Heat and Temperature Obj-1 HC Verma Solutions Vol-2 Chapter-23 Concept of Physics for Class-12. Step by Step Solutions of Objective -1 (MCQ-1) Questions of Chapter-23 Heat and Temperature (Concept of Physics) .Visit official Website CISCE for detail information about ISC Board Class-11 Physics.

## Heat and Temperature Obj-1 HC Verma Solutions Vol-2 Chapter-23 Concept of Physics

 Board ISC and other board Publications Bharti Bhawan Publishers Chapter-23 Heat and Temperature Class 12 Vol 2nd writer HC Verma Book Name Concept of Physics Topics Solution of Objective-1 (MCQ-1) Questions Page-Number 11, 12

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### HC Verma Solutions Vol-2 Chapter-23 Heat and Temperature Obj-1 (MCQ-1) Concept of Physics

(Page-11)

#### Question-1 :-

A system X is neither in thermal equilibrium with Y nor with Z. The systems Y and Z

(a) must be in thermal equilibrium

(b) cannot be in thermal equilibrium

(c) may be in thermal equilibrium

The option (c) may be in thermal equilibrium. is correct
Explanation:

The given data in the question is insufficient to specify the relation between the physical conditions of systems Y and Z. As  system X is not in thermal equilibrium with Y and Z, systems Y and Z may be at the same temperature or they may or may not be in thermal equilibrium with each other. So, the only possible option is (c).

#### Question-2 :-

Which of the curves in the following figure represents the relation between Celsius and Fahrenheit temperatures?

(a)

(b)

(c)

(d)

The option (a) . is correct

Explanation:

Celsius and Fahrenheit temperatures are related in the following way:

Here, F = temperature in Fahrenheit
C = temperature in Celsius
If this equation is plotted on the graph, then the curve will be represented by curve ‘a’ lying in the fourth quadrant with slope 5/9.

#### Question-3 :-

Which of the following pairs may give equal numerical values of the temperature of a body?

(a) Fahrenheit and Kelvin

(b) Celsius and Kelvin

(c) Kelvin and Platinum

The option (a) Fahrenheit and Kelvin is correct

Explanation:

Let θ be the temperature in Fahrenheit and Kelvin scales.

We know that the relation between the temperature in Fahrenheit and Kelvin scales is given by

5θ -160 = 9θ -2458.5

4θ = 2298.35

θ = 574.59°

If we consider the same for Celsius and Kelvin scales

t = t – 273.15

Thus, t does not exist.
The Kelvin scale uses mercury as thermometric substance, whereas the platinum scale uses platinum as thermometric substance. The scale depends on the properties of the thermometric substance used to define the scale. The platinum and Kelvin scales do not agree with each other. Therefore, there is no such temperature that has same numerical value in the platinum and Kelvin scale.

#### Question-4 :-

For a constant-volume gas thermometer, one should fill the gas at

(a)  low temperature and low pressure

(b) low temperature and high pressure

(c) high temperature and low pressure

(d) high temperature and high pressure

The option (c) high temperature and low pressure  is correct

Explanation:

A constant-volume gas thermometer should be filled with an ideal gas in which particles don’t interact with each other and are free to move anywhere, so that the thermometer functions properly. An ideal gas is only a theoretical possibility. Therefore, the gas that is filled in the thermometer should be at high temperature and low pressure, as under these conditions, a gas behaves as an ideal gas.

#### Question-5

Consider the following statements.
(A) The coefficient of linear expansion has dimension K–1.
(B) The coefficient of volume expansion has dimension K–1.

(a) A and B are correct.

(b) A is correct but B is wrong.

(c) B is correct but A is wrong.

(d) A and are both wrong.

The option (a) A and B are correct.  is correct

Explanation:

The coefficient of linear expansion,

Here, L = initial length
ΔL = change in length
ΔT = change in temperature
On the other hand, the coefficient of volume expansion,

Here, V = initial volume
Δ V = change in volume
ΔT = change in temperature
K = kelvin, the S.I. unit of temperature

(Page-12)

#### Question-6

A metal sheet with a circular hole is heated. The hole

(a) gets larger

(b) gets smaller

(c) retains its size

(d) is deformed

The option (a) gets larger  is correct

Explanation:

When a metal sheet is heated, it starts expanding and its surface area will start increasing, which will lead to an increase in the radius of the hole. Hence, the circular hole will become larger.

#### Question-7 :-

Two identical rectangular strips, one of copper and the other of steel, are riveted together to form a bimetallic strip (acopper> asteel). On heating, this strip will

(a) remain straight

(b)  bend with copper on convex side

(c) bend with steel on convex side

(d) get twisted

The option (b)  bend with copper on convex side is correct

Explanation:

We are provided with two metal strips of copper and steel. On heating, both of them will expand. Expansion coefficient of copper is more than that of steel. So, the copper metal strip will expand more, causing the bimetallic strip to bend with copper at the convex side, as it’ll have more surface area compared to the steel sheet, which will be on the concave side.

#### Question-8 :-

If the temperature of a uniform rod is slightly increased by ∆t, its moment of inertia about a perpendicular bisector increases by

(a) zero

(b) αI∆t

(c) 2αI∆t

(d)  3αI∆t.

The option (c) 2αI∆t is correct

Explanation:

The change in moment of inertia of uniform rod with change in temperature is given by,

I′ =I (1+2∝Δt)

Here, I = initial moment of inertia
I’ = new moment of inertia due to change in temperature

∝= expansion coefficient

Δt = change in temperature

So, I′ – I = 2αIΔt

#### Question-9 :-

If the temperature of a uniform rod is slightly increased by ∆t, its moment of inertia I about a line parallel to itself will increase by

(a)  zero

(b) αI∆t

(c) 2αI∆t

(d) 3αI∆t

The option (c) 2αI∆t is correct

Explanation:

The moment of inertia of a solid body of any shape changes with temperature as

I′ = I (1+2∝Δt)

Here, I = initial moment of inertia
I’ = new moment of inertia due to change in temperature

∝ = expansion coefficient

Δt =  change in temperature

So, I′ – I = 2∝IΔt

#### Question-10 :-

The temperature of water at the surface of a deep lake is 2°C. The temperature expected at the bottom is

(a) 0 °C

(b) 2 °C

(c) 4 °C

(d) 6 °C

The option (c) 4 °C is correct

Explanation:

The density of water is maximum at 4 oC, and the water at the bottom of the lake is most dense, compared to the layers of water above. Therefore, the temperature expected at the bottom is 4oC.

#### Question-11:-

An aluminium sphere is dipped into water at 10°C. If the temperature is increased, the force of buoyancy

(a) will increase

(b) will decrease

(c) will remain constant

(d) may increase or decrease depending on the radius of the sphere

The option (b) will decrease is correct

Explanation:

When an aluminium sphere is dipped in water and the temperature of water is increased, the aluminium will start expanding leading to increase in its volume. This will lead to increase in the surface area of the shell and it’ll exert less pressure on the water such that the volume of the sphere submerged in water will decrease and it’ll start float easily on water. Now, the volume of water displaced will be less compared to what was displaced initially. Therefore, the force of buoyancy will decrease, as it is directly proportional to the volume of water displaced.

—: End of Heat and Temperature Obj-1 (mcq-1) HC Verma Solutions Vol-2 Chapter-23 :–

Return to — HC Verma Solutions Vol-2 Concept of Physics

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