Thermal Properties of Matter Practice Questions - DPP for JEE

 Page 1


1. (c) In parallel combination, the equivalent thermal conductivity is given
by
For two rods of equal area,
(if A
1
 = A
2
 = A)
2. (b) Temperature of B will be higher because, due to expansion centre of
mass B will come down same heat is supplied but in B, Potential
energy is decreased therefore internal energy gain will be more.
3. (c) Since specific heat = 0.6 kcal/g × °C = 0.6 cal/g × °C
From graph it is clear that in a minute, the temperature is raised from 0°C
to 50°C
?  Heat required for a minute = 50 × 0.6 × 50 = 1500 cal.
Also from graph, Boiling point of wax is 200°C.
4. (d)
5. (a) V + ?V = (L + ?L)
3
 = (L + aL?T)
3
= L
3
 + (1 + 3a?T + 3a
2
?T
2
 +a
3
?T
3
)
? a
2 
and a
3 
terms are neglected.
? V (1 + ??T) = V (1 + 3a?T)
1 + ??T = 1 + 3a?T
? ? = 3a
6. (d) Radius of small sphere = r
Thickness of small sphere = t
Radius of bigger sphere = 2r
Thickness of bigger sphere = t/4
Mass of ice melted = (volume of sphere) × (density of ice)
Let K
1
 and K
2
 be the thermal conductivities of larger and smaller sphere.
Page 2


1. (c) In parallel combination, the equivalent thermal conductivity is given
by
For two rods of equal area,
(if A
1
 = A
2
 = A)
2. (b) Temperature of B will be higher because, due to expansion centre of
mass B will come down same heat is supplied but in B, Potential
energy is decreased therefore internal energy gain will be more.
3. (c) Since specific heat = 0.6 kcal/g × °C = 0.6 cal/g × °C
From graph it is clear that in a minute, the temperature is raised from 0°C
to 50°C
?  Heat required for a minute = 50 × 0.6 × 50 = 1500 cal.
Also from graph, Boiling point of wax is 200°C.
4. (d)
5. (a) V + ?V = (L + ?L)
3
 = (L + aL?T)
3
= L
3
 + (1 + 3a?T + 3a
2
?T
2
 +a
3
?T
3
)
? a
2 
and a
3 
terms are neglected.
? V (1 + ??T) = V (1 + 3a?T)
1 + ??T = 1 + 3a?T
? ? = 3a
6. (d) Radius of small sphere = r
Thickness of small sphere = t
Radius of bigger sphere = 2r
Thickness of bigger sphere = t/4
Mass of ice melted = (volume of sphere) × (density of ice)
Let K
1
 and K
2
 be the thermal conductivities of larger and smaller sphere.
For bigger sphere,
For smaller sphere, 
? 
7. (d) Let T be the   temperature of the interface. As the two sections are
in series, the rate of flow of heat in them will be equal.
 ,
where A is the area of cross-section.
or,     
or,     
or,    
  
8. (a) The heat flow rate is given by
where ?
1
 is the temperature of hot end and ? is temperature at a distance
x from hot end.
Page 3


1. (c) In parallel combination, the equivalent thermal conductivity is given
by
For two rods of equal area,
(if A
1
 = A
2
 = A)
2. (b) Temperature of B will be higher because, due to expansion centre of
mass B will come down same heat is supplied but in B, Potential
energy is decreased therefore internal energy gain will be more.
3. (c) Since specific heat = 0.6 kcal/g × °C = 0.6 cal/g × °C
From graph it is clear that in a minute, the temperature is raised from 0°C
to 50°C
?  Heat required for a minute = 50 × 0.6 × 50 = 1500 cal.
Also from graph, Boiling point of wax is 200°C.
4. (d)
5. (a) V + ?V = (L + ?L)
3
 = (L + aL?T)
3
= L
3
 + (1 + 3a?T + 3a
2
?T
2
 +a
3
?T
3
)
? a
2 
and a
3 
terms are neglected.
? V (1 + ??T) = V (1 + 3a?T)
1 + ??T = 1 + 3a?T
? ? = 3a
6. (d) Radius of small sphere = r
Thickness of small sphere = t
Radius of bigger sphere = 2r
Thickness of bigger sphere = t/4
Mass of ice melted = (volume of sphere) × (density of ice)
Let K
1
 and K
2
 be the thermal conductivities of larger and smaller sphere.
For bigger sphere,
For smaller sphere, 
? 
7. (d) Let T be the   temperature of the interface. As the two sections are
in series, the rate of flow of heat in them will be equal.
 ,
where A is the area of cross-section.
or,     
or,     
or,    
  
8. (a) The heat flow rate is given by
where ?
1
 is the temperature of hot end and ? is temperature at a distance
x from hot end.
The above equation can be graphically represented by option (a).
9. (d)
t' = 4 × t
3/t' = 48 s
10. (a) Suppose, height of liquid in each arm before rising the temperature
is l.
With temperature rise height of liquid in each arm increases i.e. l
1
 > l
and l
2
 > l
Also 
11. (c) The lengths of each rod increases by the same amount
?   
 
