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Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 6 Differential Equations Ex 6.2 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2

Question 1.
Obtain the differential equation by eliminating the arbitrary constants from the following equations:
(i) x³ + y³ = 4ax
Solution:
x³ + y³ = 4ax ……..(1)
Differentiating both sides w.r.t. x, we get
3x² + 3y² \(\frac{d y}{d x}\) = 4a × 1
∴ 3x² + 3y² \(\frac{d y}{d x}\) = 4a
Substituting the value of 4a in (1), we get
x³ + y³ = (3x² + 3y² \(\frac{d y}{d x}\)) x
∴ x³ + y³ = 3x³ + 3xy² \(\frac{d y}{d x}\)
∴ 2x³ + 3xy² \(\frac{d y}{d x}\) – y³ = 0
This is the required D.E.

(ii) Ax² + By² = 1
Solution:
Ax² + By² = 1
Differentiating both sides w.r.t. x, we get
A × 2x + B × 2y \(\frac{d y}{d x}\) = 0
∴ Ax + By \(\frac{d y}{d x}\) = 0 ……..(1)
Differentiating again w.r.t. x, we get

Substituting the value of A in (1), we get

This is the required D.E.

Alternative Method:
Ax² + By² = 1 ……..(1)
Differentiating both sides w.r.t. x, we get
A × 2x + B × 2y \(\frac{d y}{d x}\) = 0
∴ Ax + By \(\frac{d y}{d x}\) = 0 ……….(2)
Differentiating again w.r.t. x, we get,

The equations (1), (2) and (3) are consistent in A and B.
∴ determinant of their consistency is zero.

This is the required D.E.

(iii) y = A cos(log x) + B sin(log x)
Solution:
y = A cos(log x) + B sin (log x) ……. (1)
Differentiating w.r.t. x, we get

(iv) y² = (x + c)³
Solution:
y² = (x + c)³
Differentiating w.r.t. x, we get

This is the required D.E.

(v) y = Ae^5x + Be^-5x
Solution:
y = Ae^5x + Be^-5x ……….(1)
Differentiating twice w.r.t. x, we get

This is the required D.E.

(vi) (y – a)² = 4(x – b)
Solution:
(y – a)² = 4(x – b)
Differentiating both sides w.r.t. x, we get
2(y – a) . \(\frac{d}{d x}\)(y – a) = 4 \(\frac{d}{d x}\)(x – b)
∴ 2(y – a) . (\(\frac{d y}{d x}\) – 0) = 4(1 – 0)
∴ 2(y – a) \(\frac{d y}{d x}\) = 4
∴ (y – a) \(\frac{d y}{d x}\) = 2 ……..(1)
Differentiating w.r.t. x, we get

This is the required D.E.

(vii) y = a + \(\frac{a}{x}\)
Solution:
y = a + \(\frac{a}{x}\)
Differentiating w.r.t. x, we get

Substituting the value of a in (1), we get

This is the required D.E.

(viii) y = c₁e^2x + c₂e^5x
Solution:
y = c₁e^2x + c₂e^5x ………(1)
Differentiating twice w.r.t. x, we get
\(\frac{d y}{d x}\) = c₁e^2x × 2 + c₂e^5x × 5

The equations (1), (2) and (3) are consistent in c₁e^2x and c₂e^5x
∴ determinant of their consistency is zero.

This is the required D.E.

Alternative Method:
y = c₁e^2x + c₂e^5x
Dividing both sides by e^5x, we get

This is the required D.E.

(ix) c₁x³ + c₂y² = 5.
Solution:
c₁x³ + c₂y² = 5 ……….(1)
Differentiating w.r.t. x, we get

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

The equations (1), (2) and (3) in c₁, c₂ are consistent.
∴ determinant of their consistency is zero.

This is the required D.E.

(x) y = e^-2x(A cos x + B sin x)
Solution:
y = e^-2x(A cos x + B sin x)
∴ e^2x . y = A cos x + B sin x ………(1)
Differentiating w.r.t. x, we get

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


This is the required D.E.

Question 2.
Form the differential equation of family of lines having intercepts a and b on the coordinate axes respectively.
Solution:
The equation of the line having intercepts a and b on the coordinate axes respectively, is
\(\frac{x}{a}+\frac{y}{b}=1\) ……….(1)
where a and b are arbitrary constants.
[For different values of a and b, we get, different lines. Hence (1) is the equation of family of lines.]
Differentiating (1) w.r.t. x, we get

Differentiating again w.r.t. x, we get \(\frac{d^{2} y}{d x^{2}}=0\)
This is the required D.E.

Question 3.
Find the differential equation all parabolas having length of latus rectum 4a and axis is parallel to the X-axis.
Solution:

Let A(h, k) be the vertex of the parabola whose length of latus rectum is 4a.
Then the equation of the parabola is (y – k)² = 4a (x – h), where h and k are arbitrary constants.
Differentiating w.r.t. x, we get

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

This is the required D.E.

Question 4.
Find the differential equation of the ellipse whose major axis is twice its minor axis.
Solution:
Let 2a and 2b be lengths of major axis and minor axis of the ellipse.
Then 2a = 2(2b)
∴ a = 2b
∴ equation of the ellipse is
\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)
i.e., \(\frac{x^{2}}{(2 b)^{2}}+\frac{y^{2}}{b^{2}}=1\)
∴ \(\frac{x^{2}}{4 b^{2}}+\frac{y^{2}}{b^{2}}=1\)
∴ 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.

Question 5.
Form the differential equation of family of lines parallel to the line 2x + 3y + 4 = 0.
Solution:
The equation of the line parallel to the line 2x + 3y + 4 = 0 is 2x + 3y + c = 0, where c is an arbitrary constant.
Differentiating w.r.t. x, we get
2 × 1 + 3 \(\frac{d y}{d x}\) + 0 = 0
∴ 3 \(\frac{d y}{d x}\) + 2 = 0
This is the required D.E.

Question 6.
Find the differential equation of all circles having radius 9 and centre at point (h, k).
Solution:
Equation of the circle having radius 9 and centre at point (h, k) is
(x – h)² + (y – k)² = 81 …… (1)
where h and k are arbitrary constant.
Differentiating (1) w.r.t. x, we get

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

From (2), x – h = -(y – k) \(\frac{d y}{d x}\)
Substituting the value of (x – h) in (1), we get


This is the required D.E.

Question 7.
Form the differential equation of all parabolas whose axis is the X-axis.
Solution:

The equation of the parbola whose axis is the X-axis is
y² = 4a(x – h) …… (1)
where a and h are arbitrary constants.
Differentiating (1) w.r.t. x, we get
2y \(\frac{d y}{d x}\) = 4a(1 – 0)
∴ y \(\frac{d y}{d x}\) = 2a
Differentiating again w.r.t. x, we get

This is the required D.E.

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 4 Definite Integration Ex 4.1 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.1

I. Evaluate the following integrals as a limit of a sum.

Question 1.
\(\int_{1}^{3}(3 x-4) \cdot d x\)
Solution:
Let f(x) = 3x – 4, for 1 ≤ x ≤ 3
Divide the closed interval [1, 3] into n subintervals each of length h at the points
1, 1 + h, 1 + 2h, 1 + rh, ….., 1 + nh = 3
∴ nh = 2
∴ h = \(\frac{2}{n}\) and as n → ∞, h → 0
Here, a = 1
∴ f(a + rh) = f(1 + rh)
= 3(1 + rh) – 4
= 3rh – 1

Question 2.
\(\int_{0}^{4} x^{2} d x\)
Solution:
Let f(x) = x², for 0 ≤ x ≤ 4
Divide the closed interval [0, 4] into n subintervals each of length h at the points
0, 0 + h, 0 + 2h, ….., 0 + rh, ….., 0 + nh = 4
i.e. 0, h, 2h, ….., rh, ….., nh = 4
∴ h = \(\frac{4}{n}\) as n → ∞, h → 0
Here, a = 0

Question 3.
\(\int_{0}^{2} e^{x} d x\)
Solution:
Let f(x) = e^x, for 0 ≤ x ≤ 2
Divide the closed interval [0, 2] into n equal subntervals each of length h at the points
0, 0 + h, 0 + 2h, ….., 0 + rh, ….., 0 + nh = 2
i.e. 0, h, 2h, ….., rh, ….., nh = 2
∴ h = \(\frac{2}{n}\) and as n → ∞, h → 0
Here, a = 0

Question 4.
\(\int_{0}^{2}\left(3 x^{2}-1\right) d x\)
Solution:
Let f(x) = 3x² – 1, for 0 ≤ x ≤ 2
Divide the closed interval [0, 2] into n subintervals each of length h at the points.
0, 0 + h, 0 + 2h, ….., 0 + rh, ……, 0 + nh = 2
i.e. 0, h, 2h, ….., rh, ….., nh = 2
∴ h = \(\frac{2}{n}\) and as n → ∞, h → 0
Here, a = 0
∴ f(a + rh) = f(0 + rh)
= f(rh)
= 3(rh)² – 1
= 3r²h² – 1

Question 5.
\(\int_{1}^{3} x^{3} d x\)
Solution:
Let f(x) = x³, for 1 ≤ x ≤ 3.
Divide the closed interval [1, 3] into n equal su bintervals each of length h at the points
1, 1 + h, 1 + 2h, ……, 1 + rh, ……, 1 + nh = 3
∴ nh = 2
∴ h = \(\frac{2}{n}\) and as n → ∞, h → 0
Here a = 1

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 6 Differential Equations Ex 6.6 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.6

Question 1.
In a certain culture of bacteria, the rate of increase is proportional to the number present. If it is found that the number doubles in 4 hours, find the number of times the bacteria are increased in 12 hours.
Solution:
Let x be the number of bacteria in the culture at time t.
Then the rate of increase is \(\frac{d x}{d t}\) which is proportional to x.
∴ \(\frac{d x}{d t}\) ∝ x
∴ \(\frac{d x}{d t}\) = kx, where k is a constant
∴ \(\frac{d x}{x}\) = k dt
On integrating, we get
\(\int \frac{d x}{x}\) = k∫dt + c
∴ log x = kt + c
Initially, i.e. when t = 0, let x = x₀
log x₀ = k × 0 + c
∴ c = log x₀
∴ log x = kt + log x₀
∴ log x – log x₀ = kt
∴ log(\(\frac{x}{x_{0}}\)) = kt ………(1)
Since the number doubles in 4 hours, i.e. when t = 4, x = 2x₀

∴ the number of bacteria will be 8 times the original number in 12 hours.

Question 2.
If the population of a country doubles in 60 years; in how many years will it be triple (treble) under the assumption that the rate of increase is proportional to the number of inhabitants?
[Given log 2 = 0.6912, log 3 = 1.0986]
Solution:
Let P be the population at time t years.
Then \(\frac{d P}{d t}\), the rate of increase of population is proportional to P.
∴ \(\frac{d P}{d t}\) ∝ P
∴ \(\frac{d P}{d t}\) = kP, where k is a constant
∴ \(\frac{d P}{P}\) = k dt
On integrating, we get
\(\int \frac{d P}{P}\) = k∫dt + c
∴ log P = kt + c
Initially i.e. when t = 0, let P = P₀
∴ log P₀ = k x 0 + c
∴ c = log P₀
∴ log P = kt + log P₀
∴ log P – log P₀ = kt
∴ log(\(\frac{P}{P_{0}}\)) = kt ……(1)
Since, the population doubles in 60 years, i.e. when t = 60, P = 2P₀

∴ the population becomes triple in 95.4 years (approximately).

Question 3.
If a body cools from 80°C to 50°C at room temperature of 25°C in 30 minutes, find the temperature of the body after 1 hour.
Solution:
Let θ°C be the temperature of the body at time t minutes.
The room temperature is given to be 25°C.
Then by Newton’s law of cooling, \(\frac{d \theta}{d t}\), the rate of change of temperature, is proportional to (θ – 25).


∴ the temperature of the body will be 36.36°C after 1 hour.

Question 4.
The rate of growth of bacteria is proportional to the number present. If initially, there were 1000 bacteria and the number doubles in 1 hour, find the number of bacteria after 2½ hours. [Take √2 = 1.414]
Solution:
Let x be the number of bacteria at time t.
Then the rate of increase is \(\frac{d x}{d t}\) which is proportional to x.
∴ \(\frac{d x}{d t}\) ∝ x
∴ \(\frac{d x}{d t}\) = kx, where k is a constant
∴ \(\frac{d x}{x}\) = k dt
On integrating, we get
\(\int \frac{d x}{x}\) = k∫dt + c
∴ log x = kt + c
Initially, i.e. when t = 0, x = 1000
∴ log 1000 = k × 0 + c
∴ c = log 1000
∴ log x = kt + log 1000
∴ log x – log 1000 = kt
∴ log(\(\frac{x}{1000}\)) = kt ……(1)
Now, when t = 1, x = 2 × 1000 = 2000
∴ log(\(\frac{2000}{1000}\)) = k
∴ k = log 2

∴ the number of bacteria after 2½ hours = 5656.

Question 5.
The rate of disintegration of a radioactive element at any time t is proportional to its mass at that time. Find the time during which the original mass of 1.5 gm will disintegrate into its mass of 0.5 gm.
Solution:
Let m be the mass of the radioactive element at time t.
Then the rate of disintegration is \(\frac{d m}{d t}\) which is proportional to m.


∴ log(3)^-1 = -kt
∴ -log 3 = -kt
∴ t = \(\frac{1}{k}\) log 3
∴ the original mass will disintegrate to 0.5 gm when t = \(\frac{1}{k}\) log 3

Question 6.
The rate of decay of certain substances is directly proportional to the amount present at that instant. Initially, there is 25 gm of certain substance and two hours later it is found that 9 gm are left. Find the amount left after one more hour.
Solution:
Let x gm be the amount of the substance left at time t.
Then the rate of decay is \(\frac{d x}{d t}\), which is proportional to x.


∴ \(\frac{x}{25}=\frac{27}{125}\)
∴ x = \(\frac{27}{5}\)
∴ the amount left after 3 hours \(\frac{27}{5}\) gm.

Question 7.
Find the population of a city at any time t, given that the rate of increase of population is proportional to the population at that instant and that in a period of 40 years, the population increased from 30,000 to 40,000.
Solution:
Let P be the population of the city at time t.
Then \(\frac{d P}{d t}\), the rate of increase of population is proportional to P.
∴ \(\frac{d P}{d t}\) ∝ P
∴ \(\frac{d P}{d t}\) = kP, where k is a constant.
∴ \(\frac{d P}{P}\) = k dt
On integrating, we get
\(\int \frac{1}{P} d P\) = k∫dt + c
∴ log P = kt + c
Initially, i.e. when t = 0, P = 30000
∴ log 30000 = k × 0 + c
∴ c = log 30000
∴ log P = kt + log 30000
∴ log P – log 30000 = kt
∴ log(\(\frac{P}{30000}\)) = kt …….(1)
Now, when t = 40, P = 40000

∴ the population of the city at time t = 30000\(\left(\frac{4}{3}\right)^{\frac{t}{40}}\).

Question 8.
A body cools according to Newton’s law from 100°C to 60°C in 20 minutes. The temperature of the surroundings is 20°C. How long will it take to cool down to 30°C?
Solution:
Let θ°C be the temperature of the body at time t.
The temperature of the surrounding is given to be 20°C.
According to Newton’s law of cooling


∴ the body will cool down to 30°C in 60 minutes, i.e. in 1 hour.

Question 9.
A right circular cone has a height of 9 cm and a radius of the base of 5 cm. It is inverted and water is poured into it. If at any instant the water level rises at the rate of \(\left(\frac{\pi}{A}\right)\) cm/sec, where A is the area of the water surface
at that instant, show that the vessel will be full in 75 seconds.
Solution:

Let r be the radius of the water surface and h be the height of the water at time t.
∴ area of the water surface A = πr² sq cm.
Since height of the right circular cone is 9 cm and radius of the base is 5 cm.
\(\frac{r}{h}=\frac{5}{9}\)
∴ r = \(\frac{5}{9} h\)
∴ area of water surface, i.e. A = \(\pi\left(\frac{5}{9} h\right)^{2}\)
∴ A = \(\frac{25 \pi h^{2}}{81}\) ……..(1)
The water level, i.e. the rate of change of h is \(\frac{d h}{d t}\) rises at the rate of \(\left(\frac{\pi}{A}\right)\) cm/sec.

