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Balbharti Maharashtra State Board 11th Physics Textbook Solutions Chapter 12 Magnetism Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Physics Solutions Chapter 12 Magnetism

1. Choose the correct option.

Question 1.
Let r be the distance of a point on the axis of a bar magnet from its center. The magnetic field at r is always proportional to
(A) \(\frac {1}{r^2}\)
(B) \(\frac {1}{r^3}\)
(C) \(\frac {1}{r}\)
(D) Not necessarily \(\frac {1}{r^3}\) at all points
Answer:
(B) \(\frac {1}{r^3}\)

Question 2.
Magnetic meridian is the plane
(A) perpendicular to the magnetic axis of Earth
(B) perpendicular to geographic axis of Earth
(C) passing through the magnetic axis of Earth
(D) passing through the geographic axis
Answer:
(C) passing through the magnetic axis of Earth

Question 3.
The horizontal and vertical component of magnetic field of Earth are same at some place on the surface of Earth. The magnetic dip angle at this place will be
(A) 30°
(B) 45°
(C) 0°
(D) 90°
Answer:
(B) 45°

Question 4.
Inside a bar magnet, the magnetic field lines
(A) are not present
(B) are parallel to the cross sectional area of the magnet
(C) are in the direction from N pole to S pole
(D) are in the direction from S pole to N pole
Answer:
(D) are in the direction from S pole to N pole

Question 5.
A place where the vertical components of Earth’s magnetic field is zero has the angle of dip equal to
(A) 0°
(B) 45°
(C) 60°
(D) 90°
Answer:
(A) 0°

Question 6.
A place where the horizontal component of Earth’s magnetic field is zero lies at
(A) geographic equator
(B) geomagnetic equator
(C) one of the geographic poles
(D) one of the geomagnetic poles
Answer:
(D) one of the geomagnetic poles

Question 7.
A magnetic needle kept nonparallel to the magnetic field in a nonuniform magnetic field experiences
(A) a force but not a torque
(B) a torque but not a force
(C) both a force and a torque
(D) neither force nor a torque
Answer:
(C) both a force and a torque

2. Answer the following questions in brief.

Question 1.
What happens if a bar magnet is cut into two pieces transverse to its length/ along its length?
Answer:
i. When a magnet is cut into two pieces, then each piece behaves like an independent magnet.

ii. When a bar magnet is cut transverse to its length, the two pieces generated will behave as independent magnets of reduced magnetic length. However, the pole strength of all the four poles formed will be same as that of the original bar magnet. Thus, the new dipole moment of the smaller magnets will be,

iii. When the bar magnet is cut along its length, the two pieces generated will behave like an independent magnet with reduced pole strength. However, the magnetic length of both the new magnets will be same as that of the original bar magnet. Thus, the new dipole moment of the smaller magnets will be,

Question 2.
What could be the equation for Gauss’ law of magnetism, if a monopole of pole strength p is enclosed by a surface?
Answer:
i. According to Gauss’ law of electrostatics, the net electric flux through any Gaussian surface is proportional to net charge enclosed in it. The equation is given as,
ø_E = ∫\(\vec{E}\) . \(\vec{dS}\) = \(\frac {q}{ε_0}\)

ii. Similarly, if a monopole of a magnet of pole strength p exists, the Gauss’ law of magnetism in S.I. units will be given as,
ø_E = ∫\(\vec{B}\) . \(\vec{dS}\) = µ₀P

3. Answer the following questions in detail.

Question 1.
Explain the Gauss’ law for magnetic fields.
Answer:
i. Analogous to the Gauss’ law for electric field, the Gauss’ law for magnetism states that, the net magnetic flux (ø_B) through a closed Gaussian surface is zero. ø_B = ∫\(\vec{B}\) . \(\vec{dS}\) = 0

ii. Consider a bar magnet, a current carrying solenoid and an electric dipole. The magnetic field lines of these three are as shown in figures.

iii. The areas (P) and (Q) are the cross – sections of three dimensional closed Gaussian surfaces. The Gaussian surface (P) does not include poles while the Gaussian surface (Q) includes N-pole of bar magnet, solenoid and the positive charge in case of electric dipole.

iv. The number of lines of force entering the surface (P) is equal to the number of lines of force leaving the surface. This can be observed in all the three cases.

v. However, Gaussian surface (Q) of bar magnet, enclose north pole. As, even thin slice of a bar magnet will have both north and south poles associated with it, the number of lines of Force entering surface (Q) are equal to the number of lines of force leaving the surface.

vi. For an electric dipole, the field lines begin from positive charge and end on negative charge. For a closed surface (Q), there is a net outward flux since it does include a net (positive) charge.

vii. Thus, according to the Gauss’ law of electrostatics ø_E = ∫\(\vec{E}\) . \(\vec{dS}\) = \(\frac {q}{ε_0}\), where q is the positive charge enclosed.

viii. The situation is entirely different from magnetic lines of force. Gauss’ law of magnetism can be written as ø_B = ∫\(\vec{B}\) . \(\vec{dS}\) = 0
From this, one can conclude that for electrostatics, an isolated electric charge exists but an isolated magnetic pole does not exist.

Question 2.
What is a geographic meridian? How does the declination vary with latitude? Where is it minimum?
Answer:
A plane perpendicular to the surface of the Earth (vertical plane) and passing through geographic axis is geographic meridian.

i. Angle between the geographic and the magnetic meridian at a place is called magnetic declination (a).
ii. Magnetic declination varies with location and over time. As one moves away from the true north the declination changes depending on the latitude as well as longitude of the place. By convention, declination is positive when magnetic north is east of true north, and negative when it is to the west. The declination is small in India. It is 0° 58′ west at Mumbai and 0° 41′ east at Delhi.

Question 3.
Define the angle of dip. What happens to angle of dip as we move towards magnetic pole from magnetic equator?
Answer:
Angle made by the direction of resultant magnetic field with the horizontal at a place is inclination or angle of dip (ø) at the place.
At the magnetic pole value of ø = 90° and it goes on decreasing when we move towards equator such that at equator value of (ø) = 0°.

4. Solve the following problems.

Question 1.
A magnetic pole of bar magnet with pole strength of 100 Am is 20 cm away from the centre of a bar magnet. Bar magnet has pole strength of 200 Am and has a length 5 cm. If the magnetic pole is on the axis of the bar magnet, find the force on the magnetic pole.
Answer:
Given that, (q_m)₁ = 200 Am
and (2l) = 5 cm = 5 × 10^-2 m
∴ m = 200 × 5 × 10^-2 = 10 Am²
For a bar magnet, magnetic dipole moment is,
m = q_m (21)
For a point on the axis of a bar magnet at distance, r = 20 cm = 0.2 m,
B_a = \(\frac{\mu_{0}}{4 \pi} \times \frac{2 m}{r^{3}}\)
= 10^-7 × \(\frac{2 \times 10}{(0.2)^{3}}\)
= 0.25 × 10^-3
= 2.5 × 10^-4 Wb/m²
The force acting on the pole will be given by,
F = q_m B_a = 100 × 2.5 × 10^-4
= 2.5 × 10^-2 N

Question 2.
A magnet makes an angle of 45° with the horizontal in a plane making an angle of 30° with the magnetic meridian. Find the true value of the dip angle at the place.
Answer:
Let true value of dip be ø. When the magnet is kept 45° aligned with declination 30°, the horizontal component of Earth’s magnetic field.
B’_H = B_H cos 30° Whereas, vertical component remains unchanged.
∴ For apparent dip of 45°,
tan 45° = \(\frac{\mathrm{B}_{\mathrm{V}}^{\prime}}{\mathrm{B}_{\mathrm{H}}^{\prime}}=\frac{\mathrm{B}_{\mathrm{V}}}{\mathrm{B}_{\mathrm{H}} \cos 30^{\circ}}=\frac{\mathrm{B}_{\mathrm{v}}}{\mathrm{B}_{\mathrm{H}}} \times \frac{1}{\cos 30^{\circ}}\)
But, real value of dip is,
tan ø = \(\frac {B_V}{B_H}\)
∴ tan 45° = \(\frac {tan ø}{cos 30°}\)
∴ tan ø = tan 45° × cos 30°
= 1 × \(\frac {√3}{2}\)
∴ ø = tan^-1 (0.866)

Question 3.
Two small and similar bar magnets have magnetic dipole moment of 1.0 Am² each. They are kept in a plane in such a way that their axes are perpendicular to each other. A line drawn through the axis of one magnet passes through the centre of other magnet. If the distance between their centres is 2 m, find the magnitude of magnetic field at the midpoint of the line joining their centres.
Answer:
Let P be the midpoint of the line joining the centres of two bar magnets. As shown in figure, P is at the axis of one bar magnet and at the equator of another bar magnet. Thus, the magnetic field on the axis of the first bar magnet at distance of 1 m from the centre will be,

B_a = \(\frac{\mu_{0}}{4 \pi} \frac{2 m}{r^{3}}\)
= 10^-7 × \(\frac {2×1.0}{(1)^3}\)
= 2 × 10^-7 Wb/m²
Magnetic field on the equator of second bar magnet will be,
B_eq = \(\frac{\mu_{0}}{4 \pi} \frac{m}{r^{3}}\)
= 10^-7 × \(\frac {1.0}{(1)^3}\)
= 1 × 10^-7 Wb/m²
The net magnetic field at P,
B_net = \(\sqrt {B_a^2+B_{eq}^2}\)
= \(\sqrt {(2×10^{-7})^2+(1×10^{-7})^2}\)
= \(\sqrt {(10^{-7})^2×(4+1)}\)
= √5 × 10^-7 Wb/m²

Question 4.
A circular magnet is made with its north pole at the centre, separated from the surrounding circular south pole by an air gap. Draw the magnetic field lines in the gap. Draw a diagram to illustrate the magnetic lines of force between the south poles of two such magnets.
Answer:
i. For a circular magnet:

Question 5.
Two bar magnets are placed on a horizontal surface. Draw magnetic lines around them. Mark the position of any neutral points (points where there is no resultant magnetic field) on your diagram.
Answer:
The magnetic lines of force between two magnets will depend on their relative positions. Considering the magnets to be placed one besides the other as shown in figure, the magnetic lines of force will be as shown.

11th Physics Digest Chapter 12 Magnetism Intext Questions and Answers

Can you recall? (Textbook page no. 221)

Question 1.
What are the magnetic lines of force?
Answer:
The magnetic field around a magnet is shown by lines going from one end of the magnet to the other. These lines are named as magnetic lines of force.

Question 2.
What are the rules concerning the lines of force?
Answer:
i. Magnetic lines of force originate from the north pole and end at the south pole.
ii. The magnetic lines of force of a magnet or a solenoid form closed loops. This is in contrast to the case of an electric dipole, where the electric lines of force originate from the positive charge and end on the negative charge.
iii. The direction of the net magnetic field \(\vec {B}\) at a point is given by the tangent to the magnetic line of force at that point.
iv. The number of lines of force crossing per unit area decides the magnitude of magnetic field \(\vec {B}\)
v. The magnetic lines of force do not intersect. This is because had they intersected, the direction of magnetic field would not be unique at that point.

Question 3.
What is a bar magnet?
Answer:
Bar magnet is a magnet in the shape of bar having two poles of equal and opposite pole strengths separated by certain distance (2l).

Question 4.
If you freely hang a bar magnet horizontally, in which direction will it become stable?
Answer:
A bar magnet suspended freely in air always aligns itself along geographic N-S direction.

Try this (Textbook page no. 221)

You can take a bar magnet and a small compass needle. Place the bar magnet at a fixed position on a paper and place the needle at various positions. Noting the orientation of the needle, the magnetic field direction at various locations can be traced.
Answer:
When a small compass needle is kept at any position near a bar magnet, the needle always aligns itself in the direction parallel to the direction of magnetic lines of force.

Hence, by placing it at different positions, A, B, C, D,… as shown in the figure, the direction of magnetic lines of force can be traced. The direction of magnetic field will be a tangent at that point.

Internet my friend: (Text book page no. 227)

https://www.ngdc.noaa.gov
[Students are expected to visit above mentioned link and collect more information about Geomagnetism.]

Balbharti Maharashtra State Board 11th Physics Textbook Solutions Chapter 11 Electric Current Through Conductors Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Physics Solutions Chapter 11 Electric Current Through Conductors

1. Choose the correct Alternative.

Question 1.
You are given four bulbs of 25 W, 40 W, 60 W, and 100 W of power, all operating at 230 V. Which of them has the lowest resistance?
(A) 25 W
(B) 40 W
(C) 60 W
(D) 100 W
Answer:
(D) 100 W

Question 2.
Which of the following is an ohmic conductor?
(A) transistor
(B) vacuum tube
(C) electrolyte
(D) nichrome wire
Answer:
(D) nichrome wire

Question 3.
A rheostat is used
(A) to bring on a known change of resistance in the circuit to alter the current.
(B) to continuously change the resistance in any arbitrary manner and there by alter the current.
(C) to make and break the circuit at any instant.
(D) neither to alter the resistance nor the current.
Answer:
(B) to continuously change the resistance in any arbitrary manner and there by alter the current.

Question 4.
The wire of length L and resistance R is stretched so that its radius of cross-section is halved. What is its new resistance?
(A) 5R
(B) 8R
(C) 4R
(D) 16R
Answer:
(D) 16R

Question 5.
Masses of three pieces of wires made of the same metal are in the ratio 1 : 3 : 5 and their lengths are in the ratio 5 : 3 : 1. The ratios of their resistances are
(A) 1 : 3 : 5
(B) 5 : 3 : 1
(C) 1 : 15 : 125
(D) 125 : 15 : 1
Answer:
(D) 125 : 15 : 1

Question 6.
The internal resistance of a cell of emf 2 V is 0.1 Ω, it is connected to a resistance of 0.9 Ω. The voltage across the cell will be
(A) 0.5 V
(B) 1.8 V
(C) 1.95 V
(D) 3V
Answer:
(B) 1.8 V

Question 7.
100 cells each of emf 5 V and internal resistance 1 Ω are to be arranged so as to produce maximum current in a 25 Ω resistance. Each row contains equal number of cells. The number of rows should be
(A) 2
(B) 4
(C) 5
(D) 100
Answer:
(A) 2

Question 8.
Five dry cells each of voltage 1.5 V are connected as shown in diagram

What is the overall voltage with this arrangement?
(A) 0 V
(B) 4.5 V
(C) 6.0 V
(D) 7.5 V
Answer:
(B) 4.5 V

2. Give reasons / short answers

Question 1.
In given circuit diagram two resistors are connected to a 5V supply.

i. Calculate potential difference across the 8Q resistor.
ii. A third resistor is now connected in parallel with 6 Ω resistor. Will the potential difference across the 8 Ω resistor be larger, smaller or same as before? Explain the reason for your answer.
Answer:
Total current flowing through the circuit,
I = \(\frac {V}{R_s}\)
= \(\frac {5}{8+6}\)
= \(\frac {5}{14}\) = 0.36 A
∴ Potential difference across 8 f2 (Vi) = 0.36 × 8
= 2.88 V

ii. Potential difference across 8 Ω resistor will be larger.
Reason: As per question, the new circuit diagram will be

When any resistor is connected parallel to 6 Ω resistance. Then the resistance across that branch (6 Ω and R Ω) will become less than 6 Ω. i.e., equivalent resistance of the entire circuit will decrease and hence current will increase. Since, V = IR, the potential difference across 8 Ω resistor will be larger.

Question 2.
Prove that the current density of a metallic conductor is directly proportional to the drift speed of electrons.
Answer:
i. Consider a part of conducting wire with its free electrons having the drift speed vd in the direction opposite to the electric field \(\vec{E}\).

ii. All the electrons move with the same drift speed vd and the current I is the same throughout the cross section (A) of the wire.

iii. Let L be the length of the wire and n be the number of free electrons per unit volume of the wire. Then the total number of free electrons in the length L of the conducting wire is nAL.

iv. The total charge in the length L is,
q = nALe ………….. (1)
where, e is the charge of electron.

v. Equation (1) is total charge that moves through any cross section of the wire in a certain time interval t.
∴ t = \(\frac {L}{v_d}\) ………….. (2)

vi. Current is given by,
I = \(\frac {q}{t}\) = \(\frac {nALe}{L/v_d}\) ……………. [From Equations (1) and (2)]
= n Av_de
Hence
v_d = \(\frac {1}{nAe}\)
= \(\frac {J}{ne}\) …………. (∵ J = \(\frac {1}{A}\))
Hence for constant ‘ne’, current density of a metallic conductor is directly proportional to the drift speed of electrons, J ∝ v_d.

3. Answer the following questions.

Question 1.
Distinguish between ohmic and non ohmic substances; explain with the help of example.
Answer:

Question 2.
DC current flows in a metal piece of non uniform cross-section. Which of these quantities remains constant along the conductor: current, current density or drift speed?
Answer:
Drift velocity and current density will change as it depends upon area of cross-section whereas current will remain constant.

4. Solve the following problems.

Question 1.
What is the resistance of one of the rails of a railway track 20 km long at 20°C? The cross-section area of rail is 25 cm² and the rail is made of steel having resistivity at 20°C as 6 × 10^-8 Ω m.
Answer:
Given: l = 20 km = 20 × 10³ m,
A = 25 cm² = 25 × 10^-4 m²,
ρ = 6 × 10^-8 Ω m
To find: Resistance of rail (R)
Formula: ρ = \(\frac {RA}{l}\)
Calculation: From formula.
R = ρ\(\frac {l}{A}\)
∴ R = \(\frac {6×10^{-8}×20×10^3}{A}\) = \(\frac {6×4}{5}\) × 10^-1
= 0.48 Ω

Question 2.
A battery after a long use has an emf 24 V and an internal resistance 380 Ω. Calculate the maximum current drawn from the battery. Can this battery drive starting motor of car?
Answer:
E = 24 V, r = 380 Ω
i. Maximum current (I_max)
ii. Can battery start the motor?
Formula: I_max = \(\frac {E}{r}\)
Calculation:
From formula,
I_max = \(\frac {24}{380}\) = 0,063 A
As, the value of current is very small compared to required current to run a starting motor of a car, this battery cannot be used to drive the motor.

