Heat and Temperature HC Verma Exercise Questions Solutions Vol-2 Chapter-23

⇒ T = 373.31 K

Therefore, the temperature at the normal point (T) is 373.31 K.

Question-3 :-  (Heat and Temperature HC Verma)

Answer-3 :-

Given:
In a gas thermometer, the pressure measured at the melting point of lead, P = 2.20 × Pressure at triple point(P_​tr)
So the melting point of lead,(T) is given as:

⇒ T =2.20 × 273.16 K

=  T = 600.952 K

⇒ T ≅  601 K

Therefore, the melting point of lead is 601 K.

Question-4 :-

Answer-4 :-

Given:
Pressure measured by a constant volume gas thermometer at the triple point of water, P_tr = 40 kPa = 40 × 10³ Pa
Boiling point of water, T = 100°C = 373.16 K
Let the pressure measured at the boiling point of water be P.
​For a constant volume gas thermometer, temperature-pressure relation is given below:

Therefore, the pressure measured at the boiling point of water is 55 kPa.

Question-5 :-  (Heat and Temperature HC Verma)

The pressure of the gas in a constant volume gas thermometer is 70 kPa at the ice point.  Find the pressure at the steam point.

Answer-5 :-

Given:
​Temperature of ice point, T₁ = 273.15 K
Temperature of steam point, T₂ = 373.15 K
Pressure of the gas in a constant volume thermometer at the ice point, P_1​ = 70 kPa,
Let P_trbe the pressure at the triple point and P₂be the pressure at the steam point.
The temperature-pressure relations for ice point and steam point are given below:
For ice point,

On substituting the value of P_tr ,we get:

Therefore, the pressure at steam point is 96 kPa.

Question-6 :-

The pressures of the gas in a constant volume gas thermometer are 80 cm, 90 cm and 100 cm of mercury at the ice point, the steam point and in a heated wax bath, respectively. Find the temperature of the wax bath.

Answer-6 :-

Given:
In a constant volume gas thermometer,
Pressure of the gas at the ice point, P_​​​0 = 80 cm of Hg​
Pressure of the gas at the steam point, P₁₀₀ = 90 cm of Hg
​Pressure of the gas in a heated wax bath, P = 100 cm of Hg
The temperature of the wax bath

(T) is given by:

Therefore, the temperature of the wax bath is 200^o C.

Question-7 :-

In a Callender’s compensated  constant pressure air thermometer, the volume of the bulb is 1800 cc. When the bulb is kept immersed in a vessel, 200 cc of mercury has to be poured out. Calculate the temperature of the vessel.

Answer-7 :-

Given:
Volume of the bulb in a Callender’s compensated constant pressure air thermometer, (V) =
1800 cc
Volume of mercury that has to be poured out, V’ = 200 cc
Temperature of ice bath, T_o = 273.15 K
​So the temperature of the vessel(T’) is given by:

Therefore, the temperature of the vessel is 307 K.

Question-8 :-  (Heat and Temperature HC Verma)

A platinum resistance thermometer reads 0° when its resistance is 80 Ω and 100° when its resistance is 90 Ω.
Find the temperature at the platinum scale at which the resistance is 86 Ω.

Answer-8 :-

Given:
Resistance at 0^oC, R₀ = 80 Ω

Resistance at 100^oC, R₁₀₀ = 90 Ω

Let t be the temperature at which the resistance (R_t) is 86 Ω

Therefore, the resistance is 86 Ω at 60°C

---

Heat and Temperature HC Verma Exercise Questions Solutions

(HC Verma Ch-23 Concept of Physics Vol-2 for ISC Class-12)

(Page 13)

Question-9 :-

A resistance thermometer reads R = 20.0 Ω, 27.5 Ω, and 50.0 Ω at the ice point (0°C), the steam point (100°C) and the zinc point (420°C), respectively. Assuming that the resistance varies with temperature as R_θ = R₀ (1 + αθ + βθ²), find the values of R₀, α and β. Here θ represents the temperature on the Celsius scale.

