By how much can the upper speed limit exceed in this case?
[Ans: v_min ≅ 88 kmph, no upper limit as the road is banked for θ > 45°]
Answer:
Data : r = 72 m, θ = 78 °4′, µ_s = 0.8, g = 10 m/s² tan θ = tan 78°4′ = 5

= 24.588 m/s = 88.52 km/h
This will be the lower limit or minimum speed on this track.
Since the track is heavily banked, θ > 45 °, there is no upper limit or maximum speed on this track.

Question 16.
During a stunt, a cyclist (considered to be a particle) is undertaking horizontal circles inside a cylindrical well of radius 6.05 m. If the necessary friction coefficient is 0.5, how much minimum speed should the stunt artist maintain? Mass of the artist is 50 kg. If she/he increases the speed by 20%, how much will the force of friction be?
[Ans: v_min = 11 m/s, f_s = mg = 500 N]
Answer:
Data : r = 6.05 m, µ_s = 0.5, g = 10 m/s², m = 50 kg, ∆v = 20%

This is the required minimum speed. So long as the cyclist is not sliding, at every instant, the force of static friction is fs = mg = (50)(10) = 500 N

Question 17.
A pendulum consisting of a massless string of length 20 cm and a tiny bob of mass 100 g is set up as a conical pendulum. Its bob now performs 75 rpm. Calculate kinetic energy and increase in the gravitational potential energy of the bob. (Use π ² = 10 )
[Ans: cos θ = 0.8, K.E. = 0.45 J, ∆(P E.) = 0.04 J]
Answer:
Data : L = 0.2 m, m = 0.1 kg, n = \(\frac{75}{60}\) = \(\frac{5}{4}\) rps,
g = 10 m/s², π² = 10,

The increase in gravitational PE,
∆PE = mg(L – h)
= (0.1) (10) (0.2 – 0.16)
= 0.04 J

Question 18.
A motorcyclist (as a particle) is undergoing vertical circles inside a sphere of death. The speed of the motorcycle varies between 6 m/s and 10 m/s. Calculate diameter of the sphere of death. What are the minimum values are possible for these two speeds?
[Ans: Diameter = 3.2 m, (v₁)_min = 4 m/s, (v₂)_min = 4 \(\sqrt{5}\)/m s ]
Answer:

= 1.6 m
The diameter of the sphere of death = 3.2 m.

The required minimum values of the speeds are 4 m/s and 4\(\sqrt{5}\) m/s.

Question 19.
A metallic ring of mass 1 kg has moment of inertia 1 kg m² when rotating about one of its diameters. It is molten and remoulded into a thin uniform disc of the same radius. How much will its moment of inertia be, when rotated about its own axis.
[Ans: 1 kg m²]
Answer:
The MI of the thin ring about its diameter,
I_ring = \(\frac{1}{2}\)MR² = 1 kg.m²
Since the ring is melted and recast into a thin disc of same radius R, the mass of the disc equals the mass of the ring = M.
The MI of the thin disc about its own axis (i.e., transverse symmetry axis) is
I_disc = \(\frac{1}{2}\)MR² = I_ring
∴ I_disc = 1 kg.m²

Question 20.
A big dumb-bell is prepared by using a uniform rod of mass 60 g and length 20 cm. Two identical solid thermocol spheres of mass 25 g and radius 10 cm each are at the two ends of the rod. Calculate moment of inertia of the dumbbell when rotated about an axis passing through its centre and perpendicular to the length.
[Ans: 24000 g cm^-2]
Answer:
Data : M_sph = 50 g, R_sph = 10 cm, M_rod = 60 g, L_rod = 20 cm
The MI of a solid sphere about its diameter is
I_sph,CM = \(\frac{2}{5}\)M_sphR_sph
The distance of the rotation axis (transverse symmetry axis of the dumbbell) from the centre of sphere, h = 30 cm.
The MI of a solid sphere about the rotation axis, I_sph = I_{sph, CM} + M_sphh²
For the rod, the rotation axis is its transverse symmetry axis through CM.
The MI of a rod about this axis,
I_rod = \(\frac{1}{12}\) M_rodL²_rod
Since there are two solid spheres, the MI of the dumbbell about the rotation axis is

Question 21.
A flywheel used to prepare earthenware pots is set into rotation at 100 rpm. It is in the form of a disc of mass 10 kg and
radius 0.4 m. A lump of clay (to be taken equivalent to a particle) of mass 1.6 kg falls on it and adheres to it at a certain
distance x from the centre. Calculate x if the wheel now rotates at 80 rpm.
[Ans: x = \(\frac{1}{\sqrt{8}}\)m = 0.35 m]
Answer:
Data : f₁ = 60 rpm = 60/60 rot/s = 1 rot/s,
f₂ = 30 rpm = 30/60 rot/s = \(\frac{1}{2}\) rot/s, ∆E = — 100 J

(i)

This gives the MI of the flywheel about the given axis.

(ii) Angular momentum, L = Iω = I(2πf) = 2πIf
The change in angular momentum, ∆L
= L₂ – L₁ = 2πI(f₂ – f₁)
= 2 × 3.142 × 6.753(\(\frac{1}{2}\) – 1)
= -3.142 × 6.753 = -21.22 kg.m²/s

Question 22.
Starting from rest, an object rolls down along an incline that rises by 3 units in every 5 units (along it). The object gains a speed of \(\sqrt{10}\) m/s as it travels a distance of \(\frac{5}{3}\)m along the incline. What can be the possible shape/s of the object?
[Ans: \(\frac{K^{2}}{R^{2}}\) = 1. Thus, a ring or a hollow cylinder]
Answer:
Data : sin θ = \(\frac{3}{5}\), u = 0, v = \(\sqrt{10}\) m/s, L = \(\frac{5}{3}\)m, g = 10 m/s²


Therefore, the body rolling down is either a ring or a cylindrical shell.

12th Physics Digest Chapter 1 Rotational Dynamics Intext Questions and Answers

Activity (Textbook Page No. 3)

Question 1.
Attach a body of suitable mass to a spring balance so that it stretches by about half its capacity. Now whirl the spring balance so that the body performs a horizontal circular motion. You will notice that the balance now reads more for the same body. Can you explain this ?
Answer:
Due to outward centrifugal force.

Use your brain power (Textbook Page No. 4)

Question 1.
Obtain the condition for not toppling (rollover) for a four-wheeler. On what factors does it depend and how?
Answer:
Consider a car of mass m taking a turn of radius r along a level road. As seen from an inertial frame of reference, the forces acting on the car are :

1. the lateral limiting force of static friction \(\overrightarrow{f_{\mathrm{s}}}\) on the wheels-acting along the axis of the wheels and towards the centre of the circular path- which provides the necessary centripetal force,
2. the weight \(m \vec{g}\) acting vertically downwards at the centre of gravity (C.G.)
3. the normal reaction \(\vec{N}\) of the road on the wheels, acting vertically upwards effectively at the C.G. Since maximum centripetal force = limiting force of static friction,
   ma_r = \(\frac{m v^{2}}{r}\) = f_s…. (1)

In a simplified rigid-body vehicle model, we consider only two parameters-the height h of the C.G. above the ground and the average distance b between the left and right wheels called the track width.

The friction force \(\overrightarrow{f_{s}}\) on the wheels produces a torque \(\tau_{\mathrm{t}}\) that tends to overturn/rollover the car about the outer wheel. Rotation about the front-to-back axis is called roll.
\(\tau_{\mathrm{t}}\) = f_s.h = \(\left(\frac{m v^{2}}{r}\right)\)h … (2)

When the inner wheel just gets lifted above the ground, the normal reaction \(\vec{N}\) of the road acts on the outer wheels but the weight continues to act at the C.G. Then, the couple formed by the normal reaction and the weight produces a opposite torque \(\tau_{\mathrm{r}}\) which tends to restore the car back on all four wheels
\(\tau_{\mathrm{r}}\) = mg.\(\frac{b}{2}\) …. (3)
The car does not topple as long as the restoring torque \(\tau_{\mathrm{r}}\) counterbalances the toppling torque \(\tau_{\mathrm{t}}\). Thus, to avoid the risk of rollover, the maximum speed that the car can have is given by
\(\left(\frac{m v^{2}}{r}\right)\)h = mg.\(\frac{b}{2}\) ∴ v_max = \(\sqrt{\frac{r b g}{2 h}}\) … (4)

Thus, vehicle tends to roll when the radial acceleration reaches a point where inner wheels of the four-wheeler are lifted off of the ground and the vehicle is rotated outward. A rollover occurs when the gravitational force \(m \vec{g}\) passes through the pivot point of the outer wheels, i.e., the C.G. is above the line of contact of the outer wheels. Equation (3) shows that this maximum speed is high for a car with larger track width and lower centre of gravity.

There will be rollover (before skidding) if \(\tau_{t}\) ≥ \(\tau_{\mathrm{r}}\), that is if

The vehicle parameter ratio, \(\frac{b}{2 h}\), is called the static stability factor (SSF). Thus, the risk of a rollover is low if SSF ≤ µ_s. A vehicle will most likely skid out rather than roll if µ_s is too low, as on a wet or icy road.

Question 2.
Think about the normal reactions. Where are those and how much are those?
Answer:

In a simplified vehicle model, we assume the normal reactions to act equally on all the four wheels, i.e., mg/4 on each wheel. However, the C.G. is not at the geometric centre of a vehicle and the wheelbase (i.e., the distance L between its front and rear wheels) affects the weight distribution of the vehicle. When a vehicle is not accelerating, the normal reactions on each pair of front and rear wheels are, respectively,
N_f = \(\frac{d_{\mathrm{r}}}{L}\)mg and N_r = \(\frac{d_{\mathrm{f}}}{L}\) mg
where d_r and d_f are the distances of the rear and front axles from the C.G. [When a vehicle accelerates, additional torque acts on the axles and the normal reactions on the wheels change. So, as is common experience, a car pitches back (i.e., rear sinks and front rises) when it accelerates, and a car pitches ahead (i.e., front noses down). Rotation about the lateral axis is called pitch.]

Question 3.
What is the recommendations on loading a vehicle for not toppling easily?
Answer:
Overloading (or improper load distribution) or any load placed on the roof raises a vehicle’s centre of gravity, and increases the vehicle’s likelihood of rolling over. A roof rack should be fitted by considering weight limits.

Road accidents involving rollovers show that vehicles with higher h (such as SUVs, pickup vans and trucks) topple more easily than cars. Untripped rollovers normally occur when a top-heavy vehicle attempts to perform a panic manoeuver that it physically cannot handle.

Question 4.
If a vehicle topples while turning, which wheels leave the contact with the road? Why?
Answer:
Inner wheels.
Consider a car of mass m taking a turn of radius r along a level road. As seen from an inertial frame of reference, the forces acting on the car are :

1. the lateral limiting force of static friction \(\overrightarrow{f_{\mathrm{s}}}\) on the wheels-acting along the axis of the wheels and towards the centre of the circular path- which provides the necessary centripetal force,
2. the weight \(m \vec{g}\) acting vertically downwards at the centre of gravity (C.G.)
3. the normal reaction \(\vec{N}\) of the road on the wheels, acting vertically upwards effectively at the C.G. Since maximum centripetal force = limiting force of static friction,
   ma_r = \(\frac{m v^{2}}{r}\) = f_s…. (1)

In a simplified rigid-body vehicle model, we consider only two parameters-the height h of the C.G. above the ground and the average distance b between the left and right wheels called the track width.

The friction force \(\overrightarrow{f_{s}}\) on the wheels produces a torque \(\tau_{\mathrm{t}}\) that tends to overturn/rollover the car about the outer wheel. Rotation about the front-to-back axis is called roll.
\(\tau_{\mathrm{t}}\) = f_s.h = \(\left(\frac{m v^{2}}{r}\right)\)h … (2)

When the inner wheel just gets lifted above the ground, the normal reaction \(\vec{N}\) of the road acts on the outer wheels but the weight continues to act at the C.G. Then, the couple formed by the normal reaction and the weight produces a opposite torque \(\tau_{\mathrm{r}}\) which tends to restore the car back on all four wheels
\(\tau_{\mathrm{r}}\) = mg.\(\frac{b}{2}\) …. (3)
The car does not topple as long as the restoring torque \(\tau_{\mathrm{r}}\) counterbalances the toppling torque \(\tau_{\mathrm{t}}\). Thus, to avoid the risk of rollover, the maximum speed that the car can have is given by
\(\left(\frac{m v^{2}}{r}\right)\)h = mg.\(\frac{b}{2}\) ∴ v_max = \(\sqrt{\frac{r b g}{2 h}}\) … (4)

Thus, vehicle tends to roll when the radial acceleration reaches a point where inner wheels of the four-wheeler are lifted off of the ground and the vehicle is rotated outward. A rollover occurs when the gravitational force \(m \vec{g}\) passes through the pivot point of the outer wheels, i.e., the C.G. is above the line of contact of the outer wheels. Equation (3) shows that this maximum speed is high for a car with larger track width and lower centre of gravity.

Question 5.
How does [tendency to] toppling affect the tyres?
Answer:
While turning, shear stress acts on the tyre-road contact area. Due to this, the treads and side wall of a tyre deform. Apart from less control, this contributes to increased and uneven wear of the shoulder of the tyres.

Each wheel is placed under a small inward angle (called camber) in the vertical plane. Under severe lateral acceleration, when the car rolls, the camber angle ensures the complete contact area is in contact with the road and the wheels are now in vertical position. This improves the cornering behavior of the car. Improperly inflated and worn tyres can be especially dangerous because they inhibit the ability to maintain vehicle control. Worn tires may cause the vehicle to slide
sideways on wet or slippery pavement, sliding the vehicle off the road and increasing its risk of rolling over.

Question 6.
What is the recommendation for this?
Answer:
Because of uneven wear of the tyre shoulders, tyres should be rotated every 10000 km-12000 km. To avoid skidding, rollover and tyre-wear, the force of friction should not be relied upon to provide the necessary centripetal force during cornering. Instead, the road surface at a bend should be banked, i.e., tilted inward.

A car while taking a turn performs circular motion. If the road is level (or horizontal road), the necessary centripetal force is the force of static friction between the car tyres and the road surface. The friction depends upon the nature of the surfaces in contact and the presence of oil and water on the road. If the friction is inadequate, a speeding car may skid off the road. Since the friction changes with circumstances, it cannot be relied upon to provide the necessary centripetal force. Moreover, friction results in fast wear and tear of the tyres.

To avoid the risk of skidding as well as to reduce the wear and tear of the car tyres, the road surface at a bend is tilted inward, i.e., the outer side of the road is raised above its inner side. This is called banking of road. On a banked road, the resultant of the normal reaction and the gravitational force can act as the necessary centripetal force. Thus, every car can be safely driven on such a banked curve at certain optimum speed, without depending on friction. Hence, a road should be properly banked at a bend.

The angle of banking is the angle of inclination of the surface of a banked road at a bend with the horizontal.

When a two-wheeler takes a turn along an unbanked road, the force of friction provides the centripetal force. The two-wheeler leans inward to counteract a torque that tends to topple it outward. Firstly, friction cannot be relied upon to provide the necessary centripetal force on all road conditions. Secondly, the friction results in wear and tear of the tyres. On a banked road at a turn, any vehicle can negotiate the turn without depending on friction and without straining the tyres.

Question 7.
Determine the angle to be made with the vertical by a two-wheeler while turning on a horizontal track?
Answer:

When a bicyclist takes a turn along an unbanked road, the force of friction \(\vec{f}_{\mathrm{s}}\) provides the centripetal force; the normal reaction of the road \(\vec{N}\) is vertically up. If the bicyclist does not lean inward, there will be an unbalanced outward torque about the centre of gravity, f_s.h, due to the friction force that will topple the bicyclist outward. The bicyclist must lean inward to counteract this torque (and not to generate a centripetal force) such that the opposite inward torque of the couple formed by \(\vec{N}\) and the weight \(\vec{g}\), mg.a = f_s.h₁

Since the force of friction provides the centripetal force,
f_s = \(\frac{m v^{2}}{r}\)
If the cyclist leans from the vertical by an angle 9, the angle between \(\vec{N}\) and \(\vec{F}\) in above figure.

Hence, the cyclist must lean by an angle
θ = tan^-1\(\left(\frac{v^{2}}{g r}\right)\)

When a car takes a turn along a level road, apart from the risk of skidding off outward, it also has a tendency to roll outward due to an outward torque about the centre of gravity due to the friction force. But a car is an extended object with four wheels. So, when the inner wheels just get lifted above the ground, it can be counterbalanced by a restoring torque of the couple formed by the normal reaction (on the outer wheels) and the weight.

Question 8.
We have mentioned about ‘static friction’ between road and tyres. Why is it static friction? What about kinetic friction between road and tyres?
Answer:
When a car takes a turn on a level road, the point of contact of the wheel with the surface is instantaneously stationary if there is no slipping. Hence, the lateral force on the car is the limiting force of static friction between the tyres and road. Lateral forces allow the car to turn. As long as the wheels are rolling, there is lateral force of static friction and longitudinal force of rolling friction. Longtitudinal forces, which act in the direction of motion of the car body (or in the exact opposite direction), control the acceleration or deceleration of the car and therefore the speed of the car. These are the wheel force, rolling friction, braking force and air drag. If the car skids, the friction force is kinetic friction; more importantly, the direction of the friction force then changes abruptly from lateral to that opposite the velocity of skidding and not towards the centre of the curve, so that the car cannot continue in its curved path.