∴ t = \(\frac{81 \times 9 \times 25}{3 \times 81}\) = 75
Hence, the vessel will be full in 75 seconds.

Question 10.
Assume that a spherical raindrop evaporates at a rate proportional to its surface area. If its radius originally is 3 mm and 1 hour later has been reduced to 2 mm, find an expression for the radius of the raindrop at any time t.
Solution:
Let r be the radius, V be the volume and S be the surface area of the spherical raindrop at time t.
Then V = \(\frac{4}{3}\)πr³ and S = 4πr²
The rate at which the raindrop evaporates is \(\frac{d V}{d t}\) which is proportional to the surface area.
∴ \(\frac{d V}{d t}\) ∝ S
∴ \(\frac{d V}{d t}\) = -kS, where k > 0 ………(1)

On integrating, we get
∫dr = -k∫dt + c
∴ r = -kt + c
Initially, i.e. when t = 0, r = 3
∴ 3 = -k × 0 + c
∴ c = 3
∴ r = -kt + 3
When t = 1, r = 2
∴ 2 = -k × 1 + 3
∴ k = 1
∴ r = -t + 3
∴ r = 3 – t, where 0 ≤ t ≤ 3.
This is the required expression for the radius of the raindrop at any time t.

Question 11.
The rate of growth of the population of a city at any time t is proportional to the size of the population. For a certain city, it is found that the constant of proportionality is 0.04. Find the population of the city after 25 years, if the initial population is 10,000. [Take e = 2.7182]
Solution:
Let P be the population of the city at time t.
Then the rate of growth of population is \(\frac{d P}{d t}\) which is proportional to P.
∴ \(\frac{d P}{d t}\) ∝ P
∴ \(\frac{d P}{d t}\) = kP, where k = 0.04
∴ \(\frac{d P}{d t}\) = (0.04)P
∴ \(\frac{1}{P}\) dP = (0.04)dt
On integrating, we get
\(\int \frac{1}{P} d P\) = (0.04) ∫dt + c
∴ log P = (0.04)t + c
Initially, i.e., when t = 0, P = 10000
∴ log 10000 = (0.04) × 0 + c
∴ c = log 10000
∴ log P = (0.04)t + log 10000
∴ log P – log 10000 = (0.04)t
∴ log(\(\frac{P}{10000}\)) = (0.04)t
When t = 25, then
∴ log(\(\frac{P}{10000}\)) = 0.04 × 25 = 1
∴ log(\(\frac{P}{10000}\)) = log e ……[∵ log e = 1]
∴ \(\frac{P}{10000}\) = e = 2.7182
∴ P = 2.7182 × 10000 = 27182
∴ the population of the city after 25 years will be 27,182.

Question 12.
Radium decomposes at a rate proportional to the amount present at any time. If p percent of the amount disappears in one year, what percent of the amount of radium will be left after 2 years?
Solution:
Let x be the amount of the radium at time t.
Then the rate of decomposition is \(\frac{d x}{d t}\) which is proportional to x.


Hence, \(\left(10-\frac{p}{10}\right)^{2} \%\) of the amount will be left after 2 years.

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 3 Indefinite Integration Miscellaneous Exercise 3 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 3 Indefinite Integration Miscellaneous Exercise 3

I. Choose the correct options from the given alternatives:

Question 1.
\(\int \frac{1+x+\sqrt{x+x^{2}}}{\sqrt{x}+\sqrt{1+x}} \cdot d x=\)
(a) \(\frac{1}{2} \sqrt{x+1}+c\)
(b) \(\frac{2}{3}(x+1)^{\frac{3}{2}}+c\)
(c) \(\sqrt{x+1}+c\)
(d) \(2(x+1)^{\frac{3}{2}}+c\)
Answer:
(b) \(\frac{2}{3}(x+1)^{\frac{3}{2}}+c\)

Question 2.
\(\int \frac{1}{x+x^{5}} \cdot d x\) = f(x) + c, then \(\int \frac{x^{4}}{x+x^{5}} \cdot d x=\)
(a) log x – f(x) + c
(b) f(x) + log x + c
(c) f(x) – log x + c
(d) \(\frac{1}{5}\) x⁵ f(x) + c
Answer:
(a) log x – f(x) + c

Question 3.
\(\int \frac{\log (3 x)}{x \log (9 x)} \cdot d x=\)
(a) log(3x) – log(9x) + c
(b) log(x) – (log 3) . log(log 9x) + c
(c) log 9 – (log x) . log(log 3x) + c
(d) log(x) + log(3) . log(log 9x) + c
Answer:
(b) log(x) – (log 3) . log(log 9x) + c

Question 4.
\(\int \frac{\sin ^{m} X}{\cos ^{m+2} X} \cdot d x=\)
(a) \(\frac{\tan ^{m+1} \boldsymbol{X}}{m+1}+c\)
(b) (m + 2) tan^m+1 x + c
(c) \(\frac{\tan ^{m} \boldsymbol{X}}{m}+c\)
(d) (m + 1) tan^m+1 x + c
Answer:
(a) \(\frac{\tan ^{m+1} \boldsymbol{X}}{m+1}+c\)

Question 5.
∫tan(sin^-1 x) . dx =
(a) \(\left(1-x^{2}\right)^{-\frac{1}{2}}+c\)
(b) \(\left(1-x^{2}\right)^{\frac{1}{2}}+c\)
(c) \(\frac{\tan ^{m} \boldsymbol{X}}{\sqrt{1-x^{2}}}+c\)
(d) \(-\sqrt{1-x^{2}}+c\)
Answer:
(d) \(-\sqrt{1-x^{2}}+c\)

Hint: sin^-1 x = \(\tan ^{-1}\left(\frac{x}{\sqrt{1-x^{2}}}\right)\)

Question 6.
\(\int \frac{x-\sin x}{1-\cos x} \cdot d x=\)
(a) x cot(\(\frac{x}{2}\)) + c
(b) -x cot(\(\frac{x}{2}\)) + c
(c) cot(\(\frac{x}{2}\)) + c
(d) x tan(\(\frac{x}{2}\)) + c
Answer:
(b) -x cot(\(\frac{x}{2}\)) + c

Question 7.
If f(x) = \(\frac{\sin ^{-1} x}{\sqrt{1-x^{2}}}\), g(x) = \(e^{\sin ^{-1} x}\), then ∫f(x) . g(x) . dx =
(a) \(e^{\sin ^{-1} x} \cdot\left(\sin ^{-1} x-1\right)+c\)
(b) \(e^{\sin ^{-1} x} \cdot\left(1-\sin ^{-1} x\right)+c\)
(c) \(e^{\sin ^{-1} x} \cdot\left(\sin ^{-1} x+1\right)+c\)
(d) \(e^{\sin ^{-1} x} \cdot\left(\sin ^{-1} X-1\right)+c\)
Answer:
(a) \(e^{\sin ^{-1} x} \cdot\left(\sin ^{-1} x-1\right)+c\)

Question 8.
If ∫tan³ x . sec³ x . dx = (\(\frac{1}{m}\)) sec^m x – (\(\frac{1}{n}\)) secⁿ x + c, then (m, n) =
(a) (5, 3)
(b) (3, 5)
(c) \(\left(\frac{1}{5}, \frac{1}{3}\right)\)
(d) (4, 4)
Answer:
(a) (5, 3)

Hint: ∫tan³ x . sec³ x dx
= ∫sec² x . tan² x . sec x tan x dx
= ∫sec² x (sec² x – 1) sec x tan x dx
Put sec x = t.

Question 9.
\(\int \frac{1}{\cos x-\cos ^{2} x} \cdot d x=\)
(a) log(cosec x – cot x) + tan(\(\frac{x}{2}\)) + c
(b) sin 2x – cos x + c
(c) log(sec x + tan x) – cot(\(\frac{x}{2}\)) + c
(d) cos 2x – sin x + c
Answer:
(c) log(sec x + tan x) – cot(\(\frac{x}{2}\)) + c

Question 10.
\(\int \frac{\sqrt{\cot x}}{\sin x \cdot \cos x} \cdot d x=\)
(a) \(2 \sqrt{\cot x}+c\)
(b) \(-2 \sqrt{\cot x}+c\)
(c) \(\frac{1}{2} \sqrt{\cot x}+c\)
(d) \(\sqrt{\cot X}+c\)
Answer:
(b) \(-2 \sqrt{\cot x}+c\)

Question 11.
\(\int \frac{e^{x}(x-1)}{x^{2}} \cdot d x=\)
(a) \(\frac{e^{x}}{x}+c\)
(b) \(\frac{e^{x}}{x^{2}}+c\)
(c) \(\left(x-\frac{1}{x}\right) e^{x}+c\)
(d) x e^-x + c
Answer:
(a) \(\frac{e^{x}}{x}+c\)

Question 12.
∫sin(log x) . dx =
(a) \(\frac{x}{2}\) [sin(log x) – cos(log x)] + c
(b) \(\frac{x}{2}\) [sin(log x) + cos(log x)] + c
(c) \(\frac{x}{2}\) [cos(log x) – sin(log x)] + c
(d) \(\frac{x}{4}\) [cos(log x) – sin(log x)] + c
Answer:
(a) \(\frac{x}{2}\) [sin(log x) – cos(log x)] + c

Question 13.
∫x^x (1 + log x) . dx =
(a) \(\frac{1}{2}\) (1 + log x)² + c
(b) x^2x + c
(c) x^x log x + c
(d) x^x + c
Answer:
(d) x^x + c

Hint: \(\frac{d}{d x}\)(xx) = x^x (1 + log x)

Question 14.
\(\int \cos ^{-\frac{3}{7}} x \cdot \sin ^{-\frac{11}{7}} x \cdot d x=\)
(a) \(\log \left(\sin ^{-\frac{4}{7}} x\right)+c\)
(b) \(\frac{4}{7} \tan ^{\frac{4}{7}} x+c\)
(c) \(-\frac{7}{4} \tan ^{-\frac{4}{7}} x+c\)
(d) \(\log \left(\cos ^{\frac{3}{7}} x\right)+c\)
Answer:
(c) \(-\frac{7}{4} \tan ^{-\frac{4}{7}} x+c\)

Hint: \(\int \cos ^{-\frac{3}{7}} x \sin ^{-\frac{11}{7}} x d x\)
= \(\int \frac{\sin ^{-\frac{11}{7}} x}{\cos ^{-\frac{11}{7}} x \cdot \cos ^{2} x} d x\)
= \(\int \tan ^{-\frac{11}{7}} x \sec ^{2} x d x\)
Put tan x = t.

Question 15.
\(2 \int \frac{\cos ^{2} x-\sin ^{2} x}{\cos ^{2} x+\sin ^{2} x} \cdot d x=\)
(a) sin 2x + c
(b) cos 2x + c
(c) tan 2x + c
(d) 2 sin 2x + c
Answer:
(a) sin 2x + c

Question 16.
\(\int \frac{d x}{\cos x \sqrt{\sin ^{2} x-\cos ^{2} x}} \cdot d x=\)
(a) log(tan x – \(\sqrt{\tan ^{2} x-1}\)) + c
(b) sin^-1 (tan x) + c
(c) 1 + sin^-1 (cot x) + c
(d) log(tan x + \(\sqrt{\tan ^{2} x-1}\)) + c
Answer:
(d) log(tan x + \(\sqrt{\tan ^{2} x-1}\)) + c

Question 17.
\(\int \frac{\log x}{(\log e x)^{2}} \cdot d x=\)
(a) \(\frac{x}{1+\log x}+c\)
(b) x(1 + log x) + c
(c) \(\frac{1}{1+\log x}+c\)
(d) \(\frac{1}{1-\log x}+c\)
Answer:
(a) \(\frac{x}{1+\log x}+c\)

Question 18.
∫[sin(log x) + cos(log x)] . dx =
(a) x cos(log x) + c
(b) sin(log x) + c
(c) cos(log x) + c
(d) x sin(log x) + c
Answer:
(d) x sin(log x) + c

Question 19.
\(\int \frac{\cos 2 x-1}{\cos 2 x+1} \cdot d x=\)
(a) tan x – x + c
(b) x + tan x + c
(c) x – tan x + c
(d) -x – cot x + c
Answer:
(c) x – tan x + c

Question 20.
\(\int \frac{e^{2 x}+e^{-2 x}}{e^{x}} \cdot d x=\)
(a) \(e^{x}-\frac{1}{3 e^{3 x}}+c\)
(b) \(e^{x}+\frac{1}{3 e^{3 x}}+c\)
(c) \(e^{-x}+\frac{1}{3 e^{3 x}}+c\)
(d) \(e^{-x}-\frac{1}{3 e^{3 x}}+c\)
Answer:
(a) \(e^{x}-\frac{1}{3 e^{3 x}}+c\)

II. Integrate the following with respect to the respective variable:

Question 1.
(x – 2)² √x
Solution:

Question 2.
\(\frac{x^{7}}{x+1}\)
Solution:

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

Question 4.
\(\frac{t^{3}}{(t+1)^{2}}\)
Solution:

Question 5.
\(\frac{3-2 \sin x}{\cos ^{2} x}\)
Solution:

Question 6.
\(\frac{\sin ^{6} \theta+\cos ^{6} \theta}{\sin ^{2} \theta \cdot \cos ^{2} \theta}\)
Solution:

Question 7.
cos 3x cos 2x cos x
Solution:

Question 8.
\(\frac{\cos 7 x-\cos 8 x}{1+2 \cos 5 x}\)
Solution:

Question 9.
\(\cot ^{-1}\left(\frac{1+\sin x}{\cos x}\right)\)
Solution:
Let I = \(\int \cot ^{-1}\left(\frac{1+\sin x}{\cos x}\right) d x\)

III. Integrate the following w.r.t. x:

Question 1.
\(\frac{(1+\log x)^{3}}{x}\)
Solution:

Question 2.
cot^-1 (1 – x + x²)
Solution:

Question 3.
\(\frac{1}{x \sin ^{2}(\log x)}\)
Solution:

Question 4.
\(\sqrt{x} \sec \left(x^{\frac{3}{2}}\right) \tan \left(x^{\frac{3}{2}}\right)\)
Solution:

Question 5.
log(1 + cos x) – x tan(\(\frac{x}{2}\))
Solution:

Question 6.
\(\frac{x^{2}}{\sqrt{1-x^{6}}}\)
Solution:

Question 7.
\(\frac{1}{(1-\cos 4 x)(3-\cot 2 x)}\)
Solution:

Question 8.
log(log x) + (log x)^-2
Solution:

Question 9.
\(\frac{1}{2 \cos x+3 \sin x}\)
Solution:

Question 10.
\(\frac{1}{x^{3} \sqrt{x^{2}-1}}\)
Solution:

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

Question 12.
log(x² + 1)
Solution:

Question 13.
e^2x sin x cos x
Solution:

Question 14.
\(\frac{x^{2}}{(x-1)(3 x-1)(3 x-2)}\)
Solution:

Question 15.
\(\frac{1}{\sin x+\sin 2 x}\)
Solution:

Question 16.
\(\sec ^{2} x \sqrt{7+2 \tan x-\tan ^{2} x}\)
Solution:

Question 17.
\(\frac{x+5}{x^{3}+3 x^{2}-x-3}\)
Solution:

Question 18.
\(\frac{1}{x\left(x^{5}+1\right)}\)
Solution:

Question 19.
\(\frac{\sqrt{\tan x}}{\sin x \cdot \cos x}\)
Solution:

Question 20.
sec⁴ x cosec² x
Solution:

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 6 Differential Equations Ex 6.1 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.1

1. Determine the order and degree of each of the following differential equations:

Question (i).
\(\frac{d y}{d x^{2}}+X\left(\frac{d y}{d x}\right)+y=2 \sin x\)
Solution:
The given D.E. is \(\frac{d y}{d x^{2}}+X\left(\frac{d y}{d x}\right)+y=2 \sin x\)
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.