Question 3.
A battery of emf 12 V and internal resistance 3 O is connected to a resistor. If the current in the circuit is 0.5 A,
i. Calculate resistance of resistor.
ii. Calculate terminal voltage of the battery when the circuit is closed.
Answer:
Given: E = 12 V, r = 3 Ω, I = 0.5 A
To find:
i. Resistance (R)
ii. Terminal voltage (V)
Formulae:
i. E = I (r + R)
ii. V = IR
Calculation: From formula (i),
E = Ir + IR
∴ R = \(\frac {E-Ir}{l}\)
= \(\frac {12-0.5×3}{0.5}\)
= 21 Ω
From formula (ii),
V = 0.5 × 21
= 10.5 V

Question 4.
The magnitude of current density in a copper wire is 500 A/cm². If the number of free electrons per cm³ of copper is 8.47 × 10²², calculate the drift velocity of the electrons through the copper wire (charge on an e = 1.6 × 10^-19 C)
Answer:
Given: J = 500 A/cm² = 500 × 10⁴ A/m²,
n = 8.47 × 10²² electrons/cm³
= 8.47 × 10²⁸ electrons/m³
e = 1.6 × 10^-19 C
To Find: Drift velocity (v_d)
Formula: v_d = \(\frac {J}{ne}\)
Calculation:
From formula,
v_d = \(\frac {500×10^4}{8.47×10^{28}×1.6×10^{-19}}\)
= \(\frac {500}{8.47×1.6}\) × 10^-5
= {antilog [log 500 – log 8.47 – log 1.6]} × 10^-5
= {antilog [2.6990 – 0.9279 -0.2041]} × 10^-5
= {antilog [1.5670]} × 10^-5
= 3.690 × 10¹ × 10^-5
= 3.69 × 10^-4 m/s

Question 5.
Three resistors 10 Ω, 20 Ω and 30 Ω are connected in series combination.
i. Find equivalent resistance of series combination.
ii. When this series combination is connected to 12 V supply, by neglecting the value of internal resistance, obtain potential difference across each resistor.
Answer:
Given: R₁ = 10 Ω, R₂ = 20 Ω,
R₃ = 30 Ω, V = 12 V
To Find: i. Series equivalent resistance(R_s)
ii. Potential difference across each resistor (V₁, V₂, V₃)
Formula: i. R_s = R₁ + R₂ + R₃
ii. V = IR
Calculation:
From formula (i),
R_s = 10 + 20 + 30 = 60 Ω
From formula (ii),
I = \(\frac {V}{R}\) = \(\frac {12}{60}\) = 0.2 A
∴ Potential difference across R₁,
V₁ = I × R₁ = 0.2 × 10 = 2 V
∴ Potential difference across R₂,
V₂ = 0.2 × 20 = 4 V
∴ Potential difference across R₃,
V₃ = 0.2 × 30 = 6 V

Question 6.
Two resistors 1 Ω and 2 Ω are connected in parallel combination.
i. Find equivalent resistance of parallel combination.
ii. When this parallel combination is connected to 9 V supply, by neglecting internal resistance, calculate current through each resistor.
Answer:
R₁ = 1 kΩ = 10³ Ω,
R₂ = 2 kΩ = 2 × 10³ Ω, V = 9 V
To find:
i. Parallel equivalent resistance (R_p)
ii. Current through 1 kΩ and 2 kΩ (I₁ and I₂)
Formula:
i. \(\frac {1}{R_p}\) = \(\frac {1}{R_1}\) + \(\frac {1}{R_2}\)
ii. V = IR
Calculation: From formula (i),
\(\frac {1}{R_p}\) = \(\frac {1}{10^3}\) + \(\frac {1}{2×10^3}\)
= \(\frac {3}{2×10^3}\)
∴ R^p = \(\frac {2×10^3}{3}\) = 0.66 kΩ
From formula (ii),
I₁ = \(\frac {V}{R_1}\) + \(\frac {9}{10^3}\)
= 9 × 10^-3 A
= 3 mA
I₂ = \(\frac {V}{R_2}\) + \(\frac {9}{2×10^3}\)
= 4.5 × 10^-3 A
= 4.5 mA

Question 7.
A silver wire has a resistance of 4.2 Ω at 27°C and resistance 5.4 Ω at 100°C. Determine the temperature coefficient of resistance.
Answer:
Given: R₁ =4.2 Ω, R₂ = 5.4 Ω,
T, = 27° C, T2= 100 °C
To find: Temperature coefficient of resistance (α)
Formula: α = \(\frac {R_2-R_1}{R_1(T_2-T_1)}\)
Calculation:
From Formula
α = \(\frac {5.4-4.2}{4.2(100-27)}\) = 3.91 × 10^-3/°C

Question 8.
A 6 m long wire has diameter 0.5 mm. Its resistance is 50 Ω. Find the resistivity and conductivity.
Answer:
Given: l = 6 m, D = 0.5 mm,
r = 0.25 mm = 0.25 × 10^-3 m, R = 50 Ω
To find:
i. Resistivity (ρ)
ii. Conductivity (σ)
Formulae:
i. ρ = \(\frac {RA}{l}\) = \(\frac {Rπr^2}{l}\)
ii. σ = \(\frac {1}{ρ}\)
Calculation:
From formula (i),
ρ = \(\frac {50×3.142×(0.25×10^{-3})^2}{6}\)
= {antilog [log 50 + log 3.142 + 21og 0.25 -log 6]} × 10^-6
= {antilog [ 1.6990 + 0.4972 + 2(1.3979) -0.7782]} × 10^-6
= {antilog [2.1962 + 2 .7958 – 0.7782]} × 10^-6
= {antilog [0.9920 – 0.7782]} × 10^-6
= {antilog [0.2138]} × 10^-6
= 1.636 × 10^-6 Ω/m
From formula (ii),
σ = \(\frac {1}{1.636×10^{-6}}\)
= 0.6157 × 10⁶
….(Using reciprocal from log table)
= 6.157 × 10⁵ m/Ω

Question 9.
Find the value of resistances for the following colour code.
i. Blue Green Red Gold
ii. Brown Black Red Silver
iii. Red Red Orange Gold
iv. Orange White Red Gold
v. Yellow Violet Brown Silver
Answer:
i. Given: Blue – Green – Red – Gold
To find: Value of resistance
Formula: Value of resistance
= (xy × 10^z ± T%)Ω
Calculation:

From formula,
Value of resistance = (65 × 10² ± 5%) Ω
Value of resistance = 6.5 kΩ ± 5%

ii. Given: Brown – Black – Red – Silver
To find: Value of resistance
Formula: Value of resistance
= (xy × 10^z + T%) Ω
Calculation:

From formula,
Value of resistance = (10 × 10² ± 10%) Ω
Value of resistance = 1.0 kΩ ± 10%

iii. Given: Red – Red – Orange – Gold
To find: Value of the resistance
Formula: Value of the resistance
= (xy × 10^z ± T%)
Calculation:

From formula,
Value of resistance = (22 × 10³ ± 5%)Ω
Value of resistance = 22 kΩ ± 5%
[Note: The answer given above is presented considering correct order of magnitude.]

iv. Given: Orange – White – Red – Gold
To find: Value of the resistance
Formula: Value of the resistance
= (xy × 10^z ± T%)
Calculation:

From formula,
Value of resistance = (39 × 10² ± 5%) Ω
Value of resistance = 3.9 kΩ ± 5%

v. Given: Yellow-Violet-Brown-Silver
To find: Value of the resistance
Formula: Value of the resistance
= (xy × 10^z ± T%)
Calculation:

From formula,
Value of resistance = (47 × 10 ± 10%) Ω
Value of resistance = 470 Ω ± 10%
[Note: The answer given above is presented considering correct order of magnitude.]

Question 10.
Find the colour code for the following value of resistor having tolerance ± 10%.
i. 330 Ω
ii. 100 Ω
iii. 47 kΩ
iv. 160 Ω
v. 1 kΩ
Answer:

Question 11.
A current of 4 A flows through an automobile headlight. How many electrons flow through the headlight in a time of 2 hrs?
Answer:
Given: I = 4 A, t = 2 hrs = 2 × 60 × 60 s
To find: Number of electrons (N)
Formula: I = \(\frac {q}{t}\) = \(\frac {Ne}{t}\)
Calculation: As we know, e = 1.6 × 10^-19 C
From formula,
N = \(\frac {It}{e}\) = \(\frac {4×2×60×60}{1.6×10^{-19}}\) = 18 × 10^-23

Question 12.
The heating element connected to 230 V draws a current of 5 A. Determine the amount of heat dissipated in 1 hour (J = 4.2 J/cal).
Answer:
Given: V = 230 V, I = 5 A,
At = 1 hour = 60 × 60 sec
To find: Heat dissipated (H)
Formula: H = ∆U = I∆tV
Calculation: From formula,
H = 5 × 60 × 60 × 230
= 4.14 × 10⁶ J
Heat dissipated in calorie,
H = \(\frac {4.14×10^6}{4.2}\) = 985.7 × 10³ cal
= 985.7 kcal

11th Physics Digest Chapter 11 Electric Current Through Conductors Intext Questions and Answers

Can you recall? (Textbookpage no. 207)

An electric current in a metallic conductor such as a wire is due to the flow of electrons, the negatively charged particles in the wire. What is the role of the valence electrons which are the outermost electrons of an atom?
Answer:
i. The valence electrons become de-localized when a large number of atoms come together in a metal.
ii. These electrons become conduction electrons or free electrons constituting an electric current when a potential difference is applied across the conductor.

Internet my friend (Textbook page no. 218)

https://www.britannica.com/science/supercond activity physics

[Students are expected to visit the above-mentioned website and Collect more information about superconductivity.]

Balbharti Maharashtra State Board 11th Physics Textbook Solutions Chapter 10 Electrostatics Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Physics Solutions Chapter 10 Electrostatics

1. Choose the correct option.

Question 1.
A positively charged glass rod is brought close to a metallic rod isolated from the ground. The charge on the side of the metallic rod away from the glass rod will be
(A) same as that on the glass rod and equal in quantity
(B) opposite to that on the glass of and equal in quantity
(C) same as that on the glass rod but lesser in quantity
(D) same as that on the glass rod but more in quantity
Answer:
(A) same as that on the glass rod and equal in quantity

Question 2.
An electron is placed between two parallel plates connected to a battery. If the battery is switched on, the electron will
(A) be attracted to the +ve plate
(B) be attracted to the -ve plate
(C) remain stationary
(D) will move parallel to the plates
Answer:
(A) be attracted to the +ve plate

Question 3.
A charge of + 7 µC is placed at the centre of two concentric spheres with radius 2.0 cm and 4.0 cm respectively. The ratio of the flux through them will be
(A) 1 : 4
(B) 1 : 2
(C) 1 : 1
(D) 1 : 16
Answer:
(C) 1 : 1

Question 4.
Two charges of 1.0 C each are placed one meter apart in free space. The force between them will be
(A) 1.0 N
(B) 9 × 10⁹ N
(C) 9 × 10^-9 N
(D) 10 N
Answer:
(B) 9 × 10⁹ N

Question 5.
Two point charges of +5 µC are so placed that they experience a force of 80 × 10^-3 N. They are then moved apart, so that the force is now 2.0 × 10^-3 N. The distance between them is now
(A) 1/4 the previous distance
(B) double the previous distance
(C) four times the previous distance
(D) half the previous distance
Answer:
(B) double the previous distance

Question 6.
A metallic sphere A isolated from ground is charged to +50 µC. This sphere is brought in contact with other isolated metallic sphere B of half the radius of sphere A. The charge on the two sphere will be now in the ratio
(A) 1 : 2
(B) 2 : 1
(C) 4 : 1
(D) 1 : 1
Answer:
(D) 1 : 1

Question 7.
Which of the following produces uniform electric field?
(A) point charge
(B) linear charge
(C) two parallel plates
(D) charge distributed an circular any
Answer:
(C) two parallel plates

Question 8.
Two point charges of A = +5.0 µC and B = -5.0 µC are separated by 5.0 cm. A point charge C = 1.0 µC is placed at 3.0 cm away from the centre on the perpendicular bisector of the line joining the two point charges. The charge at C will experience a force directed towards
(A) point A
(B) point B
(C) a direction parallel to line AB
(D) a direction along the perpendicular bisector
Answer:
(C) a direction parallel to line AB

2. Answer the following questions.

Question 1.
What is the magnitude of charge on an electron?
Answer:
The magnitude of charge on an electron is 1.6 × 10^-19 C

Question 2.
State the law of conservation of charge.
Answer:
In any given physical process, charge may get transferred from one part of the system to another, but the total charge in the system remains constant”
OR
For an isolated system, total charge cannot be created nor destroyed.

Question 3.
Define a unit charge.
Answer:
Unit charge (one coulomb) is the amount of charge which, when placed at a distance of one metre from another charge of the same magnitude in vacuum, experiences a force of 9.0 × 10⁹ N.

Question 4.
Two parallel plates have a potential difference of 10V between them. If the plates are 0.5 mm apart, what will be the strength of electric charge.
Answer:
V=10V
d = 0.5 mm = 0.5 × 10^-3 m
To find: The strength of electric field (E)
Formula: E = \(\frac {V}{d}\)
Calculation: From formula,
E = \(\frac {10}{0.5×10^{-3}}\)
20 × 10³ V/m

Question 5.
What is uniform electric field?
Answer:
A uniform electric field is a field whose magnitude and direction are same at all points. For example, field between two parallel plates as shown in the diagram.

Question 6.
If two lines of force intersect of one point. What does it mean?
Answer:
If two lines of force intersect of one point, it would mean that electric field has two directions at a single point.

Question 7.
State the units of linear charge density.
Answer:
SI unit of λ is (C / m).

Question 8.
What is the unit of dipole moment?
Answer:
i. Strength of a dipole is measured in terms of a quantity called the dipole moment.

ii. Let q be the magnitude of each charge and 2\(\vec{l}\) be the distance from negative charge to positive charge. Then, the product q(2\(\vec{l}\)) is called the dipole moment \(\vec{p}\).

iii. Dipole moment is defined as \(\vec{p}\) = q(2\(\vec{l}\))

iv. A dipole moment is a vector whose magnitude is q (2\(\vec{l}\)) and the direction is from the negative to the positive charge.

v. The unit of dipole moment is coulomb-metre (C m) or debye (D).

Question 9.
What is relative permittivity?
Answer:
i. Relative permittivity or dielectric constant is the ratio of absolute permittivity of a medium to the permittivity of free space.
It is denoted as K or ε_r.
i.e., K or ε_r = \(\frac {ε}{ε_0}\)

ii. It is the ratio of the force between two point charges placed a certain distance apart in free space or vacuum to the force between the same two point charges when placed at the same distance in the given medium.
i.e., K or ε_r = \(\frac {F_{vacuum}}{F_{medium}}\)

iii. It is also called as specific inductive capacity or dielectric constant.

3. Solve numerical examples.

Question 1.
Two small spheres 18 cm apart have equal negative charges and repel each other with the force of 6 × 10^-8 N. Find the total charge on both spheres.
Solution:
Given: F = 6 × 10^-8 N, r = 18 cm = 18 × 10^-2 m
To find: Total charge (q₁ + q₂)
Formula: F = \(\frac {1}{4πε_0}\) \(\frac {q_1q_2}{r^2}\)
Calculation: From formula,

Taking square roots from log table,
∴ q = -4.648 × 10^-10 C
….(∵ the charges are negative)
Total charge = q₁ + q₂ = 2q
= 2 × (-4.648) × 10^-10
= -9.296 × 10^-10 C

Question 2.
A charge + q exerts a force of magnitude – 0.2 N on another charge -2q. If they are separated by 25.0 cm, determine the value of q.
Answer:
Given: q₁ = + q, q₂ = -2q, F = -0.2 N
r = 25 cm = 25 × 10^-2 m
To find: Charge (q)
Formula: F = \(\frac {1}{4πε_0}\) \(\frac {q_1q_2}{r^2}\)
Calculation: From formula,

[Note: The answer given above is calculated in accordance with textual method considering the given data]

Question 3.
Four charges of +6 × 10^-8 C each are placed at the corners of a square whose sides are 3 cm each. Calculate the resultant force on each charge and show in direction on a diagram drawn to scale.
Answer:
Given: q_A = q_B = q_C = q_D = 6 × 10^-8 C, a = 3 cm

∴ Resultant force on ‘A’
= F_AD cos 45 + F_AB cos 45 + F_AC
= (3.6 × 10^-2 × \(\frac {1}{√2}\)) + (3.6 × 10^-2 × \(\frac {1}{√2}\)) + 1.8 × 10^-2
= 6.89 × 10^-2 N directed along \(\vec{F_{AC}}\)

Question 4.
The electric field in a region is given by \(\vec{E}\) = 5.0 \(\hat{k}\) N/C Calculate the electric flux through a square of side 10.0 cm in the following cases
i. The square is along the XY plane
ii. The square is along XZ plane
iii. The normal to the square makes an angle of 45° with the Z axis.
Answer:
Given: \(\vec{E}\) = 5.0 \(\hat{k}\) N/C, |E| = 5 N/C
l = 10 cm = 10 × 10^-2 m = 10^-1 m
A = l² – 10^-2m²
To find: Electric flux in three cases.
(ø₁) (ø₂) (ø₃)
Formula: (ø₁) = EA cos θ
Calculation:
Case I: When square is along the XY plane,
∴ θ = 0
ø₁ = 5 × 10^-2 cos 0
= 5 × 10^-2 V m

Case II: When square is along XZ plane,
∴ θ = 90°
ø₁ = 5 × 10^-2 cos 90° = 0 V m

Case III: When normal to the square makes an angle of 45° with the Z axis.
∴ 0 = 45°
∴ ø₃ = 5 × 10^-2 × cos 45°
= 3.5 × 10^-2 V m

Question 5.
Three equal charges of 10 × 10^-8 C respectively, each located at the corners of a right triangle whose sides are 15 cm, 20 cm and 25 cm respectively. Find the force exerted on the charge located at the 90° angle.
Answer:
Given: q_A = q_B = q_C = 10 × 10^-8

Force on B due to A,

Question 6.
A potential difference of 5000 volt is applied between two parallel plates 5 cm apart. A small oil drop having a charge of 9.6 x 10-19 C falls between the plates. Find (i) electric field intensity between the plates and (ii) the force on the oil drop.
Answer:
Given: V = 5000 volt, d = 5 cm = 5 × 10^-2 m
q = 9.6 × 10^-19 C
To find:
i. Electric field intensity (E)
ii. Force (F)
Formula:
i. E = \(\frac {V}{d}\) \(\frac {q}{r}\)
ii. E = \(\frac {F}{q}\)
Calculation: From formula (i),
E = \(\frac {F}{q}\) = 10⁵ N/C
From formula (ii)
F = E x q
= 10⁵ × 9.6 × 10^-19
= 9.6 × 10^-14 N

Question 7.
Calculate the electric field due to a charge of -8.0 × 10^-8 C at a distance of 5.0 cm from it.
Answer:
Given: q = – 8 × 10^-8 C, r = 5 cm = 5 × 10^-2 m
To Find: Electric field (E)
Formula: E = \(\frac {1}{4πε_0}\) \(\frac {q}{r^2}\)
Calculation: From formula,
E = 9 × 10⁹ × \(\frac {(-8×10^{-8})}{(5×10^{-2})^2}\)
= -2.88 × 10⁵ N/C

11th Physics Digest Chapter 10 Electrostatics Intext Questions and Answers

Can you recall? (Textbookpage no. 188)

Question 1.
Have you experienced a shock while getting up from a plastic chair and shaking hand with your friend?
Answer:
Yes, sometimes a shock while getting up from a plastic chair and shaking hand with friend is experienced.

Question 2.
Ever heard a crackling sound while taking out your sweater in winter?
Answer:
Yes, sometimes while removing our sweater in winter, some crackling sound is heard and the sweater appears to stick to body.

Question 3.
Have you seen the lightning striking during pre-monsoon weather?
Answer:
Yes, sometimes lightning striking during pre-pre-monsoon weather seen.

Can you tell? (Textbook page no. 189)

i. When a petrol or a diesel tanker is emptied in a tank, it is grounded.
ii. A thick chain hangs from a petrol or a diesel tanker and it is in contact with ground when the tanker is moving.
Answer:
i. When a petrol or a diesel tanker is emptied in a tank, it is grounded so that it has an electrically conductive connection from the petrol or diesel tank to ground (Earth) to allow leakage of static and electrical charges.

ii. Metallic bodies of cars, trucks or any other big vehicles get charged because of friction between them and the air rushing past them. Hence, a thick chain is hanged from a petrol or a diesel tanker to make a contact with ground so that charge produced can leak to the ground through chain.

Can you tell? (Textbook page no. 194)

Three charges, q each, are placed at the vertices of an equilateral triangle. What will be the resultant force on charge q placed at the centroid of the triangle?
Answer:

Since AD. BE and CF meets at O, as centroid of an equilateral triangle.
∴ OA = OB = OC
∴ Let, r = OA = OB = OC
Force acting on point O due to charge on point A,
\(\overrightarrow{\mathrm{F}}_{\mathrm{OA}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}^{2}}{\mathrm{r}_{i}^{2}} \hat{\mathrm{r}}_{\mathrm{AO}}\)
Force acting on point O due to charge on point B,
\(\overrightarrow{\mathrm{F}}_{O \mathrm{~B}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}^{2}}{\mathrm{r}^{2}} \hat{\mathrm{r}}_{\mathrm{BO}}\)
Force acting on point O due to charge on point C,
\(\overrightarrow{\mathrm{F}}_{\mathrm{OC}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}^{2}}{\mathrm{r}^{2}} \hat{\mathrm{r}} \mathrm{co}\)
∴ Resultant force acting on point O,
F = \(\vec{F}\)_OA + \(\vec{F}\)_OB + \(\vec{F}\)_OC
On resolving \(\vec{F}\)_OB and \(\vec{F}\)_OC, we get –\(\vec{F}\)_OA
i.e., \(\vec{F}\)_OB + \(\vec{F}\)_OC = –\(\vec{F}\)_OA
∴ \(\vec{F}\) = \(\vec{F}\)_OA – \(\vec{F}\)_OA = 0
Hence, the resultant force on the charge placed at the centroid of the equilateral triangle is zero.

Can you tell? (Textbook page no. 197)

Why a small voltage can produce a reasonably large electric field?
Answer:

1. Electric field produced depends upon voltage as well as separation distance.
2. Electric field varies linearly with voltage and inversely with distance.
3. Hence, even if voltage is small, it can produce a reasonable large electric field when the gap between the electrode is reduced significantly.