Question 9.
What do you do if your vehicle is trapped on a slippery or sandy road? What is the physics involved?
Answer:
Driving on a country road should be attempted only with a four-wheel drive. However, if you do get stuck in deep sand or mud, avoid unnecessary panic and temptation to drive your way out of the mud or sand because excessive spinning of your tyres will most likely just dig you into a deeper hole. Momentum is the key to getting unstuck from sand or mud. One method is the rocking method-rocking your car backwards and forwards to gain momentum. Your best option is usually to gain traction and momentum by wedging a car mat (or sticks, leaves, gravel or rocks) in front and under your drive wheels. Once you start moving, keep the momentum going until you are on more solid terrain.

Use your brain power (Textbook Page No. 6)

Question 1.
As a civil engineer, you are to construct a curved road in a ghat. In order to calculate the banking angle 0, you need to decide the speed limit. How will you decide the values of speed and radius of curvature at the bend ?
Answer:
For Indian roads, Indian Road Congress (IRC), [IRC-73-1980, Table 2, p.4], specifies the design speed depending on the classification of roads (such as national and state highways, district roads and village roads) and terrain. It is the basic design parameter which determines further geometric design features. For the radius of curvature at a bend, IRC [ibid., Table 16, p.24] specifies the absolute minimum values based on the minimum design speed. However, on new roads, curves should be designed to have the largest practicable radius, generally more than the minimum values specified, to allow for ‘sight distance’ and ‘driver comfort’. To consider the motorist driving within the innermost travel lane, the radius used to design horizontal curves should be measured to the inside edge of the innermost travel lane, particularly for wide road-ways with sharp horizontal curvature.

A civil engineer refers to banking as superelevation e;e = tan θ. IRC fixes e_max = 0.07 for a non-urban road and the coefficient of lateral static friction, µ = 0.15, the friction between the vehicle tyres and the road being incredibly variable. Ignoring the product eµ, from Eq. (6)
e + µ = \(\frac{v^{2}}{g r}\) (where both v and r are in SI units)
= \(\frac{V^{2}}{127 r}\)(where V is in km / h and r is in metre) …. (1)
The sequence of design usually goes like this :

1. Knowing the design speed V and radius r, calculate the superelevation for 75% of design speed ignormg friction : e = \(\frac{(0.75 \mathrm{~V})^{2}}{127 r}\) = \(\frac{V^{2}}{225 r}\)
2. If e < 0.07, consider this calculated value of e in subsequent calculations. If e > 0.07, then take e = e_max = 0.07.
3. Use Eq. (1) above to check the value of µ for e_max = 0.07 at the full value of the design speed V : µ = \(\frac{V^{2}}{127 r}\) – 0.07
   If µ < 0.15, then e = 0.07 is safe. Otherwise, calculate the allowable speed Va as in step 4.
4. \(\frac{V_{\mathrm{a}}^{2}}{127 r}\) = e + p = 0.07 + 0.15
   If V_a > V, then the design speed V is adequate.
   If V_a < V, then speed is limited to V_a with appropriate warning sign.

Use your brain power (Textbook Page No. 7)

Question 1.
If friction is zero, can a vehicle move on the road? Why are we not considering the friction in deriving the expression for the banking angle?
Answer:
Friction is necessary for any form of locomotion. Without friction, a vehicle cannot move. The banking angle for a road at a bend is calculated for optimum speed at which every vehicle can negotiate the bend without depending on friction to provide the necessary lateral centripetal force.

Question 2.
What about the kinetic friction between the road and the lyres?
Answer:
When a car takes a turn on a level road, the point of contact of the wheel with the surface is instantaneously stationary if there is no slipping. Hence, the lateral force on the car is the limiting force of static friction between the tyres and road. Lateral forces allow the car to turn. As long as the wheels are rolling, there is lateral force of static friction and longitudinal force of rolling friction. Longtitudinal forces, which act in the direction of motion of the car body (or in the exact opposite direction), control the acceleration or deceleration of the car and therefore the speed of the car. These are the wheel force, rolling friction, braking force and air drag. If the car skids, the friction force is kinetic friction; more importantly, the direction of the friction force then changes abruptly from lateral to that opposite the velocity of skidding and not towards the centre of the curve, so that the car cannot continue in its curved path.

Use your brain power (Textbook Page No. 12)

Question 1.
What is expected to happen if one travels fast over a speed breaker? Why?
Answer:
The maximum speed with which a car can travel over a road surface, which is in the form of a convex arc of radius r, is \(\sqrt{r g}\) where g is the acceleration due to gravity. For a speed breaker, r is very small (of the order of 1 m). Hence, one must slow down considerably while going over a speed breaker. Otherwise, the car will lose contact with the road and land with a thud.

Question 2.
How does the normal force on a concave suspension bridge change when a vehicle is travelling on it with a constant speed ?
Answer:
At the lowest point, N-mg provides the centripetal force. Therefore, N-mg = \(\frac{m v^{2}}{r}\), so that N = m(g + \(\frac{v^{2}}{r}\)).
Therefore, N increases with increasing v.

Use your brain power (Textbook Page No. 15)

Question 1.
For the point P in above, we had to extend OC to Q to meet the perpendicular PQ. What will happen to the expression for I if the point P lies on OC?
Answer:
There will be no change in the expression for the MI (I) about the parallel axis through O.

Balbharti Maharashtra State Board 12th Physics Textbook Solutions Chapter 16 Semiconductor Devices Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Physics Solutions Chapter 16 Semiconductor Devices

1. Choose the correct option

i.
In a BJT, the largest current flow occurs
(A) in the emitter
(B) in the collector
(C) in the base
(D) through CB junction.
Answer:
(A) in the emitter

ii.
A series resistance is connected in the Zener diode circuit to
(A) properly reverse bias the Zener
(B) protect the Zener
(C) properly forward bias the Zener
(D) protect the load resistance.
Answer:
(A) properly reverse bias the Zener

iii.
An LED emits visible light when its
(A) junction is reverse biased
(B) depletion region widens
(C) holes and electrons recombine
(D) junction becomes hot.
Answer:
(C) holes and electrons recombine

iv.
Solar cell operates on the principle of
(A) diffusion
(B) recombination
(C) photovoltaic action
(D) carrier flow.
Answer:
(C) photovoltaic action

v.
A logic gate is an electronic circuit which
(A) makes logical decisions
(B) allows electron flow only in one direction
(C) works using binary algebra
(D) alternates between 0 and 1 value.
Answer:
(A) makes logical decisions

2 Answer in brief.

i.
Why is the base of a transistor made thin and is lightly doped?
Answer:
The base of a transistor is lightly doped than the emitter and is made narrow so that virtually all the electrons injected from the emitter (in an npn tran-sistor) diffuse right across the base to the collector junction without recombining with holes. That is, the base width is kept less than the recombination distance. Also, the emitter is much heavily doped than the base to improve emitter efficiency and common-base current gain a.

ii.
How is a Zener diode different than an ordinary diode?
Answer:
A Zener diode is heavily doped-the doping con-centrations for both p- and n-regions is greater than 1018 cm-3 while those of an ordinary diode are voltage (PIV) of an ordinary diode is higher than a Zener diode and the breakdown occurs by impact ionization (avalanche process). Their I-V characteristics are otherwise similar.

iii.
On which factors does the wavelength of light emitted by a LED depend?
Answer:
The intensity of the emitted light is directly propor-tional to the recombination rate and hence to the diode forward current. The colour of the light emitted by an LED depends on the compound semiconductor material used and its composition (and doping levels) as given below :
Table: Typical semiconductor materials and emitted colours of LEDs

Material	Emitted colour(s)
Gallium arsenide (GaAs), Indium gallium arsenide phosphide (InGaAsP)	Infrared
Aluminum gallium arsenide (AlGaAs)	Deep red, also IR laser
Indium gallium phosphide (InGaP)	Red
Gallium arsenide phosphide (GaAsP), aluminum indium gallium phosphide (AlInGaP)	Orange, red or yellow
Gallium phosphide (GaP)	Green or yellow
Aluminium gallium phosphide (AlGaP), zinc selenide (ZnSe), zinc selenide telluride (ZnSeTe), nitrogen impregnated gallium phosphide (GaP:N)	Green
Indium gallium nitride (InGaN), gallium nitride (GaN), sine sulphide (ZnS)	Blue and violet Longer wave lengths (green and yellow) are obtained by increasing the indium (In) content. Phosphor encapsulation produce white light.
Aluminium gallium nitride (AlGaN)	Ultraviolet

iv.
Why should a photodiode be operated in reverse biased mode?
Answer:
A photodiode is operated in a reverse biased mode because as photodetector or photosensor, it must conduct only when radiation is incident on it. In the reverse biased mode, the dark current for zero illumination is negligibly small—of the order of few picoamperes to nanoamperes. But when illuminated, the photocurrent is several orders of magnitude greater.

v.
State the principle and uses of a solar Cell.
Answer:
A solar cell is an unbiased pn-junction that converts the energy of sunlight directly into electricity with a high conversion efficiency.

Principle : A solar cell works on the photovoltaic effect in which an emf is produced between the two layers of a pn-junction as a result of irradiation.

Question 3.
Draw the circuit diagram of a half wave rectifier. Explain its working. What is the frequency of ripple in its output?
Answer:
A device or a circuit which rectifies only one-half of each. cycle of an alternating voltage is called a half-wave rectifier.
Electric circuit : The alternating voltage to be rectified is applied across the primary coil (P₁P₂) of a transformer. The secondary coil (S₁S₂) of the transformer is connected in series with the junction diode and a load resistance R_L, as shown in below figure. The alternating voltage across the secondary coil is the ac input voltage V_i. The dc voltage across the load resistance is called the output voltage V₀.

Working : Due to the alternating voltage V_i, the p-region of the diode becomes alternatively positive and negative with respect to the n-region.
During the half-cycle when the p-region is positive, the diode is forward biased and conducts. A current I_L passes through the load resistance RL in the direction shown.

During the next half cycle, when the p-region is negative, the diode is reverse biased and the forward current drops to zero.

Thus, the diode conducts only during one-half of the input cycle and thus acts as a half-wave rectifier. The intermittent output voltage V₀ has a fixed polarity but changes periodically with time between zero and a maximum value. I_L is unidirectional. Above figure shows the input and output voltage waveforms.

The pulsating dc output voltage of a half-wave rectifier has the same frequency as the input.

Question 4.
Why do we need filters in a power supply?
Answer:
A rectifier-half-wave or full-wave – outputs a pul-sating dc which is not directly usable in most electronic circuits. These circuits require something closer to pure dc as produced by batteries. Unlike pure dc waveform of a battery, a rectifier output has an ac ripple riding on a dc waveform.

The circuit used in a dc power supply to remove the ripple is called a filter. A filter circuit can produce a very smooth waveform that approximates the waveform produced by a battery. The most common technique used for filtering is a capacitor connected across the output of a rectifier.

Question 5.
Draw a neat diagram of a full wave rectifier and explain it’s working.
Answer:
A device or a circuit which rectifies both halves of each cycle of an alternating voltage is called a full-wave rectifier.
Electric circuit : The alternating voltage to be rectified is applied across the primary coil (P₁P₂) of a transformer with a centre-tapped secondary coil (S₁S₂). The terminals and S2 of the secondary are connected to the two p-regions of two junction diodes D₁ and D₂, respectively. The centre-tap T is connected to the ground. The load resistance RL is connected across the common n-regions and the

P₁P₂, S₁S₂ : Primary and secondary of transformer,
T : Centre-tap on secondary; D₁ D₂ : Junction diodes,
R_L : Load resistance, I_L : Load current,
V_i: AC input voltage, V₀ : DC output voltage
Above Figure : Full-wave rectifier circuit

Working : During one half cycle of the input, terminal S₁ of the secondary is positive while S₂ is negative with respect to the ground (the centre-tap T). During this half cycle, diode D₁ is forward biased and conducts, while diode D₂ is reverse biased and does not conduct. The direction of current Z_L through R_L is in the sense shown.

During the next half cycle of the input voltage, S2 becomes positive while S, is negative with respect to T. Diode D2 now conducts sending a current IL through RL in the same sense as before. Dt now does not conduct. Thus, the current through RL flows in the same direction, i.e., it is unidirectional, for both halves or the full-wave of the input. This is called full-wave rectification.

The output voltage has a fixed polarity but varies periodically with time between zero and a maximum value. Above figure shows the input and output voltage waveforms. The pulsating dc output voltage of a full-wave rectifier has twice frequency of the input.

Question 6.
Explain how a Zener diode maintains constant voltage across a load.
Answer:
Principle : In the breakdown region of a Zener diode, for widely changing Zener current, the voltage across the Zener diode remains almost constant.

Electric circuit : The circuit for regulating or stabilizing the voltage across a load resistance RL against change in load current and supply voltage is shown in above figure. The Zener diode is connected parallel to load Rh such that the current through the Zener diode is from the n to p region. The series resistance Rs limits the current through the diode below the maximum rated value.
From the circuit, I = I_Z + I_L and V = IR_s + V_Z
= (I_Z + I_L)R_s + V_Z
Working: When the input unregulated dc voltage V across the Zener diode is greater than the Zener voltage V_Z in magnitude, the diode works in the Zener breakdown region. The voltage across the diode and load Rh is then V_Z. The corresponding current in the diode is I_Z.

As the load current (I) or supply voltage (V) changes, the diode current (7Z) adjusts itself at constant V_Z. The excess voltage V-V_Zappears across the series resistance Rs.

For constant supply voltage, the supply current I and the voltage drop across R_s remain constant. If the diode is within its regulating range, an increase in load current is accompanied by a decrease in Iz at constant V_Z.
Since the voltage across R_L remains constant at V_Z, the Zener diode acts as a voltage stabilizer or voltage regulator.

Question 7.
Explain the forward and the reverse characteristic of a Zener diode.
Answer:
The forward bias region of a Zener diode is identical to that of a regular diode. There is forward current only after the barrier potential of the pn- junction is overcome. Beyond this threshold or cut in voltage, there is an exponential upward swing.

The typical forward voltage at room temperature with a current of around 1 mA is around 0.6 V.

In the reverse bias condition the Zener diode is an open circuit and only a small reverse saturation current flows as shown with change of scale. At the reverse breakdown voltage there is an abrupt rapid increase in the current-the knee is very sharp, followed by an almost vertical increase in current. The voltage across the Zener diode in the breakdown region is very nearly constant with only a small increase in voltage with increasing current. There is a minimum Zener current, I_Z (min), that places the operating point in the desired breakdown region. At some high current level, I_ZM, the power dissipation of the diode becomes excessive beyond which the diode can be damaged.

Zener diode characteristics

The I-V characteristics of a Zener diode is not totally vertical in the breakdown region. This means that for slight changes in current, there will be a small change in the voltage across the diode. The voltage change for a given change in current is the resistance R_Z of the Zener diode.

Question 8.
Explain the working of a LED.
Answer:
Working :
An LED is forward-biased with about 1.2 V to 3.6 V at 12 mA to 20 mA. Majority carriers electrons from n-type layer and holes from p-type layer are injected into the active layer. Electrons cross the junction into the p-layer. In the active p-layer, some of these excess minority carriers electrons, recombine radiatively with majority carriers, holes, thereby emitting photons. The resulting photon has an energy approximately equal to the bandgap of the active layer material. Modifying the bandgap of the active layer creates photons of diferent energies.

In the energy band diagram this recombination is equivalent to a transition of the electron from a higher energy state in the conduction band to a lower energy state in the valence band. The energy difference is emitted as a photon of energy hv.
[Note : The photons originate primarily in the p-side of the junction which has a bandgap E_Gp narrower than that of the n-side, E_Gn. Thus, with hv < E_Gn, the photons are emitted through the wide-bandgap n-region with essentially no absorption.]

Question 9.
Explain the construction and working of solar cell.
Answer:
Construction :
A simple pn-junction solar cell con-sists of a p-type semiconductor substrate backed with a metal electrode back contact. A thin n-layer (less than 2.5 pm, for silicon) is grown over the p-type substrate by doping with suitable donor impurity. Metal finger electrodes are prepared on top of the n-layer so that there is enough space between the fingers for sunlight to reach the n-layer and, subsequently, the underlying pn-junction.

Working : When exposed to sunlight, the absorption of incident radiation (in the range near-UV to infrared) creates electron-hole pairs in and near the depletion layer.

Consider light of frequency v incident on the pn-junction such that the incident photon energy hv is greater than the band gap energy EG of the semiconductor. The photons excite electrons from the valence band to the conduction band, leaving vacancies or holes in the valence band, thus generating electron-hole pairs.

The photogenerated electrons and holes move towards the n side and p side, respectively. If no external load is connected, these photogenerated charges get collected at the two sides of the junction and give rise to a forward photovoltage. In a closed- circuit, a current I passes through the external load as long as the solar cell is exposed to sunlight.