Question (ii).
\(\sqrt[3]{1+\left(\frac{d y}{d x}\right)^{2}}=\frac{d^{2} y}{d x^{2}}\)
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.

Question (iii).
\(\frac{d y}{d x}=\frac{2 \sin x+3}{\frac{d y}{d x}}\)
Solution:
The given D.E. is \(\frac{d y}{d x}=\frac{2 \sin x+3}{\frac{d y}{d x}}\)
∴ \(\left(\frac{d y}{d x}\right)^{2}\) = 2 sin x + 3
This D.E. has highest order derivative \(\frac{d y}{d x}\) with power 2.
∴ the given D.E. is of order 1 and degree 2.

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

Question (v).
\(\frac{d^{2} y}{d t^{2}}+\left(\frac{d y}{d t}\right)^{2}+7 x+5=0\)
Solution:
The given D.E. is \(\frac{d^{2} y}{d t^{2}}+\left(\frac{d y}{d t}\right)^{2}+7 x+5=0\)
This D.E. has highest order derivative \(\frac{d^{2} y}{d x^{2}}\) with power 1.
∴ the given D.E. has order 2 and degree 1.

Question (vi).
(y”‘)2 + 3y” + 3xy’ + 5y = 0
Solution:
The given D.E. is (y”‘)2 + 3y” + 3xy’ + 5y = 0
This can be written as:
\(\left(\frac{d^{3} y}{d x^{3}}\right)^{2}+3 \frac{d^{2} y}{d x^{2}}+3 x \frac{d y}{d x}+5 y=0\)
This D.E. has highest order derivative \(\frac{d^{3} y}{d x^{3}}\) with power 2.
∴ The given D.E. has order 3 and degree 2.

Question (vii).
\(\left(\frac{d^{2} y}{d x^{2}}\right)^{2}+\cos \left(\frac{d y}{d x}\right)=0\)
Solution:
The given D.E. is \(\left(\frac{d^{2} y}{d x^{2}}\right)^{2}+\cos \left(\frac{d y}{d x}\right)=0\)
This D.E. has highest order derivative \(\frac{d^{2} y}{d x^{2}}\)
∴ order = 2
Since this D.E. cannot be expressed as a polynomial in differential coefficients, the degree is not defined.

Question (viii).
\(\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{\frac{3}{2}}=8 \frac{d^{2} y}{d x^{2}}\)
Solution:
The given D.E. is \(\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{\frac{3}{2}}=8 \frac{d^{2} y}{d x^{2}}\)
On squaring both sides, we get
\(\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{3}=8^{2} \cdot\left(\frac{d^{2} y}{d x^{2}}\right)^{2}\)
This D.E. has highest order derivative \(\frac{d^{2} y}{d x^{2}}\) with power 2.
∴ the given D.E. has order 2 and degree 2.

Question (ix).
\(\left(\frac{d^{3} y}{d x^{3}}\right)^{\frac{1}{2}} \cdot\left(\frac{d y}{d x}\right)^{\frac{1}{3}}=20\)
Solution:
The given D.E. is \(\left(\frac{d^{3} y}{d x^{3}}\right)^{\frac{1}{2}} \cdot\left(\frac{d y}{d x}\right)^{\frac{1}{3}}=20\)
∴ \(\left(\frac{d^{3} y}{d x^{3}}\right)^{3} \cdot\left(\frac{d y}{d x}\right)^{2}=20^{6}\)
This D.E. has highest order derivative \(\frac{d^{3} y}{d x^{3}}\) with power 3.
∴ the given D.E. has order 3 and degree 3.

Question (x).
\(x+\frac{d^{2} y}{d x^{2}}=\sqrt{1+\left(\frac{d^{2} y}{d x^{2}}\right)^{2}}\)
Solution:
The given D.E. is \(x+\frac{d^{2} y}{d x^{2}}=\sqrt{1+\left(\frac{d^{2} y}{d x^{2}}\right)^{2}}\)
On squaring both sides, we get

This D.E. has highest order derivative \(\frac{d^{2} y}{d x^{2}}\) with power 1.
∴ the given D.E. has order 2 and degree 1.

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 5 Application of Definite Integration Ex 5.1 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 5 Application of Definite Integration Ex 5.1

1. Find the area of the region bounded by the following curves, X-axis, and the given lines:

(i) y = 2x, x = 0, x = 5.
Solution:
Required area = \(\int_{0}^{5} y d x\), where y = 2x
= \(\int_{0}^{5} 2x d x\)
= \(\left[\frac{2 x^{2}}{2}\right]_{0}^{5}\)
= 25 – 0
= 25 sq units.

(ii) x = 2y, y = 0, y = 4.
Solution:
Required area = \(\int_{0}^{4} x d y\), where x = 2y
= \(\int_{0}^{4} 2 y d y\)
= \(\left[\frac{2 y^{2}}{2}\right]_{0}^{4}\)
= 16 – 0
= 16 sq units.

(iii) x = 0, x = 5, y = 0, y = 4.
Solution:
Required area = \(\int_{0}^{5} y d x\), where y = 4
= \(\int_{0}^{5} 4 d x\)
= \([4 x]_{0}^{5}\)
= 20 – 0
= 20 sq units.

(iv) y = sin x, x = 0, x = \(\frac{\pi}{2}\)
Solution:
Required area = \(\int_{0}^{\pi / 2} y d x\), where y = sin x
= \(\int_{0}^{\pi / 2} \sin x d x\)
= \([-\cos x]_{0}^{\pi / 2}\)
= -cos \(\frac{\pi}{2}\) + cos 0
= 0 + 1
= 1 sq unit.

(v) xy = 2, x = 1, x = 4.
Solution:
For xy = 2, y = \(\frac{2}{x}\)
Required area = \(\int_{1}^{4} y d x\), where y = \(\frac{2}{x}\)
= \(\int_{1}^{4} \frac{2}{x} d x\)
= \([2 \log |x|]_{1}^{4}\)
= 2 log 4 – 2 log 1
= 2 log 4 – 0
= 2 log 4 sq units.

(vi) y² = x, x = 0, x = 4.
Solution:

The required area consists of two bounded regions A1 and A2 which are equal in areas.
For y² = x, y = √x
Required area = A₁ + A₂ = 2A₁

(vii) y² = 16x, x = 0, x = 4.
Solution:

The required area consists of two bounded regions A1 and A2 which are equal in areas.
For y² = x, y = √x
Required area = A₁ + A₂ = 2A₁

2. Find the area of the region bounded by the parabola:

(i) y² = 16x and its latus rectum.
Solution:
Comparing y² = 16x with y² = 4ax, we get
4a = 16
∴ a = 4
∴ focus is S(a, 0) = (4, 0)

For y² = 16x, y = 4√x
Required area = area of the region OBSAO
= 2 [area of the region OSAO]

(ii) y = 4 – x² and the X-axis.
Solution:
The equation of the parabola is y = 4 – x²
∴ x² = 4 – y
i.e. (x – 0)² = -(y – 4)
It has vertex at P(0, 4)
For points of intersection of the parabola with X-axis,
we put y = 0 in its equation.
∴ 0 = 4 – x²
∴ x² = 4
∴ x = ± 2
∴ the parabola intersect the X-axis at A(-2, 0) and B(2, 0)

Required area = area of the region APBOA
= 2[area of the region OPBO]

3. Find the area of the region included between:

(i) y² = 2x and y = 2x.
Solution:
The vertex of the parabola y² = 2x is at the origin O = (0, 0).

To find the points of intersection of the line and the parabola, equaling the values of 2x from both the equations we get,
y² = y
∴ y² – y = 0
∴ y = 0 or y = 1
When y = 0, x = \(\frac{0}{2}\) = 0
When y = 1, x = \(\frac{1}{2}\)
∴ the points of intersection are 0(0, 0) and B(\(\frac{1}{2}\), 1)
Required area = area of the region OABCO = area of the region OABDO – area of the region OCBDO
Now, area of the region OABDO = area under the parabola y² = 2x between x = 0 and x = \(\frac{1}{2}\)

Area of the region OCBDO = area under the line y = 2x between x = 0 and x = \(\frac{1}{2}\)

(ii) y² = 4x and y = x.
Solution:
The vertex of the parabola y² = 4x is at the origin O = (0, 0).

To find the points of intersection of the line and the parabola, equaling the values of 4x from both the equations we get,
∴ y² = y
∴ y² – y = 0
∴ y(y – 1) = 0
∴ y = 0 or y = 1
When y = 0, x = \(\frac{0}{2}\) = 0
When y = 1, x = \(\frac{1}{2}\)
∴ the points of intersection are O(0, 0) and B(\(\frac{1}{2}\), 1)
Required area = area of the region OABCO = area of the region OABDO – area of the region OCBDO
Now, area of the region OABDO = area under the parabola y² = 4x between x = 0 and x = \(\frac{1}{2}\)

Area of the region OCBDO = area under the line y = 2x between x = 0 and x = \(\frac{1}{2}\)

(iii) y = x² and the line y = 4x.
Solution:
The vertex of the parabola y = x² is at the origin 0(0, 0)
To find the points of the intersection of a line and the parabola.

Equating the values of y from the two equations, we get
x² = 4x
∴ x² – 4x = 0
∴ x(x – 4) = 0
∴ x = 0, x = 4
When x = 0, y = 4(0) = 0
When x = 4, y = 4(4) = 16
∴ the points of intersection are 0(0, 0) and B(4, 16)
Required area = area of the region OABCO = (area of the region ODBCO) – (area of the region ODBAO)
Now, area of the region ODBCO = area under the line y = 4x between x = 0 and x = 4
= \(\int_{0}^{4} y d x\), where y = 4x
= \(\int_{0}^{4} 4 x d x\)
= 4\(\int_{0}^{4} x d x\)
= 4\([latex]\int_{0}^{4} x d x\)[/latex]
= 2(16 – 0)
= 32
Area of the region ODBAO = area under the parabola y = x² between x = 0 and x = 4
= \(\int_{0}^{4} y d x\), where y = x2
= \(\int_{0}^{4} x^{2} d x\)
= \(\left[\frac{x^{3}}{3}\right]_{0}^{4}\)
= \(\frac{1}{3}\) (64 – 0)
= \(\frac{64}{3}\)
∴ required area = 32 – \(\frac{64}{3}\) = \(\frac{32}{3}\) sq units.

(iv) y² = 4ax and y = x.
Solution:
The vertex of the parabola y² = 4ax is at the origin O = (0, 0).

To find the points of intersection of the line and the parabola, equaling the values of 4ax from both the equations we get,
∴ y² = y
∴ y² – y = 0
∴ y(y – 1) = 0
∴ y = 0 or y = 1
When y = 0, x = \(\frac{0}{2}\) = 0
When y = 1, x = \(\frac{1}{2}\)
∴ the points of intersection are O(0, 0) and B(\(\frac{1}{2}\), 1)
Required area = area of the region OABCO = area of the region OABDO – area of the region OCBDO
Now, area of the region OABDO
= area under the parabola y² = 4ax between x = 0 and x = \(\frac{1}{2}\)

Area of the region OCBDO
= area under the line y
= 4ax between x = 0 and x = \(\frac{1}{4 a x}\)

(v) y = x² + 3 and y = x + 3.
Solution:
The given parabola is y = x² + 3, i.e. (x – 0)² = y – 3
∴ its vertex is P(0, 3).

To find the points of intersection of the line and the parabola.
Equating the values of y from both the equations, we get
x² + 3 = x + 3
∴ x² – x = 0
∴ x(x – 1) = 0
∴ x = 0 or x = 1
When x = 0, y = 0 + 3 = 3
When x = 1, y = 1 + 3 = 4
∴ the points of intersection are P(0, 3) and B(1, 4)
Required area = area of the region PABCP = area of the region OPABDO – area of the region OPCBDO
Now, area of the region OPABDO
= area under the line y = x + 3 between x = 0 and x = 1

Area of the region OPCBDO = area under the parabola y = x² + 3 between x = 0 and x = 1

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 6 Differential Equations Ex 6.5 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.5

Question 1.
Solve the following differential equations:
(i) \(\frac{d y}{d x}+\frac{y}{x}=x^{3}-3\)
Solution:
\(\frac{d y}{d x}+\frac{y}{x}=x^{3}-3\) …….(1)
This is the linear differential equation of the form
\(\frac{d y}{d x}\) + P . y = Q, where P = \(\frac{1}{x}\) and Q = x³ – 3

This is the general solution.

(ii) cos²x . \(\frac{d y}{d x}\) + y = tan x
Solution:

(iii) (x + 2y³) \(\frac{d y}{d x}\) = y
Solution:

(iv) \(\frac{d y}{d x}\) + y . sec x = tan x
Solution:
\(\frac{d y}{d x}\) + y sec x = tan x
∴ \(\frac{d y}{d x}\) + (sec x) . y = tan x ……..(1)
This is the linear differential equation of the form
\(\frac{d y}{d x}\) + P . y = Q, where P = sec x and Q = tan x
∴ I.F. = \(e^{\int P d x}\)
= \(e^{\int \sec x d x}\)
= \(e^{\log (\sec x+\tan x)}\)
= sec x + tan x
∴ the solution of (1) is given by
y (I.F.) = ∫Q . (I.F.) dx + c
∴ y(sec x + tan x) = ∫tan x (sec x + tan x) dx + c
∴ (sec x + tan x) . y = ∫(sec x tan x + tan²x) dx + c
∴ (sec x + tan x) . y = ∫(sec x tan x + sec²x – 1) dx + c
∴ (sec x + tan x) . y = sec x + tan x – x + c
∴ y(sec x + tan x) = sec x + tan x – x + c
This is the general solution.

(v) x \(\frac{d y}{d x}\) + 2y = x² . log x
Solution:

(vi) (x + y) \(\frac{d y}{d x}\) = 1
Solution:
(x + y) \(\frac{d y}{d x}\) = 1
∴ \(\frac{d x}{d y}\) = x + y
∴ \(\frac{d x}{d y}\) – x = y
∴ \(\frac{d x}{d y}\) + (-1) x = y ……….(1)
This is the linear differential equation of the form

This is the general solution.

(vii) (x + a) \(\frac{d y}{d x}\) – 3y = (x + a)⁵
Solution:

(viii) dr + (2r cot θ + sin 2θ) dθ = 0
Solution:
dr + (2r cot θ + sin 2θ) dθ = 0
∴ \(\frac{d r}{d \theta}\) + (2r cot θ + sin 2θ) = 0
∴ \(\frac{d r}{d \theta}\) + (2 cot θ)r = -sin 2θ ………(1)
This is the linear differential equation of the form dr
\(\frac{d r}{d \theta}\) + P . r = Q, where P = 2 cot θ and Q = -sin 2θ

This is the general solution.

(ix) y dx + (x – y²) dy = 0
Solution:
y dx + (x – y²) dy = 0
∴ y dx = -(x – y²) dy
∴ \(\frac{d x}{d y}=-\frac{\left(x-y^{2}\right)}{y}=-\frac{x}{y}+y\)
∴ \(\frac{d x}{d y}+\left(\frac{1}{y}\right) \cdot x=y\) ………(1)
This is the linear differential equation of the form

This is the general solution.