12. (a) Moment of inertia of a rod,
Differentiating w.r.t. to ?L, we get
Page 4


1. (c) In parallel combination, the equivalent thermal conductivity is given
by
For two rods of equal area,
(if A
1
 = A
2
 = A)
2. (b) Temperature of B will be higher because, due to expansion centre of
mass B will come down same heat is supplied but in B, Potential
energy is decreased therefore internal energy gain will be more.
3. (c) Since specific heat = 0.6 kcal/g × °C = 0.6 cal/g × °C
From graph it is clear that in a minute, the temperature is raised from 0°C
to 50°C
?  Heat required for a minute = 50 × 0.6 × 50 = 1500 cal.
Also from graph, Boiling point of wax is 200°C.
4. (d)
5. (a) V + ?V = (L + ?L)
3
 = (L + aL?T)
3
= L
3
 + (1 + 3a?T + 3a
2
?T
2
 +a
3
?T
3
)
? a
2 
and a
3 
terms are neglected.
? V (1 + ??T) = V (1 + 3a?T)
1 + ??T = 1 + 3a?T
? ? = 3a
6. (d) Radius of small sphere = r
Thickness of small sphere = t
Radius of bigger sphere = 2r
Thickness of bigger sphere = t/4
Mass of ice melted = (volume of sphere) × (density of ice)
Let K
1
 and K
2
 be the thermal conductivities of larger and smaller sphere.
For bigger sphere,
For smaller sphere, 
? 
7. (d) Let T be the   temperature of the interface. As the two sections are
in series, the rate of flow of heat in them will be equal.
 ,
where A is the area of cross-section.
or,     
or,     
or,    
  
8. (a) The heat flow rate is given by
where ?
1
 is the temperature of hot end and ? is temperature at a distance
x from hot end.
The above equation can be graphically represented by option (a).
9. (d)
t' = 4 × t
3/t' = 48 s
10. (a) Suppose, height of liquid in each arm before rising the temperature
is l.
With temperature rise height of liquid in each arm increases i.e. l
1
 > l
and l
2
 > l
Also 
11. (c) The lengths of each rod increases by the same amount
?   
 
12. (a) Moment of inertia of a rod,
Differentiating w.r.t. to ?L, we get
? 
As we know, ?L = La?t or 
Substituting the value , we get
13. (b) According to Wien’s law and from the figure (?
m
)
1
 < (?
m
)
3
< (?
m
)
2 
therefore T
1 
> T
3 
> T
2
.
14. (a) According to Newton’s law of cooling if temperature difference
between body & surrounding is large, then rate of cooling is also
fast hence curve A shows correct behaviour.
15. (d)
Consider a shell of thickness (dr) and of radiius (r) and let the
temperature of inner and outer surfaces of this shell be T and   (T –
dT) respectively.
 = rate of flow of heat through it
= = 
= 
To measure the radial rate of heat flow, integration technique is used,
since the area of the surface  through which heat will flow is not
Page 5


1. (c) In parallel combination, the equivalent thermal conductivity is given
by
For two rods of equal area,
(if A
1
 = A
2
 = A)
2. (b) Temperature of B will be higher because, due to expansion centre of
mass B will come down same heat is supplied but in B, Potential
energy is decreased therefore internal energy gain will be more.
3. (c) Since specific heat = 0.6 kcal/g × °C = 0.6 cal/g × °C
From graph it is clear that in a minute, the temperature is raised from 0°C
to 50°C
?  Heat required for a minute = 50 × 0.6 × 50 = 1500 cal.
Also from graph, Boiling point of wax is 200°C.
4. (d)
5. (a) V + ?V = (L + ?L)
3
 = (L + aL?T)
3
= L
3
 + (1 + 3a?T + 3a
2
?T
2
 +a
3
?T
3
)
? a
2 
and a
3 
terms are neglected.
? V (1 + ??T) = V (1 + 3a?T)
1 + ??T = 1 + 3a?T
? ? = 3a
6. (d) Radius of small sphere = r
Thickness of small sphere = t
Radius of bigger sphere = 2r
Thickness of bigger sphere = t/4
Mass of ice melted = (volume of sphere) × (density of ice)
Let K
1
 and K
2
 be the thermal conductivities of larger and smaller sphere.
For bigger sphere,
For smaller sphere, 
? 
7. (d) Let T be the   temperature of the interface. As the two sections are
in series, the rate of flow of heat in them will be equal.
 ,
where A is the area of cross-section.
or,     
or,     
or,    
  
8. (a) The heat flow rate is given by
where ?
1
 is the temperature of hot end and ? is temperature at a distance
x from hot end.
The above equation can be graphically represented by option (a).
9. (d)
t' = 4 × t
3/t' = 48 s
10. (a) Suppose, height of liquid in each arm before rising the temperature
is l.
With temperature rise height of liquid in each arm increases i.e. l
1
 > l
and l
2
 > l
Also 
11. (c) The lengths of each rod increases by the same amount
?   
 
12. (a) Moment of inertia of a rod,
Differentiating w.r.t. to ?L, we get
? 
As we know, ?L = La?t or 
Substituting the value , we get
13. (b) According to Wien’s law and from the figure (?
m
)
1
 < (?
m
)
3
< (?
m
)
2 
therefore T
1 
> T
3 
> T
2
.
14. (a) According to Newton’s law of cooling if temperature difference
between body & surrounding is large, then rate of cooling is also
fast hence curve A shows correct behaviour.
15. (d)
Consider a shell of thickness (dr) and of radiius (r) and let the
temperature of inner and outer surfaces of this shell be T and   (T –
dT) respectively.
 = rate of flow of heat through it
= = 
= 
To measure the radial rate of heat flow, integration technique is used,
since the area of the surface  through which heat will flow is not
constant.
Then, 
or 
16. (b) The wall of two layers A and B are connected in series.
Then, heat flowing per second across
both wall layers are same i.e
?  = (as K
A
 = 3K
B
)
?  = ...................(i)
Also +  = 20° C ...................(ii)
From  (i) & (ii)
 +3 = 20°C ? 4 = 20°C
?  = 5°C
17. (a) where ?l=(l’–l) = lat  so F = EAat