A solar cell module consists of several solar cells connected in series for a higher voltage output. For outdoor use with higher power output, these modules are connected in different series and parallel combinations to form a solar cell array.

[Note : Currently most of the crystalline solar cells are p-type as described above. This is because of a lower cost of production of p-type. But performance wise, n-type solar cells (a thin p-layer over an n-type substrate by doping with suitable acceptor impurity) can give much better efficiency compared to p-type solar cells.]

Question 10.
Explain the principle of operation of a photodiode.
Answer:
Construction:
A photodiode consists of an n-type silicon substrate with a metal electrode back contact. A thin p-type layer is grown over the n-type substrate by diffusing a suitable acceptor dopant.

The area of the p-layer defines the photodiode active area. An ohmic contact pad is deposited on the active area. The rest of the active area is left open with a protective antireflective coating of silicon nitride to minimize the loss of photons. The nonactive area is covered with an insulating opaque SiO₂ coating.

Depending on the required spectral sensitivity, i.e., the operating wavelength range, typical photodiode materials are silicon, germanium, indium gallium arsenide phosphide (InGaAsP) and indium gallium arsenide (InGaAs), of which silicon is the cheapest while the last two are expensive.

Working : The band gap energy of silicon is E_G = 1.12 eV at room temperature. Thus, photons or particles with energies greater than or equal to 1.12 eV, which corresponds to 110 nm, can transfer electrons from the valence band into the conduction band.

A photodiode is operated in the reverse bias mode which results in a wider depletion region. When operated in the dark (zero illumination), there is a reverse saturation current due solely to the thermally generated minority charge carriers. This is called the dark current. Depending on the minority carrier concentrations, the dark current in an Si photodiode may range from 5 pA to 10 nA.

When exposed to radiation of energy hv ≥ E_G (in the range near-UV to near-IR), electron-hole pairs are created in the depletion region. The electric field in the depletion layer accelerates these photogenerated electrons and holes towards the n side and p side, respectively, constituting a photocurrent l in the external circuit from the p side to the n side. Due to the photogeneration, more charge carriers are available for conduction and the reverse current is increased. The photocurrent is directly propor-tional to the intensity of the incident light. It is independent of the reverse bias voltage.
[Notes : Typical photodiode materials are :
(1) silicon (Si) : low dark current, high speed, good sensitivity between ~ 400 nm and 1000 run (best around 800 nm-900 nm)
(2) germanium (Ge) : high dark current, slow speed, good sensitivity between ~ 900 nm and 1600 nm (best around 1400 nm-1500 nm)
(3) indium gallium arsenide phosphide (InGaAsP) : expensive, low dark current, high speed, good sensitivity between ~ 1000 nm and 1350 nm (best around 1100 nm- 1300 nm)
(4) indium gallium arsenide (InGaAs): expensive, low dark current, high speed, good sensitivity between ~ 900 nm and 1700 nm (best around 1300 nm-1600 nm],

Question 11.
What do you mean by a logic gate, a truth table and a Boolean expression?
Answer:
A logic gate is a basic switching circuit used in digital circuits that determines when an input pulse can pass through to the output. It generates a single output from one or more inputs.

Explanation/Uses :
Any digital computation process consists in performing a sequence of arithmetical operations on the data of the problem. At each stage in the computation, the nature of the operation to be performed is decided partly by the pre-determined program and partly by the outcome of earlier stages in the process. We therefore need switches with multiple inputs to perform logical operations, i.e., the outputs of these switches are determined in specified ways by the condition (binary state) of their inputs. These arrangements are known as logic gates, and mostly they are extension of a simple transistor switch.

(1) Boolean expression : An equation expressing a logical compound statement in Boolean algebra is called a Boolean expression. A Boolean expression for a logic gate expresses the relation between input(s) and output of a logic gate.

(2) Truth table : The table which shows the truth values of a Boolean expression for a logic gate for all possible combinations of its inputs is called the truth table of logic gate.

The truth table contains one row for each input combination. Since a logical variable can assume only two possible values, 0 and 1, there are 2N combinations of N inputs so that the table has 2N rows.
[Note : Boolean algebra is a form of symbolic logic developed in 1847 by George Boole (1815-64) British mathematician.]

Question 12.
What is logic gate? Write down the truth table and Boolean expression for ‘AND’ gate.
Answer:
A logic gate is a basic switching circuit used in digital circuits that determines when an input pulse can pass through to the output. It generates a single output from one or more inputs.

The AND gate : It is a circuit with two or more inputs and one output in which the output signal is HIGH if and only if all the inputs are HIGH simultaneously.

The AND operation represents a logical multiplication.

Below figure shows the 2-input AND gate logic symbol and the Boolean expression and the truth table for the AND function.
Logic symbol:

Truth table:

Boolean expression:
Y = A ∙ B

Question 13.
What are the uses of logic gates? Why is a NOT gate known as an inverter?
Answer:
Explanation/Uses :
Any digital computation process consists in performing a sequence of arithmetical operations on the data of the problem. At each stage in the computation, the nature of the operation to be performed is decided partly by the pre-determined program and partly by the outcome of earlier stages in the process. We therefore need switches with multiple inputs to perform logical operations, i.e., the outputs of these switches are determined in specified ways by the condition (binary state) of their inputs. These arrangements are known as logic gates, and mostly they are extension of a simple transistor switch.

The NOT gate or INVERTER : It is a circuit with one input whose output is HIGH if the input is LOW and vice versa.

The NOT operation outputs an inverted version of the input. Hence, a NOT gate is also known as an INVERTER.

The small invert bubble on the output side of the inverter logic symbol, below figure and the over bar (^–) in the Boolean expression represent the invert function.

Question 14.
Write the Boolean expression for (i) OR gate, (ii) AND gate, and (iii) NAND Gate.
Answer:
(i) The OR gate : It is a circuit with two or more inputs and one output in which the output signal is HIGH if any one or more of the inputs is HIGH.
The OR operation represents a logical addition.
Below figure shows the 2-input OR gate logic sym-bol, and the Boolean expression and the truth table for the OR function.

(ii) The AND gate : It is a circuit with two or more inputs and one output in which the output signal is HIGH if and only if all the inputs are HIGH simultaneously.

The AND operation represents a logical multiplication.

Below figure shows the 2-input AND gate logic symbol and the Boolean expression and the truth table for the AND function.
Logic symbol:

Truth table:

Boolean expression:
Y = A ∙ B

(iii) The NAND gate : It is a circuit with two or more inputs and one output, whose output is HIGH if any one or more of the inputs is LOW; the output is LOW if all the inputs are HIGH.

The NAND gate is a combination of an AND gate followed by a NOT gate so that the truth table of the NAND function is obtained by inverting the outputs of the AND gate.

Question 15.
Why is the emitter, the base and the collector of a BJT doped differently?
Answer:
A BJT being a bipolar device, both electrons and holes participate in the conduction process. Under the forward-biased condition, the majority carriers injected from the emitter into the base constitute the largest current component in a BJT. For these carriers to diffuse across the base region with t negligible recombination and reach the collector junction, these must overwhelm the majority carriers of opposite charge in the base. The total emitter current has two components, that due to majority carriers in the emitter and that due to minority carriers diffused from base into emitter. The ratio of the current component due to the injected majority carriers from the emitter to the total emitter current is a measure of the emitter efficiency. To improve the emitter efficiency and the common-base current gain (a), it can be shown that’ the emitter should be much heavily doped than the base.

Also, the base width is a function of the base- collector voltage. A low doping level of the collector increases the size of the depletion region. This increases the maximum collector-base voltage and reduces the base width. Further, the large depletion region at the collector-base junction-extending mainly into the collector-corresponds to a smaller electric field and avoids avalanche breakdown of the reverse-biased collector-base junction.
[Note : Effective dopant concentrations of (a) npn transistor (b) pnp transistor are shown below.

The base doping is less than the emitter doping but greater than the collector doping. Contrary to the impression stressed in the Board’s and NCERT textbooks, collector doping is typically an order of magnitude lower than base doping. {Ref. : Semiconductor Devices Physics and Technology (3rd Edition), Simon M. Sze and M. K. Lee, p. 125}]

Question 16.
Which method of biasing is used for operating transistor as an amplifier?
Answer:
For use as an amplifier, the transistor should be in active mode. Therefore, the emitter-base junction is forward biased and the collector-base junction is reverse biased. Also, an amplifier uses an emitter bias rather than a base bias.

Question 17.
Define α and β. Derive the relation between then.
Answer:
The dc common-base current ratio or current gain (α_dc) is defined as the ratio of the collector current to emitter current.
α_dc = \(\frac{I_{C}}{I_{\mathrm{E}}}\)
The dc common-emitter current ratio or current gain (β_dc) is defined as the ratio of the collector current to base current.
β_dc = \(\frac{I_{C}}{I_{B}}\)
Since the emitter current I_E = I_B + I_C
\(\frac{I_{\mathrm{E}}}{I_{C}}=\frac{I_{\mathrm{B}}}{I_{\mathrm{C}}}+1\)
∴ \(\frac{1}{\alpha_{\mathrm{dc}}}=\frac{1}{\beta_{\mathrm{dc}}}+1\)
Therefore, the common-base current gain in terms of the common-emitter current gain is
α_dc = \(\frac{\beta_{\mathrm{dc}}}{1+\beta_{\mathrm{dc}}}\)
and the common-emitter current gain in terms of the common-base current gain is
β_dc = \(\frac{\alpha_{\mathrm{dc}}}{1-\alpha_{\mathrm{dc}}}\)
For a transistor, α_dc is close to but always less than 1 (about 0.92 to 0.98) and β_dc ranges from 20 to 200 for most general purpose transistors.

Question 18.
The common-base DC current gain of a transistor is 0.967. If the emitter current is 10mA. What is the value of base current?
Answer:
Data : α_dc = 0.967, I_E = 10 mA
α_dc = \(\frac{I_{C}}{I_{\mathrm{E}}}\) and I_E = I_B + I_C
The collector current,
I_C = α_dcI_E = 0.967 × 10 = 9.67 mA
Therefore, the base current,
I_B = I_E – I_C = 10 – 9.67 = 0.33 mA

Question 19.
In a comman-base connection, a certain transistor has an emitter current of 10mA and collector current of 9.8 mA. Calculate the value of the base current.
Answer:
DATA : I_E = 10 mA, I_C = 9.8 mA
I_E = I_B + I_C
Therefore, the base current,
I_B = I_E – I_C – 10 – 9.8 = 0.2 mA

Question 20.
In a common-base connection, the emitter current is 6.28mA and collector current is 6.20 mA. Determine the common base DC current gain.
Answer:
Data : I_E = 6.28 mA, I_C = 6.20 mA
α_dc = \(\frac{I_{C}}{I_{\mathrm{E}}}\) and β_dc = \(\frac{I_{\mathrm{C}}}{I_{\mathrm{B}}}=\frac{\alpha_{\mathrm{dc}}}{1-\alpha_{\mathrm{dc}}}\)
Common-emitter current gain, α_dc = \(\frac{6.20}{6.28}\) = 0.9873
Therefore, common-base current gain,
β_dc = \(\frac{0.9873}{1-0.9873}=\frac{0.9873}{0.0127}\) = 77.74
OR
I_E = I_B + I_C
∴ I_B = I_E – I_C = 6.28 – 6.20 = 0.08 mA
∴ β_dc = \(\frac{6.20}{0.08}\) = 77.5
[Note : The answer given in the textbook obviously refers to the common-emitter current gain.]

12th Physics Digest Chapter 16 Semiconductor Devices Intext Questions and Answers

Remember this (Textbook Page No. 346)

Question 1.
A full wave rectifier utilises both half cycles of AC input voltage to produce the DC output.
Answer:
A half-wave rectifier rectifies only one half of each cycle of the input ac wave while a full-wave rectifier rectifies both the halves. Hence the pulsating dc output voltage of a half-wave rectifier has the same frequency as the input but that of a full-wave rectifier has double the frequency of the ac input.

Do you know (Textbook Page No. 346)

Question 1.
The maximum efficiency of a full wave rectifier is 81.2% and the maximum efficiency of a half wave rectifier is 40.6%. It is observed that the maximum efficiency of a full wave rectifier is twice that of half wave rectifier.
Answer:
The ratio of dc power obtained at the output to the applied input ac power is known as rectifier efficiency. A half-wave rectifier can convert maximum 40.6% of ac power into dc power, and the remaining power of 59.4% is lost in the rectifier circuit. In fact, 50% power in the negative half cycle is not converted and the remaining 9.4% is lost in the circuit. Hence, a half wave rectifier efficiency is 40.6%. The maximum efficiency of a full-wave rectifier is 81.2%, i.e., twice that of a half-wave rectifier.

Do you know (Textbook Page No. 349)

Question 1.
The voltage stabilization is effective when there is a minimum Zener current. The Zener diode must be always operated within its breakdown region when there is a load connected in the circuit. Similarly, the supply voltage V_s must be greater than V_z.
Answer:
A Zener diode is operated in the breakdown region. There is a minimum Zener current, I_z, that places the desired operating point in the breakdown region. There is a maximum Zener current, I_zM, at which the power dissipation drives the junction temperature to the maximum allowed. Beyond that current the diode can be damaged. Hence, the supply voltage must be greater than V_z and the current-limiting resistor must limit the diode current to less than the rated maxi mum, I_zM.

Remember this (Textbook Page No. 350)

Question 1.
Zener effect occurs only if the diode is heavily doped, because when the depletion layer is thin, breakdown occurs at low reverse voltage and the field strength will be approximately 3 × 10⁷ V/m. It causes an increase in the flow of free carriers and increase in the reverse current.
Answer:
Zener breakdown occurs only in heavily doped pn junctions (doping concentrations for both p- and n-regions greater than 1018 cm3) and can take place only if the electric field in the depletion region of the reverse-biased junction is very high. It is found that the critical field at which tunneling becomes probable, i.e., at which Zener breakdown commences, is approximately 10⁶ V/cm. [“internal Field Emissiot at Narrow Silicon and Germanium PN-Junctions,” Phys. Rev., 118, 425 (1960).]

Can you tell (Textbook Page No. 350)

Question 1.
How does a cell phone charger produce a voltage of 5.0 V form the line voltage of 230V?
Answer:
A phone charger is usually a 5 V power supply. A 4-diode bridge input rectifier rectifies the ac mains voltage a provide a high voltage dc. A transistor chopper switches this on and off at high frequency. This stage is required because this high frequency allows the transformer to be smaller, lighter and much lower in cost.

A small transformer steps this down to a low voltage high-frequency ac. An output rectifier and filter convert this to low-voltage (5 V) dc and smooths out the ripple. A chopper controller provides a feedback to the chopper through an optoisohitor and adjusts the chopping cycle to maintain the output voltage at 5 V.

Question 2.
Why is a resistance connected in series with a Zener diode when used in a circuit?
Answer:
The I-V characteristics in the breakdown region of a Zener diode is almost vertical. That is, the current I_Z can rapidly increase at constant V_Z. To prevent damage due to excessive heating, the Zener current should not exceed the rated maximum current, I_ZM. Hence, a current-limiting resistor R_s is connected in series with the diode.

I_Z and the power dissipated in the Zener diode will be large for I _L = 0 (no-load condition) or when I_L is less than the rated maximum (when R_s is small and R_L is large). The current-limiting resistor R_s is so chosen that the Zener current does not exceed the rated maximum reverse current, I_ZM when there is no load or when the load is very high.
The rated maximum power of a Zener diode is
P_ZM = I_ZM = V_Z

At n-load condition, the current through R is I = I_ZM and the voltage drop across it is V – V_Z, where V is the unregulated source voltage. The diode current will be maximum when V is maxi mum at V_max and I = I_ZM. Then, the minimum value of the series resistance should be
R_{s, min} = \(\frac{V_{\max }-V_{\mathrm{Z}}}{I_{\mathrm{ZM}}}\)

Question 3.
The voltage across a Zener diode does not remain strictly constant with the changes in the Zener current. This is due to R_Z, the Zener impedance, or the internal resistance of the Zener diode. R_Z acts like a small resistance in series with the Zener. Changes in I_Z cause small changes in V_Z .
Answer:
The I-V characteristics of a Zener diode in the breakdown region is not strictly vertical. Its slope is 1/R_Z, where R_Z is the Zener impedance.

Can you know (Textbook Page No. 354)

Question 1.
What is the difference between a photo diode and a solar cell?
Answer:
Both are semiconductor photovoltaic devices. A photodiode is a reverse-biased pn-junction diode while a solar cell is an unbiased pn-junction diode. Photod iodes, however, are optimized for light detection while solar cells are optimized for energy conversion efficiency.

Question 2.
When the intensity of light incident on a photo diode increases, how is the reverse current affected?
Answer:
The photocurrent increases linearly with increasing illuminance, limited by the power dissipation of the photodiode.