Can you tell? (Textbook page no. 198)

Lines of force are imaginary; can they have any practical use?
Answer:
Yes, electric lines of force help us to visualise the nature of electric field in a region.

Can you tell? (Textbook page no. 204)

The surface charge density of Earth is σ = -1.33 nC/m². That is about 8.3 × 10⁹ electrons per square metre. If that is the case why don’t we feel it?
Answer:
The Earth along with its atmosphere acts as a neutral system. The atmosphere (ionosphere in particular) has nearly equal and opposite charge.

As a result, there exists a mechanism to replenish electric charges in the form of continual thunderstorms and lightning that occurs in different parts of the globe. This makes average charge on surface of the Earth as zero at any given time instant. Hence, we do not feel it.

Internet my friend (Textbook page no. 205)

i. https://www.physicsclassroom.com/class
ii. https://courses.lumenleaming.com/physics/
iii. https://www,khanacademy.org/science
iv. https://www.toppr.com/guides/physics/
[Students are expected to visit the above mentioned websites and collect more information about electrostatics.]

Balbharti Maharashtra State Board 11th Physics Textbook Solutions Chapter 9 Optics Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Physics Solutions Chapter 9 Optics

1. Multiple Choise Questions

Question 1.
As per the recent understanding, light consists of
(A) rays
(B) waves
(C) corpuscles
(D) photons obeying the rules of waves
Answer:
(D) photons obeying the rules of waves

Question 2.
Consider the optically denser lenses P, Q, R, and S drawn below. According to the Cartesian sign convention which of these have a positive focal length?

(A) Only P
(B) Only P and Q
(C) Only P and R
(D) Only Q and S
Answer:
(B) Only P and Q

Question 3.
Two plane mirrors are inclined at angle 40° between them. Number of images seen of a tiny object kept between them is
(A) Only 8
(B) Only 9
(C) 8 or 9
(D) 9 or 10
Answer:
(C) 8 or 9

Question 4.
A concave mirror of curvature 40 cm, used for shaving purpose produces image of double size as that of the object. Object distance must be
(A) 10 cm only
(B) 20 cm only
(C) 30 cm only
(D) 10 cm or 30 cm
Answer:
(D) 10 cm or 30 cm

Question 5.
Which of the following aberrations will NOT occur for spherical mirrors?
(A) Chromatic aberration
(B) Coma
(C) Distortion
(D) Spherical aberration
Answer:
(A) Chromatic aberration

Question 6.
There are different fish, monkeys and water of the habitable planet of the star Proxima b. A fish swimming underwater feels that there is a monkey at 2.5 m on the top of a tree. The same monkey feels that the fish is 1.6 m below the water surface. Interestingly, height of the tree and the depth at which the fish is swimming are exactly same. Refractive index of that water must be
(A) 6/5
(B) 5/4
(C) 4/3
(D) 7/5
Answer:
(B) 5/4

Question 7.
Consider following phenomena/applications: P) Mirage, Q) rainbow, R) Optical fibre and S) glittering of a diamond. Total internal reflection is involved in
(A) Only R and S
(B) Only R
(C) Only P, R and S
(D) all the four
Answer:
(A) Only R and S

Question 8.
A student uses spectacles of number -2 for seeing distant objects. Commonly used lenses for her/his spectacles are
(A) bi-concave
(B) piano concave
(C) concavo-convex
(D) convexo-concave
Answer:
(A) bi-concave

Question 9.
A spherical marble, of refractive index 1.5 and curvature 1.5 cm, contains a tiny air bubble at its centre. Where will it appear when seen from outside?
(A) 1 cm inside
(B) at the centre
(C) 5/3 cm inside
(D) 2 cm inside
Answer:
(B) at the centre

Question 10.
Select the WRONG statement.
(A) Smaller angle of prism is recommended for greater angular dispersion.
(B) Right angled isosceles glass prism is commonly used for total internal reflection.
(C) Angle of deviation is practically constant for thin prisms.
(D) For emergent ray to be possible from the second refracting surface, certain minimum angle of incidence is necessary from the first surface.
Answer:
(A) Smaller angle of prism is recommended for greater angular dispersion.

Question 11.
Angles of deviation for extreme colours are given for different prisms. Select the one having maximum dispersive power of its material.
(A) 7°, 10°
(B) 8°, 11°
(C) 12°, 16°
(D) 10°, 14°
Answer:
(A) 7°, 10°

Question 12.
Which of the following is not involved in formation of a rainbow?
(A) refraction
(B) angular dispersion
(C) angular deviation
(D) total internal reflection
Answer:
(D) total internal reflection

Question 13.
Consider following statements regarding a simple microscope:
(P) It allows us to keep the object within the least distance of distinct vision.
(Q) Image appears to be biggest if the object is at the focus.
(R) It is simply a convex lens.
(A) Only (P) is correct
(B) Only (P) and (Q) are correct
(C) Only (Q) and (R) are correct
(D) Only (P) and (R) are correct
Answer:
(D) Only (P) and (R) are correct

2. Answer the following questions.

Question 1.
As per recent development, what is the nature of light? Wave optics and particle nature of light are used to explain which phenomena of light respectively?
Answer:

1. As per recent development, it is now an established fact that light possesses dual nature. Light consists of energy carrier photons. These photons follow the rules of electromagnetic waves.
2. Wave optics explains the phenomena of light such as, interference, diffraction, polarisation, Doppler effect etc.
3. Particle nature of light can be used to explain phenomena like photoelectric effect, emission of spectral lines, Compton effect etc.

Question 2.
Which phenomena can be satisfactorily explained using ray optics?
Answer:
Ray optics or geometrical optics: Ray optics can be used for understanding phenomena like reflection, refraction, double refraction, total internal reflection, etc.

Question 3.
What is focal power of a spherical mirror or a lens? What may be the reason for using P = \(\frac {1}{f}\) its expression?
Answer:

1. Converging or diverging ability of a lens or of a mirror is defined as its focal power.
2. This implies, more the power of any spherical mirror or a lens, the more is its ability to converge or diverge the light that passes through it.
3. In case of convex lens or concave mirror, more the convergence, shorter is the focal length as shown in the figure.
4. Similarly, in case of concave lens or convex mirror, more the divergence, shorter is the focal length.
5. This explains that the focal power of any spherical lens or mirror is inversely proportional to the focal length.
6. Hence, the expression of focal power is given by the formula, P = \(\frac {1}{f}\).

Question 4.
At which positions of the objects do spherical mirrors produce (i) diminished image (ii) magnified image?
Answer:
i. Amongst the two types of spherical mirrors, convex mirror always produces a diminished image at all positions of the object.

ii. Concave mirror produces diminished image when object is placed:

• Beyond radius of curvature (i.e., u > 2f)
• At infinity (i.e., u = ∞)

iii. Concave mirror produces magnified image when object is placed:

• between centre of curvature and focus (i.e., 2f > u > f)
• between focus and pole of the mirror (i.e., u < f)

Question 5.
State the restrictions for having images produced by spherical mirrors to be appreciably clear.
Answer:
i. In order to obtain clear images, the formulae for image formation by mirrors or lens follow the given assumptions:

• Objects and images are situated close to the principal axis.
• Rays diverging from the objects are confined to a cone of very small angle.
• If there is a parallel beam of rays, it is paraxial, i.e., parallel and close to the principal axis.

ii. In case of spherical mirrors (excluding small aperture spherical mirrors), rays farther from the principle axis do not remain parallel to the principle axis. Thus, the third assumption is not followed and the focus gradually shifts towards the pole.

iii. The relation (f = \(\frac {R}{2}\)) giving a single point focus is not followed and the image does not get converged at a single point resulting into a distorted or defective image.

iv. This defect arises due to the spherical shape of the reflecting surface.

Question 6.
Explain spherical aberration for spherical mirrors. How can it be minimized? Can it be eliminated by some curved mirrors?
Answer:

1. In case of spherical mirrors (excluding small aperture spherical mirror), The rays coming from a distant object farther from principal axis do not remain parallel to the axis. Thus, the focus gradually shifts towards the pole.
2. The assumption for clear image formation namely, ‘If there is a parallel beam of rays, it is paraxial, i.e., parallel and close to the principal axis’, is not followed in this case.
3. The relation f (f = \(\frac {R}{2}\)) giving a single point focus is not followed and the image does not get converged at a single point resulting into a distorted or defective image.
4. This phenomenon is known as spherical aberration.
5. It occurs due to spherical shape of the reflecting surface, hence known as spherical aberration.
6. The rays near the edge of the mirror converge at focal point F_M Whereas, the rays near the principal axis converge at point F_P. The distance between F_M and F_P is measured as the longitudinal spherical aberration.
7. In spherical aberration, single point image is not possible at any point on the screen and the image formed is always a circle.
8. At a particular location of the screen (across AB in figure), the diameter of this circle is minimum. This is called the circle of least confusion. Radius of this circle is transverse spherical aberration.

Remedies for Spherical Aberration:

1. Spherical aberration can be minimized by reducing the aperture of the mirror.
2. Spherical aberration in curved mirrors can be completely eliminated by using a parabolic mirror.

Question 7.
Define absolute refractive index and relative refractive index. Explain in brief with an illustration for each.
Answer:
i. Absolute refractive index of a medium is defined as the ratio of speed of light in vacuum to that in the given medium.

ii. A stick or pencil kept obliquely in a glass containing water appears broken as its part in water appears to be raised.

iii. As the speed of light is different in two media, the rays of light coming from water undergo refraction at the boundary separating two media.

iv. Consider speed of light to be v in water and c in air. (Speed of light in air ~ speed of light in vacuum)
∴ refractive index of water = \(\frac {n_w}{n_s}\) = \(\frac {n_w}{n_{vacuum}}\) = \(\frac {c}{v}\)

v. Relative refractive index of a medium 2 is the refractive index of medium 2 with respect to medium 1 and it is defined as the ratio of speed of light v₁ in medium 1 to its speed v₁ in medium 2.
∴ Relative refractive index of medium 2,
¹n₂ = \(\frac {v_1}{v_2}\)

vi. Consider a beaker filled with water of absolute refractive index n₁ kept on a transparent glass slab of absolute refractive index n₂.

vii. Thus, the refractive index of water with respect to that of glass will be,
ⁿw₂ = \(\frac {n_2}{n_1}\) = \(\frac {c/v_2}{c/v_1}\) = \(\frac {v_1}{v_2}\)

Question 8.
Explain ‘mirage’ as an illustration of refraction.
Answer:
i. On a hot clear Sunny day, along a level road, there appears a pond of water ahead of the road. Flowever, if we physically reach the spot, there is nothing but the dry road and water pond again appears some distance ahead. This illusion is called mirage.

ii. Mirage results from the refraction of light through a non-uniform medium.

iii. On a hot day the air in contact with the road is hottest and as we go up, it gets gradually cooler. The refractive index of air thus decreases with height. Hot air tends to be less optically dense than cooler air which results into a non-uniform medium.

iv. Light travels in a straight line through a uniform medium but refracts when traveling through a non-uniform medium.

v. Thus, the ray of light coming from the top of an object get refracted while travelling downwards into less optically dense air and become more and more horizontal as shown in Figure.

vi. As it almost touches the road, it bends (refracts) upward. Then onwards, upward bending continues due to denser air.

vii. As a result, for an observer, it appears to be coming from below thereby giving an illusion of reflection from an (imaginary) water surface.

Question 9.
Under what conditions is total internal reflection possible? Explain it with a suitable example. Define critical angle of incidence and obtain an expression for it.
Answer:
Conditions for total internal reflection:
i. The light ray must travel from denser medium to rarer medium.

ii. The angle of incidence in the denser medium must be greater than critical angle for the given pair of media.

Total internal reflection in optical fibre:
iii. Consider an optical fibre made up of core of refractive index n₁ and cladding of refractive index n₂ such that, n₁ > n₂.

iv. When a ray of light is incident from a core (denser medium), the refracted ray is bent away from the normal.

v. At a particular angle of incidence i_c in the denser medium, the corresponding angle of refraction in the rarer medium is 90°.

vi. For angles of incidence greater than ic, the angle of refraction become larger than 90° and the ray does not enter into rarer medium at all but is reflected totally into the denser medium as shown in figure.

critical angle of incidence and obtain an expression:
i. Critical angle for a pair of refracting media can be defined as that angle of incidence in the denser medium for which the angle of refraction in the rarer medium is 90°.

ii. Let n be the relative refractive index of denser medium with respect to the rarer.

iii. Then, according to Snell’s law,
n = \(\frac {n_{denser}}{n_{rarer}}\) = \(\frac {sin r}{sin i_c}\) = \(\frac {sin 90°}{sin i_c}\)
∴ sin (i_c) = \(\frac {1}{n}\)

Question 10.
Describe construction and working of an optical fibre. What are the advantages of optical fibre communication over electronic communication?
Answer:

Construction:

1. An optical fibre consists of an extremely thin, transparent and flexible core surrounded by an optically rarer flexible cover called cladding.
2. For protection, the whole system is coated by a buffer and a jacket.
3. Entire thickness of the fibre is less than half a mm.
4. Many such fibres can be packed together in an outer cover.

Working:

1. Working of optical fibre is based on the principle of total internal reflection.
2. An optical signal (a ray of light) entering the core suffers multiple total internal reflections before emerging out after a several kilometres.
3. The optical signal travels with the highest possible speed in the material.
4. The emerged optical signal has extremely low loss in signal strength.

Advantages of optical fibre communication over electronic communication:

1. Broad bandwidth (frequency range): For TV signals, a single optical fibre can carry over 90000 independent signals (channels).
2. Immune to EM interference: Optical fibre being electrically non-conductive, does not pick up nearby EM signals.
3. Low attenuation loss: loss being lower than 0.2 dB/km, a single long cable can be used for several kilometres.
4. Electrical insulator: Optical fibres being electrical insulators, ground loops of metal wires or lightning do not cause any harm.
5. Theft prevention: Optical fibres do not use copper or other expensive material which are prone to be robbed.
6. Security of information: Internal damage is most unlikely to occur, keeping the information secure.

Question 11.
Why are prism binoculars preferred over traditional binoculars? Describe its working in brief.
Answer:

1. Traditional binoculars use only two cylinders. Distance between the two cylinders can’t be greater than that between the two eyes. This creates a limitation of field of view.
2. A prism binocular has two right angled glass prisms which apply the principle of total internal reflection.
3. The incident light rays are reflected internally twice giving the viewer a wider field of view. For this reason, prism binoculars are preferred over traditional binoculars.

Working:

1. The prism binoculars consist of 4 isosceles, right angled prisms of material having critical angle less than 45°.
2. The prism binoculars have a wider input range compared to traditional binoculars.
3. The light rays incident on the prism binoculars, first get total internally reflected by the isosceles, right angled prisms 1 and 4.
4. These reflected rays undergo another total internal reflection by prisms 2 and 3 to form the final image.

Question 12.
A spherical surface separates two transparent media. Derive an expression that relates object and image distances with the radius of curvature for a point object. Clearly state the assumptions, if any.
Answer:
i. Consider a spherical surface YPY’ of radius curvature R, separating two transparent media of refractive indices n₁ and n₂ respectively with ni₁ < n₂.

ii. P is the pole and X’PX is the principal axis. A point object O is at a distance u from the pole, in the medium of refractive index n₁.

iii. In order to minimize spherical aberration, we consider two paraxial rays.

iv. The ray OP along the principal axis travels undeviated along PX. Another ray OA strikes the surface at A.

v. As n₁ < n₂, the ray deviates towards the normal (CAN), travels along AZ and real image of point object O is formed at I.

vi. Let α, β and γ be the angles subtended by incident ray, normal and refracted ray with the principal axis.
∴ i = (α + β) and r = (β – γ)

vii. As, the rays are paraxial, all the angles can be considered to be very small.
i.e., sin i ≈ i and sin r ≈ r
Angles α, β and γ can also be expressed as,

viii. According to Snell’s law,
n₁ sin (i) = n₂ sin (r)
For small angles, Snell’s law can be written
as, n₁i = n₂r
∴ n₁ (α + β) = n₂ (β – γ)
∴ (n₂ – n₁)β = n₁α + n₂γ
Substituting values of α, β and γ, we get,
(n₂ – n₁) \(\frac {arc PA}{R}\) = n₁(\(\frac {arc PA}{-u}\)) + n₂(\(\frac {arc PA}{v}\))
∴ \(\frac {(n_2-n_1)}{R}\) = \(\frac {n_2}{v}\) – \(\frac {n_1}{u}\)

Assumptions: To derive an expression that relates object and image distances with the radius of curvature for a point object, the two rays considered are assumed to be paraxial thus making the angles subtended by incident ray, normal and refracted ray with the principal axis very small.

Question 13.
Derive lens makers’ equation. Why is it called so? Under which conditions focal length f and radii of curvature R are numerically equal for a lens?
Answer:
i. Consider a lens of radii of curvature Ri and R2 kept in a medium such that refractive index of material of the lens with respect to the medium is denoted as n.

ii. Assuming the lens to be thin, P is the common pole for both the surfaces. O is a point object on the principal axis at a distance u from P.

iii. The refracting surface facing the object is considered as first refracting surface with radii R₁.

iv. In the absence of second refracting surface, the paraxial ray OA deviates towards normal and would intersect axis at I₁. PI₁ = V₁ is the image distance for intermediate image I₁.

Before reaching I₁, the incident rays (AB and OP) strike the second refracting surface. In this case, image I₁ acts as a virtual object for second surface.

vii. For second refracting surface,
n₂ = 1, n₁ = n, R = R₂, u = v₁ and PI = v
∴ \(\frac{(1-\mathrm{n})}{\mathrm{R}_{2}}=\frac{1}{\mathrm{v}}-\frac{\mathrm{n}}{\mathrm{v}_{1}}-\frac{(\mathrm{n}-1)}{\mathrm{R}_{2}}=\frac{1}{\mathrm{v}}-\frac{\mathrm{n}}{\mathrm{v}_{1}}\) ………… (2)

viii. Adding equations (1) and (2),
(n – 1) \(\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)=\frac{1}{v}-\frac{1}{u}\)
For object at infinity, image is formed at focus, i.e., for u = ∞, v = f. Substituting this in above equation,
\(\frac{1}{\mathrm{f}}=(\mathrm{n}-1)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)\) …………. (3)
This equation in known as the lens makers’ formula.

ix. Since the equation can be used to calculate the radii of curvature for the lens, it is called the lens makers’ equation.

x. The numeric value of focal length f and radius of curvature R is same under following two conditions:
Case I:
For a thin, symmetric and double convex lens made of glass (n = 1.5), R₁ is positive and R₂ is negative but, |R₁| = |R₂|.
In this case,
\(\frac{1}{\mathrm{f}}=(1.5-1)\left(\frac{1}{\mathrm{R}}-\frac{1}{-\mathrm{R}}\right)=0.5\left(\frac{2}{\mathrm{R}}\right)=\frac{1}{\mathrm{R}}\)
∴ f = R

Case II:
Similarly, for a thin, symmetric and double concave lens made of glass (n = 1.5), R₁ is negative and R₂ is positive but, |R₁| = |R₂|.
In this case,
\(\frac{1}{\mathrm{f}}=(1.5-1)\left(\frac{1}{-\mathrm{R}}-\frac{1}{\mathrm{R}}\right)=0.5\left(-\frac{2}{\mathrm{R}}\right)=-\frac{1}{\mathrm{R}}\)
∴ f = -R or |f| = |R|

3. Answer the following questions in detail.

Question 1.
What are different types of dispersions of light? Why do they occur?
Answer:
i. There are two types of dispersions:
a. Angular dispersion
b. Lateral dispersion

ii. The refractive index of material depends on the frequency of incident light. Hence, for different colours, refractive index of material is different.

iii. For an obliquely incident ray, the angles of refraction are different for each colour and they separate as they travel along different directions resulting into angular dispersion.

iv. When a polychromatic beam of light is obliquely incident upon a plane parallel transparent slab, emergent beam consists of all component colours separated out.

v. In this case, these colours are parallel to each other and are also parallel to their initial direction resulting into lateral dispersion

Question 2
Define angular dispersion for a prism. Obtain its expression for a thin prism. Relate it with the refractive indices of the material of the prism for corresponding colours.
Answer:
i. If a polychromatic beam is incident upon a prism, the emergent beam consists of all the individual colours angularly separated. This phenomenon is known as angular dispersion for a prism.

ii. For any two component colours, angular dispersion is given by,
δ₂₁ = δ₂ – δ₁

iii. For white light, we consider two extreme colours viz., red and violet.
∴ δ_VR = δ_V – δ_R

iv. For thin prism,
δ = A(n – 1)
δ₂₁ = δ₂ – δ₁
= A(n₂ – 1) – A(n₁ – 1) = A(n₂ – n₁)
where n₁ and n₂ are refractive indices for the two colours.

v. For white light,
δ_VR = δ_V – δ_R
= A(n_V – 1) – A(n_R – 1) = A(n_V – n_R).

