CLASS : XITH SUBJECT : PHYSICS
DATE : Solutions DPP NO. : 10
Topic :- THERMAL PROPERTIES OF MATTER
1 (b)
𝑄2 𝑇2 4 2 𝑇2 4
=( ) ⇒ =( )
𝑄1 𝑇1 1 𝑇1
4 4
⇒ 𝑇2 = 2 × 𝑇1 = 2 × (273 + 727)4 ⇒ 𝑇2 = 1190𝐾
2 (b)
An ideal black body absorbs all the radiations incident upon it and has an
emissivity equal to 1. If a black body and an identical another body are kept the
same temperature, then the black body will radiate maximum power.
Hence, the black object at a temperature of 2000℃ will glow brightest.
3 (c)
The boiling point of mercury is 400℃. Therefore, the mercury thermometer can be used to
measure the temperature upto 360℃
4 (c)
Total energy radiated from a body
𝒬 = 𝐴𝜀𝜎𝑇 4 𝑡
𝒬
Or 𝑡
∝ 𝐴𝑇 4
𝒬
𝑡
∝ 𝑟 2𝑇 4 (∵ 𝐴 = 4𝜋𝑟 2 )
𝒬1 𝑟1 2 𝑇1 4 8 2 273 + 127 4
=( ) ( ) 𝑗=( ) [ ] =1
𝒬2 𝑟2 𝑇2 2 273 + 527
5 (c)
If thermal resistance of each rod is considered 𝑅 then, the given combination can be
redrawn as follows
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R 2R
R
120°C R
C
A B
20°C
R R
2R
R R
A C
120°C B 20°C
(Heat current)𝐴𝐶 = (Heat current)𝐴𝐵
(120 − 20) (120 − 𝜃)
= ⇒ 𝜃 = 70℃
2𝑅 𝑅
6 (c)
At boiling point saturation vapour pressure becomes equal to atmospheric pressure.
Therefore, at 100℃ for water. S. V. P. = 760 𝑚𝑚 of 𝐻𝑔 (atm pressure)
7 (b)
Thermal capacity = Mass × Specific heat
Due to same material both spheres will have same specific heat. Also mass = Volume (𝑉) ×
Density(𝜌)
∴ Ratio of thermal capacity
4 3
𝑚1 𝑉1 𝜌 3 𝜋𝑟1 𝑟1 3 1 3
= = = =( ) =( ) =1∶8
𝑚2 𝑉2 𝜌 4 𝜋𝑟 3 𝑟2 2
3 2
8 (d)
𝐴𝑇 16
= [Given]
𝐴2000 1
Area under 𝑒𝜆 − 𝜆 curve represents the emissive power of body and emissive power ∝ 𝑇 4
[Hence area under 𝑒𝜆 − 𝜆 curve) ∝ 𝑇 4
𝐴𝑇 𝑇 4 16 𝑇 4
⇒ =( ) ⇒ =( ) ⇒ 𝑇 = 4000𝐾
𝐴2000 2000 1 2000
9 (c)
Initial volume 𝑉1 = 47.5 units
Temperature of ice cold water 𝑇1 = 0℃ = 273 𝐾
Final volume of 𝑉2 = 67 units
𝑉1 𝑉2
Applying Charle’s law, we have =
𝑇1 𝑇2
(where temperature 𝑇2 is the boiling point)
𝑉2 67 × 273
