Class 12-science H C VERMA Solutions Physics Chapter 1 - Heat and Temperature
Heat and Temperature Exercise 12
Solution 1
For any scale
F.P.=Freezing point
B.P.=Boiling point
C=20°centigrade
Solution 2
=1.5×
Pa
P=2.05× Pa
Using
T= 
T=373.3 K
Solution 3
Pressure measured at melting point=2.2×pressure at triple point
P=2.2
Using,
T=
=
T=601 K
Solution 4
=40×
Pa; T=273+100=373 K
Using,
T=
P=
P=
P=55 kPa
Solution 5
T=
T∝P
96 kPa
Solution 6
T=
=
T=
Solution 7
T'=
=
T'≈307 K
Solution 8
=
T=
Heat and Temperature Exercise 13
Solution 9
20Ω, 
Variation of resistance with temperature
27.5=
-(i)
50=
-(ii)
Solving (i) and (ii)
Solution 10
=10.0035 m
Solution 11
cm
Solution 12
m
cm
cm
Solution 13
cm
Solution 14
(a) 
(b) 
(c) 
Solution 15
Percentage error=
(a) In Summer
%
error=
=
=0.033%
(b) In winter
%
error=
=
=-0.011%
Solution 16
Range of temperature is from 15
to 25.
Solution 17
Solution 18
Solution 19
Since, time period remains same
Solution 20
Ball will fall down when diameter of both plate and sphere becomes equal.
at T= 
at T
Solution 21
For no stress in winter final length and breadth of glass and aluminium frame must be the same.
cm
cm
Solution 22
Since, volume of air (space) remains same.
So, change in volume of Hg=change of volume of glass
cc
Solution 23
Change in volume of water
Base expansion of can is neglected
So, increase in height of water=
cm
Solution 24
(at 0)
(at 10)
1.6
Solution 25
When wood just sinks in benzene
Weight of wood=Buoyancy force
mg=V
g
Solution 26
Since, there is no opposition in expansion of length, no stress develops and hence no longitudinal strain developed.
Solution 27
Strain=
=
=
Solution 28
F=24N
Solution 29
F=384N
Heat and Temperature Exercise 14
Solution 30
Stress=Y(strain)=
Let final length of the system be l
Total strain=
=
Solution 31
Pa
Solution 32
Moment of inertia=Mass
(at 0)
(at 
)
[By binomial expansion]
Solution 33
Time period of torsional pendulum
%
change =
Solution 34
%
change in linear speed= 
