Heat Transfer Obj-1 HC Verma Solutions Vol-2 Class-12 Ch-28

Heat Transfer Obj-1 HC Verma Solutions Vol-2 Class-12 Ch-28 Concept of Physics for Class-12. Step by Step Solutions of Objective -1 (MCQ-1) Questions of Chapter-28 Heat Transfer (Concept of Physics) .Visit official Website CISCE for detail information about ISC Board Class-11 Physics.

Heat Transfer Obj-1 HC Verma Solutions Vol-2 Class-12 Ch-28

Board ISC and other board
Publications Bharti Bhawan Publishers
Chapter-28 Heat Transfer
Class 12
Vol  2nd
writer HC Verma
Book Name Concept of Physics
Topics Solution of Objective-1 (MCQ-1) Questions
Page-Number 97

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Heat Transfer Obj-1 HC Verma Solutions 

Solutions of Ch-28  Vol-2 Concept of Physics for Class-12

(Page-97)

Question-1 :-

The thermal conductivity of a rod depends on

(a) length

(b) mass

(c) area of cross section

(b) material of the rod

Answer-1 :-

The option (b) material of the rod is correct
Explanation:

The thermal conductivity of a rod depends only on the material of the rod. For example, metals are much better conductors than non-metals because metals have large number of free electron that can move freely anywhere in the body of the metal and carry thermal energy from one place to other. Also, 2 copper rods having different lengths and areas of cross-section have same thermal conductivity that depends only on the number of free electrons in copper.

Question-3 :-

A solid at temperature T1 is kept in an evacuated chamber at temperature T2 > T1. The rate of increase of temperature of the body is proportional to

(a) T2 – T1

(b) T2 – T2

(c) T32– T13

(d) T42– T14

Answer-3 :-

The option (d) T42– T14 is correct
Explanation:
From Stefan-Boltzmann law, the energy of thermal radiation emitted per unit time by a blackbody of surface area A is given by σu=σAT4
Here,  σσ is Stefan-Boltzmann constant.
Since the temperature of the solid is less than the surroundings, the temperature of the solid will increase with time until it reaches equilibrium with the surroundings. The rate of emission from the solid will be proportional to T14 and rate of emission from the surroundings will be proportional to T42.
So, the net rate of increase in temperature will be proportional to T42 – T42.

Question-4 :-

The thermal radiation emitted by a body is proportional to Tn where T is its absolute temperature. The value of n is exactly 4 for

(a) a blackbody

(b) all bodies

(c) bodies painted black only

(d) polished bodies only

Answer-4 :-

The option (b) all bodies is correct
Explanation:

From Stefan-Boltzmann law, the energy of the thermal radiation emitted per unit time by a blackbody of surface area A is given by σu=σAT4 here , σ is Stefan-Boltzmann constant.
This law holds true for all the bodies.

Question-5 :-

Two bodies A and B having equal surface areas are maintained at temperature 10°C and 20°C. The thermal radiation emitted in a given time by A and B are in the ratio

(a) 1 : 1.15

(b) 1 : 2

(c) 1 : 4

(d) 1 : 16

Answer-5 :-

The option (a) 1 : 1.15 is correct
Explanation:

From Stefan-Boltzmann law, energy of the thermal radiation emitted per unit time by a blackbody of surface area A is given by σu=σAT4
Here, σσ  is Stefan-Boltzmann constant.

hc verma heat transfer img 1

Question-6 :-

One end of a metal rod is kept in a furnace. In steady state, the temperature of the rod

(a) increases

(b) decreases

(c) remain constant

(d) is nonuniform

Answer-6 :-

The option (d) is nonuniform is correct
Explanation:

In steady state, the temperature of the rod is nonuniform maximum at the end near the furnace and minimum at the end that is away from the furnace.

A hot liquid is kept in a big room. Its temperature is plotted as a function of time. Which of the following curves may represent the plot?

A hot liquid is kept in a big room. Its temperature is plotted as a function of time. Which of the following curves may represent the plot?

(a) a

(b) b

(c) c

(d) d

Answer-8 :-

The option (a) a is correct
Explanation:
When a hot liquid is kept in a big room, the liquid will loose its temperature with time. The thermal energy emitted by the liquid will be gained by the walls of the room. As the room is big, we can assume that the temperature difference between the room and the liquid is large. From Stephen’s law, the liquid emits thermal energy in proportion to T4, where T is the initial temperature of the liquid.
As the temperature decreases, the rate of loss of thermal energy will also decrease. So, the slope of the curve will also decrease.
Therefore, the plot of temperature with time is best represented by the curve (a).

Question-9 :-

A hot liquid is kept in a big room. The logarithm of the numerical value of the temperature difference between the liquid and the room is plotted against time. The plot will be very nearly

(a) a straight line

(b) a circular are

(c) a parabola

(d) an ellipse

Answer-9 :-

The option (a) a straight line is correct
Explanation:

When a hot liquid is kept in a big room, then the liquid will loose temperature with time. The thermal energy emitted by the liquid will be gained by the walls of the room. As the room is big, we can assume that the temperature difference between the room and the liquid is large. From Stephen’s law, the liquid emits thermal energy in proportion to T4, where T is the initial temperature of the liquid. As the temperature decreases, the rate of loss will also decrease. So, the slope of the curve will also decrease. Finally, at equilibrium, the temperature of the room will become equal to the new temperature of the liquid. So, in steady state, the difference between the temperatures of the two will become zero.

A graph is plotted between the logarithm of the numerical value of the temperature difference between the liquid and the room is plotted against time.The logarithm converts the fourth power dependence into a linear dependence with some coefficient (property of log). So, the plot satisfying all the above properties will be a straight line.

Question-10 :-

A body cools down from 65°C to 60°C in minutes. It will cool down from 60°C to 55°C in

(a) 5 minutes

(b) less than 5 minutes

(c) more than 5 minutes

(d) less than or more than 5 minutes depending on whether its mass is more than or less than 1 kg.

Answer-10 :-

The option (c) more than 5 minutes is correct
Explanation:

Let the temperature of the surrounding be T°C Average temperature of the liquid in first case = 62.5°
Average temperature difference from the surroundings = (62.5-T)°C
From newton law of cooling,
1° C  min-1 = -bA ( 62.5 – T )°C

hc verma heat transfer img 3

For the second case,
Average temperature = 57.5° C
Temperature difference from the surroundings = ( 57.5 – 7 )° C
From Newton’s law of cooling and equation (i),

hc verma heat transfer img 4

—: End of Heat Transfer Obj-1 (mcq-1) HC Verma Solutions Vol-2 Chapter-28 :–


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

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