Answer-9 :-

Given:
Reading on resistance thermometer at ice point, R₀ = 20 Ω
And on resistance thermometer at steam point, R₁₀₀ = 27.5 Ω
Reading on resistance thermometer at zinc point, R₄₂₀ = 50 Ω
The variation of resistance with temperature in Celsius scale,θ, is given as:

Given:
Temperature at which the steel metre scale is calibrated, t₁ = 20^oC
Temperature at which the scale is used, t₂ = 10^oC
So, the change in temperature,

Δt = 20^o

−10^o) C

The distance to be measured by the metre scale, L_o = 51

-50) = 1 cm = 0.01 m
Coefficient of linear expansion of steel,

αsteel= 1.1 × 10^–5 °C^–1

Let the new length measured by the scale due to expansion of steel be L_​2, Change in length is given by,

ΔL = L₁ ∝_steel Δt

⇒ΔL = 1× 1.1 ×1 0-5 × 10

⇒∆L = 0.00011cm

As the temperature is decreasing, therefore length will decrease by

∆L

Therefore, ​the new length measured by the scale due to expansion of steel (L₂) will be,
L₂ = 1 cm

− 0.00011 cm = 0.99989 cm

Question-12 :-

Answer-12 :-

Given:
Length of the iron sections when there’s no effect of temperature on them, L_o  = 12.0 m
​Temperature at which the iron track is laid in winter, t_​w​ = 18 ^oC
Maximum temperature during summers, t_s = 48 ^oC
Coefficient of linear expansion of iron,

α  = 11 × 10^–6  °C^–1

Let the new lengths attained by each section due to expansion of iron in winter and summer be L_wand L_s,respectively, which can be calculated as follows:

L_w = Lο(1+αtw)

⇒ L_w = 12(1+11×10−6×18)

⇒ L_w = 12.00237m

L_s = L0(1+αts)

⇒ L_s = 12(1+11×10−6×48)

⇒ L_s = 12.006336m

∴∆L = L_s−L_w

⇒ ∆L = 12.006336−12.002376

= ∆L = 0.00396m

⇒ ∆L ≈ 0.4cm

Therefore, the gap (ΔL) that should be left between two iron sections, so that there is no compression during summer, is 0.4 cm.

Question-13 :-

Answer-13 :-

Given:
Diameter of a circular hole in an aluminium plate at 0°C, d₁ = 2 cm = 2 × 10^–2 m
Initial temperature, t₁ = 0 °C
Final temperature, t₂ = 100 °C
So, the change in temperature, (Δt) = 100°C – 0°C = 100°C
The linear expansion coefficient of aluminium, α_al​ = 2.3 × 10^–5 °C^–1
Let the diameter of the circular hole in the plate at 100^oC be d₂, which can be written as

d₂ = d₁(1+αΔt)

⇒ d₂ =2 × 10^-2(1+2.3×10^-5×10²)

⇒ d₂ = 2×10^-2(1+2.3×10^-3)

d₂ = 2×10^-2+2.3×2×10^-5

⇒ d² = 0.02 +0.000046

⇒ d² =0.020046 m

So  d² ≈ 2.0046 cm

Question-14 :-   (Heat and Temperature HC Verma)

Answer-14 :-

Given:
At 20°C, length of the metre scale made up of steel, L_st= length of the metre scale made up of aluminium, L_al​
Coefficient of linear expansion for aluminium, α_al = 2.3 × 10^–5 °C^-1
Coefficient of linear expansion for steel, α_st = 1.1 × 10^–5°C^-1
Let the length of the aluminium scale at 0°C, 40°C and 100°C be L_0al​,L​_40aland L_10​0al.
And let the length of the steel scale at 0°C, 40°C and 100°C be L_0st​,L​_40st and L_10​0st.
(a) So, L_0st(1 – α_st × 20) = L_0al(1 – α_al × 20)

Question-15 :-

Answer-15 :-

(a) Let the correct length measured by a metre scale made up of steel 16 °C be L.
Initial temperature, t₁ = 16 °C
Temperature on a hot summer day, t₂ = 46 °C
​So, change in temperature, Δθ = t₂

−t₁= 30 °C

Coefficient of linear expansion of steel,

α= 1.1 × 10^–5 °C^​-1
Therefore, change in length,
ΔL = L αΔθ = L × 1.1 × 10^–5 × 30