(x) \(\left(1-x^{2}\right) \frac{d y}{d x}+2 x y=x\left(1-x^{2}\right)^{\frac{1}{2}}\)
Solution:

(xi) \(\left(1+x^{2}\right) \frac{d y}{d x}+y=e^{\tan ^{-1} x}\)
Solution:

Question 2.
Find the equation of the curve which passes through the origin and has the slope x + 3y – 1 at any point (x, y) on it.
Solution:
Let A(x, y) be the point on the curve y = f(x).
Then slope of the tangent to the curve at the point A is \(\frac{d y}{d x}\).
According to the given condition,
\(\frac{d y}{d x}\) = x + 3y – 1
∴ \(\frac{d y}{d x}\) – 3y = x – 1 ………(1)
This is the linear differential equation of the form

This is the general equation of the curve.
But the required curve is passing through the origin (0, 0).
∴ by putting x = 0 and y = 0 in (2), we get
0 = 2 + c
∴ c = -2
∴ from (2), the equation of the required curve is 3(x + 3y) = 2 – 2e^3x i.e. 3(x + 3y) = 2 (1 – e^3x).

Question 3.
Find the equation of the curve passing through the point \(\left(\frac{3}{\sqrt{2}}, \sqrt{2}\right)\) having slope of the tangent to the curve at any point (x, y) is \(-\frac{4 x}{9 y}\).
Solution:
Let A(x, y) be the point on the curve y = f(x).
Then the slope of the tangent to the curve at point A is \(\frac{d y}{d x}\).
According to the given condition
\(\frac{d y}{d x}=-\frac{4 x}{9 y}\)
∴ y dy = \(-\frac{4}{9}\) x dx
Integrating both sides, we get
∫y dy= \(-\frac{4}{9}\) ∫x dx
∴ \(\frac{y^{2}}{2}=-\frac{4}{9} \cdot \frac{x^{2}}{2}+c_{1}\)
∴ 9y² = -4x² + 18c₁
∴ 4x² + 9y² = c where c = 18c₁
This is the general equation of the curve.
But the required curve is passing through the point \(\left(\frac{3}{\sqrt{2}}, \sqrt{2}\right)\).
∴ by putting x = \(\frac{3}{\sqrt{2}}\) and y = √2 in (1), we get
\(4\left(\frac{3}{\sqrt{2}}\right)^{2}+9(\sqrt{2})^{2}=c\)
∴ 18 + 18 = c
∴ c = 36
∴ from (1), the equation of the required curve is 4x² + 9y² = 36.

Question 4.
The curve passes through the point (0, 2). The sum of the coordinates of any point on the curve exceeds the slope of the tangent to the curve at any point by 5. Find the equation of the curve.
Solution:
Let A(x, y) be any point on the curve.
Then slope of the tangent to the curve at the point A is \(\frac{d y}{d x}\).
According to the given condition
x + y = \(\frac{d y}{d x}\) + 5
∴ \(\frac{d y}{d x}\) – y = x – 5 ………(1)
This is the linear differential equation of the form
\(\frac{d y}{d x}\) + P . y = Q, where P = -1 and Q = x – 5
∴ I.F. = \(e^{\int P d x}=e^{\int-1 d x}=e^{-x}\)
∴ the solution of (1) is given by

This is the general equation of the curve.
But the required curve is passing through the point (0, 2).
∴ by putting x = 0, y = 2 in (2), we get
2 = 4 – 0 + c
∴ c = -2
∴ from (2), the equation of the required curve is y = 4 – x – 2e^x.

Question 5.
If the slope of the tangent to the curve at each of its point is equal to the sum of abscissa and the product of the abscissa and ordinate of the point. Also, the curve passes through the point (0, 1). Find the equation of the curve.
Solution:
Let A(x, y) be the point on the curve y = f(x).
Then slope of the tangent to the curve at the point A is \(\frac{d y}{d x}\).
According to the given condition
\(\frac{d y}{d x}\) = x + xy
∴ \(\frac{d y}{d x}\) – xy = x ……….. (1)
This is the linear differential equation of the form
\(\frac{d y}{d x}\) + Py = Q, where P = -x and Q = x
∴ I.F. = \(e^{\int P d x}=e^{\int-x d x}=e^{-\frac{x^{2}}{2}}\)
∴ the solution of (1) is given by
y . (I.F.) = ∫Q . (I.F.) dx + c

This is the general equation of the curve.
But the required curve is passing through the point (0, 1).
∴ by putting x = 0 and y = 1 in (2), we get
1 + 1 = c
∴ c = 2
∴ from (2), the equation of the required curve is 1 + y = \(2 e^{\frac{x^{2}}{2}}\).

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 5 Application of Definite Integration Miscellaneous Exercise 5 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 5 Application of Definite Integration Miscellaneous Exercise 5

I. Choose the correct option from the given alternatives:

Question 1.
The area bounded by the region 1 ≤ x ≤ 5 and 2 ≤ y ≤ 5 is given by
(a) 12 sq units
(b) 8 sq units
(c) 25 sq units
(d) 32 sq units
Answer:
(a) 12 sq units

Question 2.
The area of the region enclosed by the curve y = \(\frac{1}{x}\), and the lines x = e, x = e² is given by
(a) 1 sq unit
(b) \(\frac{1}{2}\) sq units
(c) \(\frac{3}{2}\) sq units
(d) \(\frac{5}{2}\) sq units
Answer:
(a) 1 sq unit

Question 3.
The area bounded by the curve y = x³, the X-axis and the lines x = -2 and x = 1 is
(a) -9 sq units
(b) \(-\frac{15}{4}\) sq units
(c) \(\frac{15}{4}\) sq units
(d) \(\frac{17}{4}\) sq units
Answer:
(c) \(\frac{15}{4}\) sq units

Question 4.
The area enclosed between the parabola y² = 4x and line y = 2x is
(a) \(\frac{2}{3}\) sq units
(b) \(\frac{1}{3}\) sq units
(c) \(\frac{1}{4}\) sq units
(d) \(\frac{3}{4}\) sq units
Answer:
(b) \(\frac{1}{3}\) sq units

Question 5.
The area of the region bounded between the line x = 4 and the parabola y² = 16x is
(a) \(\frac{128}{3}\) sq units
(b) \(\frac{108}{3}\) sq units
(c) \(\frac{118}{3}\) sq units
(d) \(\frac{218}{3}\) sq units
Answer:
(a) \(\frac{128}{3}\) sq units

Question 6.
The area of the region bounded by y = cos x, Y-axis and the lines x = 0, x = 2π is
(a) 1 sq unit
(b) 2 sq units
(c) 3 sq units
(d) 4 sq units
Answer:
(d) 4 sq units

Question 7.
The area bounded by the parabola y² = 8x, the X-axis and the latus rectum is
(a) \(\frac{31}{3}\) sq units
(b) \(\frac{32}{3}\) sq units
(c) \(\frac{32 \sqrt{2}}{3}\) sq units
(d) \(\frac{16}{3}\) sq units
Answer:
(b) \(\frac{32}{3}\) sq units

Question 8.
The area under the curve y = 2√x, enclosed between the lines x = 0 and x = 1 is
(a) 4 sq units
(b) \(\frac{3}{4}\) sq units
(c) \(\frac{2}{3}\) sq units
(d) \(\frac{4}{3}\) sq units
Answer:
(d) \(\frac{4}{3}\) sq units

Question 9.
The area of the circle x² + y² = 25 in first quadrant is
(a) \(\frac{25 \pi}{3}\) sq units
(b) 5π sq units
(c) 5 sq units
(d) 3 sq units
Answer:
(a) \(\frac{25 \pi}{3}\) sq units

Question 10.
The area of the region bounded by the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) is
(a) ab sq units
(b) πab sq units
(c) \(\frac{\pi}{a b}\) sq units ab
(d) πa² sq units
Answer:
(b) πab sq units

Question 11.
The area bounded by the parabola y² = x and the line 2y = x is
(a) \(\frac{4}{3}\) sq units
(b) 1 sq unit
(c) \(\frac{2}{3}\) sq unit
(d) \(\frac{1}{3}\) sq unit
Answer:
(a) \(\frac{4}{3}\) sq units

Question 12.
The area enclosed between the curve y = cos 3x, 0 ≤ x ≤ \(\frac{\pi}{6}\) and the X-axis is
(a) \(\frac{1}{2}\) sq unit
(b) 1 sq unit
(c) \(\frac{2}{3}\) sq unit
(d) \(\frac{1}{3}\) sq unit
Answer:
(d) \(\frac{1}{3}\) sq unit

Question 13.
The area bounded by y = √x and line x = 2y + 3, X-axis in first quadrant is
(a) 2√3 sq units
(b) 9 sq units
(c) \(\frac{34}{3}\) sq units
(d) 18 sq units
Answer:
(b) 9 sq units

Question 14.
The area bounded by the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) and the line \(\frac{x}{a}+\frac{y}{b}=1\) is
(a) (πab – 2ab) sq units
(b) \(\frac{\pi a b}{4}-\frac{a b}{2}\) sq units
(c) (πab – ab) sq units
(d) πab sq units
Answer:
(b) \(\frac{\pi a b}{4}-\frac{a b}{2}\) sq units

Question 15.
The area bounded by the parabola y = x² and the line y = x is
(a) \(\frac{1}{2}\) sq unit
(b) \(\frac{1}{3}\) sq unit
(c) \(\frac{1}{6}\) sq unit
(d) \(\frac{1}{12}\) sq unit
Answer:
(c) \(\frac{1}{6}\) sq unit

Question 16.
The area enclosed between the two parabolas y² = 4x and y = x is
(a) \(\frac{8}{3}\) sq units
(b) \(\frac{32}{3}\) sq units
(c) \(\frac{16}{3}\) sq units
(d) \(\frac{4}{3}\) sq units
Answer:
(c) \(\frac{16}{3}\) sq units

Question 17.
The area bounded by the curve y = tan x, X-axis and the line x = \(\frac{\pi}{4}\) is
(a) \(\frac{1}{3}\) log 2 sq units
(b) log 2 sq units
(c) 2 log 2 sq units
(d) 3 log 2 sq units
Answer:
(a) \(\frac{1}{3}\) log 2 sq units

Question 18.
The area of the region bounded by x² = 16y, y = 1, y = 4 and x = 0 in the first quadrant, is
(a) \(\frac{7}{3}\) sq units
(b) \(\frac{8}{3}\) sq units
(c) \(\frac{64}{3}\) sq units
(d) \(\frac{56}{3}\) sq units
Answer:
(d) \(\frac{56}{3}\) sq units

Question 19.
The area of the region included between the parabolas y² = 4ax and x² = 4ay, (a > 0) is given by
(a) \(\frac{16 a^{2}}{3}\) sq units
(b) \(\frac{8 a^{2}}{3}\) sq units
(c) \(\frac{4 a^{2}}{3}\) sq units
(d) \(\frac{32 a^{2}}{3}\) sq units
Answer:
(a) \(\frac{16 a^{2}}{3}\) sq units

Question 20.
The area of the region included between the line x + y = 1 and the circle x² + y² = 1 is
(a) \(\frac{\pi}{2}-1\) sq units
(b) π – 2 sq units
(c) \(\frac{\pi}{4}-\frac{1}{2}\) sq units
(d) π – \(\frac{1}{2}\) sq units
Answer:
(c) \(\frac{\pi}{4}-\frac{1}{2}\) sq units

(II) Solve the following:

Question 1.
Find the area of the region bounded by the following curve, the X-axis and the given lines:
(i) 0 ≤ x ≤ 5, 0 ≤ y ≤ 2
(ii) y = sin x, x = 0, x = π
(iii) y = sin x, x = 0, x = \(\frac{\pi}{3}\)
Solution:
(i) Required area = \(\int_{0}^{5} y d x\), where y = 2
= \(\int_{0}^{5} 2 d x\)
= \([2 x]_{0}^{5}\)
= 2 × 5 – 0
= 10 sq units.

(ii) The curve y = sin x intersects the X-axis at x = 0 and x = π between x = 0 and x = π.

Two bounded regions A₁ and A₂ are obtained. Both the regions have equal areas.
∴ required area = A₁ + A₂ = 2A₁

(iii) Required area = \(\int_{0}^{\pi / 3} y d x\), where y = sin x

Question 2.
Find the area of the circle x² + y² = 9, using integration.
Solution:
By the symmetry of the circle, its area is equal to 4 times the area of the region OABO.
Clearly, for this region, the limits of integration are 0 and 3.

From the equation of the circle, y² = 9 – x².
In the first quadrant, y > 0
∴ y = \(\sqrt{9-x^{2}}\)
∴ area of the circle = 4 (area of the region OABO)

Question 3.
Find the area of the ellipse \(\frac{x^{2}}{25}+\frac{y^{2}}{16}=1\) using integration.
Solution:

By the symmetry of the ellipse, its area is equal to 4 times the area of the region OABO.
Clearly, for this region, the limits of integration are 0 and 5.
From the equation of the ellipse
\(\frac{y^{2}}{16}=1-\frac{x^{2}}{25}=\frac{25-x^{2}}{25}\)
∴ y² = \(\frac{16}{25}\) (25 – x²)
In the first quadrant y > 0
∴ y = \(\frac{4}{5} \sqrt{25-x^{2}}\)
∴ area of the ellipse = 4(area of the region OABO)

Question 4.
Find the area of the region lying between the parabolas:
(i) y² = 4x and x² = 4y
(ii) 4y² = 9x and 3x² = 16y
(iii) y² = x and x² = y.
Solution:
(i)

For finding the points of intersection of the two parabolas, we equate the values of y² from their equations.
From the equation x² = 4y, y = \(\frac{x^{2}}{4}\)
y = \(\frac{x^{4}}{16}\)
\(\frac{x^{4}}{16}\) = 4x
∴ x⁴ – 64x = 0
∴ x(x³ – 64) = 0
∴ x = 0 or x³ = 64 i.e. x = 0 or x = 4
When x = 0, y = 0
When x = 4, y = \(\frac{4^{2}}{4}\) = 4
∴ the points of intersection are 0(0, 0) and A(4, 4).
Required area = area of the region OBACO = [area of the region ODACO] – [area of the region ODABO]
Now, area of the region ODACO = area under the parabola y² = 4x, i.e. y = 2√x between x = 0 and x = 4

(ii)

For finding the points of intersection of the two parabolas, we equate the values of 4y² from their equations.
From the equation 3x² = 16y, y = \(\frac{3 x^{2}}{16}\)
∴ y = \(\frac{3 x^{4}}{256}\)
∴ \(\frac{3 x^{4}}{256}\) = 9x
∴ 3x⁴ – 2304x = 0
∴ x(x³ – 2304) = 0
∴ x = 0 or x³ = 2304 i.e. x = 0 or x = 4
When x = 0, y = 0
When x = 4, y = \(\frac{4^{2}}{4}\)
∴ the points of intersection are O(0, 0) and A(4, 4).
Required area = area of the region OBACO = [area of the region ODACO] – [area of the region ODABO]
Now, area of the region ODACO = area under the parabola y² = 4x,
i.e. y = 2√x between x = 0 and x = 4

Area of the region ODABO = area under the rabola x² = 4y,
i.e. y = \(\frac{x^{2}}{4}\) between x = 0 and x = 4

(iii)

For finding the points of intersection of the two parabolas, we equate the values of y² from their equations.
From the equation x² = y, y = \(\frac{x^{2}}{y}\)
∴ y = \(\frac{x^{2}}{y}\)
∴ \(\frac{x^{2}}{y}\) = x
∴ x² – y = 0
∴ x(x³ – y) = 0
∴ x = 0 or x³ = y
i.e. x = 0 or x = 4
When x = 0, y = 0
When x = 4, y = \(\frac{4^{2}}{4}\) = 4
∴ the points of intersection are O(0, 0) and A(4, 4).
Required area = area of the region OBACO = [area of the region ODACO] – [area of the region ODABO]
Now, area of the region ODACO = area under the parabola y² = 4x,
i.e. y = 2√x between x = 0 and x = 4

Area ofthe region ODABO = area under the rabola x² = 4y,
i.e. y = \(\frac{x^{2}}{3}\) between x = 0 and x = 4

Question 5.
Find the area of the region in the first quadrant bounded by the circle x² + y² = 4 and the X-axis and the line x = y√3.
Solution:

For finding the points of intersection of the circle and the line, we solve
x² + y² = 4 ………(1)
and x = y√3 ……..(2)
From (2), x² = 3y²
From (1), x² = 4 – y²
3y² = 4 – y²
4y² = 4
y² = 1
y = 1 in the first quadrant.
When y = 1, r = 1 × √3 = √3
∴ the circle and the line intersect at A(√3, 1) in the first quadrant
Required area = area of the region OCAEDO = area of the region OCADO + area of the region DAED
Now, area of the region OCADO = area under the line x = y√3, i.e. y = \(\frac{x}{\sqrt{3}}\) between x = 0
and x = √3

Question 6.
Find the area of the region bounded by the parabola y² = x and the line y = x in the first quadrant.
Solution:
To obtain the points of intersection of the line and the parabola, we equate the values of x from both equations.