or 𝑇2 = × 𝑇1 = = 385 𝐾 = 112℃
𝑉1 47.5
10 (a)
1 1
𝑊 = 𝐽𝑄 ⇒ ( 𝑀𝑣 2 ) = 𝐽(𝑚. 𝑐. ∆𝜃)
2 2
1
⇒ × 1 × (50)2 = 4.2[200 × 0.105 × ∆𝜃] ⇒ ∆𝜃 = 7.1℃
4
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11 (a)
According to Stefan’s law
𝐸 ∝ 𝑇4
𝐸1 𝑇 4
= [𝑇1]
𝐸2 2
𝐸1 273+627 4
0.5
= [ 273+27 ]
900 4
𝐸1 = 0.5 (300)
⇒ 𝐸1 = 40.5 J
12 (d)
Rate of cooling (here it is rate of loss of heat)
𝑑𝑄 𝑑𝜃 𝑑𝜃
= (𝑚𝑐 + 𝑊) = (𝑚𝑙 𝑐𝑙 + 𝑚𝑐 𝑐𝑐 )
𝑑𝑡 𝑑𝑡 𝑑𝑡
𝑑𝑄 60 − 55 𝐽
⇒ = (0.5 × 2400 + 0.2 × 900) ( ) = 115
𝑑𝑡 60 𝑠
13 (c)
With rise of altitude pressure decreases and boiling point decreases
14 (d)
Let final temperature of water be θ
Heat taken = Heat given
100 × 1 × (θ − 10) + 10(θ − 10) = 220 × 1(70 − θ)
⇒ θ = 48.8℃ = 50℃
15 (c)
𝐸1 𝑇1 4 7 273 + 227 4 1
𝐸 ∝ 𝑇4 ⇒ =( ) ⇒ =( ) =
𝐸2 𝑇2 𝐸2 273 + 727 16
𝑐𝑎𝑙
⇒ 𝐸2 = 112
𝑐𝑚2 × sec
16 (b)
According to Newton’s law of cooling 𝑡1 will be less than 𝑡2 .
17 (b)
Liquid having more specific heat has slow rate of cooling because for equal masses rate of
𝑑𝜃 1
cooling 𝑑𝑡 ∝ 𝑐
18 (c)
We know that 𝑃 = 𝑃0 (1 + 𝛾𝑡) and 𝑉 = 𝑉0 (1 + 𝛾𝑡)
And 𝛾 = (1/273)/℃ for 𝑡 = −273℃, we have 𝑃 = 0 and 𝑉 = 0
Hence, at absolute zero, the volume and pressure of the gas become zero
19 (b)
In series rate of flow of heat is same
𝐾𝐴 𝐴(𝜃1 − 𝜃) 𝐾𝐵 𝐴(𝜃 − 𝜃2 )
⇒ =
𝑙 𝑙
⇒ 3𝐾𝐵 (𝜃1 − 𝜃) = 𝐾𝐵 (𝜃 − 𝜃2 )
⇒ 3(𝜃1 − 𝜃) = (𝜃 − 𝜃2 )
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⇒ 3𝜃1 − 3𝜃 = 𝜃 − 𝜃2 ⇒ 4𝜃1 − 4𝜃 = 𝜃1 − 𝜃2
⇒ 4(𝜃1 − 𝜃) = (𝜃1 − 𝜃2 )
⇒ 4(𝜃1 − 𝜃) = 20 ⇒ (𝜃1 − 𝜃) = 5℃
1 2
A B
KA KB
20 (b)
𝛾𝑟 = 𝛾𝑎 + 𝛾𝑣 ; where 𝛾𝑟 = coefficient of real expansion,
𝛾𝑎 = coefficient of apparent expansion and
𝛾𝑣 = coefficient of expansion of vessel.
For copper 𝛾𝑟 = 𝐶 + 3𝛼𝐶𝑢 = 𝐶 + 3𝐴
For silver 𝛾𝑟 = 𝑆 + 3𝛼𝐴𝑔
𝐶 − 𝑆 + 3𝐴
= 𝐶 + 3𝐴 = 𝑆 + 3𝛼𝐴𝑔 ⇒ 𝛼𝐴𝑔 =
3
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ANSWER-KEY
Q. 1 2 3 4 5 6 7 8 9 10
A. B B C C C C B D C A
Q. 11 12 13 14 15 16 17 18 19 20
A. A D C D C B B C B B
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