Question 3.
Explain and define dispersive power of a transparent material. Obtain its expressions in terms of angles of deviation and refractive indices.
Answer:
Ability of an optical material to disperse constituent colours is its dispersive power.

It is measured for any two colours as the ratio of angular dispersion to the mean deviation for those two colours. Thus, for the extreme colours of white light, dispersive power is given by,
\(\omega=\frac{\delta_{\mathrm{V}}-\delta_{\mathrm{R}}}{\left(\frac{\delta_{\mathrm{V}}+\delta_{\mathrm{R}}}{2}\right)} \approx \frac{\delta_{\mathrm{V}}-\delta_{\mathrm{R}}}{\delta_{\mathrm{Y}}}=\frac{\mathrm{A}\left(\mathrm{n}_{\mathrm{V}}-\mathrm{n}_{\mathrm{R}}\right)}{\mathrm{A}\left(\mathrm{n}_{\mathrm{Y}}-1\right)}=\frac{\mathrm{n}_{\mathrm{V}}-\mathrm{n}_{\mathrm{R}}}{\mathrm{n}_{\mathrm{Y}}-1}\)

Question 4.
(i) State the conditions under which a rainbow can be seen.
Answer:
A rainbow can be observed when there is a light shower with relatively large raindrop occurring during morning or evening time with enough sunlight around.

(ii) Explain the formation of a primary rainbow. For which angular range with the horizontal is it visible?
Answer:
i. A ray AB incident from Sun (white light) strikes the upper portion of a water drop at an incident angle i.

ii. On entering into water, it deviates and disperses into constituent colours. The figure shows the extreme colours (violet and red).

iii. Refracted rays BV and BR strike the opposite inner surface of water drop and suffer internal reflection.

iv. These reflected rays finally emerge from V’ and R’ and can be seen by an observer on the ground.

v. For the observer they appear to be coming from opposite side of the Sun.

vi. Minimum deviation rays of red and violet colour are inclined to the ground level at θ_R = 42.8° ≈ 43° and θ_V = 40.8 ≈ 41° respectively. As a result, in the rainbow, the red is above and violet is below.

(iii) Explain the formation of a secondary rainbow. For which angular range with the horizontal is it visible?
Answer:
i. A ray AB incident from Sun (white light) strikes the lower portion of a water drop at an incident angle i.

ii. On entering into water, it deviates and disperses into constituent colours. The figure shows the extreme colours (violet and red).

iii. Refracted rays BV and BR finally emerge the drop from V’ and R’ after suffering two internal reflections and can be seen by an observer on the ground.

iv. Minimum deviation rays of red and violet colour are inclined to the ground level at θ_R ≈ 51° and θ_V ≈ 53° respectively. As a result, in the rainbow, the violet is above and red is below.

(iv) Is it possible to see primary and secondary rainbow simultaneously? Under what conditions?
Answer:
Yes, it is possible to see primary and secondary rainbows simultaneously. This can occur when the centres of both the rainbows coincide.

Question 5.
(i) Explain chromatic aberration for spherical lenses. State a method to minimize or eliminate it.
Answer:
Lenses are prepared by using a transparent material medium having different refractive index for different colours. Hence angular dispersion is present.
If the lens is thick, this will result into notably different foci corresponding to each colour for a polychromatic beam, like a white light. This defect is called chromatic aberration.
As violet light has maximum deviation, it is focussed closest to the pole.

Reducing/eliminating chromatic aberration:

1. Eliminating chromatic aberrations for all colours is impossible. Hence, it is minimised by eliminating aberrations for extreme colours.
2. This is achieved by using either a convex and a concave lens in contact or two thin convex lenses with proper separation. Such a combination is called achromatic combination.

(ii) What is achromatism? Derive a condition to achieve achromatism for a lens combination. State the conditions for it to be converging.
Answer:
i. To eliminate chromatic aberrations for extreme colours from a lens, either a convex and a concave lens in contact or two thin convex lenses with proper separation are used.

ii. This combination is called achromatic combination. The process of using this combination is termed as achromatism of a lens.

iii. Let ω₁ and ω₂ be the dispersive powers of materials of the two component lenses used in contact for an achromatic combination.

iv. Let V, R and Y denote the focal lengths for violet, red and yellow colours respectively.

v. For lens 1, let
K₁ = (\(\frac {1}{R_1}\)–\(\frac {1}{R_2}\))₁ and K₂ = (\(\frac {1}{R_1}\)–\(\frac {1}{R_2}\))₂

vi. For the combination to be achromatic, the resultant focal length of the combination must be the same for both the colours,

This is the condition for achromatism of a combination of lenses.

Condition for converging:
For this combination to be converging, fY must be positive.
Using equation (3), for f_Y to be positive, (f_Y)₁ < (f_Y)₂ ⇒ ω₁ < ω₂

Question 6.
Describe spherical aberration for spherical lenses. What are different ways to minimize or eliminate it?
Answer:
i. All the formulae used for image formation by lenses are based on some assumption. However, in reality these assumptions are not always true.

ii. A single point focus in case of lenses is possible only for small aperture spherical lenses and for paraxial rays.

iii. The rays coming from a distant object farther from principal axis no longer remain parallel to the axis. Thus, the focus gradually shifts towards pole.

iv. This defect arises due to spherical shape of the refracting surface, hence known as spherical aberration. It results into a blurred image with unclear boundaries.

v. As shown in figure, the rays near the edge of the lens converge at focal point F_M. Whereas, the rays near the principal axis converge at point F_P. The distance between F_M and F_P is measured as the longitudinal spherical aberration.

vi. In absence of this aberration, a single point image can be obtained on a screen. In the presence of spherical aberration, the image is always a circle.

vii. At a particular location of the screen (across AB in figure), the diameter of this circle is minimum. This is called the circle of least confusion. Radius of this circle is transverse spherical aberration.

Methods to eliminate/reduce spherical aberration in lenses:
i. Cheapest method to reduce the spherical aberration is to use a planoconvex or planoconcave lens with curved side facing the incident rays.

ii. Certain ratio of radii of curvature for a given refractive index almost eliminates the spherical aberration. For n = 1.5, the ratio is
\(\frac {R_1}{R_2}\) = \(\frac {1}{6}\) and for n = 2, \(\frac {R_1}{R_2}\) = \(\frac {1}{5}\)

iii. Use of two thin converging lenses separated by distance equal to difference between their focal lengths with lens of larger focal length facing the incident rays considerably reduces spherical aberration.

iv. Spherical aberration of a convex lens is positive (for real image), while that of a concave lens is negative. Thus, a suitable combination of them can completely eliminate spherical aberration.

Question 7.
Define and describe magnifying power of an optical instrument. How does it differ from linear or lateral magnification?
Answer:
i. Angular magnification or magnifying power of an optical instrument is defined as the ratio of the visual angle made by the image formed by that optical instrument (β) to the visual angle subtended by the object when kept at the least distance of distinct vision (α).

ii. The linear magnification is the ratio of the size of the image to the size of the object.

iii. When the distances of the object and image formed are very large as compared to the focal lengths of the instruments used, the magnification becomes infinite. Whereas, the magnifying power being the ratio of angle subtended by the object and image, gives the finite value.

iv. For example, in case of a compound microscope,
M_min = \(\frac {D}{f}\) = \(\frac {25}{5}\) = 5 and M_max = 1 + \(\frac {D}{f}\) = 6
Hence image appears to be only 5 to 6 times bigger for a lens of focal length 5 cm.
For M_min = \(\frac {D}{f}\) = 5, V = ∞
∴ Lateral magnification (m) = \(\frac {v}{u}\) = ∞
Thus, the image size is infinite times that of the object, but appears only 5 times bigger.

Question 8.
Derive an expression for magnifying power of a simple microscope. Obtain its minimum and maximum values in terms of its focal length.
Answer:
i. Figure (a) shows visual angle a made by an object, when kept at the least distance of distinct vision (D = 25 cm). Without an optical instrument this is the greatest possible visual angle as we cannot get the object closer than this.

ii. Figure (b) shows a convex lens forming erect, virtual and magnified image of the same object, when placed within the focus.

iii. The visual angle p of the object and the image in this case are the same. However, this time the viewer is looking at the image which is not closer than D. Hence the same object is now at a distance smaller than D.

iv. Angular magnification or magnifying power, in this case, is given by
M = \(\frac {Visual angle of theimage}{Visual angle of the object at D}\) = \(\frac {β}{α}\)
For small angles,
M = \(\frac {β}{α}\) ≈ \(\frac {tan(β)}{tan(α)}\) = \(\frac {BA/PA}{BA/D}\) = \(\frac {D}{u}\)

v. For maximum magnifying power, the image should be at D. For thin lens, considering thin lens formula.

Question 9.
Derive the expressions for the magnifying power and the length of a compound microscope using two convex lenses.
Answer:
i. The final image formed in compound microscope (A” B”) as shown in figure, makes a visual angle β at the eye.
Visual angle made by the object from distance D is α.

From figure,
tan β = \(\frac {A”B”}{v_c}\) = \(\frac {A’B’}{u_c}\)
and tan α = \(\frac {AB}{D}\)

Question 10.
What is a terrestrial telescope and an astronomical telescope?
Answer:

1. Telescopes used to see the objects on the Earth, like mountains, trees, players playing a match in a stadium, etc. are called terrestrial telescopes.
2. In such case, the final image must be erect. Eye lens used for this purpose must be concave and such a telescope is popularly called a binocular.
3. Most of the binoculars use three convex lenses with proper separation. The image formed by second lens is inverted with respect to object. The third lens again inverts this image and makes final image erect with respect to the object.
4. An astronomical telescope is the telescope used to see the objects like planets, stars, galaxies, etc. In this case there is no necessity of erect image. Such telescopes use convex lens as eye lens.

Question 11.
Obtain the expressions for magnifying power and the length of an astronomical telescope under normal adjustments.
Answer:
i. For telescopes, a is the visual angle of the object from its own position, which is practically at infinity.

ii. Visual angle of the final image is p and its position can be adjusted to be at D. However, under normal adjustments, the final image is also at infinity making a greater visual angle than that of the object.

iii. The parallax at the cross wires can be avoided by using the telescopes in normal adjustments.

iv. Objective of focal length f₀ focusses the parallel incident beam at a distance f₀ from the objective giving an inverted image AB.

v. For normal adjustment, the intermediate image AB forms at the focus of the eye lens. Rays refracted beyond the eye lens form a parallel beam inclined at an angle β with the principal axis.

vi. Angular magnification or magnifying power for telescope is given by,
M = \(\frac {β}{α}\) ≈ \(\frac {tan(β)}{tan(α)}\) = \(\frac {BA/P_cB}{BA/P_0B}\) = \(\frac {f_0}{f_e}\)

vii. Length of the telescope for normal adjustment is, L = f₀ + f_e.

Question 12.
What are the limitations in increasing the magnifying powers of (i) simple microscope (ii) compound microscope (iii) astronomical telescope?
Answer:
i. In case of simple microscope
\(\mathrm{M}_{\max }=\frac{\mathrm{D}}{\mathrm{u}}=1+\frac{\mathrm{D}}{\mathrm{f}}\)
Thus, the limitation in increasing the magnifying power is determined by the value of focal length and the closeness with which the lens can be held near the eye.

ii. In case of compound microscope,
M = \(\mathrm{m}_{0} \times \mathrm{M}_{\mathrm{e}}=\frac{\mathrm{v}_{0}}{\mathrm{u}_{\mathrm{o}}} \times \frac{\mathrm{D}}{\mathrm{u}_{\mathrm{e}}}\)
Thus, in order to increase m₀, we need to decrease u₀. Thereby, the object comes closer and closer to the focus of the objective. This increases v₀ and hence length of the microscope. Therefore, mQ can be increased only within the limitation of length of the microscope.

iii. In case of telescopes,
M = \(\frac {f_0}{f_e}\)
Where f₀ = focal length of the objective
f_e = focal length of the eye-piece
Length of the telescope for normal adjustment is, L = f₀ + f_e.
Thus, magnifying power of telescope can be increased only within the limitations of length of the telescope.

4. Solve the following numerical examples

Question 1.
A monochromatic ray of light strikes the water (n = 4/3) surface in a cylindrical vessel at angle of incidence 53°. Depth of water is 36 cm. After striking the water surface, how long will the light take to reach the bottom of the vessel? [Angles of the most popular Pythagorean triangle of sides in the ratio 3 : 4 : 5 are nearly 37°, 53° and 90°]
Answer:
From figure, ray PO = incident ray
ray OA = refracted ray
QOB = Normal to the water surface.
Given that,
∠POQ = angle of incidence (θ₁) = 53°
Seg OB = 36 cm and n_water = \(\frac {4}{3}\)
From Snell’s law,
n₁ sin θ₁ = n₂ sin θ₂
∴ n_water = \(\frac {sinθ_1}{sinθ_2}\)
Or sin θ₂ = \(\frac {sinθ_1}{n_{water}}\) = \(\frac {sin(53°)×3}{4}\)
∴ θ₂ ~ 37°
ΔOBA forms a Pythagorean triangle with angles 53°, 37° and 90°.
Thus, sides of ΔOBA will be in ratio 3 : 4 : 5 Such that OA forms the hypotenuse. From figure, we can infer that,
Seg OB = 4x = 36 cm
∴ x = 9
∴ seg OA = 5x = 45 cm and
seg AB = 3x = 27 cm.
This means the light has to travel 45 cm to reach the bottom of the vessel.
The speed of the light in water is given by,
v = \(\frac {c}{n}\)
∴ v = \(\frac {3×10^8}{4/3}\) = \(\frac {9}{4}\) × 10⁸ m/s
∴ Time taken by light to reach the bottom of vessel is,
t = \(\frac {s}{v}\) = \(\frac {45×10^{-2}}{\frac {9}{4} × 10^8}\) = 20 × 10^-10 = 2 ns or 0.002 µs

Question 2.
Estimate the number of images produced if a tiny object is kept in between two plane mirrors inclined at 35°, 36°, 40° and 45°.
Answer:
i. For θ₁ =35°
n₁ = \(\frac {360}{θ_1}\) = \(\frac {360}{35}\) = 10.28
As ni is non-integer, N₁ = integral part of n₁ = 10

ii. For θ₂ = 36°
n₂ = \(\frac {360}{36}\) = 10
As n₂ is even integer, N₂ = (n₂ – 1) = 9

iii. For θ₃ = 40°
n₃ = \(\frac {360}{36}\) = 9
As n₃ is odd integer.
Number of images seen (N₃) = n₃ – 1 = 8
(if the object is placed at the angle bisector) or Number of images seen (N₃) = n₃ = 9
(if the object is placed off the angle bisector)

iv. For θ₄ = 45°
n₄ = \(\frac {360}{45}\) = 8
As n₄ is even integer,
N₄ = n₄ – 1 = 7

Question 3.
A rectangular sheet of length 30 cm and breadth 3 cm is kept on the principal axis of a concave mirror of focal length 30 cm. Draw the image formed by the mirror on the same diagram, as far as possible on scale.

Answer:

[Note: The question has been modified and the ray digram is inserted in question in order to find the correct position of the image.]

Question 4.
A car uses a convex mirror of curvature 1.2 m as its rear-view mirror. A minibus of cross section 2.2 m × 2.2 m is 6.6 m away from the mirror. Estimate the image size.
Answer:
For a convex mirror,
f = +\(\frac {R}{2}\) = \(\frac {1.2}{2}\) = +0.6m
Given that, a minibus, approximately of a shape of square is at distance 6.6 m from mirror.
i.e., u = -6.6 m

∴ h₂ = 0.183 m
i.e., h₂ 0.2 m

Question 5.
A glass slab of thickness 2.5 cm having refractive index 5/3 is kept on an ink spot. A transparent beaker of very thin bottom, containing water of refractive index 4/3 up to 8 cm, is kept on the glass block. Calculate apparent depth of the ink spot when seen from the outside air.
Answer:
When observed from the outside air, the light coming from ink spot gets refracted twice; once through glass and once through water.
∴ When observed from water,

∴ Apparent depth = 2 cm
Now when observed from outside air, the total real depth of ink spot can be taken as (8 + 2) cm = 10 cm.
∴ \(\frac {n_w}{n_{air}}\) = \(\frac {Real depth}{Apparent depth}\)
∴ Apparent depth = \(\frac {10}{4/3}\)
= \(\frac {10×3}{4}\) = 7.5 cm

Question 64.
A convex lens held some distance above a 6 cm long pencil produces its image of SOME size. On shifting the lens by a distance equal to its focal length, it again produces the image of the SAME size as earlier. Determine the image size.
Answer:
For a convex lens, it is given that the image size remains unchanged after shifting the lens through distance equal to its focal length. From given conditions, it can be inferred that the object distance should be u = –\(\frac {f}{2}\)
Also, h₁ = 6 cm, v₁ = v₂
From formula for thin lenses,

Question 7.
Figure below shows the section ABCD of a transparent slab. There is a tiny green LED light source at the bottom left corner B. A certain ray of light from B suffers total internal reflection at nearest point P on the surface AD and strikes the surface CD at point Question Determine refractive index of the material of the slab and distance DQ. At Q, the ray PQ will suffer partial or total internal reflection?