=[1.1 × 10^-5 × 30 ×100]%

=3.3 × 10^-2 %

(b) Temperature on a winter day, t₂ = 6 °C​
​So, change in temperature, Δθ = t₁

−t₂= 10 °C​
ΔL = L_​2

− L₁ = L αΔθ = L × 1.1 × 10^–5 × 10​

Question-16 :-  (Heat and Temperature HC Verma)

Answer-16 :-

Given:
Temperature at which a metre scale gives an accurate reading, T₁ = 20 °C
The value of variation admissible, ΔL = 0.055 mm = 0.055 × 10^–3 m, in the length, L₀ = 1 m
Coefficient of linear expansion of steel, α  = 11 × 10^–6  °C^–1
Let the range of temperature in which the experiment can be performed be T₂.
We know:  ΔL = L₀ αΔT

Either T₂ = 20 + 5 = 25°C

or T₂= 20 – 5 = 15°C

Hence, the experiment can be performed in the temperature range of 15 °C

Question-17 :-

Answer-17 :-

Given:
Density of water at 0°C, ( f₀)= 0.998 g cm^-3
Density of water at 4°C,  ​(f₄) = 1.000 g cm-3
Change in temperature, (Δt) = 4^oC
Let the average coefficient of volume expansion of water in the temperature range of 0 to 4°C be γ.

As the density decreases,

Therefore,the average coefficient of volume expansion of water in the temperature range of 0 to 4°C will be

Question-18 :-

Answer-18 :-

Let the original length of iron rod be L_Fe and L^‘​​​_Fe be its length when temperature is increased by ΔT.
Let the original length of aluminium rod be L_Al and L^‘​​​_Al be its length when temperature is increased by ΔT.

Coefficient of linear expansion of iron,

αFe = 12 × 10^–6 °C​

− 1

Coefficient of linear expansion of aluminium, α_Al = 23 × 10^–6 °C^​​

− 1

Since the difference in length is independent of temperature, the difference is always constant.

The ratio of the lengths of the iron to the aluminium rod is 23:12.

Question-19 :-  (Heat and Temperature HC Verma)

A pendulum clock shows correct time at 20°C at a place where g = 9.800 m s^–2. The pendulum consists of a light steel rod connected to a heavy ball. It is taken to a different place where g = 9.788 m s^–1. At what temperature will the clock show correct time? Coefficient of linear expansion of steel = 12 × 10^–6 °C^–1.

Answer-19 :-

Given:
The temperature at which the pendulum shows the correct time, T₁ = 20 °C
Coefficient of linear expansion of steel,

α = 12 × 10^–6 °C^–1
Let T₂ be the temperature at which the value of g is 9.788 ms^–2 and

ΔΔT be  the change in temperature.
^​So, the time periods of pendulum at different values of g will be t₁ and t₂ , such that

Therefore, for a pendulum clock to give correct time, the temperature at which the value of g is 9.788 ms^–2 should be

− 82^oC.

Question-20 :-  (Heat and Temperature HC Verma)

An aluminium plate fixed in a horizontal position has a hole of diameter 2.000 cm. A steel sphere of diameter 2.005 cm rests on this hole. All the lengths refer to a temperature of 10 °C. The temperature of the entire system is slowly increased. At what temperature will the ball fall down? Coefficient of linear expansion of aluminium is 23 × 10^–6 °C^–1and that of steel is 11 × 10^–6 °C^–1.

Answer-20 :-

Given:
Diameter of the steel sphere at temperature (T₁= 10 °C) , d_st = 2.005 cm
Diameter of the aluminium sphere, d_Al = 2.000 cm
Coefficient of linear expansion of steel, α_st  = 11 × 10

−6  °C

−1

Coefficient of linear expansion of aluminium, α_Al  = 23 × 10

− 6 °C

− 1

Let the temperature at which the ball will fall be T_2 , so that change in temperature be ΔT​.
d‘_st= 2.005(1 + α_stΔT)      

⇒ d’_st  = 2.005+2.005×11×10^–6×∆T

d’_st = 2(1+α_Al×∆T)

⇒d”_Al = 2+2×23×10^–6×∆T

The steel ball will fall when both the diameters become equal.
So, d’‘_st= d’‘_Al