∴ y² = y
∴ y² – y = 0
∴ y(y – 1) = 0
∴ y = 0 or y = 1
When y = 0, x = 0
When y = 1, x = 1
∴ the points of intersection are O(0, 0) and A(1, 1).
Required area = area of the region OCABO = area of the region OCADO – area of the region OBADO
Now, area of the region OCADO = area under the parabola y² = x i.e. y = +√x (in the first quadrant) between x = 0 and x = 1

Area of the region OBADO = area under the line y = x between x = 0 and x = 1

Question 7.
Find the area enclosed between the circle x² + y² = 1 and the line x + y = 1, lying in the first quadrant.
Solution:

Required area = area of the region ACBPA = (area of the region OACBO) – (area of the region OADBO)
Now, area of the region OACBO = area under the circle x² + y² = 1 between x = 0 and x = 1

Area of the region OADBO = area under the line x + y = 1 between x = 0 and x = 1

∴ required area = \(\left(\frac{\pi}{4}-\frac{1}{2}\right)\) sq units.

Question 8.
Find the area of the region bounded by the curve (y – 1)² = 4(x + 1) and the line y = (x – 1).
Solution:
The equation of the curve is (y – 1)² = 4(x + 1)
This is a parabola with vertex at A (-1, 1).
To find the points of intersection of the line y = x – 1 and the parabola.
Put y = x – 1 in the equation of the parabola, we get
(x – 1 – 1)² = 4(x + 1)
∴ x² – 4x + 4 = 4x + 4
∴ x² – 8x = 0
∴ x(x – 8) = 0
∴ x = 0, x = 8
When x = 0, y = 0 – 1 = -1
When x = 8, y = 8 – 1 = 7
∴ the points of intersection are B (0, -1) and C (8, 7).

To find the points where the parabola (y – 1)² = 4(x + 1) cuts the Y-axis.
Put x = 0 in the equation of the parabola, we get
(y – 1)² = 4(0 + 1) = 4
∴ y – 1 = ±2
∴ y – 1 = 2 or y – 1 = -2
∴ y = 3 or y = -1
∴ the parabola cuts the Y-axis at the points B(0, -1) and F(0, 3).
To find the point where the line y = x – 1 cuts the X-axis.
Put y = 0 in the equation of the line, we get
x – 1 = 0
∴ x = 1
∴ the line cuts the X-axis at the point G (1, 0).
Required area = area of the region BFAB + area of the region OGDCEFO + area of the region OBGO
Now, area of the region BFAB = area under the parabola (y – 1)² = 4(x + 1), Y-axis from y = -1 to y = 3

Since, the area cannot be negative,
Area of the region BFAB = \(\left|-\frac{8}{3}\right|=\frac{8}{3}\) sq units.
Area of the region OGDCEFO = area of the region OPCEFO – area of the region GPCDG

Since, area cannot be negative,
area of the region = \(\left|-\frac{1}{2}\right|=\frac{1}{2}\) sq units.
∴ required area = \(\frac{8}{3}+\frac{109}{6}+\frac{1}{2}\)
= \(\frac{16+109+3}{6}\)
= \(\frac{128}{6}\)
= \(\frac{64}{3}\) sq units.

Question 9.
Find the area of the region bounded by the straight line 2y = 5x + 7, X-axis and x = 2, x = 5.
Solution:
The equation of the line is
2y = 5x + 7, i.e., y = \(\frac{5}{2} x+\frac{7}{2}\)
Required area = area of the region ABCDA = area under the line y = \(\frac{5}{2} x+\frac{7}{2}\) between x = 2 and x = 5

Question 10.
Find the area of the region bounded by the curve y = 4x², Y-axis and the lines y = 1, y = 4.
Solution:

By symmetry of the parabola, the required area is 2 times the area of the region ABCD.
From the equation of the parabola, x² = \(\frac{y}{4}\)
In the first quadrant, x > 0
∴ x = \(\frac{1}{2} \sqrt{y}\)
∴ required area = \(\int_{1}^{4} x d y\)

Balbharti Maharashtra State Board 11th Physics Textbook Solutions Chapter 7 Thermal Properties of Matter Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Physics Solutions Chapter 7 Thermal Properties of Matter

1. Choose the correct option.

Question 1.
The range of temperature in a clinical thermometer, which measures the temperature of the human body, is
(A) 70 ºC to 100 ºC
(B) 34 ºC to 42 ºC
(C) 0 ºF to 100 ºF
(D) 34 ºF to 80 ºF
Answer:
(B) 34 ºC to 42 ºC

Question 2.
A glass bottle completely filled with water is kept in the freezer. Why does it crack?
(A) Bottle gets contracted
(B) Bottle is expanded
(C) Water expands on freezing
(D) Water contracts on freezing
Answer:
(C) Water expands on freezing

Question 3.
If two temperatures differ by 25 °C on Celsius scale, the difference in temperature on Fahrenheit scale is
(A) 65°
(B) 45°
(C) 38°
(D) 25°
Answer:
(B) 45°

Question 4.
If α, β and γ are coefficients of linear, area l and volume expansion of a solid then
(A) α: β:γ 1:3:2
(B) α:β:γ 1:2:3
(C) α:β:γ 2:3:1
(D) α:β:γ 3:1:2
Answer:
(B) α:β:γ 1:2:3

Question 5.
Consider the following statements-
(I) The coefficient of linear expansion has dimension K^-1
(II) The coefficient of volume expansion has dimension K^-1
(A) I and II are both correct
(B) I is correct but II is wrong
(C) II is correct but I is wrong
(D) I and II are both wrong
Answer:
(A) I and II are both correct

Question 6.
Water falls from a height of 200 m. What is the difference in temperature between the water at the top and bottom of a water fall given that specific heat of water is 4200 J kg^-1 °C^-1?
(A) 0.96 °C
(B) 1.02 °C
(C) 0.46 °C
(D) 1.16 °C
Answer:
(C) 0.46 °C

2. Answer the following questions.

Question 1.
Clearly state the difference between heat and temperature?
Answer:

Question 2.
How a thermometer is calibrated?
Answer:

1. For the calibration of a thermometer, a standard temperature interval is selected between two easily reproducible fixed temperatures.
2. The fact that substances change state from solid to liquid to gas at fixed temperatures is used to define reference temperature called fixed point.
3. The two fixed temperatures selected for this purpose are the melting point of ice or freezing point of water and the boiling point of water.
4. This standard temperature interval is divided into sub-intervals by utilizing some physical property that changes with temperature.
5. Each sub-interval is called as a degree of temperature. Thus, an empirical scale for temperature is set up.

Question 3.
What are different scales of temperature? What is the relation between them?
Answer:

1. Celsius scale:
   • The ice point (melting point of pure ice) is marked as O °C (lower point) and steam point (boiling point of water) is marked as 100 °C (higher point).
   • Both are taken at one atmospheric pressure.
   • The interval between these points is divided into two equal parts. Each of these parts is called as one degree celsius and it is ‘written as 1 °C.
2. Fahrenheit scale:
   • The ice point (melting point of pure ice) is marked as 32 °F and steam point (boiling point of water) is marked as 212 °F.
   • The interval between these two reference points is divided into 180 equal parts. Each part is called as degree fahrenheit and is written as 1 °F.
3. Kelvin scale:
   • The temperature scale that has its zero at -273.15 °C and temperature intervals are same as that on the Celsius scale is called as kelvin scale or absolute scale.
   • The absolute temperature, T and celsius temperature, T_C are related as, T = T_C + 273.15
     eg.: when T_C = 27 °C,
     T = 27+273.15 K = 300.15 K

Relation between different scales of temperature:
\(\frac{\mathrm{T}_{\mathrm{F}}-32}{180}=\frac{\mathrm{T}_{\mathrm{C}}-0}{100}=\frac{\mathrm{T}_{\mathrm{K}}-273.15}{100}\)
where,
T_F = temperature in fahrenheit scale,
T_C = temperature in celsius scale,
T_K = temperature in kelvin scale,
[Note: At zero of the kelvin scale, every substance in nature has the least possible activity.]

Question 4.
What is absolute zero?
Answer:

1. When the graph of pressure (P) against temperature T (°C) at constant volume for three ideal gases A, B and C is plotted, in each case, P -T graph is straight line indicating direct proportion between them. The slopes of these graphs are different.
2. The individual straight lines intersect the pressure axis at different values of pressure at O °C. but each line intersects the temperature axis at the same point, i.e., at absolute temperature (-273.15 °C).
3. Similarly graph at constant pressure for three different ideal gases A, B and C extrapolate to the same temperature intercept -273.15 °C i.e., absolute zero temperature.
4. It is seen that all the lines for different gases Cut the temperature axis at the same point at -273.15 °C.
5. This point is termed as the absolute zero of temperature.
6. It is not possible to attain a temperature lower than this value. Even to achieve absolute zero temperature is not possible in practice.
   [Note: The point of zero pressure or zero volume does not depend on am specific gas.]

Question 5.
Derive the relation between three coefficients of thermal expansion.
Answer:
Consider a square plate of side l₀ at 0 °C and h at T °C.

1. l_T = l₀ (1 + αT)
   If area of plate at 0 °C is A₀, A₀ = \(l_{0}^{2}\)
   If area of plate at T °C is A_T,
   A_T = \(l_{\mathrm{T}}^{2}=l_{0}^{2}\) (1 + αT)²
   or A_T = A₀ (1 + αT)² …………… (1)
   Also,
   A_T = A₀(1 + βT)² …………… (2)
   ……………. [∵ β = \(\frac{\mathrm{A}_{\mathrm{T}}-\mathrm{A}_{0}}{\mathrm{~A}_{0}\left(\mathrm{~T}-\mathrm{T}_{0}\right)}\)]
2. Using Equations (1) and (2),
   A₀ (1 + αT)² = A₀(1 + βT)
   ∴ 1 + 2αT + α²T² = 1 + βT
3. Since the values of a are very small, the term α²T² is very small and may be neglected,
   ∴ β = 2a
4. The result is general because any solid can be regarded as a collection of small squares.

Relation between coefficient of linear expansion (α) and coefficient of cubical expansion (γ).

1. Consider a cube of side l₀ at 0 °C and l_T at T °C.
   ∴ l_T = l₀(1 + αT)
   If volume of the cube at 0 °C is V₀, V₀ = \(l_{0}^{3}\)
   If volume of the cube at T °C is
   V_T, V_T = \(l_{\mathrm{T}}^{3}=l_{0}^{3}\) (1 + αT)³
   V_T = V₀ (1 + αT)³ ………. (1)
   Also,
   V_T = V₀(1 + γT) …………. (2)
   …………. [∵ γ = \(\frac{\mathrm{V}_{\mathrm{T}}-\mathrm{V}_{0}}{\mathrm{~V}_{0}\left(\mathrm{~T}-\mathrm{T}_{0}\right)}\)]
2. Using Equations (1) and (2),
   V₀(1 + αT)³ = V₀(1 + γT)
   ∴ 1 + 3αT + 3α²T² + α³T³ = 1 + γT
3. Since the values of a are very small, the terms with higher powers of a may be neglected,
   ∴ γ = 3α
4. The result is general because any solid can be regarded as a collection of small cubes.

Relation between α, β and γ is given by,
α = \(\frac{\beta}{2}=\frac{\gamma}{3}\)
where, α = coefficient of linear expansion.
β = coefficient of superficial expansion,
γ = coefficient of cubical expansion.

Question 6.
State applications of thermal expansion.
Answer:
Applications of thermal expansion:

• The steel wheel is heated to expand. This expanded wheel can easily fit over axle. The wheel is then cooled quickly. Upon cooling, wheel contracts and fits tightly upon the axle.
• An electric light bulb gets hot quickly when in use. The wire leads to the filament are sealed into the glass. 1f the glass of the bulbs has significantly different thermal expansivity from the wire leads, the glass and the wire would separate, breaking down the vacuum. To prevent this, wires are made of platinum or some suitable alloy with the same expansivity as ordinary glass.

Question 7.
Why do we generally consider two specific heats for a gas?
Answer:

• A slight change in temperature causes considerable change in pressure as well as volume of the gas.
• Therefore, two principal specific heats are defined for a gas viz., specific heat capacity at constant volume (S_V) and specific heat capacity at constant pressure (S_p).

Question 8.
Are freezing point and melting point same with respect to change of state ? Comment.
Answer:
Though freezing point and melting point mark same temperature (0°C or 32° F), state of change is different for the two points. At freezing point liquid gets converted into solid, whereas at melting point solid gets converted into its liquid state.

Question 9.
Define
(i) Sublimation
(ii) Triple point.
Answer:

1. The change from solid state to vapour stale without passing through the liquid state is called sublimation and the substance is said to sublime.
   Examples: Dry ice (solid CO₂) and iodine.
2. The triple point of water is that point where water in a solid, liquid and gas state co-exists in equilibrium and this occurs only at a unique temperature and a pressure.

Question 10.
Explain the term ‘steady state’.
Answer:

1. When one end of a metal rod is heated, the heat flows by conduction from hot end to the cold end.
   
2. As a result, the temperature of every section of the rod starts increasing.
3. Under this condition, the rod is said to be in a variable temperature state.
4. After some time, the temperature at each section of the rod becomes steady i.e., does not change.
5. Temperature of each cross-section of the rod now becomes constant though not the same. This is called steady state condition.

Question 11.
Define coefficient of thermal conductivity. Derive its expression.
Answer:
Coefficient of thermal conductivity of a material is defined as the quantity of heat that flows in one second between the opposite faces of a cube of side 1 m, the faces being kept at a temperature difference of 1°C (or 1 K).

Expression for coefficient of thermal conductivity:

1. Under steady state condition, the quantity of heat ‘Q’ that flows from the hot face at temperature T₁ to the cold face at temperature T₂ of a cube with side x and area of cross-section A is
   • directly proportional to the cross-sectional area A of the face. i.e.. Q ∝ A
   • directly proportional to the temperature difference between the two faces i.e., Q ∝ (T₁ – T₂)
   • directly proportional to time t (in seconds) for which heat flows i.e.. Q ∝ t
   • inversely proportional to the perpendicular distance x between hot and cold faces i.e., Q ∝ 1/x
2. Combining the above four factors, we have the quantity of heat
   Q ∝ \(\frac{\mathrm{A}\left(\mathrm{T}_{1}-\mathrm{T}_{2}\right) \mathrm{t}}{\mathrm{x}}\)
   ∴ Q = \(\)
   where k is a constant of proportionality and is called coefficient of thermal conductivity. Its value depends upon the nature of the material.

Question 12.
Give any four applications of thermal conductivity in every day life.
Answer:
Answer: Applications of thermal conductivity:

1. • Thick walls are used in the construction of cold storage rooms.
   • Brick being a bad conductor of heat is used to reduce the flow of heat from the surroundings to the rooms.
   • Better heat insulation is obtained by using hollow bricks.
   • Air being a poorer conductor than a brick, it further avoids the conduction of heat from outside.
2. Street vendors keep ice blocks packed in saw dust to prevent them from melting rapidly.
3. The handle of a cooking utensil is made of a bad conductor of heat, such as ebonite, to protect our hand from the hot utensil.
4. Two bedsheets used together to cover the body help retain body heat better than a single bedsheet of double the thickness. Trapped air being a bad conductor of heat, the layer of air between the two sheets reduces thermal conduction better than a sheet of double the thickness. Similarly, a blanket coupled with a bedsheet is a cheaper alternative to using two blankets.