Answer:

As, the light ray undergo total internal reflection at P, the ray BP may be incident at critical angle.
For a Pythagorean triangle with sides in ratio 3 : 4 : 5 the angle opposite to side 3 units is 37° and that opposite to 4 units is 53°.
Thus, from figure, we can say, in ΔBAP
∠ABP = 53°
∠BPN = i_c = 53°
∴ n_glass = \(\frac {1}{sin_c}\) = \(\frac {1}{sin(53°)}\) ≈ \(\frac {1}{0.8}\) = \(\frac {5}{4}\)
∴ Refractive index (n) of the slab is \(\frac {5}{4}\)
From symmetry, ∆PDQ is also a Pythagorean triangle with sides in ratio QD : PD : PQ = 3 : 4 : 5.
PD = 2 cm ⇒ QD = 1.5 cm.
As critical angle is i_c = 53° and angle of incidence at Q, ∠PQN = 37° is less than critical angle, there will be partial internal reflection at Question

Question 8.
A point object is kept 10 cm away from a double convex lens of refractive index 1.5 and radii of curvature 10 cm and 8 cm. Determine location of the final image considering paraxial rays only.
Answer:
Given that, R₁ = 10 cm, R₂ = -8 cm,
u = -10 cm and n = 1.5
From lens maker’s equation,

Question 9.
A monochromatic ray of light is incident at 37° on an equilateral prism of refractive index 3/2. Determine angle of emergence and angle of deviation. If angle of prism is adjustable, what should its value be for emergent ray to be just possible for the same angle of incidence.
Answer:
By Snell’s law, in case of prism,

For equilateral prism, A = 60°
Also, A= r₁ + r₂
∴ r₂ = A – r₁ = 60° – 23°39′ = 36°21′
Applying snell’s law on the second surface of

= sin^-1 (0.889)
= 62°44′
≈ 63°
For any prism,
i + e = A + δ
∴ δ = (i + e) – A
= (37 + 63) – 60
= 40°
For an emergent ray to just emerge, the angle r’₂ acts as a critical angle.
∴ r’₂ = sin^-1 (\(\frac {1}{n}\))
= sin^-1 (\(\frac {2}{3}\))
= 41°48′
As, A = r’₁ + r’₂ and i to be kept the same.
⇒ A’ = r’₁ + r’₂‘
= 23°39′ + 41°48′
= 65°27’

Question 10.
From the given data set, determine angular dispersion by the prism and dispersive power of its material for extreme colours. n_R = 1.62 nv = 1.66, δ_R = 3.1°
Answer:
Given: n_R = 1.62, n_V = 1.66, δ_R = 3.1°
To find:
i. Angular dispersion (δ_vr)
ii. Dispersive power (ω_VR)
Formula:
i. δ = A (n – 1)
ii. δ_VR = δ_V – δ_R
(iii) ω = \(\frac{\delta_{\mathrm{V}}-\delta_{\mathrm{R}}}{\left(\frac{\delta_{\mathrm{V}}+\delta_{\mathrm{R}}}{2}\right)}\)
Calculation: From formula (i),
δ_R = A(n_R – 1)
∴ A = \(\frac{\delta_{R}}{\left(n_{R}-1\right)}=\frac{3.1}{(1.62-1)}=\frac{3.1}{0.62}\)
= 5
δV = A(n_v – 1) = 5 × (1.66 – 1) = 3.3C
From formula (ii),
δ_VR = 3.3 – 3.1 = 0.2°
From formula (iii),
ω_VR = \(\frac{3.3-3.1}{\left(\frac{3.3+3.1}{2}\right)}=\frac{0.2}{6.4} \times 2=\frac{0.2}{3.2}=\frac{1}{16}\)
= 0.0625

Question 11.
Refractive index of a flint glass varies from 1.60 to 1.66 for visible range. Radii of curvature of a thin convex lens are 10 cm and 15 cm. Calculate the chromatic aberration between extreme colours.
Answer:
Given the refractive indices for extreme colours. As, n_R < n_V
n_R = 1.60 and n_V = 1.66
For convex lens,
R₁ = 10 cm and R₁₀ = -15 cm

= 0.11
∴ f_V = 11 cm
∴ Longitudinal chromatic aberration
= f_V – f_R = 11 – 10 = 1 cm

Question 12.
A person uses spectacles of ‘number’ 2.00 for reading. Determine the range of magnifying power (angular magnification) possible. It is a concave convex lens (n = 1.5) having curvature of one of its surfaces to be 10 cm. Estimate that of the other.
Answer:
For a single concavo-convex lens, the magnifying power will be same as that for simple microscope As, the number represents the power of the lens,
P = \(\frac {1}{f}\) = 2 ⇒ f = 0.5 m.
∴ Range of magnifying power of a lens will be,
M_min = \(\frac {D}{f}\) = \(\frac {0.25}{0.5}\) = 0.5
and M_min = 1 + \(\frac {D}{f}\) = 1 + 0.5 = 1.5
Given that, n = 1.5, |R₁| = 10 cm
f = 0.5 m = 50 cm
From lens maker’s equation,

Question 13.
Focal power of the eye lens of a compound microscope is 6 dioptre. The microscope is to be used for maximum magnifying power (angular magnification) of at least 12.5. The packing instructions demand that length of the microscope should be 25 cm. Determine minimum focal power of the objective. How much will its radius of curvature be if it is a biconvex lens of n = 1.5.
Answer:
Focal power of the eye lens,
P_e = \(\frac {1}{f_e}\) = 6D
∴ f_e = \(\frac {1}{6}\) = 0.1667 m = 16.67 cm
Now, as the magnifying power is maximum,
v_e = 25 cm,
Also (M_e)_max = 1 + \(\frac {D}{f_e}\) = 1 + \(\frac {25}{16.67}\) ≈ 2.5
Given that,
M = m₀ × M_e = 12.5
∴ m₀ × 2.5 = 12.5
∴ m₀ = \(\frac {v_0}{u_0}\) = 5 ……….. (1)
From thin lens formula,

Length of a compound microscope,
L = |v₀| +|u₀|
∴ 25 = |v₀| + 10
∴ |v₀|= 15 cm
∴ |u₀| = \(\frac {v_0}{5}\) = 3 cm …………… (from 1)
From lens formula for objective,
\(\frac {1}{f_0}\) = \(\frac {1}{v_0}\) – \(\frac {1}{u_0}\)
= \(\frac {1}{15}\) – \(\frac {1}{-3}\)
= \(\frac {2}{5}\)
∴ f₀ = 2.5 cm = 0.025 m
Thus, focal power of objective,
P = \(\frac {1}{f_0(m)}\)
= \(\frac {1}{0.025}\) = 40 D
Using lens maker’s equation for a biconvex lens,
\(\frac{1}{f_{o}}=(n-1)\left(\frac{1}{R}-\frac{1}{-R}\right)\)
∴ \(\frac{1}{2.5}=(1.5-1)\left(\frac{2}{R}\right)=\frac{1}{R}\)
∴ R = 2.5 cm

11th Physics Digest Chapter 9 Optics Intext Questions and Answers

Can you recall? (Textbook rage no 159)

What are laws of reflection and refraction?
Answer:
Laws of reflection:
a. Reflected ray lies in the plane formed by incident ray and the normal drawn at the point of incidence and the two rays are on either side of the normal.
b. Angles of incidence and reflection are equal (i = r).

Laws of refraction:
a. Refracted ray lies in the plane formed by incident ray and the normal drawn at the point of incidence; and the two rays are on either side of the normal.

b. Angle of incidence (θ₁) and angle of refraction (θ₂) are related by Snell’s law, given by, n₁ sin θ₁ = n₂ sin θ₂ where, n₁, n₂ = refractive indices of medium 1 and medium 2 respectively.

Can you recall? (Textbook page no. 159)

Question 1.
What is refractive index?
Answer:
The ratio of velocity of light in vacuum to the velocity’ of light in a medium is called the refractive index of the medium.

Question 2.
What is total internal reflection?
Answer:
For angles of incidence larger than the critical angle, the angle of refraction is larger than 90°. Thus, all the incident light gets reflected back into the denser medium. This is called total internal reflection.

Question 3.
How does a rainbow form?
Answer:

1. The rainbow appears in the sky after a rainfall.
2. Water droplets present in the atmosphere act as small prism.
3. When sunlight enters these water droplets it gets refracted and dispersed.
4. This dispersed light gets totally reflected inside the droplet and again is refracted while coming out of the droplet.
5. As a combined effect of all these phenomena, the seven coloured rainbow is observed.

Question 4.
What is dispersion of light?
Answer:
Splitting of a white light into its constituent colours is known as dispersion of light.

Balbharti Maharashtra State Board 11th Physics Textbook Solutions Chapter 8 Sound Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Physics Solutions Chapter 8 Sound

1. Choose the correct alternatives

Question 1.
A sound carried by the air from a sitar to a listener is a wave of the following type.
(A) Longitudinal stationary
(B)Transverse progressive
(C) Transverse stationery
(D) Longitudinal progressive
Answer:
(D) Longitudinal progressive

Question 2.
When sound waves travel from air to water, which of these remains constant?
(A) Velocity
(B) Frequency
(C) Wavelength
(D) All of above
Answer:
(B) Frequency

Question 3.
The Laplace’s correction in the expression for velocity of sound given by Newton is needed because sound waves
(A) are longitudinal
(B) propagate isothermally
(C) propagate adiabatically
(D) are of long wavelength
Answer:
(C) propagate adiabatically

Question 4.
Speed of sound is maximum in
(A) air
(B) water
(C) vacuum
(D) solid
Answer:
(D) solid

Question 5.
The walls of the hall built for music concerns should
(A) amplify sound
(B) Reflect sound
(C) transmit sound
(D) Absorb sound
Answer:
(D) Absorb sound

2. Answer briefly.

Question 1.
Wave motion is doubly periodic. Explain.
Answer:
i. A wave particle repeats its motion after a definite interval of time at every location, making it periodic in time.
ii. Similarly, at any given instant, the form of a wave repeats itself at equal distances making it periodic in space.
iii. Thus, wave motion is a doubly periodic phenomenon, i.e., periodic in time as well as periodic in space.

Question 2.
What is Doppler effect?
Answer:
The apparent change in the frequency of sound heard by a listener, due to relative motion between the source of sound and the listener is called Doppler effect in sound.

Question 3.
Describe a transverse wave.
Answer:
Transverse wave:
A wave in which particles of the medium vibrate in a direction perpendicular to the direction of propagation of the wave is called transverse wave.
Example: Ripples on the surface of water, light waves.

Characteristics of transverse waves:

1. All the particles of medium in the path of wave vibrate in a direction perpendicular to the direction of propagation of wave with same period and amplitude.
2. When transverse wave passes through the medium, the medium is divided into alternate crests i.e., regions of positive displacements and troughs i.e., regions of negative displacement, that are periodic in time.
3. A crest and an adjacent trough form one cycle of a transverse wave. The distance between any two successive crests or troughs is called wavelength ‘λ’ of the wave.
4. Crests and troughs advance in the medium and are responsible for transfer of energy.
5. Transverse waves can travel only through solids and not through liquids and gases. Electromagnetic waves are transverse waves, but they do not require material medium for propagation.
6. When transverse waves advance through a medium, there is no change of pressure and density at any point of the medium, but the shape changes periodically.
7. Transverse wave can be polarised.
8. Medium conveying a transverse wave must possess elasticity of shape, i.e., modulus of rigidity.

Question 4.
Define a longitudinal wave.
Answer:
A wave in which particles of medium vibrate in a direction parallel to the direction of propagation of the wave is called longitudinal wave. Example: Sound waves.

Question 5.
State Newton’s formula for velocity of sound.
Answer:
Newton’s formula for velocity of sound:
i. Sound wave travels through a medium in the form of compression and rarefaction. At compression, the density of medium is greater while at rarefaction density is smaller. This is possible only in elastic medium.

ii. Thus, the velocity of sound depends upon density and elasticity of medium. It is given by
v = \(\sqrt{\frac {E}{ρ}}\) ….(1)
Where, E is the modulus of elasticity of medium and ρ is density of medium.

Assumptions:
1. Newton assumed that during propagation of sound wave in air, average temperature of the medium remains constant. Hence, propagation of sound wave in air is an isothermal process and isothermal elasticity should be considered.

2. The volume elasticity of air determined under isothermal change is called isothermal bulk modulus.

Calculations:
1. For a gas or air, the isothermal elasticity E is equal to the atmospheric pressure P.
Substituting this value in equation (1), the velocity of sound in air or a gas is given by
v = \(\sqrt{\frac {P}{ρ}}\) ….(∵ E = P)
This is the Newton’s formula for velocity of sound in air.

2. But atmospheric pressure is given by,
P = hdg
∴ v = \(\sqrt{\frac {hdg}{ρ}\) ….(2)

3. At N.T.P., h = 0.76 m of mercury, density of mercury d = 13600 kg/m³ and acceleration due to gravity, g = 9.8 m/s², density of air ρ = 1.293 kg/m³

4. From equation (2) we have velocity of sound,
v = \(\sqrt{\frac {0.76×13600×9.8}{1.293}}\) = 279.9 m/s at N.T.P

Question 6.
What is the effect of pressure on velocity of sound?
Answer:
Effect of pressure:
i. Let v be the velocity of sound in air when the pressure is P and density is ρ.

ii. Using Laplace’s formula, we can write,
v = \(\sqrt{\frac {γP}{ρ}}\) ….(1)

iii. If V be the volume of a gas having mass M then, ρ = \(\frac {M}{V}\)

iv. Substituting ρ in equation (1), we get,
v = \(\sqrt{\frac {γPV}{M}}\) ….(2)

v. But according to Boyle’s law,
PV = constant (at constant temperature)
Also, M and γ are constant.
∴ v = constant

vi. Hence, the velocity of sound does not depend upon the change in pressure, as long as the temperature remains constant.

vii. For a gaseous medium, PV= nRT.
Substituting in equation (2), we get,
v = \(\sqrt{\frac {γnRT}{M}}\)

viii. Thus, even for a gaseous medium obeying ideal gas equation, the velocity of sound does not depend upon the change in pressure, as long as the temperature remains constant.

Question 7.
What is the effect of humidity of air on velocity of sound?
Answer:
Effect of humidity:
i. Let v_m and v_d be the velocities of sound in moist air and dry air respectively.

ii. Humid air contains a large proportion of water vapour. Density of water vapour at 0 °C is 0.81 kg/m³ while that of dry air at 0°C is 1.29 kg/m³. So, the density ρ_m of moist air is less than the density ρ_d of dry air i.e., ρ_m < ρ_d.

iii. Thus \(\frac {v_m}{v_d}\) > 1
∴ v_m > v_d

iv. Hence, sound travels faster in moist air than in dry air. It means that velocity of sound increases with increase in moistness (humidity) of air.

Question 8.
What do you mean by an echo?
Answer:
An echo is the repetition of the original sound because of reflection from some rigid surface at a distance from the source of sound.

Question 9.
State any two applications of acoustics.
Answer:
Application of acoustics in nature:
i. Bats apply the principle of acoustics to locate objects. They emit short ultrasonic pulses of frequency 30 kHz to 150 kHz. The resulting echoes give them information about location of the obstacle. This helps the bats to fly in even in total darkness of caves.

ii. Dolphins navigate underwater with the help of an analogous system. They emit subsonic frequencies which can be about 100 Hz. They can sense an object about 1.4 m or larger.

Medical applications of acoustics:
i. High pressure and high amplitude shock waves are used to split kidney stones into smaller pieces without invasive surgery. A reflector or acoustic lens is used to focus a produced shock wave so that as much of its energy as possible converges on the stone. The resulting stresses in the stone causes the stone to break into small pieces which can then be removed easily.

ii. Ultrasonic imaging uses reflection of ultrasonic waves from regions in the interior of body. It is used for prenatal (before the birth) examination, detection of anomalous conditions like tumour etc. and the study of heart valve action.

iii. Ultrasound at a very high-power level, destroys selective pathological tissues which is helpful in treatment of arthritis and certain type of cancer.

Underwater applications of acoustics:
i. SONAR (Sound Navigational Ranging) is a technique for locating objects underwater by transmitting a pulse of ultrasonic sound and detecting the reflected pulse.
ii. The time delay between transmission of a pulse and the reception of reflected pulse indicates the depth of the object.
iii. Motion and position of submerged objects like submarine can be measured with the help of this system.

Applications of acoustics in environmental and geological studies:
i. Acoustic principle has important application to environmental problems like noise control. The quiet mass transit vehicle is designed by studying the generation and propagation of sound in the motor’s wheels and supporting structures.

Reflected and refracted elastic waves passing through the Earth’s interior can be measured by applying the principles of acoustics. This is useful in studying the properties of the Earth.

Principles of acoustics are applied to detect local anomalies like oil deposits etc. making it useful for geological studies.

Question 10.
Define amplitude and wavelength of a wave.
Answer:
i. Amplitude (A): The largest displacement of a particle of a medium through which the wave is propagating, from its rest position, is called amplitude of that wave.
SI unit: (m)

ii. Wavelength (λ): The distance between two successive particles which are in the same state of vibration is called wavelength of the wave.
SI unit: (m)

Question 11.
Draw a wave and indicate points which are (i) in phase (ii) out of phase (iii) have a phase difference of π/2.
Answer:

i. In phase point: A and F; B and H; C and I; D and J
ii. Out of phase points: A and B, B and D, FI and J, E and F,
iii. Point having phase difference of π/2: A and B; B and C; D; D and F; F and H; H and I; J and I

Question 12.
Define the relation between velocity, wavelength and frequency of wave.
Answer:
i. A wave covers a distance equal to the wavelength (λ) during one period (T).
Therefore, the magnitude of the velocity (v) is given by,
Magnitude of velocity = \(\frac {Distance covered}{Corresponding time}\)

ii. v = \(\frac {22}{7}\) i.e., v = λ × (\(\frac {1}{T}\)) …………….. (1)

iii. But reciprocal of the period is equal to the frequency (n) of the waves.
∴ \(\frac {1}{T}\) = n …………… (2)

iv. From equations (1) and (2), we get
v = nλ
i.e., wave velocity = frequency × wavelength.

Question 13.
State and explain principle of superposition of waves.
Answer:
Principle:
As waves don’t repulse each other, they overlap in the same region of the space without affecting each other. When two waves overlap, their displacements add vectorially.

Explanation:
i. Consider two waves travelling through a medium arriving at a point simultaneously.

ii. Let each wave produce its own displacement at that point independent of the others. This displacement can be given as,
y₁ = displacement due to first wave.
y₂ = displacement due to second wave.

iii. Then according to superposition of waves, the resultant displacement at that point is equal to the vector sum of the displacements due to all the waves.
∴\(\vec{y}\) = \(\vec{y_1}\) + \(\vec{y_2}\)

Question 14.
State the expression for apparent frequency when source of sound and listener are
i) moving towards each other
ii) moving away from each other
Answer:
i. Let,
n = actual frequency of the source.
n₀ = apparent frequency of the source,
v = velocity of sound in air.
v_s = velocity of the source.
v_l = velocity of the listener.

ii. Apparent frequency heard by the listener is given by,
n = n₀(\(\frac {v±v_L}{v±v_s}\))
Where upper signs (+ ve in numerator and -ve in denominator) indicate that source and observer move towards each other. Lower signs (-ve in numerator and +ve in denominator) indicate that source and listener move away from each other.

iii. If source and listener are moving towards each other, then apparent frequency is given by,
n = n₀(\(\frac {v+v_L}{v-v_s}\)) i.e., apparent frequency increases.

iv. If source and listener are moving away from each other, then apparent frequency is given by,
n = n₀(\(\frac {v-v_L}{v+v_s}\)) i.e., apparent frequency decreases.

Question 15.
State the expression for apparent frequency when source is stationary and listener is
1) moving towards the source
2) moving away from the source
Answer:
Let,
n = actual frequency of the source.
n₀ = apparent frequency of the source,
v = velocity of sound in air.
v_s = velocity of the source.
v_l = velocity of the listener.

i. If listener is moving towards source then apparent frequency is given by,
n = n₀(\(\frac {v+v_L}{v}\)) i.e., apparent frequency increases.

ii. If listener is receding away from source then apparent frequency is given by,
n = n₀(\(\frac {v-v_L}{v}\)) i.e., apparent frequency decreases.

Question 16.
State the expression for apparent frequency when listener is stationary and source is.

(i) moving towards the listener
(ii) moving away from the listener
Answer:
Let,
n = actual frequency of the source.
n₀ = apparent frequency of the source,
v = velocity of sound in air.
v_s = velocity of the source.
v_l = velocity of the listener.

i. If source is moving towards observer then apparent frequency is given by,
n = n₀(\(\frac {v}{v-v_s}\)) i.e., apparent frequency increases.

ii. If source is receding away from observer then apparent frequency is given by,
n = n₀(\(\frac {v}{v+v_s}\)) i.e., apparent frequency decreases.

Question 17.
Explain what is meant by phase of a wave.
Answer:

i. The state of oscillation of a particle is called the phase of the particle.

ii. The displacement, direction of velocity and oscillation number of the particle describe the phase of the particle at a place.

iii. Particles r and t (q and u or v and s) have same displacements but the directions of their velocities are opposite.

iv. Particles having same magnitude of displacements and same direction of velocity are said to be in phase during their respective oscillations. Example: particles v and p.

v. Separation between two particles which are in phase is wavelength (λ).

vi. The two successive particles differ by ‘1’ in their oscillation number i.e., if particle v is at its n^th oscillation, particle p will be at its (n + 1)^th oscillation as the wave is travelling along + X direction.

vii. In the given graph, if the disturbance (energy) has just reached the particle w, the phase angle corresponding to particle is 0°. At this instant, particle v has completed quarter oscillation and reached its positive maximum (sin θ = +1). The phase angle θ of this particle v is \(\frac {π^c}{2}\) = 90° at this instant.

viii. Phase angles of particles u and q are π^c (180°) and 2rcc (360°) respectively.

ix. Particle p has completed one oscillation and is at its positive maximum during its second oscillation.
∴ phase angle = 2π^c + \(\frac {π^c}{2}\)
= \(\frac {5π^c}{2}\)

x. v and p are the successive particles in the same state (same displacement and same direction of velocity) during their respective oscillations. Phase angle between these two differs by 2π^c.

Question 18.
Define progressive wave. State any four properties.
Answer:
i. Waves in which a disturbance created at one place travels to distant points and keeps travelling unless stopped by an external force are known as travelling or progressive waves.
Properties of progressive waves are:
Amplitude, wavelength, period, double periodicity, frequency and velocity.

Question 19.
Distinguish between traverse waves and longitudinal waves.
Answer:

Question 20.
Explain Newtons formula for velocity of sound. What is its limitation?
Answer:
Newton’s formula for velocity of sound:
i. Sound wave travels through a medium in the form of compression and rarefaction. At compression, the density of medium is greater while at rarefaction density is smaller. This is possible only in elastic medium.

ii. Thus, the velocity of sound depends upon density and elasticity of medium. It is given by
v = \(\sqrt{\frac {E}{ρ}}\) ….(1)
Where, E is the modulus of elasticity of medium and ρ is density of medium.