⇒ 2.005+2.005×11×10^–6∆T=2+2×23×10^–6∆T

=  (46−22.055)×10^–6∆T=0.005

Question-23 :- (Heat and Temperature HC Verma)

An aluminium can of cylindrical shape contains 500 cm³ of water. The area of the inner cross section of the can is 125 cm². All measurements refer to 10°C.
Find the rise in the water level if the temperature increases to 80°C. The coefficient of linear expansion of aluminium is 23 × 10^–6 °C^–1 and the average coefficient of the volume expansion of water is 3.2 × 10^–4 °C^–1   respectively

Answer-23 :-

Given:
Volume of water contained in the aluminium can, V₀ = 500 cm³
Area of inner cross-section of the can, A = 125 cm^2
​Coefficient of volume expansion of water, γ​ = 3.2 × 10^–4 °C^–1
Coefficient of linear expansion of aluminium,

α_Al  = 23 × 10^–6 °C^–1
If  ∆θ  is the change in temperature, then final volume of water

(V) due to expansion,
V = V₀(1 + γΔθ)
= 500 [1 + 3.2 × 10^–4 × (80 – 10)]
and 500 [1 + 3.2 × 10^–4 × 70]
= 511.2 cm³
The aluminium vessel expands in its length only.
So, area of expansion of the base can be neglected.
Increase in volume of water = 11.2 cm³ 
Consider a cylinder of volume 11.2 cm³
∴ Increase in height of the water

= 11.2/125 = 0.0896

= 0.089 cm

Question-24 :-

Answer-24 :-

Given: At 0^o​C, volume of glass vessel, V_g = 10 × 10 × 10 = 1000 cc = volume of mercury, V_Hg
Let the volume of mercury at 10°C be V’_Hg and that of glass be V’_g.
At 10^​oC, the additional volume of mercury than glass, due to heating, V’_Hg – V’_g = 1.6 cm³
So change in temperature, ΔT = 10°C
Coefficient of linear expansion of glass, α_g = 6.5 × 10^–6 °C^–1
Therefore, the coefficient of volume expansion of glass, γ_g = 3 × 6.5 × 10^–6°C^–1​
Let the coefficient of volume expansion of mercury be γ_Hg.
We know that
V’_Hg = V_Hg (1 + γ_Hg ΔT)         …(1)
V’_g = V_g (1 + γ_g ΔT)              …(2)
Subtracting (2) from (1) we get,
V’_Hg – V’_g = V_Hg – V_g + V_Hg γ_Hg ΔT – V_g γ_g ΔT (as V_Hg = V_g)

⇒ 1.6 = 1000 × γ_Hg × 10 − 1000 × 6.5 × 3 × 10^–6 × 10

⇒ γ_Hg ≅ 1.8 × 10^-4°C^-1

Therefore, the coefficient of volume expansion of mercury is 1.8× 10^–4 °C^–1.

Question-25 :-

Answer-25 :-

Given:
Density of wood at 0 °C, f_w = 880 kgm^​-3
Density of benzene at 0 °C, f_b = 900 kgm^–​3
Coefficient of volume expansion for wood, γ_w = Coefficient of volume expansion for benzene, γ_b = 1.5 × 10^–3 °C^–1
So, initial temperature, T₁ = 0 °C
Let T₂ be the temperature at which the piece of wood will just sink in benzene and ΔT = T_2-T₁.
The piece of wood begins to sink when its weight is equal to the weight of the benzene displaced.
Mass = volume × density

Therefore, Vf_wg = Vf_bg

⇒ 880+880 × 1.5 × 10^-3 ΔT =900+900 ×1.2×10^-3 ΔT

= (1320 -1080)× 10^-3 ΔT =20

⇒ ΔT = 83.3°C

= T₂ -T₁ ≅ 83°

⇒ T₂ -0° ≅ 83°

= T₂ ≅ 83°C

Therefore, the piece of wood will just sink in benzene at 83 ^oC.

Question-26 :-  (Heat and Temperature HC Verma)

Answer-26 :-

The steel rod is resting on a horizontal base at 0 °C. When the temperature is increased to 100 °C, it will lead to an increase in the length of the steel due to expansion on heating. Since, there is no opposition in expansion of length, no longitudinal strain will be developed.

Question-27 :-

A steel rod is clamped at its two ends and rests on a fixed horizontal base. The rod is unstrained at 20°C. Find the longitudinal strain developed in the rod if the temperature rises to 50°C. Coefficient of linear expansion of steel = 1.2 × 10^–5 °C^–1.

Answer-27 :-

Given:
Temperature at which rod is resting on a fixed horizontal base without any strain, T₁=20 °C. Then the rod is heated to temperature, T₂ = 50 °C

​So change in temperature,ΔT =T₂-T₁=30°C

Coefficient of linear expansion of steel, α = 1.2 × 10^–5 °C^$-$1​ 
Let L be the length of the rod without heating and L’ be the length of the rod on heating.
Let longitudinal strain developed in the rod be S.
We know that

⇒ S =1.2 × 10^-5 ×(50-20)

=1.2 × 10^-5 ×30

=1.2 × 10^-5 × 30

=36 × 10^-5

S = 3.6 × 10^-4

The strain of 3.6 × 10^-4 will be opposite to the direction of expansion.

Question-28 :-  (Heat and Temperature HC Verma)

Answer-28 :-

Given:
Cross-sectional area of the steel wire, A = 0.5 mm² = 0.5 × 10^–6 m²
The wire is taut at a temperature, T₁ = 20 °C,
After this, the temperature is reduced to T₂ = 0 °C
​So, the change in temperature, Δθ = T_1-T₂ = 20 °C
Coefficient of linear expansion of steel, α = 1.2 ×10^–5 °C^​-1
Young’s modulus, γ = 2 ×10¹¹ Nm^​-2
Let L be the initial length of the steel wire and L’ be the length of the steel wire when temperature is reduced to 0°C.

Decrease in length due to compression, ΔL =L’ – L = LαΔθ …(1)

Let the tension applied be F.

Change in length due to tension produced is given by (1) and (2).
So, on equating (1) and (2), we get:

⇒ F = αΔθAγ

= 1.2 × 10^-5 × (20-0) × 0.5 ×10^-6 × 2 ×10¹¹

= 1.2 × 20

⇒ F = 24 N

Therefore, the tension produced when the temperature falls to 0°C

Question-29 :-

Answer-29 :-

Given:
Temperature of the rod at zero tension, T₁ = 20 °C
Final temperature, T₂ = 100 °C

Change in temperature, Δθ =80°C

Cross-sectional area of the rod, A = 2 mm² = 2 × 10^-6m²

Coefficient of linear expansion of steel, α = 12 ×10^–6 °C^​-1
Young’s modulus of steel, Y = 2 × 10¹¹ Nm^-2
Let L be the length of the steel rod at 20 °C and L’ be the length of steel rod at 100 °C.

Change of length of the rod, ΔL =L’ – L

If F be the force exerted by the rod on one of the clamps due to rise in temperature, then

F =YAαΔθ

= 2 × 10¹¹× 2 × 10^-6 × 12 × 10^-6 × 80

=48 × 80 × 10^-1

So, F = 384 N

Therefore, the rod will exert a force of 384 N on one of the clamps.

Heat and Temperature HC Verma Exercise Questions Solutions

(HC Verma Ch-23 Concept of Physics Vol-2 for ISC Class-12)

(Page 14)

Question-30 :-

Two steel rods and an aluminium rod of equal length l₀ and equal cross-section are joined rigidly at their ends, as shown in the figure below. All the rods are in a state of zero tension at 0°C. Find the length of the system when the temperature is raised to θ. Coefficient of linear expansion of aluminium and steel are α_a and α_s, respectively. Young’s modulus of aluminium is Y_a and of steel is Y_s.

Answer-30 :-

Given :
Initial length of the system (two identical steel rods + an aluminium rod) at 0 °C = l₀
Coefficient of linear expansion of steel and aluminium are α_s and α_Al​, respectively.
Temperature is raised by θ.