Question 13.
Explain the term thermal resistance. State its SI unit and dimensions.
Answer:

1. Consider expression for conduction rate,
   P_cond = kA \(\frac{\left(\mathrm{T}_{1}-\mathrm{T}_{2}\right)}{\mathrm{x}}\)
   ⇒ \(\frac{\mathrm{T}_{1}-\mathrm{T}_{2}}{\mathrm{P}_{\text {cond }}}=\frac{\mathrm{x}}{\mathrm{kA}}\) ……………. (1)
2. Ratio \(\frac{\mathrm{T}_{1}-\mathrm{T}_{2}}{\mathrm{P}_{\text {cond }}}\) is called as thermal resistance (R_T) of material.

The SI unit of thermal resistance is °C s/kcal or °C s/J and its dimensional formula is [L^-2M^-1T³K¹].

Question 14.
How heat transfer occurs through radiation in absence of a medium?
Answer:

1. All objects possess thermal energy due to their temperature T(T > 0 K).
2. The rapidly moving molecules of a hot body emit EM waves travelling with the velocity of light. These are called thermal radiations.
3. These carry energy with them and transfer it to the low-speed molecules of a cold body on which they fall.
4. This results in an increase in the molecular motion of the cold body and its temperature rises.
5. Thus transfer of heat by radiation is a two fold process-the conversion of thermal energy into waves and reconversion of waves into thermal energy by the body on which they fall.

Question 15.
State Newton’s law of cooling and explain how it can be experimentally verified.
Answer:
The rate of loss of heat dT/dt of the both’ is directly proportional to the difference of temperature (T – T₀) of the body and the surroundings provided the difference in temperatures is small.

Mathematically, Newton’s law of cooling can be expressed as:
\(\frac{\mathrm{dT}}{\mathrm{dt}}\) ∝ (T – T₀)
∴ \(\frac{\mathrm{dT}}{\mathrm{dt}}\) ∝ C(T – T₀)
where, C is constant of proportionality. Experimental verification of Newton’s law of cooling:

1. Fill a calorimeter upto \(\frac{2}{3}\) of its capacity with a boiling water. Cover it with lid with a hole for passing the thermometer.
2. Insert the thermometer through the hole and adjust it so that the bulb of the thermometer is fully immersed in hot water.
3. Keep calorimeter vessel in constant temperature enclosure or just in open air since room temperature will not change much during the experiment.
4. Note down the temperature (T) on the thermometer at every one minute interval until the temperature of water decreases by about 25 °C.
5. Plot a graph of temperature (T) on Y-axis against time (t) on X-axis. This graph is called cooling curve as shown in figure (a).
   
6. Draw tangents to the curve at suitable points on the curve. The slope of each tangent is \(\lim _{\Delta t \rightarrow 0} \frac{\Delta \mathrm{T}}{\Delta \mathrm{t}}\) and gives the rate of fall of temperature at that temperature (T).
7. Now the graph of \(\left|\frac{\mathrm{dT}}{\mathrm{dt}}\right|\) on Y-axis against (T – T₀) on X-axis is plotted with (0, 0) origin. The graph is straight line and passes through origin as shown in figure (b), which verities Newton’s law of cooling.
   
   (b) Graphical verification of Newton’s law of cooling

Question 16.
What is thermal stress? Give an example of disadvantages of thermal stress in practical use?
Answer:

1. Consider a metallic rod of length l₀ fixed between two rigid supports at T °C.
   If the temperature of rod is increased by ∆T, length of rod would become,
   l = l₀(1 + α∆T)
   Where, α is the coefficient of linear expansion of material of the rod.
   But the supports prevent expansion of rod. As a result, rod exerts stress on the supports. Such stress is termed as thermal stress.
2. Disadvantage: Thermal stress can lead to fracture or deformation in substance under certain conditions.
3. Railway tracks are made up of metals which expand upon heating. If no gap is kept between tracks, in hot weather, expansion of metal tracks may exert thermal stress on track. This may lead to bending of tracks which would be dangerous. Hence, railway track is not a continuous piece but is made up of segments separated by gaps.

Question 17.
Which materials can be used as thermal insulators and why?
Answer:

1. Substances such as glass, wood, rubber, plastic, etc. can be used as thermal insulators.
2. These substances do not have free electrons to conduct heat freely throughout the body. Hence, they arc poor conductors of heat.

3. Solve the following problems.

Question 1.
A glass flask has volume 1 × 10^-4 m³. It is filled with a liquid at 30 ºC. If the temperature of the system is raised to 100 ºC, how much of the liquid will overflow. (Coefficient of volume expansion of glass is 1.2 × 10^-5 (ºC)^-1 while that of the liquid is 75 × 10^-5 ºC^-1)
Solution:
Given: V₁ = 1 × 10^-4 m³ = 10^-4 m³, T₁ = 30°C,
T₂ = 100 °C
To find: Volume of liquid that overflows
Formula: γ = \(\frac{V_{2}-V_{1}}{V_{1}\left(T_{2}-T_{1}\right)}\)
Calculation: From formula,
Increase is volume = V₂ – V₁
= γV₁(T₂ – T₁)
increase in volume of beaker
= γ_glass × V₁ (T₂ – T₁)
= 1.2 × 10^-5 × 10^-4 × (100 – 30)
= 1.2 × 70 × 10^-9
= 4 × 10^-9 m³
∴ Increase in volume of beaker
= 84 × 10^-9 m³
Increase in volume of liquid
= γ_liquid × V₁ (T₂ – T₁)
= 75 × 10^-5 × 10 × (100 – 30)
= 75 × 70 × 10
= 5250 × 10^-9 m³
∴ Increase in volume of liquid = 5250 × 10^-9 m³
∴ Volume of liquid which overflows
= (5250 – 84) × 10^-9 m³
= 5166 × 10^-9 m³
= 0.5166 × 10^-7 m³
Volume of liquid that overflows is 0.5166 × 10^-7 m³.
[Note: The answer given above is presented considering standard conventions of writing number with its correct order of magnitude.]

Question 2.
Which will require more energy, heating a 2.0 kg block of lead by 30 K or heating a 4.0 kg block of copper by 5 K? (s_lead = 128 J kg^-1 K^-1, s_copper = 387 J kg^-1 K^-1)
Solution:
Given: m_lead = 2 kg, ∆T_lead = 30 K,
s_lead = 128 J/kg K,
m_Cu =4 kg, ∆T_Cu = 5 K,
s_Cu = 387 J/kg K
To find: Substance requiring more heat energy.
Formula: Q = ms ∆T
Calculation: From formula,
For lead, Q_lead = 2 × 128 × 30 = 7680J
For Copper, Q_Cu = 4 × 387 × 5 = 7740 J
Q_Cu > Q_lead, copper will require more heat energy.
Copper will require more heat energy.

Question 3.
Specific latent heat of vaporization of water is 2.26 × 10⁶ J/kg. Calculate the energy needed to change 5.0 g of water into steam at 100 ºC.
Solution:
Given: L_vap = 2.26 × 10⁶ J/kg
m = 5g = 5 × 10^-3 kg
In this case, no temperature change takes place only change of state occurs.
To find: Heat required to convert water into steam.
Formula: Heat required = mL_vap
Calculation: From formula,
Heat required = 5 × 10^-3 × 2.26 × 10⁶
= 11300J
= 1.13 × 10⁴ J
Heat required to convert water into steam is 1.13 × 10⁴ J
[Note: The answer given above is presented considering standard conventions of writing number with its correct order of magnitude.]

Question 4.
A metal sphere cools at the rate of 0.05 ºC/s when its temperature is 70ºC and at the rate of 0.025 ºC/s when its temperature is 50 ºC. Determine the temperature of the surroundings and find the rate of cooling when the temperature of the metal sphere is 40 ºC.
Solution:

∴ 2(50 – T₀) = 70 – T₀
∴ T₀ = 30 πC
Substituting value of T₀.
0.05 = C (70 – 30)
∴ C = \(\frac{0.05}{40}\) = 0.00125/s.
For T₃ = 40 °C
\(\left(\frac{\mathrm{d} \mathrm{T}}{\mathrm{dt}}\right)_{3}\) = C(T₃ – T₀)
= 0.00125 (40 – 30)
= 0.00125 × 10
= 0.0125°C/s.
i) Temperature of surrounding is 30 °C.
ii) Rate of cooling at 40 °C is 0.0125 °C/s.

Question 5.
The volume of a gas varied linearly with absolute temperature if its pressure is held constant. Suppose the gas does not liquefy even at very low temperatures, at what temperature the volume of the gas will be ideally zero?
Answer:
At temperature of -273.15 °C, the volume of the gas will be ideally zero.

Question 6.
In olden days, while laying the rails for trains, small gaps used to be left between the rail sections to allow for thermal expansion. Suppose the rails are laid at room temperature 27 ºC. If maximum temperature in the region is 45 ºC and the length of each rail section is 10 m, what should be the gap left given that α = 1.2 × 10^-5 K^-1 for the material of the rail section?
Solution:
Given. T₁ = 27 °C, T₂ = 45 °C,
L₁ = 10m.
α = 1.2 × 10^-5 K^-1
To find: Gap that should be left (L₂ – L₁)
Formula: L₂ – L₁ = L₁ α(T₂ – T₁)
Calculation: From formula,
L₂ – L₁ = 10 × 1.2 × 10^-5 × (45 – 27)
= 2.16 × 10^-3 m
= 2.16 mm
The gap that should be left between rail sections is 2.16 mm.

Question 7.
A blacksmith fixes iron ring on the rim of the wooden wheel of a bullock cart. The diameter of the wooden rim and the iron ring are 1.5 m and 1.47 m respectively at room temperature of 27 ºC. To what temperature the iron ring should be heated so that it can fit the rim of the wheel (α_iron = 1.2 × 10^-5 K^-1).
Solution:
Given: d_w = 1.5 m, d = 1.47 m, T₁ = 27 °C.
α_i = 1.2 × 10^-5/ K
To find: Temperature (T₂)
Formula. α = \(\frac{\mathrm{d}_{\mathrm{w}}-\mathrm{d}_{\mathrm{i}}}{\mathrm{d}_{\mathrm{i}}\left(\mathrm{T}_{2}-\mathrm{T}_{1}\right)}\)
Calculation: From formula,
T₂ = \(\frac{\mathrm{d}_{\mathrm{w}}-\mathrm{d}_{\mathrm{i}}}{\mathrm{d}_{\mathrm{i}} \alpha}\) + T₁
= \(\frac{1.5-1.47}{1.47 \times 1.2 \times 10^{-5}}\) + 27
= 1700.7 + 27
= 1727.7 °C
Iron ring should be heated to temperature of 1727.7 °C.

Question 8.
In a random temperature scale X, water boils at 200 °X and freezes at 20 °X. Find the boiling point of a liquid in this scale if it boils at 62 °C.
Solution:
Here thermometric property P is temperature at some random scale X.
Using equation,
T = \(\frac{100\left(P_{T}-P_{1}\right)}{\left(P_{2}-P_{1}\right)}\)
For P₁ = 20 °X,
P₂ = 200 °X,
T = 62°C
∴ 62 = \(\frac{100\left(\mathrm{P}_{\mathrm{T}}-20\right)}{(200-20)}\)
∴ P_T = \(\frac{62 \times(200-20)}{100}\) + 20 = 111.6 + 20
= 131.6 °X
The boiling point of a liquid in this scale is 131.6 °X.

Question 9.
A gas at 900°C is cooled until both its pressure and volume are halved. Calculate its final temperature.
Solution:
Given: T₁ = 900 °C = 900 + 273.15 = 1173.15 K
V₂ = \(\frac{\mathrm{V}_{1}}{2}\), P₂ = \(\frac{\mathrm{P}_{1}}{2}\)
To find: Final temperature (T₂)
Formula: \(\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{~T}_{\mathrm{I}}}=\frac{\mathrm{P}_{2} \mathrm{~V}_{2}}{\mathrm{~T}_{2}}\)
Calculation: From formula.
\(\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{1173.15}=\frac{\mathrm{P}_{1} \mathrm{~V}_{\mathrm{l}}}{4 \mathrm{~T}_{2}}\)
∴ T₂ = \(\frac{1173.15}{4}\) = 293.29 K
Final temperature of gas is 293.29 K.

Question 10.
An aluminium rod and iron rod show 1.5 m difference in their lengths when heated at all temperature. What are their lengths at 0 °C if coefficient of linear expansion for aluminium is 24.5 × 10^-6 /°C and for iron is 11.9 × 10^-6 /°C
Solution:
Given: (L_T)_i – (L_T)_al = 1.5 m, T₀ = 0 °C
α_al = 24.5 × 10^-6/°C
α_i = 11.9 × 10^-6 /°C
To find: Lengths of aluminium and iron rod (L₀)_al and (L₀)_i
Formula: L_T = L₀[(1 + α(T – T₀)]
Calculation: For T₀ = 0 °C
From formula,
L_T = L₀(1 + αT)
For aluminium,
(L₀)_al = (L₀)_al(1 + α_alT) ……………. (1)
For iron,
(L_T)_i = (L₀)_i (1 + α_iT) ………….. (2)
Subtracting equation (2) by (1),
(L_T)_i – (L_T)_al = [(L₀)_i + (L₀)_i α_iT] – [(L₀)_al + (L₀)_alα_alT]
= (L₀)_i – (L₀)_al + [(L₀)_i α_i – (L₀)_al α_al]T
∴ 1.5 = 1.5 + [(L₀)_i α_i – (L₀)_al α_al)]T
⇒ [(L₀)_iα_i – (L₀)_alα_al] T = 0
∴ (L₀)_alα_al = (L₀)_iα_i

Length of aluminium rod at 0 °C is 1.417 m and that of iron rod is 2.917 m.

Question 11.
What is the specific heat of a metal if 50 cal of heat is needed to raise 6 kg of the metal from 20°C to 62 °C ?
Solution:
Given: Q = 50 cal, m =6 kg,
∆T = 62 – 20 = 42 °C
To find: Specific heat (s)
Formula: Q = ms ∆T
Calculation: From formula,
s = \(\frac{\mathrm{Q}}{\mathrm{m} \Delta \mathrm{T}}=\frac{50}{6 \times 42}\) = 0.198 cal/kg °C
Specific heat of metal is copper 0.198 cal/kg °C.

Question 12.
The rate of flow of heat through a copper rod with temperature difference 30 °C is 1500 cal/s. Find the thermal resistance of copper rod.
Solution:
Given: ∆T = 30 °C, P_cond = 1500 cal/s
To find: Thermal resistance (R_T)
Formula: R_T = \(\frac{\Delta \mathrm{T}}{\mathrm{P}_{\text {cond }}}\)
Calculation: From formula,
R_T = \(\frac{30}{1500}\)
= 0.02 °C s/cal.
Thermal resistance of copper rod is 0.02 °C s/cal.

Question 13.
An electric kettle takes 20 minutes to heat a certain quantity of water from 0°C to its boiling point. It requires 90 minutes to turn all the water at 100°C into steam. Find the latent heat of vaporisation. (Specific heat of water = 1cal/g°C)
Solution:
Let heat supplied by kettle in 20 minutes be Q₁ and that in 90 min. be Q₂.
Using heat temperature of water is raised from O °C to 100 °C.
If mass of water in the kettle is ‘m’ then.
Q₁ = ms_water∆T m × 1 × (100 – 0)
= 100 m ………….. (i)
…………. (∵ S_water = 1 cal/g °C)
Similarly using heat Q₂ water is converted from liquid to gas,
∴ Q₂ = mL_vap ……………. (ii)
Given that heat Q₁, Q₂ are supplied to water in 20 min. (t₁) and 90 min (t₂) respectively.
Kettle being same its conduction rate (P_cond) is same.
Using P_cond = \(\frac{\mathrm{Q}_{1}}{\mathrm{t}_{1}}=\frac{\mathrm{Q}_{2}}{\mathrm{t}_{2}}\) …………… (iii)
From (i), (ii) and (iii),
\(\frac{100 \mathrm{~m}}{20}=\frac{\mathrm{mL}_{\text {vap }}}{90}\),
∴ L_vap = 5 × 90 = 450 cal/g
Latent heat of vaporisation for water is 450 cal/g.

Question 14.
Find the temperature difference between two sides of a steel plate 4 cm thick, when heat is transmitted through the plate at the rate of 400 k cal per minute per square metre at steady state. Thermal conductivity of steel is 0.026 kcal/m s K.
Solution:

Temperature difference between two sides is 10.26 K.
[Note: Above answer is expressed in K (‘kelvin considering that thermal conductivity is expressed in units of kcal / ms K, and not as kcal / m s °C. As 1 °C equivalent to 1 K. conceptually temperature difference of 10.26 K will correspond to 10.26 t]

Question 15.
A metal sphere cools from 80 °C to 60 °C in 6 min. How much time with it take to cool from 60 °C to 40 °C if the room temperature is 30°C?
Solution:
Given: T₁ = 80 °C, T₂ = 60 °C, T₃ = 40 °C, T₀ = 30 °C, (dt)₁ = 6 min.
To find: Time taken in cooling (dt)₂

Time taken in cooling is 10 min.

11th Physics Digest Chapter 7 Thermal Properties of Matter Intext Questions and Answers

Can you tell? (Textbook Page No. 125)

Question 1.
i) Why the metal wires for electrical transmission lines sag?
ii) Why a railway track is not a continuous piece but is made up of segments separated by gaps?
iii) How a steel wheel is mounted on an axle to fit exactly?
Answer:

1. In hot weather, metal wires get heated due to increased temperature of surrounding. As a result, they expand increasing the slack between transmission line structure, causing them to sag.
2. Railway tracks are made up of metals which expand upon heating. If no gap is kept between tracks, in hot weather, expansion of metal tracks may exert thermal stress on track. This may lead to bending of tracks which would be dangerous. Hence, railway track is not a continuous piece but is made up of segments separated by gaps.
3. The steel wheel is heated to expand. This expanded wheel can easily fit over axle. The wheel is then cooled quickly. Upon cooling, wheel contracts and fits tightly upon the axle.

Intext question. (Textbook Page No 124)

Question 1.
Can you now tell why the balloon bursts sometimes when you try to fill air in it?
Answer:

1. When balloon is blown, air that is blown inside makes the balloon expand.
2. A given size of balloon can expand upto certain limit.
3. Once that limit is reached and air is still blown inside the balloon, balloon cannot expand further.
4. As a result, air causes additional pressure on inner surface of balloon.
5. Since, pressure inside balloon is now greater than pressure outside balloon, balloon bursts equalizing the two pressures.

Can you tell? (Textbook Page No. 125)

Question 1.
Why lakes freeze first at the surface?
Answer:

1. In cold climate, temperature of water in ponds and lakes starts falling.
2. On getting colder, water contracts. As a result, density of water increases and it goes down. To replace it, warmer water from below rises up. This process continues till temperature of water at the bottom of pond becomes 4 °C.
3. Water, due to its anomalous behaviour possesses maximum density at 4 °C.
4. If the temperature lowers further, ice is formed at the surface of pond with water below it.
5. Ice being poor conductor of heat blocks the further heat exchange between atmosphere and water in the pond and maintains water below surface in liquid state.

Activity (Textbook Page No. 129)

Question 1.
To understand the process of change of state:
Take some cubes of ice in a beaker. Note the temperature of ice (0 °C). Start heating it slowly on a constant heat source. Note the temperature after every minute. Continuously stir the mixture of water and ice. Observe the change in temperature. Continue heating even after the whole of ice gets converted into water. Observe the change in temperature as before till vapours start coming out. Plot the graph of temperature (along Y-axis) versus time (along X-axis). Obtain a graph of temperature versus time.
Answer:
[Students are expected to attempt the activity on their own.]

Can you tell? (Textbook Page No. 130)

Question 1.
What is observed after point D in graph? Can steam be hotter than 100 °C?
Answer:
Beyond point D, thermometer again shows rise in temperature. This means, steam can be hotter than 100 °C and is termed as superheated steam.

Question 2.
Why steam at 100 °C causes more harm to our skin than water at 100 °C?
Answer:

1. Though steam and boiling water have same temperature, the heat contained in steam is more than that in boiling water.
2. Steam is formed when boiling water absorbs specific latent heat of vaporisation i.e.. 22.6 × 10⁵ J/kg.
3. As a result, when steam comes in contact with the skin of a person, it gives off additional 22.6 × 10⁵ joule per kilogram causing severe (more serious) burns.
   Hence, burns caused from steam are more serious than those caused from boiling water at same temperature.

Activity (Textbook Page No. 130)

Activity to understand the dependence of boiling point on pressure:
Take a round bottom flask, more than half filled with water. Keep it over a burner and fix a thermometer and steam outlet through the cork of the flask as shown in figure. As water in the flask gets heated, note that first the air, which was dissolved in the water comes out as small bubbles. Later bubbles of steam form at the bottom but as they rise to the cooler water near the top, they condense and disappear. Finally, as the temperature of the entire mass of the water reaches 100 oc, bubbles of steam reach the surface and boiling is said to occur. The steam in the flask may not be visible hut as it comes out of the flask, It condenses as tiny droplets of water giving a foggy appearance.

If now the steam outlet is closed for a few seconds to increase the pressure in the flask, you will notice that boiling stops. More heat would be required to raise the temperature (depending on the increase in pressure) before boiling starts again. Thus, boiling point increases with increase in pressure. Let us now remove the burner. Allow water to cool to about 80°C. Remove the thermometers and steam outlet. Close the flask with a air tight cork. Keep the flask turned upside down on a stand. Pour icecold water on the flask. Water vapours in the flask condense reducing the pressure on the water surface inside the flask. Water begins to boil again, now at a lower temperature. Thus boiling point decreases with decrease in pressure and increases with increase in pressure.
Answer:
[Students are expected to attempt the activity an their own.]

Can you tell? (Textbook Page No. 131)

Question 1.
i) Why is cooking difficult at high altitude?
ii) Why is cooking faster in pressure cooker?
Answer:

1. • At high altitude density of air is low which causes reduction in atmospheric pressure.
   • As pressure is less, boiling point of water lowers.
   • Water, at high altitude, starts boiling below 100 OC.
   • As food is cooked mostly through the water boiling, cooking of food becomes difficult.
2. • Pressure cooker operates by expelling air within the cooker and trapping steam produced from the liquid. (mostly water) boiling inside.
   • Due to high internal pressure, boiling point of liquid increases and liquid boils at temperature higher than its boiling point.
   • The increased boiling point allows more absorption of heat by liquid and steam formed is superheated.
   • As a result, food gets cooked quickly.

Internet my friend (Textbook Page No. 139)

i) https ://hyperphysics. phy-astr.gsu.edul/base/hframe.html
ii) https://youtu.be/7ZKHc5J6R5Q
iii) https://physics. info/expansion
Answer:
[Students are expected to visit the above mentioned webs it es and collect more information about the thermal properties of matter.]

Balbharti Maharashtra State Board 11th Physics Textbook Solutions Chapter 6 Mechanical Properties of Solids Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Physics Solutions Chapter 6 Mechanical Properties of Solids

1. Choose the correct answer:

Question 1.
Change in dimensions is known as …………..
(A) deformation
(B) formation
(C) contraction
(D) strain.
Answer:
(A) deformation

Question 2.
The point on stress-strain curve at which strain begins to increase even without increase in stress is called…………
(A) elastic point
(B) yield point
(C) breaking point
(D) neck point
Answer:
(B) yield point

Question 3.
Strain energy of a stretched wire is 18 × 10^-3 J and strain energy per unit volume of the same wire and same cross section is 6 × 10^-3 J/m³. Its volume will be………….
(A) 3cm³
(B) 3 m³
(C) 6 m³
(D) 6 cm³
Answer:
(B) 3 m³

Question 4.
……………. is the property of a material which enables it to resist plastic deformation.
(A) elasticity
(B) plasticity
(C) hardness
(D) ductility
Answer:
(C) hardness

Question 5.
The ability of a material to resist fracturing when a force is applied to it, is called……………
(A) toughness
(B) hardness
(C) elasticity
(D) plasticity.
Answer:
(A) toughness

2. Answer in one sentence:

Question 1.
Define elasticity.
Answer:
If a body regains its original shape and size after removal of the deforming force, it is called an elastic body and the property is called elasticity.

Question 2.
What do you mean by deformation?
Answer:
The change in shape or size or both of u body due to an external force is called deformation.

Question 3.
State the SI unit and dimensions of stress.
Answer:

1. SI unit: N m^-2 or pascal (Pa)
2. Dimensions: [L^-1M¹T^-2]

Question 4.
Define strain.
Answer:
Strain:

1. Strain is defined as the ratio of change in dimensions of the body to its original dimensions.
   Strain = \(\frac{\text { change in dimensions }}{\text { original dimensions }}\)
2. Types of strain:
   • Longitudinal strain,
   • Volume strain,
   • Shearing strain.

Question 5.
What is Young’s modulus of a rigid body?
Answer:
Young’s modulus (Y): It is the modulus of elasticity related to change in length of an object like a metal wire, rod, beam, etc., due to the applied deforming force.

Question 6.
Why bridges are unsafe after a very long use?
Answer:
A bridge during its use undergoes recurring stress depending upon the movement of vehicles on it. When bridge is used for long time, it loses its elastic strength and ultimately may collapse. Hence, the bridges are declared unsafe after long use.

Question 7.
How should be a force applied on a body to produce shearing stress?
Answer:
A tangential force which is parallel to the top and the bottom surface of the body should be applied to produce shearing stress.

Question 8.
State the conditions under which Hooke’s law holds good.
Answer:
Hooke’s Taw holds good only when a wire/body is loaded within its elastic limit.

Question 9.
Define Poisson’s ratio.
Answer:
Within elastic limit, the ratio of lateral strain to the linear strain is called the Poisson‘s ratio.

Question 10.
What is an elastomer?
Answer:
A material that can be elastically stretched to a larger value of strain is called an elastomer.

Question 11.
What do you mean by elastic hysteresis?
Answer:

1. In case of some materials like vulcanized rubber, when the stress applied on a body decreases to zero, the strain does not return to zero immediately. The strain lags behind the stress. This lagging of strain behind the stress is called elastic hysteresis.
2. Below figure shows the stress-strain curve for increasing and decreasing load. It encloses a loop. Area of loop gives the energy dissipated during deformation of a material.
   

Question 12.
State the names of the hardest material and the softest material.
Answer:
Hardest material: Diamond
Softest material: Aluminium
[Note: Material with highest strength is steel whereas material with lowest strength is plasticine clay.]

Question 13.
Define friction.
Answer:
The property which resists the relative motion between two surfaces in contact is called friction.

Question 14.
Why force of static friction is known as ‘self-adjusting force?
Answer:
The force of static friction varies in accordance with applied force. Hence, it is called as self adjusting force.

Question 15.
Name two factors on which the coefficient of friction depends.
Answer:
Coefficient of friction depends upon:

1. the materials of the surfaces in contact.
2. the nature of the surfaces.

3. Answer in short:

Question 1.
Distinguish between elasticity and plasticity.
Answer:

Question 2.
State any four methods to reduce friction.
Answer:
Friction can be reduced by using polished surfaces, using lubricants, using grease and using ball bearings.

Question 3.
What is rolling friction? How does it arise?
Answer:

1. Friction between two bodies in contact when one body is rolling over the other, is called rolling friction.
2. Rolling friction arises as the point of contact of the body with the surface keep changing continuously.

Question 4.
Explain how lubricants help in reducing friction?
Answer:

1. The friction between lubricant to surface is much less than the friction between two same surfaces. Hence using lubricants reduces the friction between the two surfaces.
2. When lubricant is applied to machine parts, it fills the depression present on the surface in contact. Thus, less friction is occurred between machine parts.
3. Application of lubricants also reduces wear and tear of machine parts which in turn reduces friction.
4. Advantage: Reduction in function reduces dissipation of energy in machines due to which efficiency of machines increases.

Question 5.
State the laws of static friction.
Answer:
Laws of static friction:

1. First law: The limiting force of static friction (F_L) is directly proportional to the normal reaction (N) between the two surfaces in contact.
   F_L ∝ N
   ∴ F_L = µ_s N
   where, µ_s = constant called coefficient of static friction.
2. Second law: The limiting force of friction is
   independent of the apparent area between the surfaces in contact, so long as the normal reaction remains the same.
3. Third law: The limiting force of friction depends upon materials in contact and the nature of their surfaces.

Question 6.
State the laws of kinetic friction.
Answer:
Laws of kinetic friction:

1. First law: The force of kinetic friction (Fk) is directly proportional to the normal reaction (N) between two surfaces in contact.
   F_k ∝ N
   ∴ F_k = µ_kN
   where, µ_k = constant called coefficient of kinetic friction.
2. Second law: Force of kinetic friction is independent of shape and apparent area of the surfaces in contact.
3. Third law: Force of kinetic friction depends upon the nature and material of the surfaces in contact.
4. Fourth law: The magnitude of the force of kinetic friction is independent of the relative velocity between the object and the surface provided that the relative velocity is neither too large nor too small.

Question 7.
State advantages of friction.
Answer:
Advantages of friction:

1. We can walk due to friction between ground and feet.
2. We can hold object in hand due to static friction.
3. Brakes of vehicles work due to friction; hence we can reduce speed or stop vehicles.
4. Climbing on a tree is possible due to friction.

Question 8.
State disadvantages of friction.
Answer:
Disadvantages of friction:

1. Friction opposes motion.
2. Friction produces heat in different parts of machines. It also produces noise.
3. Automobile engines consume more fuel due to friction.

Question 9.
What do you mean by a brittle substance? Give any two examples.
Answer:

1. Substances which breaks within the elastic limit are called brittle substances.
2. Examples: Glass, ceramics.

4. Long answer type questions:

Question 1.
Distinguish between Young’s modulus, bulk modulus and modulus of rigidity.
Answer:

Question 2.
Define stress and strain. What are their different types?
Answer:
i) Stress:

1. The internal restoring force per unit area of a both is called stress.
   Stress = \(\frac{\text { deforming force }}{\text { area }}=\frac{|\vec{F}|}{\mathrm{A}}\)
   where \(\vec{F}\) is internal restoring force or external applied deforming force.
2. Types of stress:
   • Longitudinal stress,
   • Volume stress,
   • Shearing stress.

ii. Strain:

1. Strain is defined as the ratio of change in dimensions of the body to its original dimensions.
   Strain = \(\frac{\text { change in dimensions }}{\text { original dimensions }}\)
2. Types of strain:
   • Longitudinal strain,
   • Volume strain,
   • Shearing strain.