Assumptions:
1. Newton assumed that during propagation of sound wave in air, average temperature of the medium remains constant. Hence, propagation of sound wave in air is an isothermal process and isothermal elasticity should be considered.

2. The volume elasticity of air determined under isothermal change is called isothermal bulk modulus.

Calculations:
1. For a gas or air, the isothermal elasticity E is equal to the atmospheric pressure P.
Substituting this value in equation (1), the velocity of sound in air or a gas is given by
v = \(\sqrt{\frac {P}{ρ}}\) ….(∵ E = P)
This is the Newton’s formula for velocity of sound in air.

2. But atmospheric pressure is given by,
P = hdg
∴ v = \(\sqrt{\frac {hdg}{ρ}}\) ….(2)

3. At N.T.P., h = 0.76 m of mercury, density of mercury d = 13600 kg/m³ and acceleration due to gravity, g = 9.8 m/s², density of air ρ = 1.293 kg/m³

4. From equation (2) we have velocity of sound,
v = \(\sqrt{\frac {0.76×13600×9.8}{1.293}}\) = 279.9 m/s at N.T.P

Limitations:
1. Experimentally, it is found that the velocity of sound in air at N. T. P is 332 m/s. Thus, there is considerable difference between the value predicted by Newton’s formula and the experimental value.

2. Experimental value is 16% greater than the value given by the formula. Newton failed to provide a satisfactory explanation for the difference.

3. Solve the following problems.

Question 1.
A certain sound wave in air has a speed 340 m/s and wavelength 1.7 m for this wave, calculate
(i) the frequency
(ii) the period.
Answer:
Given: v = 340 m/s, λ = 1.7 m
To find: frequency (n), period (T)
Formulae:
i. n = \(\frac {v}{λ}\)
ii. T = \(\frac {1}{n}\)
Calculation: From formula, (i)
n = \(\frac {340}{1.7}\)
∴ n = 200 Hz
From formula, (ii)
T = \(\frac {1}{n}\) = \(\frac {1}{2×10^2}\)
= 5 × 10^-3
…….. (using reciprocal Table)
∴ T = 0.005 s

Question 2.
A tuning fork of frequency 170 Hz produces sound waves of wavelength 2m. Calculate speed of sound.
Answer:
Given: n = 170 Hz, λ = 2 m
To find: velocity of sound (v)
Formula: v = nλ
Calculation: From formula,
v = 170 × 2
∴ v = 340 m/s

Question 3.
An echo-sounder in a fishing boat receives an echo from a shoal of fish 0.45s after it was sent. If the speed of sound in water is 1500 m/s, how deep is the shoal?
Answer:
Given: t = 0.45 s, v = 1500 m/s,
To Find: depth (d)
Formula: speed (v) = \(\frac {distance}{time}\)
Calculation:
For an echo distance travelled by the sound wave = 2 × (distance between echo sounder and shoal) (d)
v = \(\frac {2 × d}{t}\)
∴ d = \(\frac {1500 × 0.45}{2}\) = 337.5 m

Question 4.
A girl stands 170 m away from a high wall and claps her hands at a steady rateso that each clap coincides with the echo of the one before.
a) If she makes 60 claps in 1 minute, what value should be the speed of sound in air?
b) Now, she moves to another location and finds that she should now make 45 claps in 1 minute to coincide with successive echoes. Calculate her distance for the new position from the wall.
Answer:
i. When the girl makes 60 claps in 1 minute, the value of speed of is 340 m/s.

ii. The girl is at a distance of 226.67 m from the wall when she produces 45 claps per minute.
[Note: The answer given above is calculated in accordance with textual method considering the given data]

Question 5.
Sound wave A has period 0.015 s, sound wave B has period 0.025. Which sound has greater frequency?
Answer:
Given: T_A = 0.015 s, T_B = 0.025 s
To find: greater frequency (n)
Formula: n = \(\frac {1}{T}\)
Calculation: From formula,
n_A = \(\frac {1}{T_A}\) = \(\frac {1}{0.025}\) = \(\frac {1}{2.5 ×10^{-2}}\)
∴ n_A = 66.67
…. (using reciprocal table)
n_B = \(\frac {1}{T_B}\) = \(\frac {1}{0.025}\) = \(\frac {1}{2.5 ×10^{-2}}\)
∴ n_B = 40 Hz
…. (using reciprocal table)
∴ n_A > n_B

Question 6.
At what temperature will the speed of sound in air be 1.75 times its speed at N.T.P?
Answer:
Given:
v_air = 1.75 V_S.T.P = \(\frac {7}{4}\) v_S.T.P
T_S.T.P = 273 K
To find: temperature T_air
Formula: v ∝ √T
Calculation: From formula,

Question 7.
A man standing between 2 parallel eliffs fires a gun. He hearns two echos one after 3 seconds and other after 5 seconds. The separation between the two cliffs is 1360 m, what is the speed of sound?
Answer:
distance (s) = 1360 m,
time for first echo = 3 s,
time for second echo = 5 s
To Find : speed of sound (v)
Formula : speed = \(\frac {distence}{time}\)
Calculation:
Time for first echo = 3 s
∴ time taken by sound to travel given distance t₁
= \(\frac {3}{2}\) = 1.5 s
Time for second echo = 5 s
∴ time taken by sound to travel given distance t₂
= \(\frac {5}{2}\) = 2.5 s
∴Total time taken by sound to travel given distance, T = 1.5 + 2.5 = 4 s
From formula,
v = \(\frac {1360}{4}\)
∴v = 340 m/s

Question 8.
If the velocity of sound in air at a given place on two different days of a given week are in the ratio of 1 : 1.1. Assuming the temperatures on the two days to be same what quantitative conclusion can your draw about the condition on the two days?
Answer:
Let v₁ and v₂ be the velocity of sound on day 1 and day 2 respectively.
\(\frac {v_1}{v_2}\) = \(\frac {1}{1.1}\)
We know, v ∝ \(\frac {1}{√ρ}\)
Let ρ₁ and ρ₂ be the density of air on day 1 and day 2 respectively.
∴ \(\sqrt{\frac {ρ_2}{ρ_1}}\) = \(\frac {1}{1.1}\)
∴ \(\frac {ρ_2}{ρ_1}\) = (\(\frac {1}{1.1}\))²
∴ ρ₁ = 1.1² ρ₂ = 1.21 ρ²
From above equation, we can conclude,
ρ₁ > ρ₂
∴ v₂ > v₁ i.e., the velocity of sound is greater on the second day than on the first day.
We know, speed of sound in moist air (v_m) is greater than speed of sound in dry air (v_d).
∴ We can conclude, air is moist on second day and dry on the first day.

Question 9.
A police car travels towards a stationary observer at a speed of 15 m/s. The siren on the car emits a sound of frequency 250 Hz. Calculate the recorded frequency. The speed of sound is 340 m/s.
Answer:
Given: v_s = 15 m/s, n0 = 250 Hz, v = 340 m/s
To find: Frequency (n)
Formula: n = n₀(\(\frac {v}{v-v_s}\))
Calculation: As the source approaches listener, apparent frequency is given by,
n = 250 (\(\frac {340}{340-15}\)) = \(\frac {3400}{13}\)
∴ n = 261.54 Hz

Question 10.
The sound emitted from the siren of an ambulance has frequency of 1500 Hz. The speed of sound is 340 m/s. Calculate the difference in frequencies heard by a stationary observer if the ambulance initially travels towards and then away from the observer at a speed of 30 m/s.
Answer:
Given: v_s = 30 m/s, n₀ = 1500 Hz, v = 340 m/s
To find: Difference in apparent frequencies (n_A – n’_A)
Formulae:
i. When the ambulance moves towards he stationary observer then n_A = n₀(\(\frac {v}{v-v_s}\))

ii. When the ambulance moves away from the stationary observer then, n’_A = n₀(\(\frac {v}{v+v_s}\))

Calculation:
From formula (i), icon’ 340
n_A = 1500(\(\frac {340}{340-30}\))
∴ n_A = 1645 Hz
From (ii)
n’_A = 1500(\(\frac {340}{340+30}\))
∴ n_A = 1378 Hz
Difference between n_A and n’_A
= n_A – n’_A = 1645 – 1378 = 267 Hz

11th Physics Digest Chapter 8 Sound Intext Questions and Answers

Can you recall? (Textbook page no. 142)

i. What type of wave is a sound wave?
ii. Can sound travel in vacuum?
iii. What are reverberation and echo?
iv. What is meant by pitch of a sound?
Answer:
i. Sound wave is a longitudinal wave.

ii. Sound cannot travel in vacuum.

iii. a. Reverberation is the phenomenon in which sound waves are reflected multiple times causing a single sound to be heard more than once.
b. An echo is the repetition of the original sound because of reflection by some surface.

iv. The characteristic of sound which is determined by the value of frequency is called as the pitch of the sound.

Activity (Textbook page no. 144)

i. Using axes of displacement and distance, sketch two waves A and B such that A has twice the wavelength and half the amplitude of B.
ii. Determine the wavelength and amplitude of each of the two waves P and Q shown in figure below.

Answer:

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 6 Differential Equations Ex 6.4 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.4

I. Solve the following differential equations:

Question 1.
\(x \sin \left(\frac{y}{x}\right) d y=\left[y \sin \left(\frac{y}{x}\right)-x\right] d x\)
Solution:

Question 2.
(x² + y²) dx – 2xy . dy = 0
Solution:
(x² + y²) dx – 2xy dy = 0
∴ 2xy dy = (x² + y²) dx
∴ \(\frac{d y}{d x}=\frac{x^{2}+y^{2}}{2 x y}\) ………(1)

Question 3.
\(\left(1+2 e^{\frac{x}{y}}\right)+2 e^{\frac{x}{y}}\left(1-\frac{x}{y}\right) \frac{d y}{d x}=0\)
Solution:

Question 4.
y² dx + (xy + x²) dy = 0
Solution:
y² dx + (xy + x²) dy = 0
∴ (xy + x²) dy = -y² dx
∴ \(\frac{d y}{d x}=\frac{-y^{2}}{x y+x^{2}}\) ……..(1)
Put y = vx
∴ \(\frac{d y}{d x}=v+x \frac{d v}{d x}\)
Substituting these values in (1), we get

Question 5.
(x² – y²) dx + 2xy dy = 0
Solution:

Question 6.
\(\frac{d y}{d x}+\frac{x-2 y}{2 x-y}=0\)
Solution:

Question 7.
\(x \frac{d y}{d x}-y+x \sin \left(\frac{y}{x}\right)=0\)
Solution:

Question 8.
\(\left(1+e^{\frac{x}{y}}\right) d x+e^{\frac{x}{y}}\left(1-\frac{X}{y}\right) d y=0\)
Solution:

Question 9.
\(y^{2}-x^{2} \frac{d y}{d x}=x y \frac{d y}{d x}\)
Solution:

Question 10.
xy \(\frac{d y}{d x}\) = x² + 2y², y(1) = 0
Solution:

Question 11.
x dy + 2y · dx = 0, when x = 2, y = 1
Solution:
∴ x dy + 2y · dx = 0
∴ x dy = -2y dx
∴ \(\frac{1}{y} d y=\frac{-2}{x} d x\)
Integrating, we get

This is the general solution.
When x = 2, y = 1, we get
4(1) = c
∴ c = 4
∴ the particular solution is x²y = 4.

Question 12.
x² \(\frac{d y}{d x}\) = x² + xy + y²
Solution:
x² \(\frac{d y}{d x}\) = x² + xy + y²
∴ \(\frac{d y}{d x}=\frac{x^{2}+x y+y^{2}}{x^{2}}\) ………(1)
Put y = vx
∴ \(\frac{d y}{d x}=v+x \frac{d v}{d x}\)

Question 13.
(9x + 5y) dy + (15x + 11y) dx = 0
Solution:
(9x + 5y) dy + (15x + 11y) dx = 0
∴ (9x + 5y) dy = -(15x + 11y) dx
∴ \(\frac{d y}{d x}=\frac{-(15 x+11 y)}{9 x+5 y}\) ………(1)
Put y = vx
∴ \(\frac{d y}{d x}=v+x \frac{d v}{d x}\)

Question 14.
(x² + 3xy + y²) dx – x² dy = 0
Solution:
(x² + 3xy + y²) dx – x² dy = 0
∴ x² dy = (x² + 3xy + y²) dx
∴ \(\frac{d y}{d x}=\frac{x^{2}+3 x y+y^{2}}{x^{2}}\) ………(1)

Question 15.
(x² + y²) dx – 2xy dy = 0.
Solution:

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 6 Differential Equations Miscellaneous Exercise 6 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6

(I) Choose the correct option from the given alternatives:

Question 1.
The order and degree of the differential equation \(\sqrt{1+\left(\frac{d y}{d x}\right)^{2}}=\left(\frac{d^{2} y}{d x^{2}}\right)^{\frac{3}{2}}\) are respectively……..
(a) 2, 1
(b) 1, 2
(c) 3, 2
(d) 2, 3
Answer:
(d) 2, 3

Question 2.
The differential equation of y = c² + \(\frac{c}{x}\) is…….
(a) \(x^{4}\left(\frac{d y}{d x}\right)^{2}-x \frac{d y}{d x}=y\)
(b) \(\frac{d y}{d x^{2}}+x \frac{d y}{d x}+y=0\)
(c) \(x^{3}\left(\frac{d y}{d x}\right)^{2}+x \frac{d y}{d x}=y\)
(d) \(\frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}-y=0\)
Answer:
(a) \(x^{4}\left(\frac{d y}{d x}\right)^{2}-x \frac{d y}{d x}=y\)

Question 3.
x² + y² = a² is a solution of ………
(a) \(\frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}-y=0\)
(b) \(y=x \sqrt{1+\left(\frac{d y}{d x}\right)^{2}}+a^{2} y\)
(c) \(y=x \frac{d y}{d x}+a \sqrt{1+\left(\frac{d y}{d x}\right)^{2}}\)
(d) \(\frac{d^{2} y}{d x^{2}}=(x+1) \frac{d y}{d x}\)
Answer:
(c) \(y=x \frac{d y}{d x}+a \sqrt{1+\left(\frac{d y}{d x}\right)^{2}}\)

Question 4.
The differential equation of all circles having their centres on the line y = 5 and touching the X-axis is
(a) \(y^{2}\left(1+\frac{d y}{d x}\right)=25\)
(b) \((y-5)^{2}\left[1+\left(\frac{d y}{d x}\right)^{2}\right]=25\)
(c) \((y-5)^{2}+\left[1+\left(\frac{d y}{d x}\right)^{2}\right]=25\)
(d) \((y-5)^{2}\left[1-\left(\frac{d y}{d x}\right)^{2}\right]=25\)
Answer:
(b) \((y-5)^{2}\left[1+\left(\frac{d y}{d x}\right)^{2}\right]=25\)

Question 5.
The differential equation y \(\frac{d y}{d x}\) + x = 0 represents family of ………
(a) circles
(b) parabolas
(c) ellipses
(d) hyperbolas
Answer:
(a) circles

Hint:
y \(\frac{d y}{d x}\) + x = 0
∴ ∫y dy + ∫x dx = c
∴ \(\frac{y^{2}}{2}+\frac{x^{2}}{2}=c\)
∴ x² + y² = 2c which is a circle.

Question 6.
The solution of \(\frac{1}{x} \cdot \frac{d y}{d x}=\tan ^{-1} x\) is……
(a) \(\frac{x^{2} \tan ^{-1} x}{2}+c=0\)
(b) x tan^-1x + c = 0
(c) x – tan^-1x = c
(d) \(y=\frac{x^{2} \tan ^{-1} x}{2}-\frac{1}{2}\left(x-\tan ^{-1} x\right)+c\)
Answer:
(d) \(y=\frac{x^{2} \tan ^{-1} x}{2}-\frac{1}{2}\left(x-\tan ^{-1} x\right)+c\)

Question 7.
The solution of (x + y)² \(\frac{d y}{d x}\) = 1 is…….
(a) x = tan^-1(x + y) + c
(b) y tan^-1(\(\frac{x}{y}\)) = c
(c) y = tan^-1(x + y) + c
(d) y + tan^-1(x + y) = c
Answer:
(c) y = tan^-1(x + y) + c

Question 8.
The Solution of \(\frac{d y}{d x}=\frac{y+\sqrt{x^{2}-y^{2}}}{2}\) is……
(a) sin^-1(\(\frac{y}{x}\)) = 2 log |x| + c
(b) sin^-1(\(\frac{y}{x}\)) = log |x| + c
(c) sin(\(\frac{x}{y}\)) = log |x| + c
(d) sin(\(\frac{y}{x}\)) = log |y| + c
Answer:
(b) sin^-1(\(\frac{y}{x}\)) = log |x| + c

Question 9.
The solution of \(\frac{d y}{d x}\) + y = cos x – sin x is……
(a) y e^x = cos x + c
(b) y e^x + e^x cos x = c
(c) y e^x = e^x cos x + c
(d) y² e^x = e^x cos x + c
Answer:
(c) y e^x = e^x cos x + c
Hint:
\(\frac{d y}{d x}\) + y = cos x – sin x
I.F. = \(e^{\int 1 d x}=e^{x}\)
∴ the solution is y . e^x = ∫(cos x – sin x) e^x + c
∴ y . e^x = e^x cos x + c

Question 10.
The integrating factor of linear differential equation x \(\frac{d y}{d x}\) + 2y = x² log x is……..
(a) \(\frac{1}{x}\)
(b) k
(c) \(\frac{1}{n^{2}}\)
(d) x²
Answer:
(d) x²
Hint:
I.F. = \(e^{\int \frac{2}{x} d x}\)
= e^{2 log x}
= x²

Question 11.
The solution of the differential equation \(\frac{d y}{d x}\) = sec x – y tan x is…….
(a) y sec x + tan x = c
(b) y sec x = tan x + c
(c) sec x + y tan x = c
(d) sec x = y tan x + c
Answer:
(b) y sec x = tan x + c

Hint:
\(\frac{d y}{d x}\) = sec x – y tan x
∴ \(\frac{d y}{d x}\) + y tan x = sec x
I.F. = \(e^{\int \tan x d x}=e^{\log \sec x}\) = sec x
∴ the solution is
y . sec x = ∫sec x . sec x dx + c
∴ y sec x = tan x + c

Question 12.
The particular solution of \(\frac{d y}{d x}=x e^{y-x}\), when x = y = 0 is……
(a) e^x-y = x + 1
(b) e^x+y = x + 1
(c) e^x + e^y = x + 1
(d) e^y-x = x – 1
Answer:
(a) e^x-y = x + 1

Question 13.
\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) is a solution of……..
(a) \(\frac{d^{2} y}{d x^{2}}+y x+\left(\frac{d y}{d x}\right)^{2}=0\)
(b) \(x y \frac{d^{2} y}{d x^{2}}+2\left(\frac{d y}{d x}\right)^{2}-y \frac{d y}{d x}=0\)
(c) \(y \frac{d^{2} y}{d x^{2}}+2\left(\frac{d y}{d x}\right)^{2}+y=0\)
(d) \(x y \frac{d y}{d x}+y \frac{d^{2} y}{d x^{2}}=0\)
Answer:
(b) \(x y \frac{d^{2} y}{d x^{2}}+2\left(\frac{d y}{d x}\right)^{2}-y \frac{d y}{d x}=0\)

Question 14.
The decay rate of certain substances is directly proportional to the amount present at that instant. Initially, there are 27 grams of substance and 3 hours later it is found that 8 grams left. The amount left after one more hour is……
(a) 5\(\frac{2}{3}\) grams
(b) 5\(\frac{1}{3}\) grams
(c) 5.1 grams
(d) 5 grams
Answer:
(b) 5\(\frac{1}{3}\) grams

Question 15.
If the surrounding air is kept at 20°C and the body cools from 80°C to 70°C in 5 minutes, the temperature of the body after 15 minutes will be…..
(a) 51.7°C
(b) 54.7°C
(c) 52.7°C
(d) 50.7°C
Answer:
(b) 54.7°C

(II) Solve the following:

Question 1.
Determine the order and degree of the following differential equations:
(i) \(\frac{d^{2} y}{d x^{2}}+5 \frac{d y}{d x}+y=x^{3}\)
Solution:
The given D.E. is \(\frac{d^{2} y}{d x^{2}}+5 \frac{d y}{d x}+y=x^{3}\)
This D.E. has highest order derivative \(\frac{d^{2} y}{d x^{2}}\) with power 1.
∴ the given D.E. is of order 2 and degree 1.