_​So, the change in temperature, Δθ = θ – 0 ° C = θ

Young’s modulus of steel and aluminium are γ_s and γ_Al, respectively.
If l be the final length of the system at temperature θ,

Young’s modulus of the system,

Y = ϒ_s +ϒ_s+ϒ_Al = 2ϒ_s + ϒ_Al  …(3)

Using (1), (2) and (3), we get:

Therefore, the final length of the system will be l_{0        where l0 is its  initial length.}

Question-31 :-

A steel ball that is initially at a pressure of 1.0 × 10⁵ Pa is heated from 20°C to 120°C, keeping its volume constant. Find the pressure inside the ball. Coefficient of linear expansion of steel = 12 × 10^–6 °C^–1and bulk modulus of steel = 1.6 × 10¹¹ Nm^–2.

Answer-31 :-

Given:
Initial pressure on the steel ball = 1.0 × 10⁵ Pa
The ball is heated from 20 °C to 120 °C.
So, change in temperature, Δθ = 100°C.

Coefficient of linear expansion of steel, α = 12 ×10^-6°C^-1

Bulk modulus of steel ,B = 1.6 × 10¹¹ Nm^–2

Pressure is given as,

⇒ P = B × γ Δθ

P =B × 3 αΔθ  (∵ γ = 3 α )

⇒ P = 1.6 × 10¹¹ ×3 × 12 × 10^-6 ×(120-20)

=1.6 ×  3 × 12 ×10¹¹ ×10^-6 ×10²

= 57.6 × 10⁷

⇒ P =5.8 × 10⁸ Pa

Therefore, the pressure inside the ball is 5.8 × 10⁸ Pa.

Question-32 :-

Answer-32 :-

Given:
Coefficient of linear expansion of solid = α
Moment of inertia at 0 °C = I₀
If temperature changes to θ from 0 °C, then change in temperature, (ΔT) =θ
Let I be the new moment of inertia attained due to rise in temperature.
Let R₀ be the radius of gyration at 0 °C.
We know that on heating, radius of gyration will change as
R = R₀(1 + αθ)
Here, R is the radius of gyration after heating.
I₀ = MR₀² , where M = mass of the body
Now, I = MR² = MR₀²(1 + αθ)²
Expanding binomially and neglecting the higher terms of order (αθ) that will be very small, we get
I = MR₀²(1 + 2 αθ)
So, I = I₀(1 + 2 αθ)
Hence, proved.

Question-33 :-

A torsional pendulum consists of a solid  disc connected to a thin wire (α = 2.4 × 10^–5°C^–1) at its centre. Find the percentage change in the time period between peak winter (5°C) and peak summer (45°C).

Answer-33 :-

Given:
Coefficient of linear expansion of the wire, α = 2.4 × 10^–5 °C^–1
Let I₀ be the moment of inertia of the torsional pendulum at 0 °C.
If K is the torsional constant of the wire, then time period of torsional pendulum (T):

Here, I = moment of inertia after change in temperature
When the temperature is changed by Δθ, moment of inertia (I),

l = l₀(1+2αΔθ)

On substituting the value of I  in equation(1), we get:

In winter ,  Δθ = 5 °C

∴ Time period , (T₁)

In summer , Δθ = 45 °C

Time period (T₂)

So,

= 0.0959 %

⇒ % change in time period ≈ 9.6 ×  10^-2 %

Therefore, the percentage change in time period of a torsional pendulum between peak winters and peak summers is 9.6 × 10^–2  % .

Question-34 :-  (Heat and Temperature HC Verma)

Answer-34 :-

Let initial radius of the circular disc at 20 ^ₒC = r₂₀
and Let final radius of the circular disc at 50 ^ₒC = r₅₀
Coefficient of linear expansion of iron, α = 1.2 × 10^–5 °C^–1.

change in temperature,ΔT = 30°C

Let R’ and R be the radius of the paricle at 50 ^ₒC and 20 ^ₒC respectively.
If v and v’ be the linear speed of the particle at 50 ^ₒC and 20 ^ₒC respectively, as the angular velocity remains (ω) constant.
Therefore,

Now,

R’ = R (1+αΔT)

⇒ R’ =R +R × 1.2 ×10^-5 ° C^-1 × Δ T

⇒ R’ = 1.00036R

Using equation(1) we have,

⇒ v’ = 1.00036v

Percentage change in linear speed will be,

—: End of Heat and Temperature HC Verma Exercise Solutions Concept of Physics:–

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