Question 3.
What is Young’s modulus? Describe an experiment to find out Young’s modulus of material in the form of a long straight wire.
Answer:
Definition: Young ‘s modulus is the ratio of longitudinal stress to longitudinal strain.
It is denoted by Y.
Unit: N/m² or Pa in SI system.
Dimensions: [L^-1M¹T^-2]

Experimental description to find Young’s modulus:

i. Consider a metal wire suspended from a rigid support. A load is attached to the free end of the wire. Due to this, deforming force gets applied to the free end of wire in downward direction and it produces a change in length.
Let,
L = original length of wire,
Mg = weight suspended to wire,
l = extension or elongation,
(L + l) = new length of wire.
r = radius of the cross section of wire

ii. In its equilibrium position,

Question 4.
Derive an expression for strain energy per unit volume of the material of a wire.
Answer:
Expression for strain energy per unit volume;

i. Consider a wire of original length L and cross sectional area A stretched by a force F acting along its length. The wire gets stretched and elongation l is produced in it

ii. If the wire is perfectly elastic then,
Longitudinal stress = \(\frac{F}{A}\)
Longitudinal strain = \(\frac{l}{L}\)

iii. The magnitude of stretching force increases from zero to F during elongation of wire.
Let ‘f’ be the restoring force and ‘x’ be its corresponding extension at certain instant during the process of extension.
∴ f = \(\frac{\text { YAx }}{\mathrm{L}}\) ……………. (2)

iv. Let ‘dW’ be the work done for the further small extension ‘dx’.
Work = force × displacement
∴ dW = fdx
∴ dW= \(\frac{\text { YAx }}{L}\) dx …………..(3) [From (2)]

v. The total amount of work done in stretching the wire from x = 0 to x = l can be found out by integrating equation (3).

∴ Work done in stretching a wire,
W = \(\frac{1}{2}\) × load × extension

vi. Work done by stretching force is equal to strain energy gained by the wire.
∴ Strain energy = \(\frac{1}{2}\) × load × extension

vii. Work done per unit volume

∴ Strain energy per unit volume = \(\frac{1}{2}\) × stress × strain

viii. Other forms:

Question 5.
What is friction? Define coefficient of static friction and coefficient of kinetic friction. Give the necessary formula for each.
Answer:

1. The property which resists the relative motion between two surfaces in contact is called friction.
2. The coefficient of static friction is defined as the ratio of limiting force of friction to the normal reaction.
   Formula: \(\mu_{\mathrm{S}}=\frac{\mathrm{F}_{\mathrm{L}}}{\mathrm{N}}\)
3. The coefficient of kinetic friction is defined as the ratio of force of kinetic friction to the normal reaction between the two surfaces in contact.
   Formula: \(\mu_{\mathrm{k}}=\frac{\mathrm{F}_{\mathrm{K}}}{\mathrm{N}}\)

Question 6.
State Hooke’s law. Draw a labelled graph of tensile stress against tensile strain for a metal wire up to the breaking point. In this graph show the region in which Hooke’s law is obeyed.
Answer:
i) Statement: Within elastic limit, stress is directly proportional to strain.
Explanation;

1. According to Hooke’s law,
   Stress ∝ Strain
   
   This constant of proportionality is called modulus of elasticity.
2. Modulus of elasticity of a material is the slope of stress-strain curve in elastic deformation region and depends on the nature of the material.
3. The graph of strain (on X-axis) and stress (on Y-axis) within elastic limit is shown in the figure.

ii)

iii) Hooke’s law is completely obeyed in the region OA.

5. Answer the following

Question 1.
Calculate the coefficient of static friction for an object of mass 50 kg placed on horizontal table pulled by attaching a spring balance. The force is increased gradually it is observed that the object just moves when spring balance shows 50N.
[Answer: µs = 0.102]
Solution:
Given: m = 50 kg, F_L = 50 N, g = 9.8 m/s²
To find: Coefficient of static friction (µ_s)
Formula: µ_s = \(\frac{\mathrm{F}_{\mathrm{L}}}{\mathrm{N}}=\frac{\mathrm{F}_{\mathrm{L}}}{\mathrm{mg}}\)
µ_s = \(\frac{50}{50 \times 9.8}\) = 0.102
Answer:
The coefficient of static friction is 0.102.

Question 2.
A block of mass 37 kg rests on a rough horizontal plane having coefficient of static friction 0.3. Find out the least force required to just move the block horizontally.
[Answer: F= 108.8N]
Solution:
Given: m = 37 kg, µ_s = 0.3, g = 9.8 m /s²
To find: Limiting force (F_L)
Formula: F_L = µ_SN = µ_S mg
Calculation: From formula,
F_L = 0.3 × 37 × 9.8 = 108.8 N
Answer:
The force required to move the block is 108.8 N.

Question 3.
A body of mass 37 kg rests on a rough horizontal surface. The minimum horizontal force required to just start the motion is 68.5 N. In order to keep the body moving with constant velocity, a force of 43 N is needed. What is the value of
a) coefficient of static friction? and
b) coefficient of kinetic friction?
Asw:
a) µ_s = 0.188
b) µ_k = 0.118]
Solution:
Given:
F_L = 68.5 N, F_k = 43 N,
m = 37 kg, g = 9.8 m/s²

To find:

i. Coefficient of static friction (µ_s)
ii. Coefficient of kinetic friction (µ_k)

Formulae:

i. µ_s = \(\frac{F_{L}}{N}\) = \(\frac{F_{L}}{m g}\)
ii. µ_k = \(\frac{F_{k}}{N}\) = \(\frac{\mathrm{F}_{\mathrm{k}}}{\mathrm{mg}}\)

Calculation:
From formula (i),
∴ µ_s = \(\frac{F_{S}}{N}=\frac{68.5}{37 \times 9.8}\) = 0.1889
From formula (ii),
∴ µ_k = \(\frac{F_{k}}{N}=\frac{43}{37 \times 9.8}\) = 0.1186
Answer:

1. The coefficient of static friction is 0.1889.
2. The coefficient of kinetic friction is 0.1186.

[Note: Answers calculated above are in accordance with textual methods of calculation.]

Question 4.
A wire gets stretched by 4mm due to a certain load. If the same load is applied to a wire of same material with half the length and double the diameter of the first wire. What will be the change in its length?
Solution:
Given. l₁ = 4mm = 4 × 10^-3 m
L₂ = \(\frac{\mathrm{L}_{1}}{2}\), D₂ = 2D, r₂ = 2r₁
To find: Change in length (l₂)
Formula: Y = \(\frac{\mathrm{FL}}{\mathrm{Al}}=\frac{\mathrm{FL}}{\pi \mathrm{r}^{2} l}\)
Calculation: From formula,

= 0.5 × 10^-3 m
= 0.5 mm
The new change in length of the wire is 0.5 mm.

Question 5.
Calculate the work done in stretching a steel wire of length 2m and cross sectional area 0.0225mm² when a load of 100 N is slowly applied to its free end. [Young’s modulus of steel= 2 × 10¹¹ N/m²]
Solution:
Given. L = 2m, F = 100 N,
A = 0.0225 mm² = 2.25 × 10^-8 m²,
Y = 2 × 10^-11 N/m²,
To find: Work (W)
Formula: W = \(\frac{1}{2}\) × F × l
Claculation:

= antilog [log 10 – log 4.5]
= antilog [1.0000 – 0.6532 ]
= antilog [0.3468]
∴ W = 2.222 J
Answer:
The work done in stretching the steel wire is 2.222 J.

Question 6.
A solid metal sphere of volume 0.31m³ is dropped in an ocean where water pressure is 2 × 10⁷ N/m². Calculate change in volume of the sphere if bulk modulus of the metal is 6.1 × 10¹⁰ N/m²
Solution:
Given: V= 0.31 m³, dP = 2 × 10⁷ N/m²,
K = 6.1 × 10¹⁰ N/m²
To find: Change in volume (dV)
Formula: K = V × \(\frac{\mathrm{dP}}{\mathrm{dV}}\)
Calculation: From formula,
dV = \(\frac{\mathrm{V} \times \mathrm{dP}}{\mathrm{K}}\)
∴ dV = \(\frac{0.31 \times 2 \times 10^{7}}{6.1 \times 10^{10}}\) ≈ 10^-4 m³
The change in volume of the sphere is 10^-4 m³.

Question 7.
A wire of mild steel has initial length 1.5 m and diameter 0.60 mm is extended by 6.3 mm when a certain force is applied to it. If Young’s modulus of mild steel is 2.1 × 10¹¹ N/m², calculate the force applied.
Solution:
Given:
L = 1.5m, d = 0.60 mm,
r = \(\frac{d}{2}\) = 0.30 mm = 3 × 10^-4 m,
Y = 2.1 × 10¹¹ N/m²,
l = 6.3 mm = 6.3 × 10^-3 m
To find: Force (F)
Calculation:
From formula,

= 2.1 × 3.142 × 6 × 6.3
= antilog [log 2.1 + log 3.142 + log 6 + log 6.3]
= antilog [0.3222 + 0.4972 + 0.7782 + 0.7993]
= antilog [2.3969]
= 2.494 × 10²
≈ 250 N
The force applied on wire is 250 N.

Question 8.
A composite wire is prepared by joining a tungsten wire and steel wire end to end. Both the wires are of the same length and the same area of cross section. If this composite wire is suspended to a rigid support and a force is applied to its free end, it gets extended by 3.25mm. Calculate the increase in length of tungsten wire and steel wire separately.
[Given: Y_steel = 2 × 10¹¹ Pa, Y_Tungsten = 4.11 × 10¹¹ Pa]
Solution:
Given: l_s + l_T = 3.25 mm,
Y_T = 4.11 × 10¹¹ Pa
Y_s = 2 × 10¹¹ Pa
To find: Extension in tungsten wire (l_T)
Extension in steel wire (l_s)

But l_s + l_T = 3.25
l_s + 0.487 l_s = 3.25
l_s(1 + 0.487) = 3.25
l_s = 2.186 mm
∴ l_T = 3.25 – 2.186
= 1.064 mm
The extension in tungsten wire is 1.064 mm and the extension in steel wire is 2.186 mm.

[Note: Values of Young’s modulus of tungsten and steel considered above are standard values. Using them, calculation is carried out ¡n accordance with textual method.]

Question 9.
A steel wire having cross sectional area 1.2 mm² is stretched by a force of 120 N. If a lateral strain of 1.455 mm is produced in the wire, calculate the Poisson’s ratio.
Solution:
Given: A = 1.2 mm² = 1.2 × 10^-6 m²,
F = 120 N, Y_steel = 2 × 10¹¹ N/m²,
Lateral strain = 1.455 × 10^-4
To find: Poisson’s ratio (σ)

The Poisson’s ratio of steel is 0.291.
[Note: Lateral strain being ratio of two same physical quantities, is unitless. hence, value given in question ¡s modified to 1.455 × 10^-4 to reach the answer given in textbook.]

Question 10.
A telephone wire 125m long and 1mm in radius is stretched to a length 125.25m when a force of 800N is applied. What is the value of Young’s modulus for material of wire?
Solution:
Given: L = 125m,
r = 1 mm= 1 × 10^-3 m
l = 125.25 – 125 = 0.25 m,
F = 800N
To find: Young’s modulus (Y)
Formula: Y \(\frac{\mathrm{FL}}{\mathrm{Al}}=\frac{\mathrm{FL}}{\pi \mathrm{r}^{2} l}\)
Calculation: From formula,
Y = \(\frac{800 \times 125}{3.142 \times 10^{-6} \times 0.25}\)
= {antilog [log 800 + log 125 – log 3.142 – log 0.25 ]} × 10⁶
= {antilog [2.9031 + 2.0969 – 0.4972 – \(\overline{1}\) .3979]} × 10⁶
= {antilog[5.1049]} × 10⁶
= 1.274 × 10⁵
= 1.274 × 10¹¹ N/m²
The Young’s modulus of telephone wire is 1.274 × 10¹¹ N/m².

Question 11.
A rubber band originally 30cm long is stretched to a length of 32cm by certain load. What is the strain produced?
Solution:
Given: L = 30 cm = 30 × 10 ^-2 m,
∆l = 32 cm – 30 cm = 2cm = 2 × 10 ^-2 m
To find. Strain
Formula: Strain = \(\frac{\Delta l}{\mathrm{~L}}\)
Calculation: From formula,
Strain = \(\frac{2 \times 10^{-2}}{30 \times 10^{-2}}\) = 6.667 × 10 ^-2
The strain produced in the wire is 6.667 × 10 ^-2.

Question 12.
What is the stress in a wire which is 50m long and 0.01cm² in cross section, if the wire bears a load of 100kg?
Solution:
Given: M = 100 kg, L 50 m, A = 0.01 × 10^-4 m
To find: Stress
Formula: Stress = \(\frac{\mathrm{F}}{\mathrm{A}}=\frac{\mathrm{Mg}}{\mathrm{A}}\)
Calculation: From formula,
Stress = \(\frac{100 \times 9.8}{0.01 \times 10^{-4}}\) = 9.8 × 10⁸ N/m²
The stress in the wire is 9.8 × 10⁸ N/m².

Question 13.
What is the strain in a cable of original length 50m whose length increases by 2.5cm when a load is lifted?
Solution:
Given: L = 50m, ∆l = 2.5cm = 2.5 × 10 ^-2 m
To find: Strain
Formula: Strain = \(\frac{\Delta l}{\mathrm{~L}}\)
Calculation: From formula,
Strain = \(\frac{2.5 \times 10^{-2}}{50}\) = 5 × 10^-4
The Strain produced in wire is 5 × 10^-4 .

11th Physics Digest Chapter 6 Mechanical Properties of Solids Intext Questions and Answers

Can you recall? (Textbook Page No. 100)

Question 1.

1. Can you name a few objects which change their shape and size on application of a force and regain their original shape and size when the force is removed?
2. Can you name objects which do not regain their original shape and size when the external force is removed?

Answer:

1. Objects such as rubber, metals, quartz, etc. change their shape and size on application of a force (within specific limit) and regain their original shape and size when the force is removed.
2. Objects such as putty, clay, thick mud. etc. do not regain their original shape and size when the external force is removed.

Can you tell? (Textbook Page No. 107)

Question 1.
Why does a rubber band become loose after repeated use?
Answer:

1. After repeated use of rubber band, its stress-strain curve does not remain linear.
2. In such case, since rubber crosses its elastic limit, there is a permanent set formed on the rubber due to which it becomes loose.

Can you tell? (Textbook Page No.111)

Question 1.
i. It is difficult to run fast on sand.
ii. It is easy to roll than pull a barrel along a road.
iii. An inflated tyre rolls easily than a flat tyre.
iv. Friction is a necessary evil.
Answer:
i.

1. The intermolecular space between crystals of sand is very large as compared to that in a rigid surface.
2. Thus, there are number of depressions at the points of contact of feet and sand surface.
3. Projections and depressions between sand and feet are not completely interlocked.
4. Thus, action and reaction force become unbalanced. The horizontal component of force helps to move forward and vertical component of the force resist to move.
   Hence, it becomes difficult to run fast on sand.

ii.

1. When a barrel is pulled along a road, the friction between the tyres and road is kinetic friction, but when its rolls along the road it undergoes rolling friction.
2. The force of kinetic friction is greater than force of rolling friction.
   Hence, it is easy to roll than pull a barrel along a road.

iii.

1. When the tyre is inflated, the pressure inside the tyre is reducing the normal force between tyre and the ground, and thus reducing the friction between the tyre and the road.
2. When the tyre gets deflated, it gets deformed during rolling, the supplied energy is used up in changing the shape and not overcoming the friction, and thus due to deformation, friction increases.
   Hence, an inflated tyre rolls easily than a flat tyre.

iv.

1. Friction helps us to walk, hold objects in hand, lift objects and without friction we cannot walk, we cannot grip or hold objects with our hands,
2. Friction is responsible for wear and tear of various part of machines, it produces heat in different parts of machine and also produces noise but it also helps in ball bearing or connecting screws.
   Hence, friction is said to be a necessary evil because it is useful as well as harmful.

Internet my friend (Textbook Page No. 111)

Question 1.
i. https ://opentextbc. ca/physicstestbook2/ chapter/friction/
ii. https://www.livescience.com/
iii. https://www.khanacademy.org/science/physics
iv. https://courses.lumenleaming.com/physics/ chapter/5-3-elasticity-stress-and-strain/
v. https://www.toppr.com/guides/physics/
Answer:
[Students are expected to visit the above mentioned websites and collect more information about mechanical properties of solid.]