(ii) \(\left(\frac{d^{3} y}{d x^{3}}\right)^{2}=\sqrt[5]{1+\frac{d y}{d x}}\)
Solution:
The given D.E. is \(\left(\frac{d^{3} y}{d x^{3}}\right)^{2}=\sqrt[5]{1+\frac{d y}{d x}}\)
\(\left(\frac{d^{3} y}{d x^{3}}\right)^{2 \times 5}=1+\frac{d y}{d x}\)
\(\left(\frac{d^{3} y}{d x^{3}}\right)^{10}=1+\frac{d y}{d x}\)
This D.E. has highest order derivative \(\frac{d^{3} y}{d x^{3}}\) with power 10.
∴ the given D.E. is of order 3 and degree 10.

(iii) \(\sqrt[3]{1+\left(\frac{d y}{d x}\right)^{2}}=\frac{d^{2} y}{d x^{2}}\)
Solution:
The given D.E. is \(\sqrt[3]{1+\left(\frac{d y}{d x}\right)^{2}}=\frac{d^{2} y}{d x^{2}}\)
On cubing both sides, we get
\(1+\left(\frac{d y}{d x}\right)^{2}=\left(\frac{d^{2} y}{d x^{2}}\right)^{3}\)
This D.E. has highest order derivative \(\frac{d^{2} y}{d x^{2}}\) with power 3.
∴ the given D.E. is of order 2 and degree 3.

(iv) \(\frac{d y}{d x}=3 y+\sqrt[4]{1+5\left(\frac{d y}{d x}\right)^{2}}\)
Solution:
The given D.E. is

This D.E. has the highest order derivative \(\frac{d y}{d x}\) with power 4.
∴ the given D.E. is of order 1 and degree 4.

(v) \(\frac{d^{4} y}{d x^{4}}+\sin \left(\frac{d y}{d x}\right)=0\)
Solution:
The given D.E. is \(\frac{d^{4} y}{d x^{4}}+\sin \left(\frac{d y}{d x}\right)=0\)
This D.E. has highest order derivative \(\frac{d^{4} y}{d x^{4}}\).
∴ order = 4
Since this D.E. cannot be expressed as a polynomial in differential coefficient, the degree is not defined.

Question 2.
In each of the following examples verify that the given function is a solution of the differential equation.
(i) \(x^{2}+y^{2}=r^{2} ; x \frac{d y}{d x}+r \sqrt{1+\left(\frac{d y}{d x}\right)^{2}}=y\)
Solution:
x² + y² = r² ……. (1)
Differentiating both sides w.r.t. x, we get

Hence, x² + y² = r² is a solution of the D.E.
\(x \frac{d y}{d x}+r \sqrt{1+\left(\frac{d y}{d x}\right)^{2}}=y\)

(ii) y = e^ax sin bx; \(\frac{d^{2} y}{d x^{2}}-2 a \frac{d y}{d x}+\left(a^{2}+b^{2}\right) y=0\)
Solution:

(iii) y = 3 cos(log x) + 4 sin(log x); \(x^{2} \frac{d^{2} y}{d x^{2}}+x \frac{d y}{d x}+y=0\)
Solution:
y = 3 cos(log x) + 4 sin (log x) …… (1)
Differentiating both sides w.r.t. x, we get

(iv) xy = ae^x + be^-x + x²; \(x \frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}+x^{2}=x y+2\)
Solution:

(v) x² = 2y² log y, x² + y² = xy \(\frac{d x}{d y}\)
Solution:
x² = 2y² log y ……(1)
Differentiating both sides w.r.t. y, we get

∴ x² + y² = xy \(\frac{d x}{d y}\)
Hence, x² = 2y² log y is a solution of the D.E.
x² + y² = xy \(\frac{d x}{d y}\)

Question 3.
Obtain the differential equation by eliminating the arbitrary constants from the following equations:
(i) y² = a(b – x)(b + x)
Solution:
y² = a(b – x)(b + x) = a(b² – x²)
Differentiating both sides w.r.t. x, we get
2y \(\frac{d y}{d x}\) = a(0 – 2x) = -2ax
∴ y \(\frac{d y}{d x}\) = -ax …….(1)
Differentiating again w.r.t. x, we get

This is the required D.E.

(ii) y = a sin(x + b)
Solution:
y = a sin(x + b)

This is the required D.E.

(iii) (y – a)² = b(x + 4)
Solution:
(y – a)² = b(x + 4) …….(1)
Differentiating both sides w.r.t. x, we get
\(2(y-a) \cdot \frac{d}{d x}(y-a)=b \frac{d}{d x}(x+4)\)

(iv) y = \(\sqrt{a \cos (\log x)+b \sin (\log x)}\)
Solution:
y = \(\sqrt{a \cos (\log x)+b \sin (\log x)}\)
∴ y² = a cos (log x) + b sin (log x) …….(1)
Differentiating both sides w.r.t. x, we get

(v) y = Ae^3x+1 + Be^-3x+1
Solution:
y = Ae^3x+1 + Be^-3x+1 …… (1)
Differentiating twice w.r.t. x, we get

This is the required D.E.

Question 4.
Form the differential equation of:
(i) all circles which pass through the origin and whose centres lie on X-axis.
Solution:

Let C (h, 0) be the centre of the circle which pass through the origin. Then radius of the circle is h.
∴ equation of the circle is (x – h)² + (y – 0)² = h²
∴ x² – 2hx + h² + y² = h²
∴ x² + y² = 2hx ……..(1)
Differentiating both sides w.r.t. x, we get
2x + 2y \(\frac{d y}{d x}\) = 2h
Substituting the value of 2h in equation (1), we get
x² + y² = (2x + 2y \(\frac{d y}{d x}\)) x
∴ x² + y² = 2x² + 2xy \(\frac{d y}{d x}\)
∴ 2xy \(\frac{d y}{d x}\) + x² – y² = 0
This is the required D.E.

(ii) all parabolas which have 4b as latus rectum and whose axis is parallel to Y-axis.
Solution:
Let A(h, k) be the vertex of the parabola which has 4b as latus rectum and whose axis is parallel to the Y-axis.
Then equation of the parabola is
(x – h)² = 4b(y – k) ……. (1)
where h and k are arbitrary constants.

Differentiating both sides of (1) w.r.t. x, we get
2(x – h). \(\frac{d}{d x}\)(x – h) = 4b . \(\frac{d}{d x}\)(y – k)
∴ 2(x – h) x (1 – 0) = 4b(\(\frac{d y}{d x}\) – 0)
∴ (x – h) = 2b \(\frac{d y}{d x}\)
Differentiating again w.r.t. x, we get
1 – 0 = 2b \(\frac{d^{2} y}{d x^{2}}\)
∴ 2b \(\frac{d^{2} y}{d x^{2}}\) – 1 = 0
This is the required D.E.

(iii) an ellipse whose major axis is twice its minor axis.
Solution:
Let 2a and 2b be lengths of the major axis and minor axis of the ellipse.
Then 2a = 2(2b)
∴ a = 2b
∴ equation of the ellipse is
\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)
∴ \(\frac{x^{2}}{(2 b)^{2}}+\frac{y^{2}}{b^{2}}=1\)
∴ \(\frac{x^{2}}{4 b^{2}}+\frac{y^{2}}{b^{2}}=1\)
∴ x² + 4y² = 4b²
Differentiating w.r.t. x, we get
2x + 4 × 2y \(\frac{d y}{d x}\) = 0
∴ x + 4y \(\frac{d y}{d x}\) = 0
This is the required D.E.

(iv) all the lines which are normal to the line 3x + 2y + 7 = 0.
Solution:
Slope of the line 3x – 2y + 7 = 0 is \(\frac{-3}{-2}=\frac{3}{2}\).
∴ slope of normal to this line is \(-\frac{2}{3}\)
Then the equation of the normal is
y = \(-\frac{2}{3}\)x + k, where k is an arbitrary constant.
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}=-\frac{2}{3} \times 1+0\)
∴ 3\(\frac{d y}{d x}\) + 2 = 0
This is the required D.E.

(v) the hyperbola whose length of transverse and conjugate axes are half of that of the given hyperbola \(\frac{x^{2}}{16}-\frac{y^{2}}{36}=k\).
Solution:
The equation of the hyperbola is \(\frac{x^{2}}{16}-\frac{y^{2}}{36}=k\)
i.e., \(\frac{x^{2}}{16 k}-\frac{y^{2}}{36 k}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\), we get
a² = 16k, b² = 36k
∴ a = 4√k, b = 6√k
∴ l(transverse axis) = 2a = 8√k
and l(conjugate axis) = 2b = 12√k
Let 2A and 2B be the lengths of the transverse and conjugate axes of the required hyperbola.
Then according to the given condition
2A = a = 4√k and 2B = b = 6√k
∴ A = 2√k and B = 3√k
∴ equation of the required hyperbola is
\(\frac{x^{2}}{A^{2}}-\frac{y^{2}}{B^{2}}=1\)
i.e., \(\frac{x^{2}}{4 k}-\frac{y^{2}}{9 k}=1\)
∴ 9x² – 4y² = 36k, where k is an arbitrary constant.
Differentiating w.r.t. x, we get
9 × 2x – 4 × 2y \(\frac{d y}{d x}\) = 0
∴ 9x – 4y \(\frac{d y}{d x}\) = 0
This is the required D.E.

Question 5.
Solve the following differential equations:
(i) log(\(\frac{d y}{d x}\)) = 2x + 3y
Solution:

(ii) \(\frac{d y}{d x}\) = x²y + y
Solution:

(iii) \(\frac{d y}{d x}=\frac{2 y-x}{2 y+x}\)
Solution:

(iv) x dy = (x + y + 1) dx
Solution:

(v) \(\frac{d y}{d x}\) + y cot x = x² cot x + 2x
Solution:
\(\frac{d y}{d x}\) + y cot x = x cot x + 2x ……..(1)
This is the linear differential equation of the form
\(\frac{d y}{d x}\) + Py = Q, where P = cot x and Q = x² cot x + 2x
∴ I.F. = \(e^{\int P d x}\)
= \(e^{\int \cot x d x}\)
= \(e^{\log (\sin x)}\)
= sin x
∴ the solution of (1) is given by
y(I.F.) = ∫Q . (I.F.) dx + c
∴ y sin x = ∫(x² cot x + 2x) sin x dx + c
∴ y sinx = ∫(x² cot x . sin x + 2x sin x) dx + c
∴ y sinx = ∫x² cos x dx + 2∫x sin x dx + c
∴ y sinx = x² ∫cos x dx – ∫[\(\frac{d}{d x}\left(x^{2}\right)\) ∫cos x dx] dx + 2∫x sin x dx + c
∴ y sin x = x² (sin x) – ∫2x(sin x) dx + 2∫x sin x dx + c
∴ y sin x = x² sin x – 2∫x sin x dx + 2∫x sin x dx + c
∴ y sin x = x² sin x + c
∴ y = x² + c cosec x
This is the general solution.

(vi) y log y = (log y² – x) \(\frac{d y}{d x}\)
Solution:

(vii) 4 \(\frac{d x}{d y}\) + 8x = 5e^-3y
Solution:

Question 6.
Find the particular solution of the following differential equations:
(i) y(1 + log x) = (log xx) \(\frac{d y}{d x}\), when y(e) = e²
Solution:

(ii) (x + 2y²) \(\frac{d y}{d x}\) = y, when x = 2, y = 1
Solution:


This is the general solution.
When x = 2, y = 1, we have
2 = 2(1)² + c(1)
∴ c = 0
∴ the particular solution is x = 2y².

(iii) \(\frac{d y}{d x}\) – 3y cot x = sin 2x, when y(\(\frac{\pi}{2}\)) = 2
Solution:
\(\frac{d y}{d x}\) – 3y cot x = sin 2x
\(\frac{d y}{d x}\) = (3 cot x) y = sin 2x ……..(1)
This is the linear differential equation of the form

(iv) (x + y) dy + (x – y) dx = 0; when x = 1 = y
Solution:

(v) \(2 e^{\frac{x}{y}} d x+\left(y-2 x e^{\frac{x}{y}}\right) d y=0\), when y(0) = 1
Solution:

Question 7.
Show that the general solution of defferential equation \(\frac{d y}{d x}+\frac{y^{2}+y+1}{x^{2}+x+1}=0\) is given by (x + y + 1) = c(1 – x – y – 2xy).
Solution:

Question 8.
The normal lines to a given curve at each point (x, y) on the curve pass through (2, 0). The curve passes through (2, 3). Find the equation of the curve.
Solution:
Let P(x, y) be a point on the curve y = f(x).
Then slope of the normal to the curve is \(-\frac{1}{\left(\frac{d y}{d x}\right)}\)
∴ equation of the normal is

This is the general equation of the curve.
Since, the required curve passed through the point (2, 3), we get
2² + 3² = 4(2) + c
∴ c = 5
∴ equation of the required curve is x² + y² = 4x + 5.

Question 9.
The volume of a spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of the balloon after t seconds.
Solution:
Let r be the radius and V be the volume of the spherical balloon at any time t.
Then the rate of change in volume of the spherical balloon is \(\frac{d V}{d t}\) which is a constant.


Hence, the radius of the spherical balloon after t seconds is \((63 t+27)^{\frac{1}{3}}\) units.

Question 10.
A person’s assets start reducing in such a way that the rate of reduction of assets is proportional to the square root of the assets existing at that moment. If the assets at the beginning are ₹ 10 lakhs and they dwindle down to ₹ 10,000 after 2 years, show that the person will be bankrupt in 2\(\frac{2}{9}\) years from the start.
Solution:
Let x be the assets of the presort at time t years.
Then the rate of reduction is \(\frac{d x}{d t}\) which is proportional to √x.
∴ \(\frac{d x}{d t}\) ∝ √x
∴ \(\frac{d x}{d t}\) = -k√x, where k > 0
∴ \(\frac{d x}{\sqrt{x}}\) = -k dt
Integrating both sides, we get
\(\int x^{-\frac{1}{2}} d x\) = -k∫dt
∴ \(\frac{x^{\frac{1}{2}}}{\left(\frac{1}{2}\right)}\) = -kt + c
∴ 2√x = -kt + c
At the beginning, i.e. at t = 0, x = 10,00,000
2√10,00,000 = -k(0) + c
∴ c = 2000
∴ 2√x = -kt + 2000 ……..(1)
Also, when t = 2, x = 10,000
∴ 2√10000 = -k × 2 + 2000
∴ 2k = 1800
∴ k = 900
∴ (1) becomes,
∴ 2√x = -900t + 2000
When the person will be bankrupt, x = 0
∴ 0 = -900t + 2000
∴ 900t = 2000
∴ t = \(\frac{20}{9}=2 \frac{2}{9}\)
Hence, the person will be bankrupt in \(2 \frac{2}{9}\) years.

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 4 Definite Integration Miscellaneous Exercise 4 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 4 Definite Integration Miscellaneous Exercise 4

I. Choose the correct option from the given alternatives:

Question 1.
\(\int_{2}^{3} \frac{d x}{x\left(x^{3}-1\right)}=\)
(a) \(\frac{1}{3} \log \left(\frac{208}{189}\right)\)
(b) \(\frac{1}{3} \log \left(\frac{189}{208}\right)\)
(c) \(\log \left(\frac{208}{189}\right)\)
(d) \(\log \left(\frac{189}{208}\right)\)
Answer:
(a) \(\frac{1}{3} \log \left(\frac{208}{189}\right)\)

Question 2.
\(\int_{0}^{\pi / 2} \frac{\sin ^{2} x \cdot d x}{(1+\cos x)^{2}}=\)
(a) \(\frac{4-\pi}{2}\)
(b) \(\frac{\pi-4}{2}\)
(c) 4 – \(\frac{\pi}{2}\)
(d) \(\frac{4+\pi}{2}\)
Answer:
(a) \(\frac{4-\pi}{2}\)

Question 3.
\(\int_{0}^{\log 5} \frac{e^{x} \sqrt{e^{x}-1}}{e^{x}+3} \cdot d x=\)
(a) 3 + 2π
(b) 4 – π
(c) 2 + π
(d) 4 + π
Answer:
(b) 4 – π

Question 4.
\(\int_{0}^{\pi / 2} \sin ^{6} x \cos ^{2} x \cdot d x=\)
(a) \(\frac{7 \pi}{256}\)
(b) \(\frac{3 \pi}{256}\)
(c) \(\frac{5 \pi}{256}\)
(d) \(\frac{-5 \pi}{256}\)
Answer:
(c) \(\frac{5 \pi}{256}\)

Question 5.
If \(\int_{0}^{1} \frac{d x}{\sqrt{1+x}-\sqrt{X}}=\frac{k}{3}\), then k is equal to
(a) √2(2√2 – 2)
(b) \(\frac{\sqrt{2}}{3}\)(2 – 2√2)
(c) \(\frac{2 \sqrt{2}-2}{3}\)
(d) 4√2
Answer:
(d) 4√2

Question 6.
\(\int_{1}^{2} \frac{1}{x^{2}} e^{\frac{1}{x}} \cdot d x=\)
(a) √e + 1
(b) √e − 1
(c) √e(√e − 1)
(d) \(\frac{\sqrt{e}-1}{e}\)
Answer:
(c) √e(√e − 1)

Question 7.
If \(\int_{2}^{e}\left[\frac{1}{\log x}-\frac{1}{(\log x)^{2}}\right] \cdot d x=a+\frac{b}{\log 2}\), then
(a) a = e, b = -2
(b) a = e, b = 2
(c) a = -e, b = 2
(d) a = -e, b = -2
Answer:
(a) a = e, b = -2

Question 8.
Let \(\mathrm{I}_{1}=\int_{e}^{e^{2}} \frac{d x}{\log x}\) and \(\mathrm{I}_{2}=\int_{1}^{2} \frac{e^{x}}{\boldsymbol{X}} \cdot d x\), then
(a) I₁ = \(\frac{1}{3}\) I₂
(b) I₁ + I₂ = 0
(c) I₁ = 2I₂
(d) I₁ = I₂
Answer:
(d) I₁ = I₂

Question 9.
\(\int_{0}^{9} \frac{\sqrt{X}}{\sqrt{X}+\sqrt{9-X}} \cdot d x=\)
(a) 9
(b) \(\frac{9}{2}\)
(c) 0
(d) 1
Answer:
(b) \(\frac{9}{2}\)

Question 10.
The value of \(\int_{-\pi / 4}^{\pi / 4} \log \left(\frac{2+\sin \theta}{2-\sin \theta}\right) \cdot d \theta\) is
(a) 0
(b) 1
(c) 2
(d) π
Answer:
(a) 0

II. Evaluate the following:

Question 1.
\(\int_{0}^{\pi / 2} \frac{\cos x}{3 \cos x+\sin x} d x\)
Solution:
Let I = \(\int_{0}^{\pi / 2} \frac{\cos x}{3 \cos x+\sin x} d x\)
Put Numerator = A(Denominator) + B[\(\frac{d}{d x}\)(Denominator)]
∴ cos x = A(3 cos x + sin x) + B[\(\frac{d}{d x}\)(3 cos x + sin x)]
= A(3 cos x + sin x) + B(-3 sin x + cos x)
∴ cos x + 0 . sin x = (3A + B) cos x + (A – 3B) sin x
Comparing the coefficients of sinx and cos x on both the sides, we get
3A + B = 1 ………. (1)
A – 3B = 0 ………. (2)
Multiplying equation (1) by 3, we get
9A + 3B = 3 ………(3)
Adding (2) and (3), we get
10A = 3
∴ A = \(\frac{3}{10}\)

Question 2.
\(\int_{\pi / 4}^{\pi / 2} \frac{\cos \theta}{\left[\cos \frac{\theta}{2}+\sin \frac{\theta}{2}\right]^{3}} d \theta\)
Solution:

Question 3.
\(\int_{0}^{1} \frac{1}{1+\sqrt{x}} d x\)
Solution:
Let I = \(\int_{0}^{1} \frac{1}{1+\sqrt{x}} d x\)
Put √x = t
∴ x = t² and dx = 2t . dt
When x = 0, t = 0
When x = 1, t = 1