Do you know (Textbook Page No. 355)

Question 1.
LED junction does not actually emit that much light so the epoxy resin body is constructed in such a way that the photons emitted by the junction are reflected away from the surrounding substrate base to which the diode is attached and are focused upwards through the domed top of the LED, which itself acts like a lens concentrating the light. This is why the emitted light appears to be brightest at the top of the LED.
Answer:
The pn-junction of an LED is encased in a transparent, hard plastic (epoxy resin), not only for shock protection but also for enhancing the brightness in one direction. Light emitted by the pn-junction is not directional. The hemispherical epoxy lens focuses the light in the direction of the hemispherical part. This is why the emitted light appears to be brightest at the top of the LED.

Question 2.
The current rating of LED is of a few tens of milli-amps. Hence it is necessary to connect a high resistance in series with it. The forward voltage drop of an LED is much larger than an ordinary diode and is around 1.5 to 3.5 volts.
Answer:
Most common LEDs require a forward operating voltage of between approximately 1.2 V (for a standard red LED) to 3.6 V (for a blue LED) with a forward current rating of about 10 mA to 30 mA, with 12 mA to 20 mA being the most common range. Like any diode, the forward current is approximately an exponential function of voltage and the forward resistance is very small. A small voltage change may result in a large change in current. If the current exceeds the rated maximum, an LED may overheat and get destroyed. LEDs are current driven devices and a current-limiting series resistor is required to prevent burning up the LED.

Do you know (Textbook Page No. 356)

Question 1.
White Light LEDs or White LED Lamps:
Shuji Nakamura, a Japanese – born American electronic engineer invented the blue LED. He was awarded the Nobel prize for physics for 2014. He was also awarded the global energy prize in the year 2015. His invention of blue LED made the fabrication of white LED possible.
LED lamps, bulbs, street lighting are becoming very popular these days because of the very high efficiency of LEDs in terms of light output per unit input power(in milliWatts), as compared to the incandescent bulbs. So for general purpose lightings, white light is preferred.
Commercially available white LEDs are normally manufactured by using the technique of wavelength conversion. It is a process which partly or completely converts the radiation of a LED into white light. There are many ways of wavelength conversion. One of these methods uses blue LED and yellow phosphor. In this method of wavelength conversion, a LED which emits blue colour is used to excite a yellow colour phosphor. This results in the emission of yellow and blue light and this mixture of blue and yellow light gives the appearance of white light. This method is the least expensive method for producing white light.
Answer:

The all important blue LEDs
The development of LEDs has made more efficient light sources possible. Creating white light that can be used for lighting requires a combination of red, green and blue light. Blue LEDs proved to be much more difficult to create than red and green LEDs. During the 1980s and 1990s Isamu Akasaki, Hiroshi Amano, and Shuji Nakamura successfully used the difficult-to-handle semiconductor gallium nitride to create efficient blue LEDs. Isamu Akasaki is known for invent ing the bright gallium nitride (CaN) pn-junction blue LED in 1989 and subsequently the high-brightness CaN blue LED.

Using blue LEDs, highly efficient white light sources. became possible by converting part of the blue light emitted from an LED to yellow using a phosphor. To the human eye, the combination of blue and yellow light is perceived as white. A white LED can be created by embedding phosphors in the plastic cap which surrounds a blue LED. Higher quality white light can also be created by mixing blue light with other colors as well, including red and green

Isamu Akasaki, together with Shuji Nakamura and Hiroshi Amano, received the 2014 Nobel Prize in Physics for the invention of efficient blue light-emitting diodes which has enabled bright and energy saving white light sources.

Use your brain power (Textbook Page No. 357)

Question 1.
What would happen if both junctions of a BJT are forward biased or reverse biased?
Answer:
A BJT has four regimes of operation, depending on the four combinations of the applied biases (voltage polarities) to the emitter-base junction and the collector-base junction, as shown in the following table; ‘F’ and ‘R’ indicate forward bias and reverse bias, respectively.

Remember This (Textbook Page No. 358)

Question 1.
The lightly doped, thin base region sandwiched between the heavily doped emitter region and the intermediate doped collector region plays a crucial role in the transistor action.
Answer:
If the two junctions in a BJT are physically close compared with the minority carrier diffusion length (i.e., the distance within which recombination will take place), the careers injected from the emitter can diffuse through the base to reach the base-collector junction. The narrow width of the base is thus crucial for transistor action.

Use your brain power (Textbook Page No. 361)

Question 1.
If a transistor amplifies power, explain why it is not used to generate power.
The term ‘amplification’ is used as an abstraction of the transistor properties so that we have few equations which are useful for a large number of practical problems. Transistors use a small power to control a power supply which can output a huge power. The large output comes from the power supply, while the input signal valves the transistor on and off. The increased power comes from the power supply so that a transistor does not violate the law of conservation of energy.

Balbharti Maharashtra State Board Class 11 Sociology Solutions Chapter 8 Social Change Textbook Exercise Questions and Answers.

Maharashtra State Board Class 11 Sociology Solutions Chapter 8 Social Change

1. (A) Choose the correct alternative and complete the statements.

Question 1.
Social change as a term is ………………
(value loaded / ethically neutral / prejudiced)
Answer:
Ethically neutral

Question 2.
The effects of an earthquake on people is a ………………. factor of change.
(geographical / biological / cultural)
Answer:
geographical

Question 3.
The study of sex ratio is a ………………. factor of change.
(biological / technological / natural)
Answer:
biological

Question 4.
The slum rehabilitation programme within a city is an example of ………………… social change.
(planned / unplanned / revolutionary)
Ans.
planned

1. (B) Correct the incorrect pair.

Question 1.
(a) Earthquake – Biological
(b) Fundamentalism – Economic
(c) Growing Urbanization – Technological
(d) E-governance – Physical
Answer:
(b) Fundamentalism – Socio – cultural factor

1. (C) Identify the appropriate term from the given options.

(Physical Factor, Educational Factor, Economic Factor)
Question 1.
Impact of rising sea water level on coastal regions.
Answer:
Physical factor

Question 2.
Creating awareness about the problem of sexual abuse.
Answer:
Educational factor

1. (D) Correct the underlined words and complete the sentence.

Question 1.
Social change is a linear process.
Answer:
Social change is a continuous process.

Question 2.
All teachers are expected to think about how they will teach a unit in the classroom. This is an example of unplanned change.
Answer:
All teachers are expected to think about how they will teach a unit in the classroom. This is an example of planned change.

2. Differentiate between.

Question 1.
Planned change and unplanned change.
Answer:

Planned change	Unplanned change
(i) Planned change occurs when purposeful changes are promoted by the government or other agencies.	(i) Unplanned change is a type of changes that is not planned. It happens suddenly.
(ii) In the case of planned cities in India, they have definite spaces marked for residence, parks, grounds, places of worship so on; the five years plans, educational plans, tribal welfare programmes, etc.	(ii) In the case of natural disaster, there is a loss of human and animal lives as well as property. Rehabilitation programmes have to be immediately designed and implemented for the affected persons.
(iii) Planned change occurs when deliberate decisions are taken to bring change.	(iii) Unplanned change is a result of unforeseen occurrences.
(iv) Planned social change is based on directions and goals.	(iv) Unplanned social change occurs without any directions or goals.

Question 2.
Short-term change and Long-term change.
Answer:

Short-term change	Long-term change
(i) Some social changes which may bring about immediate results are known as short-term change	(i) Some social change which may take years or decades to produce results are known as long-term change.
(ii) The purchase of new gadgets like home theatre for the purpose of entertainment within the home is rapid.	(ii) Giving up social evils like dowry, early marriage or domestic violence take decades to get rid of.
(iii) Short term change is change in material culture.	(iii) Long term change is change in non¬material culture.
(iv) Technological changes such as inventions and discovery play important role in bringing short term changes.	(iv) Social movements and revolutions play important role in bringing long term changes.

3. Explain the following concept with an example.

Question 1.
Social change is interactional chain reaction
Answer:
1. A single factor may trigger a particular change, but it is almost associated with other factors like physical, biological, technological, cultural, social, economic, which may together bring about a social change.

2. This is due to mutual interdependence of social phenomenon.
Example : A huge increase in school fees will have an impact on student enrolment. It may further result in higher dropouts especially for the girl child from the system of school. Increase in school fees is an economic factor which may give rise to social factor like problems of girls dropout.

Technological factor of social change:

1. Today, as we live in a digitalized world, we have been increasingly loaded with technology from our homes to our workplace.
2. Technological changes have affected our social, economic, religious, political, and cultural life.
3. Technological development creates new conditions of life and new conditions for adaptation. It continues to be an index of the overall progress of society.

Example : During the British period in India, systems of transportation and communication were laid. These may have served the needs of colonizers then, but we still continue to benefit from the systems.

Dysfunctional of social system:

1. The social system may become dysfunctional at times.
2. Hence, human beings have to make conscious efforts to help bring stability, balance and equilibrium in society.

Example : Emile Durkheim makes reference to anomic suicide where there is a state of normlessness or chaos, which can trigger off suicidal feelings that makes the social system dysfunctional.

Change in performance of social roles of individuals is also social change:

1. The social system comprises of social institutions like education, government, economy, etc., they regulate human contact, allocate roles and provide resources.
2. Social change also refers to change in performance of social roles of individuals according to changing times.

Example : In today’s Information Age, the role of a teacher in school is radically different than it was during the early Vedic period. There was marked differences in terms of the size of the school, learners, content of education, educational philosophy, methods of teaching and evaluation, etc.

Question 2.
Long term change

4. (A) Complete the concept maps.

Identify the significant factor of change for each.
Question 1.

inventions	——–
Effects of earthquake	——–
Declining sex ratio	——–
Student exchange programme	——-
Cultural diffusion	——–
Materialism	——–

Answer:

inventions	Technological factor
Effects of earthquake	Physical factor
Declining sex ratio	Biological factor
Student exchange programme	Educational factor
Cultural diffusion	Socio – cultural factor
Materialism	Economic factor

4. (B) State whether the following statements are true or false with reasons.

Question 1.
Prejudice and fear of the unknown is an obstacle to change.
Answer:
This statement is True.

1. Sometimes people are not open to change as they are too comfortable within their life.
2. Sometimes people don’t perceive the need to change prejudice or attitude towards a change also becomes obstacles.
3. Fear of unknown leads people to avoid difference.
   Hence, prejudice and fear of unknown is an obstacle to social change.

Question 2.
Social changes can be predicted accurately.
Answer:
This statement is False.

1. The concept of social change involves a transition in society from one state to another through time. The change depends upon complex factors. Hence social change cannot be predicted accurately.
2. Social change is not instant; it takes place over time. There is no inherent law of social change.
3. The forces of social change may not remain the same and the process of social change does not remain uniform.

5. Give your personal response.

Question 1.
Do you think people do not accept change easily? Why?
Answer:
Yes, I think people do not accept change easily. Customs and traditions which are embedded in society do not allow people to accept new ideas and acts as an obstacle to social change. Sometimes lack of motivation or interest also causes hindrance to social change. Even though social change is universal, there are more often some quarters of resistance to change.

Question 2.
Do you think the Swachh Bharat Abhiyan has had a positive impact on society? Justify your response.
Answer:
The physical environment has also been adversely affected by human behaviour in the name of development. In this era of global warming and climate change, everyone is striving towards a clean and safe India. The campaign of clean India, i.e., the Swachh Bharat Abhiyan is the biggest step taken over as a cleanliness drive and has a huge possible impact on society.

11th Sociology Digest Chapter 8 Social Change Intext Questions and Answers

ACTIVITY (Textbook Page No. 83)

Question 1.
Do a Google search for ‘Punk Hairstyle’.
Answer:
Relate ‘Punk Hairstyle’ to cultural change in society. The inspiration for the hairstyle came from the punk rock music in the 70’s. People have long been in the practice of using hair dyes to change the colour of their hair as a means of making themselves more attractive. Punks use hair dyes to make themselves appear different from mainstream society. One of the most common punk hairstyles is the Mohawk and use of bright colours on the hair.

Question 2.
You have learnt about the physical factor of social change. Now, write one page about how the natural calamities affect the life of people and society by giving some suitable examples. (Textbook Page No. 86)
Answer:
Natural disaster in India, cause massive losses of life and property. Droughts, cyclones, landslides pose greatest threat. Landslides are common in the lower Himalayas. Parts of Western Ghats also suffer from low intensity landslides. Floods are the most common natural disaster in India. The heavy southwest monsoon rain causes the Brahmaputra and other rivers to over-cross their banks, often flooding the surrounding areas. The floods kill and displace many. Temperatures in three Indian cities of Chennai, Mumbai and Delhi in the last five decades have seen a steady rise. This rise in temperature has led to a higher incidence of natural disaster storms, floods and drought, which have increased. The cost of damages has gone up. The latest cyclone Vayu in Gujrat have led to widespread devastation along parts of the eastern coast of India.

Question 3.
Try to understand the meaning of globalisation and observe changes brought about by globalisation in the world around you. (Textbook Page No. 88)
Answer:
Globalization is a process of integrating a country’s economy with the world economy with a view to exploit global opportunities for local growth. Globalization has resulted in both advantage and disadvantage for the Indian society. On one hand it has promoted the process of industrialization but on the other small-scale industries are the worst affected by the entry of large-scale multinational companies. Though globalization has increased the export of Indian industrial and agricultural products, there are lot of hindrance in path of export.

Globalization has led to new and better employment opportunities but there has been also a negative impact of globalization on the employment situation in India, since it has to shift many of its workers from the organized sector to the unorganized sector of Indian economy. It has promoted international travel and tourism leading to cultural exchange.

Question 4.
You have understood the importance of technological factor of social change. Now, try to collect the data from ten families in your neighbourhood, about the use of modern technology in their day-to-day life. (Textbook Page No. 89)
Answer:
Students should attempt this question on their own.

Question 5.
Study the educational transformation in the last 10 years e.g., Teach for India campaign (Textbook Page No. 90)
Answer:
Teach for India campaign have transformed the lives of children in low-income communities.
They have re-imagined education by being holistic and differentiated such that every single child learns and grows to his or her fullest potential. 37,920 children have learned across Teach for India classrooms; They are committed to a singular goal – an excellent education for all children. Teach for India is striving to end the problem of educational inequality in India.

Question 6.
Trace changes in fashions and eating habits followed by teenagers in the past decade. Make a pictorial album or photo essay to show the changes. (Textbook Page No. 91)
Answer:
Students should attempt this question of their own.

Balbharti Maharashtra State Board Class 11 Sociology Solutions Chapter 7 Social Stratification Textbook Exercise Questions and Answers.

Maharashtra State Board Class 11 Sociology Solutions Chapter 7 Social Stratification

1. (A) Choose the correct alternative and complete the statements.

Question 1.
Social stratification is ……………….
(local / national / universal)
Answer:
universal

Question 2.
Class is a ………………. form of stratification.
(open / closed / rigid)
Answer:
open

Question 3.
Gender based stratification has led to ………………. in society.
(justice / exploitation / equality)
Answer:
exploitation

Question 4.
Social stratification of ………………. is based on the principle of purity and pollution.
(class / gender / caste)
Answer:
caste

1. (B) Correct the incorrect pair.

Question 1.
(a) Ownership of wealth – Economic Capital
(b) Membership and involvement in social network – Social Capital
(c) Gained through education – Cultural Capital
(d) Prestige, status and social honour – Economic Capital
Answer:
(d) Prestige, status, social honour – Symbolic Capital

1. (C) Correct underlined words and complete the sentence.

Question 1.
Caste is based on wealth.
Answer:
Class is based on wealth.

Question 2.
A hierarchical system where women are given a lower social status is stratification based on class.
Answer:
A hierarchical system where women are given a lower social status is stratification based on gender.

2. Write short notes.

Question 1.
Principles of social stratification.
Answer:

1. Stratification is social: Social stratification is not determined by biological differences but it is governed by social norms and sanctions.
2. Social stratification persists over generations : In all society’s parents confer their social status on their children. Thus, the pattern of inequality stays same from generation to generation.
3. Social stratification is universal but variable : Social stratification is found everywhere. At the same time the nature of inequality varies. ‘What’ is unequal and ‘how’ unequal, changes within the context of a society.
4. Social stratification involves inequality : Any stratified system not only gives people more resources but also justifies this arrangement and defines them as fair.
5. Social stratification is consequential : Stratification affects every aspect of life of all individuals. Social life is affected because of the position of an individual in the social hierarchy. Some experience positive consequences, while others face negative consequence of the hierarchy in a particular society.

Question 2.
Characteristics of caste according to Dr. G.S. Ghurye.
Answer:
Dr. G.S. Ghurye a well known Sociologist and Indologist defines caste in terms of its essential characteristics. They are as follows:
1.Segmental division by society : Society is divided into various castes. The membership of castes are determined by birth. Therefore, mobility from one caste to another is impossible.

2. Hierarchy: Castes or segments are arranged in terms of hierarchy. According to Dr. Ghurye, castes are graded and arranged into a hierarchy on the basis of the concept of ‘purity and pollution’.

3. Restriction on feeding and social intercourse : This fact of separation is reinforced by the notion of ‘purity and pollution’. Each caste imposes restrictions on its members with regard to food and social intercourse.

4. Differential civil and religious privileges and disabilities : In a caste society there is an unequal distribution of privileges and disabilities among its members. The higher castes enjoy all privileges and lower caste suffer from all kinds of disabilities.

5. Lack of unrestricted choice of occupation : Choice of occupation is not free under caste system. Occupations are hereditary and the members of the caste are expected to follow their traditional occupation.

6. Endogamy : Endogamy is the essence of caste system. Every caste of sub-caste insists that its member should marry within the group.

Question 3.
Types of mobility.
Answer:
1. Horizontal Mobility : It refers to change of residence or job without status change. Under this type of social mobility, a person changes one’s occupation but the overall social standing remains the same. Certain occupation like doctor, engineer and teacher may enjoy the same status but when an engineer changes one’s occupation from engineer to teaching engineering there is a horizontal shift from one occupational category to another but no change has taken place in the system of social stratification.

2. Vertical Mobility : Vertical mobility refers to any change in the occupational, economic, political status of an individual or a group which leads to change of their position. Vertical Mobility stands for change of social position, either upward or downward.

3. Intergenerational Mobility : This type of mobility means that one generation changes its social status in contrast to the previous generation. However, this mobility may be upward or downward. For e.g., people of lower caste or class may provide facilities to their children to get higher education, training and skills, with the help of which the younger generation may get employment in higher position.

4. Intragenerational Mobility : This type of mobility takes place in the lifespan of one generation. A person may start one’s career as a clerk and after acquiring more education, becomes an IFS Officer. Here the individual moves up and occupies a higher social position than previously.

3. Differentiate between.

Question 1.
Caste and Class.
Answer:

Caste	Class
(i) Different castes form a hierarchy of social preference and each position in the caste structure is defined in terms of its ‘purity and pollution’.	(i) A social class is made up of similar social status who regard one another as social equals.
(ii) In a caste stratification system, an individual’s position depends on the status attributes ascribed by birth.	(ii) In a class stratification system and individual’s position depends on the possession of substantial amounts of wealth, occupation, education and prestige which is achieved.
(iii) Caste is an example of closed stratification.	(iii) Class system is an example of open stratification.
(iv) In this type of social stratification there is no scope for social mobility.	(iv) In this type of social stratification there is scope for social mobility.

Question 2.
Intragenerational Mobility and Intergenerational Mobility.
Answer:

Intragenerational Mobility	Intergenerational Mobility
(i) This type of mobility takes place in the lifespan of one generation.	(i) This type of mobility means that one generation changes its social status in contrast to the previous generation.
(ii) This mobility is upward.	(ii) The mobility may be upward or downward.
(iii) A person may start one’s own career as a clerk. He / she acquires more education and over a period of time becomes an IFS Officer. Here the individual moves up and occupies a higher social position than previously.	(iii) People of lower caste or class may provide facilities to their children to get higher education, training and skills. With the help of these skills the younger generation may get employment in higher position.
(iv) It refers to advancement in one’s social level during the course of one’s lifetime.	(iv) It refers to a change in the status of family members, one generation to the next.

4. Explain the following concept with suitable examples.

Question 1.
Vertical Mobility
Answer:

1. Vertical mobility refers to any change in the occupational economic or political status of an individual or a group which leads to change of their position.
2. Vertical mobility stands for change of social position either upward or downward, which can be labelled as ascending or descending type of mobility.

Example : A person who works as a customer assistant, works hard and starts his own business successfully. In such a position there is a clear change in the position of the individual.

Question 2.
Intergenerational Mobility
Answer:

1. This type of mobility means that one generation changes its social status in contrast to the previous generation.
2. However, this mobility may be upward or downward.

Example : People of lower caste or class may provide facilities to their children to get higher education, training and skills, with the help of which the younger generation may get employment in higher position.

5. (A) Complete the concept maps.

Question 1.

Answer:

5. (B) State whether the following statements are true or false with reasons.

Question 1.
There is no mobility in the class system.
Answer:
This statement is False.

1. Class system is an example of open stratification in which individuals or groups enjoy the freedom of changing their social strata, i.e., in class system there is scope for social mobility. Individuals or groups move from one strata to another.
2. The class system in modern industrial society (Upper class, middle class and lower class) is an example of open stratification.
3. The criteria of open stratification i.e., class system are power, property, intelligence, skills, etc.

Question 2.
Education had led to women’s empowerment.
Answer:
This statement is True.

1. Education is a milestone of women empowerment because it enables them to respond to challenges, to confront their traditional role and change their life.
2. Education creates occupational achievement, self-awareness, satisfaction etc.
3. Education is one of the main levers of social class which has helped women empower and change their status in society.

6. Answer the following in detail (About 150-200 words).

Question 1.
Discuss class and gender as forms of social stratification with suitable examples of your own.
Answer:
Class as a form of social stratification:
A social class is made up of people of similar social status who regard one another as social equals.
Each class has a set of values, attitudes, beliefs and behaviour norms which differ from those of the other classes. A social class is essentially a status group which is achieved. Class is almost a universal phenomenon. Each social class has its own status in the society. Status is associated with prestige. A social class is relatively a stable group. Social class represent an open social system. An open class system in one in which vertical social mobility is possible.

Example : Within this system, individuals can move from one class to another through hard work, education and skills. Ownership of wealth and occupation are the chief criteria of class differences but education, hereditary, prestige, group participation, self identification and recognition by others, also play an important role in class distinction.

Gender as a form of social stratification:
Gender stratification refers to social ranking, where men typically inhabit higher statuses than women. A common general definition of gender stratification refers to the unequal distribution of wealth, power and privilege between the two sexes. Throughout the world, most societies allocate fewer resources to women than men. Almost all societies are characterized by sexism. Sexism is the belief that one sex is superior than the other. Although, societies have been believing in the superiority of men over women and therefore have been dominating women. This male dominance is supported further by patriarchy. The process of socialization is gendered and creates gender hierarchy. Example : Boys are given toy cars or lego sets or bat and ball to play whereas girls are given household sets, medical sets, dolls, etc.

11th Sociology Digest Chapter 7 Social Stratification Intext Questions and Answers

ACTIVITY (Textbook Page No. 75)

Question 1.
Watch the Marathi movie, ‘Fandry’ and write a film review describing the social, cultural and economic obstacles created by caste barriers.
Answer:
Review of the Marathi Movie ‘Fandry’. The film powerfully busts the myth of individual merit in a caste-decided society. In a small village in Maharashtra Jabya portrays friend Pirya are the only two boys from a so-called untouchable caste. Jabya doesn’t want to consider his caste an obstacle to his aspirations. These hopes of wanting to move out of the confines of his caste are shown through Jabya’s love for his classmate Shalu, an upper caste by birth. Jabya and Pirya, meanwhile want to hunt down the exclusive black sparrow which Jabya believes would help him to win Shalu’s love. His father Kachru wants him to continue their tradition. From being called blacky to being made to feel ashamed of his mother’s occupation. When she comes to school Jabya’s trials indicate the prejudices that make the promise of equality sound like unreal.

In theory, Jabya’s school is supposed to uplift him to a modern and caste-less society where he should be able to choose the work he wants to do. Yet we see how modern education itself is not free from caste. In caste system social set up everything is pre-decided by one’s caste, whom one can love and be friends with, the occupation he has to choose etc. Fandry makes visible the way in which caste is so central to all our relatives.

Question 2.
In today’s world the characteristics of caste are changing. Find out which of the characteristics are changing and which are remaining constant. Conduct a group discussion on the same. (Textbook Page No. 75)
Answer:
In the modern age, many changes happen in the features and functions of caste system. A group discussion can be conducted on the following changes within the caste system.

1. Decline in the superiority of upper caste.
2. Changes in the restrictions regarding social habits.
3. Changes in the restrictions regarding marriage.
4. Changes in the restrictions regarding occupation.
5. Changes in the disabilities of lower castes.
6. Loss of faith in the ascribed status.
7. Changes in lifestyle.
8. Changes in inter-caste relations.
9. Changes in the lower of caste Panchayats.
10. Restrictions on education removed
11. Changes in the philosophical basis.

Question 3.
Divide the class into groups. Each group can select one of the issue mentioned and collect information on it. The group should present their findings to the class. (Textbook Page No. 78)
Answer:
Present findings on any one of the issues to the class.
1. The Economy : Explains how women are being paid low for some amount of work done by men in various unorganized sectors. Also, dual role played by women and unpaid work.

2. The Polity : Explains about women exercising the power of right to vote, in spite of reservation for women, the number of women in official positions of power are less as world leaders, less number of women at war and peace movements.

3. Crime : Explains the crime committed by women, increase in number of women coming in conflict with the law; women prisons in India are relatively less crowded, women commit fewer and different crimes compared to men.

4. Religion : Most religions elevate the status of men over women and have striker sanctions against women and require them to be submissive.

5. Family : In spite of women sharing the economic role, the role of men in raising children is still minimum or negligible. Traditional sexual division of labour where women looked after the house and men played the role of economic provider is still prevalent in the society. Women are expected to balance between home and work.

6. Health : Women neglect their health and nutrition. The frequency of women to visit a doctor is very less as most of the time they manage with home remedies.

Question 4.
Find out examples of intergenerational and intragenerational mobility from your surroundings and present it in your classroom.
Answer:
1. Intergenerational mobility means one generation changes its social status in contrast to the previous generation.
Example : Eminent personality like Dr. B. R. Ambedkar.

2. Intragenerational mobility this type of mobility takes place in the life span of one generation. Example : A person may start one’s career as a clerk and after acquiring more education over a period of time he becomes an IAS officer. Students should find out similar examples of intergenerational and intragenerational mobility from their surroundings.

Question 5.
Arrange the group reading of any two of the following books and conduct a group discussion on the caste and gender discrimination/inequality Baburao Bagul-Jevha Mi Jaat chorli Hoti, Daya Pawar- Baluta, Urmila Pawar- Aaydaan, Omprakash Valmiki- Jhootan, Kishor Shantabai Kale- Against all Odds. (Textbook Page No. 81)
Answer:
Baburao Bagul – ‘Jevha Mi Jaat Chorli Hoti’: This most poignant story recites about an educated Dalit trying to escape his caste profession of scavenging, is an ethnography of caste oppression, description of gender roles shaped by caste, the way Dalit women are oppressed, critique of the political economy of a caste society.

Daya Pawar – ‘Baluta’ : It generalizes the status of rural untouchables. Baluta is a collection of memories of life trapped within the framework of India’s caste system. The frustration and helplessness of being born as a Dalit and the inner conflict in the writer’s mind. He thinks of education as a means to escape from his downtrodden life but ends up being the agent of his lifelong distress.

Urmila Pawar – ‘Aaydaan’: The lives of different members of the family are woven together in a narrative that gradually reveals different aspects of the everyday life of Dalits the manifold ways in which caste assets itself and grinds them down.

Omprakash Valmiki – ‘Jhootan’ : An autobiography by Omprakash Valmiki in which he has explored the issues of Dalits. Being socially segregated for centuries the Dalits are obliged to live a helpless life.

Kishor Shantabai Kale – ‘Against all Odds: The book raises many questions about the exploitation life of women in Kolhati community.

Balbharti Maharashtra State Board Class 11 Sociology Solutions Chapter 6 Socialization Textbook Exercise Questions and Answers.

Maharashtra State Board Class 11 Sociology Solutions Chapter 6 Socialization

1. (A) Choose the correct alternative and complete the statements.

Question 1.
The process whereby an individual learns to conform to the norms of society is called …………………..
(assimilation / socialization / co-operation)
Answer:
socialization

Question 2.
Family is a ………………….. agency of socialization.
(primary / secondary / tertiary)
Answer:
primary

Question 3.
School is an ………………….. agency of socialization.
(primary / secondary / tertiary)
Answer:
secondary

Question 4.
Television is a / an ………………….. medium of communication.
(audio / visual / audio visual)
Answer:
audio-visual

1. (B) Correct the incorrect pair.

Question 1.
(a) Language, behaviour – Family
(b) Social values like friendship – Peer Group
(c) Teamwork, discipline – Neighbourhood
(d) To build opinion – Mass media
Answer:
(c) Team work, discipline – Workplace

1. (C) Identify the appropriate term from the given options.

(Internet, Peer Group, Childhood, Socialization)
Question 1.
Takes place in the early years of life.
Answer:
Socialization

Question 2.
Global impact in today’s world.
Answer:
Internet

1. (D) Correct the underlined words and complete the sentence.

Question 1.
Radio is an audiovisual medium.
Answer:
Radio is an audio medium.

Question 2.
Peer group is an example of an authoritarian agency.
Answer:
Family is an example of an authoritarian agency.

2. Write short notes.

Question 1.
Formation of ‘self ’ according to Mead.
Answer:
George Mead has elaborated on the process of building social self which does not exist at birth. According to Mead, formation of self occurs in three distinct stages.
Stage 1 – Imitation : In this stage, children imitate behaviour of adults without understanding it. Example : A little boy might drive his mother to her office by driving his toy car or help his parents clean the floor by pushing a broom.

Stage 2 – Play stage : A child plays, sometimes as being a mother or a teacher, at times a postal worker, a police officer etc. In this stage, responses are not organized. A child internalises the attitudes of others who are significant to her/his through enacting the roles of others. A significant other is someone whose opinions matter to us and who is in a position to influence our thinking.

Stage 3 – Game stage : As a child matures, and as the self gradually develops, one internalises the expectations of a large number of people. Children learn to behave according to the impressions of others. They understand that role play in each situation involves following a consistent set of rules and expectations. For example, a child at this stage is likely to be aware of the different responsibilities of people in a restaurant who together, make for a smooth dining experience. Thus, the self is mainly formed through our interactions with others and our understanding of others responses. Socialization, in this sense is a process of self-awareness.

Question 2.
Agencies of socialization.
Answer:
There are different social groups which can be seen as agencies of socialization.
1. Family : Family is the main agent of socialization. The child learns language and other basic behavioural patterns in family. Socialization through family is varied because there is no single, uniform pattern to do so. A child brought up in nuclear family will undergo different pattern of socialization. Patterns of child rearing vary across families with different caste, class, and ethnic backgrounds.

2. Peer groups : Peer groups are friendship groups made up of people of similar age. In peer groups, the interactions are reasonably egalitarian as there is a greater amount of give and take, when compared to family or school. Peer groups use informal sanctions including positive sanctions like approving gestures or laughing at your jokes, and negative sanctions like disapproving jokes, labelling or rejecting your company.

3. Schools : Schooling and education are considered as secondary agencies. School involves learning values and norms at a step higher than those learnt in a family. Skills and values like team work, discipline, conformity to authority are learnt in schools and this helps prepare students for the adult world.

4. Mass Media : One of the significant forces of socialization in modern culture is mass media. Mass media are the means for delivering impersonal communication directed to a vast audience. Mass media includes traditional print media like newspapers and magazines, electronic media like radio and television and current IT enabled media and social media. Television has an influence on children from a very young age and affects their cognitive and social development. Modern technological advancements have strengthened and changed the role of mass media. Technology has certainly increased the spread of mass media.

5. Neighbourhood : A neighbourhood community is an important agency of socialization. A neighbourhood is a geographically localized community within a larger city, town or suburb. Neighbourhoods are formed through considerable face to face interaction among members often living near one another. A neighbourhood community provides the base for an individual to extend social relations and interactions beyond the narrow limits of the home.

6. Workplace : Socialization is a life long process. Adult socialization indicates this continuous process of learning. One of the significant agents of adult socialization is the workplace.

Adult individuals spend significant amount of time at the workplace. Socialization through work place involves acquiring new skills, knowledge and behaviour patterns suitable to the requirements of the job.

Question 3.
Resocialization.
Answer:
The process of unlearning old norms, roles, values and behavioural patterns and learning new patterns is called re-socialization. Sometimes an individual is caught in a situation where one has to break away from past experience and internalise different norms and values. Re-socialization can also be defined as a process which subjects an individual to new values, attitudes and skills according to the norms of a particular institution and the person has to completely re-engineer one’s sense of social values and norms.

The person may be in a jail, hospital, in religious organization, police, army etc. In such institutions there is total break up from the normal social life outside. A prison sentence is a good example. The individual not only has to change and rehabilitate one’s behaviour in order to return to society but must also accommodate the new norms required for living, while in prison.

3. Explain the following concept with an example.

Question 1.
Primary socialization
Answer:

1. The most critical process of socialization happens in the early years.
2. This learning in the early years is termed as primary socialization.
3. Primary socialization takes place in infancy and childhood and involves intense cultural learning.
4. A child gets acquainted with values, customs, behavioural norms and manners. It is an informal process.

Example : Family is the main agent of primary socialization. Peer group and neighbourhood is also seen as a primary socializing agency.

Question 2.
Secondary socialization
Answer:

1. Socialization as a process is lifelong.
2. The learning which extends over the entire life of a person is known as secondary socialization. It is a formal process of socialization.

Example : Schooling and education are considered as secondary agencies of socialization. What we learn through a formal curriculum with specific subjects and skills. Schooling involves learning values and norms at a step higher than those learnt in family.

4. (A) Complete the concept maps.

Question 1.

Answer:

4. (B) State whether the following statements are true or false with reasons.

Question 1.
Socialization is a life-long process.
Answer:
This statement is True.
(i) The process of learning attitudes, norms and behaviour patterns and becoming members of different social groups like family, kin network, peer group and later, formal groups like school, professional networks etc., is a life long process.

(ii) Socialization is an ongoing process of continuous learning The birth of a child is a new experience of parenting for a couple. Similarly, older people become grandparents thus creating another set of relationships connecting different generations with each other.

(iii) Thus, socialization as a learning process is life long even though the most critical process happens in the early years but secondary socialization extends over the entire life of a person.

Question 2.
Advertisements influence consumer behaviour.
Answer:
This statement is True.

1. Mass media has become an integral part of our day to day life. Advertisements through mass media are the means for delivering impersonal communication directed to a vast audience.
2. Advertisements transmit information and messages which influence the behaviour of the consumer to a great extent.
3. The use of colours, words, music, images, videos influence our behaviour and persuades us to take action. Advertisements through mass media has wider approach.

5. Give your personal response.

Question 1.
‘Breaking News’ tends to create panic or emotional responses. Why do you think this happens? Give relevant examples to illustrate.
Answer:
Many newspapers as well as some private news channels very frequently transmit news of murders, accidents, stealing, dacoity, beating, rape, economic cheating, fraud, scams, etc., as breaking news. Constant hearing of such news affects the minds of the people and it weakens the faith in ideals and values of life. This happens because breaking news get much more viewers than normal news.

Question 2.
The use of ‘unacceptable language’ is often picked up by children even if this kind of language is not used within the home. Explain how this might happen.
Answer:
Even though the new born is initiated with this learning process in family it is not the only agency of socialization. School, peer groups, neighbourhood, mass media are different social groups and social contexts which can be seen as agencies of socialization. Children pick up unacceptable language from variety of other sources like television which has strong influence on viewers. The child might hear one of his friends or someone in neighbourhood using slang words or abusing language.

6. Answer the following question in detail (About 150-200 words).

Question 1.
You belong to a generation that has been exposed to internet. Discuss how internet has brought about positive and negative results.
Answer:
Modern technological advancements have strengthened and changed the role of mass media as an agent of socialization. Technology like internet has certainly increased the spread of mass media. People spend most of their time in touch with the world. Internet has enhanced communication and social connection. It has also increased political and civic participations. Social media allow students to learn outside of their class rooms. ‘School in the cloud’ is yet another example of how the internet and social media can help to improve global education.

Internet has helped to transmit information and create awareness about a wide range of issues and events among members of the society. It influences attitudes, values and moulds public opinion and acts as an effective way to change the society. Through the internet we can access online educational courses or training. In fact, any type of information from any part of the world can be accessed through the internet.

There is also negative impact of internet on society as – Youth access the internet and indulge in chatting, emailing, watching restricted site that leads to cyber crimes instead of creating interest in reading and creative activities. Sometimes internet may not give accurate information hence the validity and accuracy of the messages must be considered. Internet reaches the masses in developing countries, but there are many tribal, rural and poor urban people having no access to any kind of information. Communication technologies are expensive and need maintenance. Thus, internet may help to develop knowledge and spread information but it also has adverse effects on the society and have promoted values like individualism and materialism.

11th Sociology Digest Chapter 6 Socialization Intext Questions and Answers

ACTIVITY (Textbook Page No. 68)

Question 1.
Conduct a group discussion on the threatening challenge of online games like ‘Blue Whale’. Try to find answers to issues like why do children even consider participating in such games? Are parents to be blamed? What is the role of Law?
Answer:
Games like ‘Blue Whale’ has the challenges of self-harm. It exploits vulnerable people. It blocks the boundary between virtual and real world. There’s a constant competition, level up, which drive the children to perform their best amongst others.

Most games are addictive become of the challenges involved. Once the children are engrossed in it, there is no coming back and they strive hard to achieve the next level, the next goal. This sense of achievement targets the brain’s reward system and compels the gamer to perform the act again and again.

Are parents to be blamed?
Children are becoming addictive to online games because they are designed to be addictive and not because parents allow them to play too much.

What is the role of Law?
With dangerous online games like ‘Blue Whale’ claiming several innocent lives in the recent past, the supreme court has directed the centre to constitute a panel of experts to block such life-threatening games.

Question 2.
Watch advertisements or messages on T.V. and see how effective mass media is in creating awareness against corruption, drug addiction, smoking or any other relevant social issue. (Textbook Page No. 68)
Answer:
The mass media has potential to create awareness against various issues like corruption, drug addiction, smoking etc., by propagandise simple and focused messages to large audiences repeatedly, overtime, at a low cost. They are able to reach a large heterogeneous population. Media campaigns can help in the reduction of smoking and drug addiction and have shown positive results in number of other relevant social areas. Techniques of mass media can be effectively used to counter corruption as well.

Question 3.
Do you think resocialization requires total institutions? Why? Why not? (Textbook Page No. 70)
Answer:
In the process of resocialization old behaviours are removed because they are of no use. Resocialization is necessary when a person moves to a senior care centre, goes to a boarding school or serves time in jail. I think, resocialization requires ‘total institutions’ in a new environment as they can learn new norms and unlearn existing behaviours. The most common way of resocialization occurs in a total institution where people are isolated from society and made to follow new rules and behaviours. A ship at sea military camps, religious convents, prisons or some cult organizations. They are cut off from a larger society. Members entering an institution have to leave behind their old identify to be socialized.

Question 4.
Collect data from five students regarding their experience with social networking sites (example Facebook, Snapchat, Twitter). Find out about how much time they spend online, what kinds of people they interact with, what topics are usually discussed, the uses and problems of social networking sites. Write a 100-word Report on your findings. (Textbook Page No. 71)
Answer:
With respect to overall media consumption, most students spend hours on social networking sites using mobile phones, tablets, laptops, desktops, etc. This age group restricts watching television and is considered as the largest part of change in the media landscape. Example: More three to eleven years of age group are online than in 2016, with much of this growth coming from increased use of tablets. Unsurprisingly, tablets and other portable, connected devices are also playing an important role.

Uses:
Social networking sites allow users living at distant places within their network to connect to another thus increasing social connection, share ideas, photographs, videos, information and other happenings around the world.

Problem:

1. Untrustworthy Member Data.
2. Users submit inaccurate information on their profile.
3. Leaving social networking is difficult; there are saved accounts, and ways to continue to reconnect to the site, even after an individual uninstall the account.
4. Less time for face to face connections with family members.
5. Being too much online diminishes our skills and can have serious side effects. These side effects are becoming more and more frequent amongst the waves of generations.

Balbharti Maharashtra State Board Class 11 Sociology Solutions Chapter 5 Culture Textbook Exercise Questions and Answers.

Maharashtra State Board Class 11 Sociology Solutions Chapter 5 Culture

1. (A) Choose the correct alternative and complete the statements.

Question 1.
Culture is ……………….
(natural / personal / adaptive)
Answer:
adaptive

Question 2.
Material culture is ……………….
(concrete / abstract / intangible)
Answer:
concrete

Question 3.
Bollywood music is an example of ……………….
(high culture / popular culture / folk culture)
Answer:
popular culture

1. (B) Correct the incorrect pair.

Question 1.
(a) Classical singing of Bhimsen Joshi – High culture
(b) Shakespeare’s literature – Folk culture
(c) Harry Potter books – Popular culture
(d) Religious group – Sub-culture
Answer:
(b) Shakespeare’s literature – High culture

1. (C) Identify the appropriate term from the given options.

(Folk Culture, Material Culture, Popular Culture)
Question 1.
Songs transmitted from one generation to the next.
Answer:
Folk Culture

Question 2.
Use of mobile phones today.
Answer:
Material Culture

1. (D) Correct underlined words and complete the sentence.

Question 1.
Belief in superstition is an example of material culture.
Answer:
Belief in superstition is an example of non-material culture.

Question 2.
E-commerce is an example of popular culture.
Answer:
E-commerce is an example of mass culture.

2. Write short notes.

Question 1.
Characteristics of Culture.
Answer:
The term culture refers to the way of life of a member of various societies or groups. Culture has the following characteristics:

1. Culture is acquired : Culture is learnt by each member through socialization. Cultural learning takes place through experience and symbolic interactions. Culture is propagated through generations.
2. Culture is abstract : Culture exist in the minds or habits of the members in a society. We cannot see culture but can see human behaviour.
3. Culture is shared : Culture is shared by a group of people belonging to the same community. They share same values, beliefs and traditions. These aspects develop a sense of unity.
4. Culture is man-made : Culture is a human product and does nothing on its own.
5. Culture is idealistic : Culture embodies the ideas and norms of a group. It consists of intellectual, artistic and social ideas which are followed by members of the society.
6. Culture is transmitted among the members of the society : The cultural ways are learned by persons from persons and many of them are handed down by one’s elders, parents, teachers and others.

Question 2.
Social Benefits of Culture.
Answer:
Culture has many social benefits:

1. Fundamental benefits : Cultural experiences are opportunities for leisure, entertainment, learning and sharing experiences with others. These benefits are intrinsic to culture. They are what attracts us and the reason why we participate.
2. Improved, learning and valuable skills for the future : In children and youth, participation in culture helps to develop thinking skills and build self-esteem, which enhance educational outcomes.
3. Better health and well-being : Participation in culture contributes and cultural engagement improves both mental and physical health.
4. Social solidarity and cohesion : Culture helps build social capital – the bond that holds communities together. Cultural activities such as festivals, bring people together and build social solidarity. Our diverse cultural heritage develops a feeling of pride and a sense of belonging to a wider community.

3. Differentiate between.

Question 1.
Material Culture and Non-Material Culture.
Answer:

Material Culture	Non-Material Culture
(i) Material culture refers to the physical objects which are man-made.	(i) Non-material culture refers to non¬physical ideas created by human beings.
(ii) Material culture is concrete and tangible in nature.	(ii) Non-material culture is abstract and intangible in nature.
(iii) It consists of manufactured objects like clothing, roads, jewellery, computers, airplanes etc.	(iii) It consists of norms, regulations, values, signs, symbols, knowledge, beliefs, etc.
(iv) The material aspect of culture changes very fast.	(iv) Change in non-material culture is difficult and not readily accepted by society.

Question 2.
Folkways and Mores.
Answer:

Folkways	Mores
(i) Folkways are mildly enforced social expectations.	(i) Mores are strictly held beliefs about behaviours.
(ii) Violation of folkways is not seen as a serious threat to social order.	(ii) Violation of mores is seen as a serious threat to social order.
(iii) Folkways are less deeply rooted in society and change more rapidly.	(iii) Mores are more deeply rooted and change less frequently.
(iv) Folkways are customary, normal and habitual ways of a group, to meet certain needs or solving day to day problems.	(iv) Mores are more serious norms and have serious binding on groups.
(v) The manner of speech, dressing, the time of meals and numerous other practices of daily life are some examples of customary practices to which an individual confirm in their personal habits.	(v) Murder, stealing, lying, incest are examples of social Mores in almost all cultures.

4. Explain the following concept with suitable examples.

Question 1.
Norms
Answer:

1. Norms are rules and behavioural expectations by which a society guides the behaviours of its members.
2. Some norms are prescriptive and some are prescriptive norms Most norms apply universally but some norms are culture specific.
3. Social norms are further divided into folkways and mores. Folkways are mildly enforced social expectations, while mores are strictly held beliefs about behaviours.
   Example : Folkways – the concept of appropriate dress. Mores – Religious doctrines, taboos, customs, laws, etc.

Question 2.
Folk Culture
Answer:
Folk culture refers to the culture of ordinary people particularly those living in pre-industrial societies. It is an authentic culture. It never aspire to be an art but its distinctiveness is accepted and respected.

Example : Parents expect obedience from children, the time of meals, the number of meals per day, the manner of taking meals the manner of speech; dressing; forms of etiquette and numerous other practices of daily life.

5. Complete the concept maps.

Question 1.

Answer:

6. Give your personal response.

Question 1.
Very few people make an effort to learn classical music today.
Answer:
Classical music is not popular among people today because like any other form of music one requires exposure over a period of time to become familiar. While pop music is appreciated by a large number of people with no cultural expertise.

Question 2.
It is not easy to give up superstitious beliefs.
Answer:
Superstitious beliefs are form of non-material culture which are rooted in society for many decades and centuries. Change in this aspect is not readily accepted by the society or certain sections of the society. It is rooted in society for many decades and centuries. Hence, change in these aspects is not easy

11th Sociology Digest Chapter 5 Culture Intext Questions and Answers

ACTIVITY (Textbook Page No. 61)

Question 1.
Presentation : students make groups of 5 in class and present an aspect of culture, (e.g., language, dialect, dress, folklore, dances, music, art, food habits, architecture, literature, tribal life, rural life, urban life) of any state in India.
Answer:
Students should conduct a presentation in the classroom with the help of using power point, charts to explain the various cultural elements of any one state in India.

Question 2.
Culture varies from society to society. Each Society or a group will have different culture. These cultures are sometimes overlapping and sometimes exclusive. Give examples. (Textbook Page No. 51)
Answer:
Compare culture of different states in India, how they are different in their lifestyles, food habits, dressing styles etc. Also explain by giving examples how certain elements of culture overlap or have similarities. Many cultural elements of different states are also exclusive in nature, peculiar to that region to maintain the ethnicity.

Both Gujarat and Maharashtra were created on May 1, 1960. The dialects spoken in each state are also different. In Maharashtra, the majority of the people speak the Marathi language. The same is true for the Gujarati language in Gujarat this is an example of exclusive culture.
Hindi, one of the official language of India, is a common language. This is an example of overlapping culture.

Question 3.
Observe cultural change around you and list examples of cultural lag in society. (Textbook Page No. 52)
Answer:
Make a note of cultural changes around you.
Example:

1. Younger generations have become more independent.
2. Indian culture today allows young men and woman to have more freedom of choice with respect to marriage partner.
3. Impact of internet similarly, list examples of cultural lag in society.

Example of cultural lag
For example, expectant parents can use genetic engineering to select their unborn child’s eye colour or sex. However, many people view this type of genetic engineering as unethical and believe it could lead to unintended social consequences. This an example of cultural lag.

Question 4.
Look at your surrounding and list out the examples of cultural hybridisation in the areas of food, toys, religious practices, festivals, celebrations. (Textbook Page No. 59)
Answer:

1. Burger and pizza with a pinch of Indian spices, Indianisation of Chinese food.
2. Celebration of Valentine’s Day.
3. Hybrid version of Barbie, fusion music, formation of new language after blending different languages etc.

Balbharti Maharashtra State Board Class 11 Sociology Solutions Chapter 4 Social Institutions Textbook Exercise Questions and Answers.

Maharashtra State Board Class 11 Sociology Solutions Chapter 4 Social Institutions

1. (A) Choose the correct alternative and complete the statements.

Question 1.
A rule that prescribes marriage within a group is called ……………….
(exogamy / endogamy / polygamy)
Answer:
endogamy

Question 2.
A family that comprises at least three generation is a ………………… family.
(nuclear / joint / cohabitation)
Answer:
joint

Question 3.
The 10 + 2 + 3 system is part of the ………………. system of education.
(informal / formal / universal)
Answer:
formal

1. (B) Correct the incorrect pair.

Question 1.
(a) School – Distance Education
(b) Peer group – Informal Education
(c) Health care training – Non-formal Education
(d) Family – Informal Education
Answer:
(a) School – Formal education

1. (C) Identify the appropriate term from the given options.

(Homosexual Marriage, Matrilocal, Heterosexual Marriage)
Question 1.
A married couple is expected to live at the wife’s residence.
Answer:
Matrilocal

Question 2.
A form of marriage between persons of the same sex.
Answer:
Homosexual Marriage

1. (D) Correct underlined words and complete the sentence.

Question 1.
X married Y. Later she divorced her spouse and married Z. This is an example of polygamy.
Answer:
X married Y. Later she divorced her spouse and married Z. This is an example of monogamy.

Question 2.
Mass production and factory system are found in agricultural society.
Answer:
Mass production and factory system are found in industrial society.

2. Write short notes.

Question 1.
Types of family based on authority.
Answer:
On the basis of authority within family, two broad forms of family can be identified. Matriarchal Family : In matriarchal family authority rests at the hands of the mother and she is considered the head of the family. In matriarchal family descent is traced through mother’s line, known as matrilineal system. Residence of the marriage is matrilocal i.e., the bridegroom lives with the wife’s family and the name follows the mother’s line. It is matronymic. The Garo and Khasi tribes of North East India and the Nair’s of South India follow the matriarchal system.

Patriarchal Family : In patriarchal family all authority rests at the hands of the male head and the eldest male is regarded as the head of the patriarchal family. It is patrilineal i.e., descent is traced through father’s line. Residence after marriage is patrilocal. The married couples live with the husband’s family, and the name of the father’s side continues to the next generation. It is patronymic. Patriarchal family is the most widespread form of family in India and most parts of the world.

Question 2.
Stages of Economy.
Answer:
Human society has evolved through various stages, depending upon the way it evolved mechanisms to adapt to changing material needs.

With respect to the changing nature of work and economy three broad stage of economy are as follows: Agriculture Revolution : From hunting-gathering and then pastoral phases, human society entered into a new stage with discovery of agriculture. With development of agriculture, food production increased. Complex division of labour, advanced trade and permanent settlement are the result of economy expansion through agricultural technology.

Industrial Revolution : Industrial revolution which happened around the middle of the 18th century, first in England transformed social and economic life like never before. It brought four notable economic changes.

1. New forms of energy : With the pioneering invention of the steam engine in 1765 by James Watt, the use of animal and human muscle energy was significantly reduced.
2. Centralisation of work in factories : Use of machinery gave rise to a new workplace called factory.
3. Mass production : Industrial economy developed the manufacturing sector very fast and produced goods on a large scale.
4. Division of labour : The factory system reduced the importance of human skills and resulted to specialised division of labour.
5. Industrialization transformed the whole society and also created huge economic gaps in the society.

The Information Revolution : By the middle of the 20th century the nature of production started to change. The use of automated machinery drastically reduced the role of human labour in manufacturing. Three key changes were seen due to the development of computer and information technology.

1. Tangible products to idea : The tertiary or service sector providing services and dealing with production of ideas and symbols is expanding very fast.
2. Mechanical skills to literacy skills : The information revolution demands a new set of skills such as literacy skills – the ability to communicate, to write, to present and use computer technology.
3. Work from anywhere : Computer technology is allowing for decentralisation of work.

Question 3.
Importance of Education.
Answer:
Education is broad process of learning that broadens an individual’s cognitive horizons and develops in them the skills to choose, evaluate and add to existing knowledge.
The functions of education can be as follows:
1. Inculcate common values and moral beliefs : Children coming from diverse family and community backgrounds gain an understanding of the common values and moral beliefs in society through school education.

2. Fosters self-discipline : Through education children learn self-discipline. They internalise the social rules that contribute to the smooth function of society.

3. Teaches specialised skills: The education system teaches specialised skills needed for a complex modern economy. For example, technical schools are specially designed to provide technical / vocational training.

4. Instils the value of achievement in children : Schools like wider society, largely operate on a meritocratic basis. Those with ability and talent achieve their just rewards.

3. Differentiate between.

Question 1.
Matriarchal Family and Patriarchal Family.
Answer:

Matriarchal Family	Patriarchal Family
(i) Authority rests at the hands of the mother in a matriarchal family. She is considered the head of the family.	(i) The eldest male is regarded as the head of the patriarchal family. All authority rests at the hands of the male head.
(ii) Matriarchal families follow a matrilineal system of lineage i.e., descent through mother’s line.	(ii) Patriarchal family is also patrilineal i.e., descent is traced through father’s line.
(iii) Residence after marriage is matrilocal i.e., the bridegroom lives with the wife’s family.	(iii) Residence after marriage is patrilocal. The married couple lives with the husband’s family.
(iv) The name follows the mother’s line. It is matronymic.	(iv) The name of the father’s side continues to the next generation. It is patronymic.
(v) The Garo and Khasi tribes of North East India and the Nair’s of South India follow the matriarchal system.	(v) The most widespread form of family in India and most parts of the world is patriarchal family.

Question 2.
Agriculture Revolution and Industrial Revolution.
Answer:

Agriculture Revolution	Industrial Revolution
(i) From hunting – gathering and then pastoral phases, human society entered into a new stage with discovery of agriculture.	(i) Industrial revolution happened around the middle of the eighteenth century and transformed social and economic life.
(ii) Agricultures involved using technology of large -scale farming using ploughs harnessed to animals.	(ii) Industrialisation involved using of machinery and new forms of energy.
(iii) This increased the productive power of hunting and gathering more than tenfold.	(iii) This increased more mass production which turned raw materials into a wide range of goods.
(iv) griculture revolution resulted into expansion of economy through agricultural technology, complex division of labours permanent settlement and advanced trade.	(iv) Industrial revolution resulted into centralisation of work in factories and specialized division of labour.

4. Explain the following concept with suitable examples.

Question 1.
Cohabitation
Answer:

1. Cohabitation is the sharing of a household by an unmarried couple.
2. Live in relations or cohabitation may or may not lead to marriage.
   Example : Younger generations specially in many parts of Europe and in some urban areas in India are preferring cohabitation as family relation. This is especially true among same sex couples.

Question 2.
Formal Education
Answer:

1. Formal education is planned with a specific end in view.
2. It involves direct schooling and instruction. Since it is provided to fulfill specific ends, formal education is limited to a specific period. Formal education has a well-defined and systematic curriculum based on aims designed according to the needs of society.

Example : Schooling and education are considered as a formal education based on formal curriculum with specific subject and skills.

5. State whether the following statements are true or false with reasons.

Question 1.
In a modernizing society the role of parents are changing.
Answer:
This statement is True.

1. Rapid growth of divorce and changing roles of woman give rise to a new family form and relation which has led to changes in the role of parents.
2. The average age at which people get married is also increasing. There is also an increasing trend of individuals not getting married. There is a changing role of women due to education and employment. All these changes are affecting family as an institution.
3. In our contemporary society, the traditional belief of fathers are the breadwinners and mothers only do household chores is no longer true. With the advent of globalization, their roles are changing which is vastly different from their previous generations.

Question 2.
India is in the stage of industrial revolution.
Answer:
This statement is False.

1. India is in the stage of information revolution. The development of computer and information technology have changed the nature of work in India.
2. The use of automated machinery has reduced the role of human labour in manufacturing. Service industries like public relations, banking and sales, media, advertising have expanded.
3. The introduction of computer and information have changed the character of work in India.

6. Give your personal response.

Question 1.
What are your views about love marriage? Do you support it? Explain.
Answer:
The basic concept of love marriage lies in the fact that the girl or boy chooses his or her life partner. The restrictions of caste, religion, physical appearance does not apply when a person falls in love. In India these restrictions are seriously enforced which limit the number of desirable matches for a person. Also, it curbs the practice of dowry as one does not need to prove one’s worth. Therefore, I support love marriages.

Question 2.
How has the information revolution impacted your life?
Answer:
Information revolution has made our life easier. Information revolution has made it possible to do many things from the comforts of one’s own home. I can take virtual tour of museums, buildings etc. It has exposed one to different cultures. One can do all of shopping online. It has created abundance of data on every possible subject or interest.

7. Answer the following in detail (About 150-200 words).

Question 1.
Show how the role of family has changed in the present times. Illustrate with your own examples.
Answer:
Families are the nexus of activities that include parenting , employment and leisure. As society changes, families must adapt to the new structures and processes resulting from this change. One of the most striking features of modern societies has been rapid growth of divorce.

The average age at which people get married is also increasing along with an increasing trend of an individuals not getting married. The changing roles of woman through increased education and employment, has put additional pressures on the family to adapt to dual-earner, households and changing needs of child care. All these changes are affecting family as an institution.

It has given rise to new family relationship:
(i) Single-parent Family : Majority of single parent families are headed by single mothers. This may result from divorce, separation, death or by choice. Sometimes growing up in a single parent family can be a disadvantage for children.

(ii) Cohabitation : Cohabitation is the sharing of a household by an unmarried couple. Younger generations, especially in many parts of Europe and in some urban areas in India are preferring cohabitation as family relation. This is especially true among some same sex couples.

(iii) Step-parenting : As rates of divorce and remarriage is steadily on an increase, it gives rise to a new family form and relation of step-parenting. The extent of children staying in step families is increasing.

In the traditional family living, the wife had no voice in family decision making but in contemporary family she has equal power role to play. The authority has shifted from patriarchal to parents who consult their children an all-important issues before taking any decisions about them. Younger generations now claim more individuality. Change in family does not mean the complete erosion of previous norms and structure.

11th Sociology Digest Chapter 4 Social Institutions Intext Questions and Answers

ACTIVITY (Textbook Page No. 41)

Question 1.
Discuss why hypogamous marriages are resisted in society. Take help of newspapers, magazines and cinema.
Answer:
Reasons for resisting hypogamous marriage. Hypergamy is a term used for the practice of a person marrying a spouse of higher caste or social status than themselves. Young women generally marry older men of higher status, with general rule that older men have more time to create wealth and status than younger men. Today such marriages are on decline, as most people marry their approximate social equals and in some parts of the world hypergamy has decreased. Also, it is becoming less common for women to marry older men though hypergamy does not require the man to be older but only of higher status.

Question 2.
Look at the matrimonial advertisements in newspapers, magazines and internet and discuss the findings. Do you think endogamy is still the prevalent norm in India? (Textbook Page No. 42)
Answer:
Religion and Caste endogamy still dominates while choosing a life partner. Women have the freedom to select a spouse in urban areas, yet, the choice can be restricted. So, in general, marriage is often seen as socially determined institutions.

Question 3.
Watch the classic movie ‘Modern Times’ made by Charlie Chaplin and have a discussion on the effects of mechanisation on human labour. (Textbook Page No. 44)
Answer:
The film ‘Modern Times’ is a comment on the desperate employment and financial conditions that people faced during the great depression conditions created, by the modern industrialization in the view of Chaplin. Modern Times’ portray Chaplin as a factory worker where he is subjected to such indignities as being forced fed by a malfunctioning ‘feeding machine’. In the movie, industry is portrayed as something bigger and more valuable than the people who work within it. Chaplin tries to point out the fact at the industrialized world functions in a way that suppresses human agency and creativity.

Question 4.
Eminent personalities having achieved social prestige and position on the basis of their educational qualification. (Textbook Page No. 48)
Answer:
(i) Bhimrao Ramji Ambedkar : Dr. Bhimrao Ambedkar was born on 14th April 1891. During his childhood he was subjected to socio-economic discrimination and faced severe humiliation. Dr. Ambedkar earned his doctorates in Economics from both Columbia University and the London School of Economics. He gained reputation as a scholar for his research in Law, economics and political science. In his initial career, he worked as an economist, professor and lawyer. He was posthumous awarded the Bharat Ratna, India’s highest civilian award, in 1990.

(ii) A. P. J. Abdul Kalam : A. P. J. Abdul Kalam was born on 15th October 1931. His father was an owner of a boat, an imam of local mosque. Due to destruction in business and loss of the family fortune, Kalam’s family suffered poverty. To help his family, he started selling newspapers at an early age. Abdul Kalam completed his graduation from the Madras Institute of Technology and joined the Aeronautical Development Establishment of the Defence Research and Development Organisation (DRDO) as a scientist after becoming a member of Defence Research and Development Service (DRDS). He came to be known as the Missile Man of India’. He was the 11th President of India.

Balbharti Maharashtra State Board Class 11 Sociology Solutions Chapter 3 Basic Concepts in Sociology Textbook Exercise Questions and Answers.

Maharashtra State Board Class 11 Sociology Solutions Chapter 3 Basic Concepts in Sociology

1. (A) Choose the correct alternative and complete the statements.

Question 1.
Society is based on ……………….
(co-operation / competition / conflict)
Answer:
Co-operation

Question 2.
Villages are a part of ……………….. community.
(rural / city / metropolitan)
Answer:
rural

Question 3.
Family is an example of a ………………. group.
(large / primary / formal)
Answer:
primary

Question 4.
Social status refers to ……………… in a group.
(function / position / purpose)
Answer:
position

1. (B) Correct the incorrect pair.

Question 1.
(a) Family – Voluntary Group
(b) Caste – Involuntary Group
(c) Village – Primary Group
(d) Workplace – Secondary Group
Answer:
(a) Family – Involuntary Group

1. (C) Identify the appropriate term from the given options.

(Role, Folkways, Society)
Question 1.
Punctuality in class is the duty of a student.
Answer:
Role

Question 2.
In many places in India, eating with hands is a common practice.
Answer:
Folkways

1. (D) Correct underlined words and complete the sentence.

Question 1.
Norms which can be applied in daily life are called law.
Answer:
Norms which can be applied in daily life are called standard norms.

Question 2.
Family is an example of a/an voluntary group.
Answer:
Family is an example of a/an involuntary group.

2. Write short notes.

Question 1.
Characteristics of Society.
Answer:
(i) Society includes likeness : According to Maclver ‘society’ means likeness. Social relationships exist among those who are alike in body and mind. People have similarities with regard to their needs aim, ideals, values, outlook towards life and so on.

(ii) Society includes difference : A society together with likeness is also based on differences. Differences is an important factor for a healthy society. Different types of personalities are essential in our society in order to satisfy the different needs. These differences are based on sex, age, physical strength, intelligence, talent, personality and unequal possessions of material objects and wealth.

(iii) Interdependence : Individuals in a society depend upon one another for the satisfaction of needs. One group, one nation, one community depends upon the other for its development, t Interdependence is seen in family groups as well. The existence of society depends upon different types of social relationships.

(iv) Co-operation : Every society is characterized by co-operation and division of labour. It is necessary for survival.

(v) Normative nature : In a society, social behaviour is evaluated by social control. These controlling means are known as norms. Norms control the misbehaviour or the acts which are harmful to society thus protecting the society.

(vi) Society is dynamic : No society is static. It changes continuously; old customs, traditions, I values, norms and institutions replace the new customs, traditions, values, norms and institutions.

Question 2.
Characteristics of Primary Group.
Answer:

1. Physical proximity : This is essential for a primary group as people have close relations with each other. Hence, they experience physical proximity.
2. Smallness of the group : These social groups are small in size and bring out the close relations among its members.
3. Permanence of relationship : Close relations create more unity among the members.
4. Face-to-face relationship : These social groups are small in size and are based on greater closeness. It is useful for maintaining group stability.
5. Similar objectives and goals : This group is homogeneous in nature. Hence, there is uniformity of objectives and goals among its members.
6. The relationship is an end in itself: The relationships in this group are very natural.
7. Informal control: This group is conventional and based on emotional bonds.

Question 3.
Types of Norms.
Answer:
Folkways : The term was first used by William Sumner. According to him, folkways are the recognized ways of behaving and acting in society. Example, eating using one’s fingers, eating with chopsticks, eating with fork and spoon; different ways of wearing a sari.

Mores : Mores are more rigid than folkways. They are instruments of social control and deals with higher values of people. Example, sexual relations before marriage are not permitted.

Law : Law is an important and essential element of society.
They are deliberately formulated rules of behaviour. It is universal in nature and common for all to follow. There are two types of law- Customary Law and Enacted Law.

1. Customary law – It is not in a written form but orally transmitted. It is followed in tribal as well as in rural society.
2. Enacted law – This law is in a written form. It is important as well as obligatory in a modern, complex and dynamic society. For example, The Hindu Marriage Act, 1955; The Domestic Violence Act, 2005.

3. Differentiate between.

Question 1.
Primary Group and Secondary Group.
Answer:

Primary Group	Secondary Group
(i) Primary group is a group in which relationships are personal, informal, face to face and intimate in nature.	(i) Secondary group is a group in which relationships are impersonal, formal and contractual in nature.
(ii) This group is small in size.	(ii) This group is large in size.
(iii) Relationships are personal, so physical proximity is more.	(iii) Due to formal relationships there is no physical proximity between the members.
(iv) Relations are permanent.	(iv) Relations are temporary.
(v) Behaviour is controlled in an informal way.	(v) Behaviour is controlled in a formal way.
(vi) Example Family, peer group, neighbourhood, etc.	(vi) Example Nation, State, Labour unions, etc.

Question 2.
Voluntary Group and Involuntary Group.
Answer:

Voluntary Group	Involuntary Group
(i) Membership is based on choice.	(i) Membership is based on birth.
(ii) It may be temporary or permanent in nature.	(ii) It is mostly permanent in nature.
(iii) Individual has choice to continue or to leave the membership.	(iii) One cannot leave the membership or it may be difficult to leave the group.
(iv) Example : Political parties, youth organisation, cultural association, etc.	(iv) Example : Family, caste, race, religion, etc.

4. Explain the following concept with suitable examples.

Question 1.
Reference Group
Answer:
The concept of reference group was introduced by Robert Merton. A reference group is a group to which an individual or another group is compared. We use reference groups in order to guide our behaviour and attitudes that help us to identify social norms.

Reference Groups are of two types Informal and Formal.

1. Informal reference groups are formed by the individual as per their likes, interests and attitudes.
   Example : Family, peer groups, teachers, siblings, associates, etc.
2. Formal Reference Group are the ones that have a specific goal or mission.
   Example : Labour unions, supreme court, military units, corporations, worship place, universities, etc.

Question 2.
Role Conflict
Answer:

1. Role conflict refers to a clash between roles.
2. When one faces incompatibility between two or more roles at the same time or in a given situation. Role conflict is inevitable.

Example: A boss will suffer role conflict if forced to fine an employee who is also a close friend. A newly married bride experiences role conflict because she has to adjust with her in-laws, household work and office work.

5. (A) Complete the concept maps.

Question 1.

Answer:

5. (B) State whether the following statements are true or false with reasons.

Question 1.
Society never changes.
Answer:
This statement is False.
1. Society is a web of social relationships. These relationships and interactions are based upon different institutions, traditions, customs, values and norms of the society.

2. Changes in any one of this is reflected in the changes in relationship and interaction among the individuals. Change is a universal phenomenon. No single society is static. Factors like, westernisation, modernisation, industrialisation, education have brought various changes in society.

3. Today, educational system, family patterns, norms, values, needs, etc., are changed. People have become more self-centred. Individual freedom, consciousness towards one’s own rights are increasing. Hence, society is changing continuously.

Question 2.
People living in a community must have awareness of sharing a way of life.
Answer:
This statement is True.

1. Individuals are emotionally attached to their community. People those who live in a particular community occupy a definite territorial area. They share common objectives and needs.
2. They have love and affection towards each other.
3. This develops a sense of belongingness. Therefore, people living in a community must have awareness of sharing a way of life.

6. Give your personal response.

Question 1.
Show how folkways, mores and laws may clash with each other.
Answer:
Folkways is one of the essential elements of culture. Folkways govern our daily routine and ordinary contacts with other people. Whereas mores are considered as vital to the welfare of the group. Folkways and mores are the customary way of life and standards of right and wrong. Example: wearing clothes are mores and wearing clothes of different styles are folkways.

Laws are deliberately formulated rules of behaviour that are enforced by a special authority, e.g., there are laws that can punish people for marrying more than one person. Hindu Marriage Act, 1955.

Mores, folkways and laws are taught through the process of socialisation by various sources like family, friends, peer groups, schools, etc. However, these three clash each other in various ways. In today’s complex, competitive society, one finds it difficult to cope up with these three elements of society. Example: there is a man who belongs to a poor family and has the responsibility of marriage of his sister. Being born in a poor family, he doesn’t have enough money for his sister’s marriage. As per the laws, asking for dowry is illegal, but due to his financial condition, he has to ask for dowry from his in-laws to get his sister married. In such situation folkways, mores and law clash each other.

Question 2.
Do you think that role conflict is inevitable in social life?
Answer:
Yes, the role conflict is inevitable in social life.
When an individual has to play several roles at a time, then it is not possible to perform one role appropriately, it is called as role conflict. In today’s society, everyone wants to achieve high status in society. Everyone wants to achieve a life of luxury and comfort. People want to earn more and more money. Expectations have been increased. In order to adjust with such competitive situation, one has to play different role in the society. That creates role conflict.

Example: A woman working as a nurse in the hospital, has to look after her sick in-laws at home, because of which she can’t pay proper attention towards her duty. A husband who has to attend urgent meeting as a manager, can’t give time to his family. One has to perform multiple roles in a complex society. That is why role conflict is inevitable in social life.

7. Answer the following in detail (About 150-200 words).

Question 1.
Identify any two secondary groups of which you are a member. Discuss any four characteristics with reference to the groups that you have named.
Answer:
Secondary groups have impersonal, formal, indirect, temporary and goal oriented relationships with large number of members. It is not possible to have regular, permanent, close and intimate relations with everyone. People meet each other only for serving specific purposes. Relations are maintained only when the specific purposes are fulfilled and people are satisfied.

People are always very busy and it is necessary to have appointments before meeting. Secondary group are heterogeneous and formal in natural. People are given time to meet but for that purpose only. People do not meet for long term goals. Free expressions of emotions, feelings and intimacy is not allowed. In secondary group, members have to follow rules and regulations strictly.

Example : (i) Student of a college
(ii) Member of NSUI – National Students Union of India (or) ABVP – Akhil Bhartiya Vidhyarthi Parishad
Characteristics of secondary group with reference to the above mentioned group. Large Size: In college or in NSUI number of students and members is large. Here, the membership is unlimited as compared to primary group.

Indirect relations : Due to large size, direct relations among all the students and members are not possible. To convey any message, we take help of modern means of communication, e.g., mobile phones, e-mail, WhatsApp, etc.

Impersonal relations : As the members are large in size personal interactions is very rarely seen. Here, we do not know each other personally. That is why, relationship is not personal.

Formal relations : Informal relations is not possible because face to face contact can’t take place. People are more goal oriented. Membership is strictly based upon laws and regulations.

11th Sociology Digest Chapter 3 Basic Concepts in Sociology Intext Questions and Answers

ACTIVITY (Textbook Page No. 25)

Question 1.
Visit a rural, tribal or urban community and speak to 3-5 people and find out about their work, families, customs, beliefs etc. Write a report or make a short film on your mobile phone.
Answer:
Tribal, Rural and Urban Community – Work, Families, Customs, Beliefs-
(i) The Tribal Community : The tribal community is small in size. Each tribe has its own culture. They are close to nature and their life is influenced by religion. Therefore, they worship natural forces and seek assistance from the shaman and magician for solving their problems. Their economy and usage of technology is simple. This is known as subsistence economy.

(ii) Rural Community : The rural community is largely homogeneous. The main occupation of the people in this community is agriculture and non-agricultural occupations such as dairy farming, poultry, etc. This community is influenced by nature and they worship it as God or Goddess.

The rural community is influenced by the institution of family and characterized by primary relations. Joint family continues to exist in rural areas though some modification has occurred.

(iii) Urban Community : The urban community is large in size and consists of mostly nuclear families. It is a settlement of socially heterogeneous people. Occupations of the urban people are non agricultural. There is more scope for division of labour. This community is greatly impacted by artificial or man-made environment rather than natural environment on the urban community.

Question 2.
Prepare a photo essay regarding interactions observed within primary and secondary groups. Students should go around in their neighbouring localities and capture with their devices (like mobiles), the interaction patterns based on characteristics of various types of groups. How to write a photo essay? (Textbook Page No. 29)
Answer:
(Students are required to perform this activity themselves by visiting the link http://www. collectivelens.com/blog/creating-photoessay/ as a guideline.)

Question 3.
Conduct a role play exercise to understand the concept of role conflict and role strain. For example: Working women performing various roles inside and outside the home. (Textbook Page No. 32)
Answer:
Understanding 1: Women consider that the household affairs and care for the family is their primary duty. Women are thus confronted with the challenge of playing the dual role, to excel at home, as well as at the workplace. To prove her competence on both the fronts, women are facing the problem of overwork and are often being stressed or drained of their energy. Yet, they find themselves unable to match the expectations at any of the two places, which gives rise to role conflict and role strain.

Understanding 2:
Role Conflict:
Role conflict occurs when conflicting expectations arise from two or more statuses than an individual occupies.

Role Strain:
Role strain arises when conflicting expectations are built into a single status.

Question 4.
List out various folkways and mores that we find in our day-to-day lives. (Textbook Page No. 33)
Answer:
Folkways: The folkways are the recognized ways of behaving and acting in society. Folkways are norms that stem from and organize casual interactions and emerge out of repetition and routines. We engage in them to satisfy our daily needs and they are most often unconscious in operation, though they are quite useful for the ordered functioning of the society.

1. Waiting in a queue
2. Appropriate dressing.
3. Practice of raising one’s hand to take turns while speaking in a group.
4. While eating using one’s fingers, eating with chopsticks, eating with fork and spoon.
5. Different ways of wearing a sari.

Mores : “More are the popular habits and traditions when they include a judgment that are conductive to social welfare and when they exert a coercion on an individual to conform to them”.

More are stricter than folkways, as they determine what is considered moral and ethical behaviour. Mores structure the difference between right and wrong. Violation of mores results in disapproval or ostracizing. As such mores play a key role in shaping our value, beliefs, behaviour, and interactions than folkways.

1. Religious doctrine.
2. Sexual relations before marriage are not permitted.
3. It is not acceptable to use drugs such as heroin and cocaine.
4. It is expected that one will be one time for work.
5. Talking to oneself in public is not considered normal behaviour.
6. Nudity in public is not acceptable.

Balbharti Maharashtra State Board Class 11 Sociology Solutions Chapter 2 Contribution of Western and Indian Sociologists Textbook Exercise Questions and Answers.

Maharashtra State Board Class 11 Sociology Solutions Chapter 2 Contribution of Western and Indian Sociologists

1. (A) Choose the correct alternative and complete the statements.

Question 1.
The Industrial Revolution took place in ………………..
(North America / Europe / Australia)
Answer:
Europe

Question 2.
Science is based on …………………
(beliefs / facts / guess work)
Answer:
facts

Question 3.
Significant work has been done on kinship organisation in India, by female sociologist ………………
(Iravati Karve / Suma Chitnis / Neera Desai)
Answer:
Iravati Karve

Question 4.
The ……………….. stage of societal growth is characterised by explanations which are abstract but not God-centred.
(Theological / Metaphysical / Positive)
Answer:
Metaphysical

1. (B) Correct the incorrect pair.

Question 1.
(a) Abdul Rahman Ibn-Khaldun – North Africa
(b) Auguste Comte – France
(c) Harriet Martineau – England
(d) Karl Marx – Russia
Answer:
(d) Karl Marx – Germany

1. (C) Identify the appropriate term from the given options.

(Polarization. M. N. Srinivas, R. N. Mukherjee)
Question 1.
Marx argues that classes will become hostile towards each other.
Answer:
Polarization

Question 2.
The Indian sociologist who coined the term westernisation.
Answer:
M. N. Srinivas

1. (D) Correct the underlined words and complete the sentence.

Question 1.
The stage of society where empirical evidence forms the basis for explanation is the theological stage.
Answer:
The stage of society where empirical evidence forms the basis for explanation is the positive / scientific stage.

Question 2.
The book ‘Le Suicide’ was written by Hobbes.
Answer:
The book ‘Le Suicide’ was written by Emile Durkheim

2. Write short notes.

Question 1.
Contribution of G. S. Ghurye to Indian Sociology.
Answer:
1. Govind Sadashiv Ghurye was the first who introduced sociology in India after independence. Therefore, he is considered as the ‘Father of Indian Sociology’. He established ‘Indian Sociological Society’ and started the journal ‘Sociological Bulletin’.

2. Ghurye’s work on scheduled tribes, were based on the historical, administrative and social dimensions of Indian tribes. He wanted the tribals to be integrated with the wider Indian society. He has written on tribes like Mahadev Kolis in Maharashtra. His book, ‘Caste and Race in India’ is one of the most important contribution to Indian sociology.

3. Ghurye referred to the long process of Hinduisation of the tribes from different parts of India. He considered cultural unity between tribes and caste as the only means to promote integration in Indian society.

4. His book, ‘Caste and Race in India’, published in 1932 combines historical, anthropological and sociological perspectives to understand caste and kinship system in India.

Question 2.
Types of suicide according to Durkheim.
Answer:
Suicide is a type of death carried by an individual intentionally or deliberately. According to Durkheim suicide is a social phenomenon. Following are the types of suicide.
1. Egoistic suicide : It takes place when an individual is very much self-centred and least bothered about the other members of society, e.g., it is committed by people who become introvert and have less desires to live in the company of others, suicide rates are higher for those widowed, single and divorced.

2. Anomic suicide : When situations go against the norms and values of a person. In such condition individual finds it difficult to face the situation and adjust with that, e.g., suicide of farmers.

3. Altruistic suicide : This type of suicide is different from egoistic suicide. This type of suicide means sacrificing own life for the betterment of people or society, e.g., Sati system, a patient commits suicide for the sake of family.

4. Fatalistic suicide : When an individual is under excessive control from the outside factors fatalistic suicide takes place e.g., Dowry death. According to Durkheim social situations, circumstances, factors are responsible for suicide and become a ‘social fact.’

3. Differentiate between.

Question 1.
Theological Stage and Positive Stage.
Answer:

Theological Stage	Positive Stage
(i) Theological stage is the first stage of Comte’s Law of three stages.	(i) Positive stage is last or third stage of Comte’s law of three stages.
(ii) Human thinking, ideas and views were influenced by spiritual and supernatural factors.	(ii) This is an improved rational, scientific form of human thinking.
(iii) All societies believed that God controls all events in the world.	(iii) Instead of imagination and superstitious belief, societies turn towards empirical scientific approach.
(iv) Example : Natural calamities like flood, drought etc., were the expressions of God’s anger.	(iv) Example : Discoveries of Newton and Galileo, enlightement.

Question 2.
Anomic Suicide and Altruistic Suicide.
Answer:

Anomic Suicide	Altruistic Suicide
(i) Anomic suicide takes place in situations where one is not able to adjust with circumstances which arise unexpectedly.	(i) Altruistic suicide means, an individual commits suicide with the object of doing well for others.
(ii) This is due to economic instability and personal struggle.	(ii) This is due to fanatical love for one’s own community.
(iii) It reflects a normlessness.	(iii) It is value-oriented.
(iv) Example : Businessman committed suicide due economic depression or extreme prosperity.	(iv) Example : Sati system (committed by Indian women in the past), Hara-kiri (committed by Japanese)

4. (A) Complete the concept maps.

Question 1.

Answer:

4. (B) State whether the following statements are true or false with reasons.

Question 1.
According to Marx, capitalism gives rise to we feeling among workers.
Answer:
This statement is True.

1. Capitalists accumulate profit through the exploitation of labour. The poverty of the workers’ class grows with increasing exploitation of labour.
2. Economic exploitation and inhuman conditions lead to the increasing alienation of workers.
3. The classes tend to become internally homogeneous and class struggle, more intensified and creates class solidarity and we feeling among the workers.

Question 2.
Iravati Karve has made significant contribution to the study of Kinship in India.
Answer:
This statement is True.

1. Iravati Karve has contributed in the field of Sociology as well as Anthropology. She wrote a book ‘Kinship Organization in India’ in which she describes major kinship systems in India.
2. Kinship is one of the concepts which is socially and culturally related with factors like family, caste and languages in India.
3. According to her, kinship system is based on the geographical and linguistic group differences.
   Thus, Iravati Karve has made significant contribution to the study of Kinship in India

5. Give your personal response.

Question 1.
Do you think globalization has led to polarization of classes? Discuss with relevant examples of your own.
Answer:
Yes, globalization has led to polarization of classes. This has led to segregation of people in the society that may emerge from income inequality, economic restructuring etc. It leads to differentiation of groups on the basis of high income and low income. Skilled people manage to get high paying jobs while the less educated/skilled people receive low wages.

Question 2.
Do you think Kinship bonds are weakening? Give reasons for your response?
Answer:
Yes, I think as we Indians are progressing and trying to establish our nation as a developed nation, we on the other side are losing importance of our family ties. Kinship bonds means ties based on blood and marriage.
Following are some of the reasons which are axing Kinship bonds.

1. Urbanisation, one of the major reasons of migration of people from rural to urban resulting into formation of a nuclear family.
2. Modernisation, making individuals self-centred, career oriented simultaneously detach from family.
3. Change in the medium of recreation, earlier it was family get together or celebration of festivals or religious ceremony now it is Cinema Hall or Visit to Mall.
4. Increasing control of electronic gadgets like T.V., Mobile Internet, etc., which make virtual world closer but taxing to kinship bond, care and attachment.
5. Dominance of materialistic world results into abundance of needs. Above all, todays smart and modern generation is chopping their own roots, meaning, weakening kinship bonds.

6. Answer the following in detail (About 150-200 words).

Question 1.
You have studied about Comte’s Law of Three Stages of human thought. With reference to the first and third stage, comment on the challenges it poses for Indian society.
Answer:
Auguste Comte believed that the evolution of human mind had taken place along with the evolution of the individual mind. Comte’s social philosophy is based on the concept of three stage of human thought.

Theological stage : In this stage human beings believed in supernatural and spiritual factors. It was believed that three was a control of God on all events in the world. The human mind, at this level, supposed that all phenomena was produced by the immediate action of supernatural beings. For example, all natural calamities were the expressions of God’s anger. This stage is dominated by priests and ruled by military men.
Following are the challenges which poses for Indian society:

1. During this stage, human mind is dominated by sentiments, feelings and emotions.
2. Certain section of Indian society believed that all actions/events happened due to power of God/ Supernatural beings.
   E.g. drought, flood, earthquake etc., natural calamities were nothing but anger of God.
3. Explanation for all happening events are the form of myths concerning spirits and supernatural beings.
4. During this stage, military society was basically dying and priests were dominated.
5. In short, due to effect of theological stage there was lack of logical and orderly thinking in the Indian society.

Positive/Scientific stage : Comte says, scientific stage is an improved and scientific form of human thinking. Now, instead of imagination, we turn to observation. Since Reason and examination were basic planks of knowledge at this stage hence, one seeks to establish laws which link facts and which govern social life.

Effect of scientific stage on Indian Society:

1. Finally, in this stage, Indian society turned towards the scientific way of thinking.
2. Instead of imaginations, society turned towards observation, reason and examination all these were basic planks of knowledge.
3. The concept of God started vanishing from human mind.
4. Society adopted the scientific rational way of thinking and there is no place for any belief or superstition in it.
5. In this way, Comte’s first and third stage affected the Indian Society.

11th Sociology Digest Chapter 2 Contribution of Western and Indian Sociologists Intext Questions and Answers

ACTIVITY (Textbook Page No. 16)

Question 1.
Discuss how ‘polarization of classes’ and ‘class solidarity’ are relevant in the present times.
Answer:
In present times, polarization of classes and class solidarity is associated with the segregation within a society which are immediate cause of emergence of income inequality, economic displacements, formation of various social groups from high income to low income, economic restructuring particularly in cities, economic inequality etc.

Question 2.
Read newspaper articles related to caste issues (e.g. representation, atrocities, reservation) and share the findings with your class. (Textbook Page No. 19)
Answer:
Reference : Times of India (March 29, 2019)
The Indian Express (June 6, 2019)
Times of India (June 3, 2019)