Question 4.
\(\int_{0}^{\pi / 4} \frac{\tan ^{3} x}{1+\cos 2 x} d x\)
Solution:

Question 5.
\(\int_{0}^{1} t^{5} \sqrt{1-t^{2}} d t\)
Solution:

Question 6.
\(\int_{0}^{1}\left(\cos ^{-1} x\right)^{2} d x\)
Solution:

Question 7.
\(\int_{-1}^{1} \frac{1+x^{3}}{9-x^{2}} d x\)
Solution:

Question 8.
\(\int_{0}^{\pi} x \cdot \sin x \cdot \cos ^{4} x d x\)
Solution:

Question 9.
\(\int_{0}^{\pi} \frac{x}{1+\sin ^{2} x} d x\)
Solution:

Question 10.
\(\int_{1}^{\infty} \frac{1}{\sqrt{x}(1+x)} d x\)
Solution:

III. Evaluate the following:

Question 1.
\(\int_{0}^{1}\left(\frac{1}{1+x^{2}}\right) \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) d x\)
Solution:

Question 2.
\(\int_{0}^{\pi / 2} \frac{1}{6-\cos x} d x\)
Solution:

Question 3.
\(\int_{0}^{a} \frac{1}{a^{2}+a x-x^{2}} d x\)
Solution:

Question 4.
\(\int_{\pi / 5}^{3 \pi / 10} \frac{\sin x}{\sin x+\cos x} d x\)
Solution:

Question 5.
\(\int_{0}^{1} \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) d x\)
Solution:
Let I = \(\int_{0}^{1} \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) d x\)

Question 6.
\(\int_{0}^{\pi / 4} \frac{\cos 2 x}{1+\cos 2 x+\sin 2 x} d x\)
Solution:

Question 7.
\(\int_{0}^{\pi / 2}[2 \log (\sin x)-\log (\sin 2 x)] d x\)
Solution:

Question 8.
\(\int_{0}^{\pi}\left(\sin ^{-1} x+\cos ^{-1} x\right)^{3} \sin ^{3} x d x\)
Solution:

Question 9.
\(\int_{0}^{4}\left[\sqrt{x^{2}+2 x+3}\right]^{-1} d x\)
Solution:

Question 10.
\(\int_{-2}^{3}|x-2| d x\)
Solution:
|x – 2|= 2 – x, if x < 2
= x – 2, if x ≥ 2

IV. Evaluate the following:

Question 1.
If \(\int_{a}^{a} \sqrt{x} d x=2 a \int_{0}^{\pi / 2} \sin ^{3} x d x\), find the value of \(\int_{a}^{a+1} x d x\).
Solution:

Question 2.
If \(\int_{0}^{k} \frac{1}{2+8 x^{2}} \cdot d x=\frac{\pi}{16}\), find k.
Solution:

Question 3.
If f(x) = a + bx + cx², show that \(\int_{0}^{1} f(x) d x=\frac{1}{6}\left[f(0)+4 f\left(\frac{1}{2}\right)+f(1)\right]\)
Solution:

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 6 Differential Equations Ex 6.3 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3

Question 1.
In each of the following examples verify that the given expression is a solution of the corresponding differential equation.
(i) xy = log y + c; \(\frac{d y}{d x}=\frac{y^{2}}{1-x y}\)
Solution:
xy = log y + c
Differentiating w.r.t. x, we get

Hence, xy = log y + c is a solution of the D.E.
\(\frac{d y}{d x}=\frac{y^{2}}{1-x y^{\prime}}, x y \neq 1\)

(ii) y = (sin^-1x)² + c; (1 – x²) \(\frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}=2\)
Solution:
y = (sin^-1 x)² + c …….(1)
Differentiating w.r.t. x, we get

Differentiating again w.r.t. x, we get

(iii) y = e^-x + Ax + B; \(e^{x} \frac{d^{2} y}{d x^{2}}=1\)
Solution:
y = e^-x + Ax + B
Differentiating w.r.t. x, we get

∴ \(e^{x} \frac{d^{2} y}{d x^{2}}=1\)
Hence, y = e^-x + Ax + B is a solution of the D.E.
\(e^{x} \frac{d^{2} y}{d x^{2}}=1\)

(iv) y = x^m; \(x^{2} \frac{d^{2} y}{d x^{2}}-m x \frac{d y}{d x}+m y=0\)
Solution:
y = x^m
Differentiating twice w.r.t. x, we get

This shows that y = x^m is a solution of the D.E.
\(x^{2} \frac{d^{2} y}{d x^{2}}-m x \frac{d y}{d x}+m y=0\)

(v) y = a + \(\frac{b}{x}\); \(x \frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}=0\)
Solution:
y = a + \(\frac{b}{x}\)
Differentiating w.r.t. x, we get

Differentiating again w.r.t. x, we get

Hence, y = a + \(\frac{b}{x}\) is a solution of the D.E.
\(x \frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}=0\)

(vi) y = e^ax; x \(\frac{d y}{d x}\) = y log y
Solution:
y = e^ax
log y = log e^ax = ax log e
log y = ax …….(1) ……..[∵ log e = 1]
Differentiating w.r.t. x, we get
\(\frac{1}{y} \cdot \frac{d y}{d x}\) = a × 1
∴ \(\frac{d y}{d x}\) = ay
∴ x \(\frac{d y}{d x}\) = (ax)y
∴ x \(\frac{d y}{d x}\) = y log y ………[By (1)]
Hence, y = e^ax is a solution of the D.E.
x \(\frac{d y}{d x}\) = y log y.

Question 2.
Solve the following differential equations.
(i) \(\frac{d y}{d x}=\frac{1+y^{2}}{1+x^{2}}\)
Solution:

(ii) log(\(\frac{d y}{d x}\)) = 2x + 3y
Solution:

(iii) y – x \(\frac{d y}{d x}\) = 0
Solution:
y – x \(\frac{d y}{d x}\) = 0
∴ x \(\frac{d y}{d x}\) = y
∴ \(\frac{1}{x} d x=\frac{1}{y} d y\)
Integrating both sides, we get
\(\int \frac{1}{x} d x=\int \frac{1}{y} d y\)
∴ log |x| = log |y| + log c
∴ log |x| = log |cy|
∴ x = cy
This is the general solution.

(iv) sec²x . tan y dx + sec²y . tan x dy = 0
Solution:
sec²x . tan y dx + sec²y . tan x dy = 0
∴ \(\frac{\sec ^{2} x}{\tan x} d x+\frac{\sec ^{2} y}{\tan y} d y=0\)
Integrating both sides, we get
\(\int \frac{\sec ^{2} x}{\tan x} d x+\int \frac{\sec ^{2} y}{\tan y} d y=c_{1}\)
Each of these integrals is of the type
\(\int \frac{f^{\prime}(x)}{f(x)} d x\) = log |f(x)| + c
∴ the general solution is
∴ log|tan x| + log|tan y | = log c, where c₁ = log c
∴ log |tan x . tan y| = log c
∴ tan x . tan y = c
This is the general solution.

(v) cos x . cos y dy – sin x . sin y dx = 0
Solution:
cos x . cos y dy – sin x . sin y dx = 0
\(\frac{\cos y}{\sin y} d y-\frac{\sin x}{\cos x} d x=0\)
Integrating both sides, we get
∫cot y dy – ∫tan x dx = c₁
∴ log|sin y| – [-log|cos x|] = log c, where c₁ = log c
∴ log |sin y| + log|cos x| = log c
∴ log|sin y . cos x| = log c
∴ sin y . cos x = c
This is the general solution.

(vi) \(\frac{d y}{d x}\) = -k, where k is a constant.
Solution:
\(\frac{d y}{d x}\) = -k
∴ dy = -k dx
Integrating both sides, we get
∫dy = -k∫dx
∴ y = -kx + c
This is the general solution.

(vii) \(\frac{\cos ^{2} y}{x} d y+\frac{\cos ^{2} x}{y} d x=0\)
Solution:
\(\frac{\cos ^{2} y}{x} d y+\frac{\cos ^{2} x}{y} d x=0\)
∴ y cos²y dy + x cos²x dx = 0
∴ \(x\left(\frac{1+\cos 2 x}{2}\right) d x+y\left(1+\frac{\cos 2 y}{2}\right) d y=0\)
∴ x(1 + cos 2x) dx + y(1 + cos 2y) dy = 0
∴ x dx + x cos 2x dx + y dy+ y cos 2y dy = 0
Integrating both sides, we get
∫x dx + ∫y dy + ∫x cos 2x dx + ∫y cos 2y dy = c₁ ……..(1)
Using integration by parts

Multiplying throughout by 4, this becomes
2x² + 2y² + 2x sin 2x + cos 2x + 2y sin 2y + cos 2y = 4c₁
∴ 2(x² + y²) + 2(x sin 2x + y sin 2y) + cos 2y + cos 2x + c = 0, where c = -4c₁
This is the general solution.

(viii) \(y^{3}-\frac{d y}{d x}=x^{2} \frac{d y}{d x}\)
Solution:

(ix) 2e^x+2y dx – 3 dy = 0
Solution:

(x) \(\frac{d y}{d x}\) = e^x+y + x² e^y
Solution:

∴ 3e^x + 3e^-y + x³ = -3c₁
∴ 3e^x + 3e^-y + x³ = c, where c = -3c₁
This is the general solution.

Question 3.
For each of the following differential equations, find the particular solution satisfying the given condition:
(i) 3e^x tan y dx + (1 + e^x) sec²y dy = 0, when x = 0, y = π
Solution:
3e^x tan y dx + (1 + e^x) sec²y dy = 0

(ii) (x – y²x) dx – (y + x²y) dy = 0, when x = 2, y = 0
Solution:
(x – y²x) dx – (y + x²y) dy = 0
∴ x(1 – y²) dx – y(1 + x²) dy = 0

When x = 2, y = 0, we have
(1 + 4)(1 – 0) = c
∴ c = 5
∴ the particular solution is (1 + x²)(1 – y²) = 5.

(iii) y(1 + log x) \(\frac{d x}{d y}\) – x log x = 0, y = e², when x = e
Solution:
y(1 + log x) \(\frac{d x}{d y}\) – x log x = 0

(iv) (e^y + 1) cos x + e^y sin x \(\frac{d y}{d x}\) = 0, when x = \(\frac{\pi}{6}\), y = 0
Solution:
(e^y + 1) cos x + e^y sin x \(\frac{d y}{d x}\) = 0

\(\int \frac{f^{\prime}(x)}{f(x)} d x\) = log|f(x)| + c
∴ from (1), the general solution is
log|sin x| + log|e^y + 1| = log c, where c₁ = log c
∴ log|sin x . (e^y + 1)| = log c
∴ sin x . (e^y + 1) = c
When x = \(\frac{\pi}{4}\), y = 0, we get
\(\left(\sin \frac{\pi}{4}\right)\left(e^{0}+1\right)=c\)
∴ c = \(\frac{1}{\sqrt{2}}\)(1 + 1) = √2
∴ the particular solution is sin x . (e^y + 1) = √2

(v) (x + 1) \(\frac{d y}{d x}\) – 1 = 2e^-y, y = 0, when x = 1
Solution:

This is the general solution.
Now, y = 0, when x = 1
∴ 2 + e⁰ = c(1 + 1)
∴ 3 = 2c
∴ c = \(\frac{3}{2}\)
∴ the particular solution is 2 + e^y = \(\frac{3}{2}\) (x + 1)
∴ 2(2 + e^y) = 3(x + 1).

(vi) cos(\(\frac{d y}{d x}\)) = a, a ∈ R, y (0) = 2
Solution:
cos(\(\frac{d y}{d x}\)) = a
∴ \(\frac{d y}{d x}\) = cos^-1 a
∴ dy = (cos^-1 a) dx
Integrating both sides, we get
∫dy = (cos^-1 a) ∫dx
∴ y = (cos^-1 a) x + c
∴ y = x cos^-1 a + c
This is the general solution.
Now, y(0) = 2, i.e. y = 2,
when x = 0, 2 = 0 + c
∴ c = 2
∴ the particular solution is
∴ y = x cos^-1 a + 2
∴ y – 2 = x cos^-1 a
∴ \(\frac{y-2}{x}\) = cos^-1a
∴ cos(\(\frac{y-2}{x}\)) = a

Question 4.
Reduce each of the following differential equations to the variable separable form and hence solve:
(i) \(\frac{d y}{d x}\) = cos(x + y)
Solution:

(ii) (x – y)² \(\frac{d y}{d x}\) = a²
Solution:

(iii) x + y \(\frac{d y}{d x}\) = sec(x² + y²)
Solution:

Integrating both sides, we get
∫cos u du = 2 ∫dx
∴ sin u = 2x + c
∴ sin(x² + y²) = 2x + c
This is the general solution.

(iv) cos²(x – 2y) = 1 – 2 \(\frac{d y}{d x}\)
Solution:

Integrating both sides, we get
∫dx = ∫sec²u du
∴ x = tan u + c
∴ x = tan(x – 2y) + c
This is the general solution.

(v) (2x – 2y + 3) dx – (x – y + 1) dy = 0, when x = 0, y = 1
Solution:
(2x – 2y + 3) dx – (x – y + 1) dy = 0
∴ (x – y + 1) dy = (2x – 2y + 3) dx
∴ \(\frac{d y}{d x}=\frac{2(x-y)+3}{(x-y)+1}\) ………(1)
Put x – y = u, Then \(1-\frac{d y}{d x}=\frac{d u}{d x}\)

∴ u – log|u + 2| = -x + c
∴ x – y – log|x – y + 2| = -x + c
∴ (2x – y) – log|x – y + 2| = c
This is the general solution.
Now, y = 1, when x = 0.
∴ (0 – 1) – log|0 – 1 + 2| = c
∴ -1 – o = c
∴ c = -1
∴ the particular solution is
(2x – y) – log|x – y + 2| = -1
∴ (2x – y) – log|x – y + 2| + 1 = 0

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 4 Definite Integration Ex 4.2 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2

I. Evaluate:

Question 1.
\(\int_{1}^{9} \frac{x+1}{\sqrt{x}} \cdot d x\)
Solution:

Question 2.
\(\int_{2}^{3} \frac{1}{x^{2}+5 x+6} \cdot d x\)
Solution:

Question 3.
\(\int_{0}^{\pi / 4} \cot ^{2} \cdot d x\)
Solution:

The integral does not exist since cot 0 is not defined.

Question 4.
\(\int_{-\pi / 4}^{\pi / 4} \frac{1}{1-\sin x} \cdot d x\)
Solution:

Question 5.
\(\int_{3}^{5} \frac{1}{\sqrt{2 x+3}-\sqrt{2 x-3}} \cdot d x\)
Solution:

Question 6.
\(\int_{0}^{1} \frac{x^{2}-2}{x^{2}+1} \cdot d x\)
Solution:

Question 7.
\(\int_{0}^{\pi / 4} \sin 4 x \sin 3 x \cdot d x\)
Solution:

Question 8.
\(\int_{0}^{\pi / 4} \sqrt{1+\sin 2 x} \cdot d x\)
Solution:

Question 9.
\(\int_{0}^{\pi / 4} \sin ^{4} x \cdot d x\)
Solution:

Question 10.
\(\int_{-4}^{2} \frac{1}{x^{2}+4 x+13} \cdot d x\)
Solution:

Question 11.
\(\int_{0}^{4} \frac{1}{\sqrt{4 x-x^{2}}} \cdot d x\)
Solution:

Question 12.
\(\int_{0}^{1} \frac{1}{\sqrt{3+2 x-x^{2}}} \cdot d x\)
Solution:

Question 13.
\(\int_{0}^{\pi / 2} x \cdot \sin x \cdot d x\)
Solution:

Question 14.
\(\int_{0}^{1} x \cdot \tan ^{-1} x \cdot d x\)
Solution:

Question 15.
\(\int_{0}^{\infty} x \cdot e^{-x} \cdot d x\)
Solution:

II. Evaluate:

Question 1.
\(\int_{0}^{\frac{1}{\sqrt{2}}} \frac{\sin ^{-1} x}{\left(1-x^{2}\right)^{\frac{3}{2}}} \cdot d x\)
Solution:

Question 2.
\(\int_{0}^{\pi / 4} \frac{\sec ^{2} x}{3 \tan ^{2} x+4 \tan x+1} \cdot d x\)
Solution:

Question 3.
\(\int_{0}^{4 \pi} \frac{\sin 2 x}{\sin ^{4} x+\cos ^{4} x} \cdot d x\)
Solution:

Question 4.
\(\int_{0}^{2 \pi} \sqrt{\cos x} \cdot \sin ^{3} x \cdot d x\)
Solution:

Question 5.
\(\int_{0}^{\pi / 2} \frac{1}{5+4 \cos x} \cdot d x\)
Solution:

Question 6.
\(\int_{0}^{\pi / 4} \frac{\cos x}{4-\sin ^{2} x} \cdot d x\)
Solution:

Question 7.
\(\int_{0}^{\pi / 2} \frac{\cos X}{(1+\sin x)(2+\sin x)} \cdot d x\)
Solution:

Question 8.
\(\int_{-1}^{1} \frac{1}{a^{2} e^{x}+b^{2} e^{-x}} \cdot d x\)
Solution:

Question 9.
\(\int_{0}^{\pi} \frac{1}{3+2 \sin x+\cos x} \cdot d x\)
Solution:

Question 10.
\(\int_{0}^{\pi / 4} \sec ^{4} x \cdot d x\)
Solution:

Question 11.
\(\int_{0}^{1} \sqrt{\frac{1-x}{1+x}} \cdot d x\)
Solution:

Question 12.
\(\int_{0}^{\pi} \sin ^{3} x(1+2 \cos x)(1+\cos x)^{2} \cdot d x\)
Solution:

Question 13.
\(\int_{0}^{\pi / 2} \sin 2 x \cdot \tan ^{-1}(\sin x) \cdot d x\)
Solution:

Question 14.
\(\int_{\frac{1}{\sqrt{2}}}^{1} \frac{\left(e^{\cos ^{-1} x}\right)\left(\sin ^{-1} x\right)}{\sqrt{1-x^{2}}} \cdot d x\)
Solution:

Question 15.
\(\int_{2}^{3} \frac{\cos (\log x)}{x} \cdot d x\)
Solution:

III. Evaluate:

Question 1.
\(\int_{0}^{a} \frac{1}{x+\sqrt{a^{2}-x^{2}}} \cdot d x\)
Solution:

Question 2.
\(\int_{0}^{\pi / 2} \log \tan x \cdot d x\)
Solution:

Question 3.
\(\int_{0}^{1} \log \left(\frac{1}{x}-1\right) \cdot d x\)
Solution:

Question 4.
\(\int_{0}^{\pi / 2} \frac{\sin x-\cos x}{1+\sin x \cdot \cos x} \cdot d x\)
Solution:

Question 5.
\(\int_{0}^{3} x^{2}(3-x)^{\frac{5}{2}} \cdot d x\)
Solution:

Question 6.
\(\int_{-3}^{3} \frac{x^{3}}{9-x^{2}} \cdot d x\)
Solution:

Question 7.
\(\int_{-\pi / 2}^{\pi / 2} \log \left(\frac{2+\sin x}{2-\sin x}\right) \cdot d x\)
Solution:

Question 8.
\(\int_{-\pi / 4}^{\pi / 4} \frac{x+\frac{\pi}{4}}{2-\cos 2 x} \cdot d x\)
Solution:

Question 9.
\(\int_{-\pi / 4}^{\pi / 4} x^{3} \cdot \sin ^{4} x \cdot d x\)
Solution:

Question 10.
\(\int_{0}^{1} \frac{\log (x+1)}{x^{2}+1} \cdot d x\)
Solution:

Question 11.
\(\int_{-1}^{1} \frac{x^{3}+2}{\sqrt{x^{2}+4}} \cdot d x\)
Solution:

Question 12.
\(\int_{-a}^{a} \frac{x+x^{3}}{16-x^{2}} \cdot d x\)
Solution:

Question 13.
\(\int_{0}^{1} t^{2} \sqrt{1-t} \cdot d t\)
Solution:

Question 14.
\(\int_{0}^{\pi} x \cdot \sin x \cdot \cos ^{2} x \cdot d x\)
Solution:

Question 15.
\(\int_{0}^{1} \frac{\log x}{\sqrt{1-x^{2}}} \cdot d x\)
Solution: