Thermal Properties of Matter Test

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Thermal Properties of Matter Test - 65

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Question 1
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Two metallic sphere $$S_1$$ and $$S_2$$ are made of the same material and have got identical surface finish.The mass of $$S_1$$ is thrice that of $$S_2$$.Both the sphere are heated to the same high temperature and placed in the same room having lower temperature but are thermally insulated from each other.The ratio of the initial rate of cooling of $$S_1$$ to that $$S_2$$ is :

A
$$\dfrac{1}{3}$$

B
$$\dfrac{1}{\sqrt{3}}$$

C
$$\dfrac{\sqrt{3}}{1}$$

D
$$\left(\dfrac{1}{3}\right)^{\dfrac{1}{3}}$$

Solution
Question 2
1 / -0

The pressure of an ideal gas veries according to the law $$P=P_{0}-AV^{2}$$ where $$P_{0}$$ and A are positive constants. What is the highest temperature that can be attained by the gas?

A
$$\dfrac{P_{0}}{nR}\sqrt{\dfrac{P_{0}}{A}}$$

B
$$\dfrac{P_{0}}{nR}\sqrt{\dfrac{P_{0}}{2A}}$$

C
$$\dfrac{2P_{0}}{nR}\sqrt{\dfrac{P_{0}}{2A}}$$

D
$$\dfrac{2P_{0}}{3nR}\sqrt{\dfrac{P_{0}}{3A}}$$

Solution
Question 3
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Consider the situation as shown in fig. Calculate the amount of heat flowing per second through a cross section of the bent part, if the total heat taken out per second from the end at $$100^{o}C$$ is $$130\ J$$.

A
$$130\ J$$

B
$$60\ J$$

C
$$70\ J$$

D
$$80\ J$$

Solution
Question 4
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$$10g$$ of ice at $$-{20}^{o}C$$ is added to $$10g$$ of water at $${50}^{o}C$$. Specific heat of water $$=1cal/g ^{o}C$$, specific heat of ice$$=0.5cal/g^{o}C$$. Latent heat of ice $$=80cal/g$$. Then resulting temperature is -

A
$$-{20}^{o}C$$

B
$${15}^{o}C$$

C
$${0}^{o}C$$

D
$${50}^{o}C$$

Solution
Question 5
1 / -0

A hot body is placed in cooler surrounding. When the hot body temperature $$75^{\circ}C$$, the rate of cooling is $$4^{\circ}C/min$$. When it is $$55^{\circ}C$$, the rate of cooling is $$2^{\circ}C/min$$. The temperature of the surroundings is

A
$$20^{\circ}C$$

B
$$25^{\circ}C$$

C
$$30^{\circ}C$$

D
$$35^{\circ}C$$

Solution

We know that,
cooling rate is directly proportional to difference of temperature of body and temperature of surroundings
$$\dfrac { d\theta }{ dt } \propto ({ \theta }_{ t }-{ \theta }_{ s })\\ $$
$$\dfrac { d\theta }{ dt } =k({ \theta }_{ t }-{ \theta }_{ s })$$ --------------------------(1)
Given,
$$ At\quad { 75 }^{ \circ }C\quad ,\dfrac { d\theta }{ dt } ={ 4 }^{ \circ }C/min$$
And
$$ At\quad { 55 }^{ \circ }C\quad ,\dfrac { d\theta }{ dt } ={ 2 }^{ \circ }C/min$$
Substitute all value in equation(1)
$$4=k(75-{ \theta }_{ s })$$ ---------------------------------------(2)
And
$$2=k(55-{ \theta }_{ s })$$ ----------------------------------------(3)
By Dividing both the above equation we will get
$$\dfrac { 4 }{ 2 } =\dfrac { k(75-{ \theta }_{ s }) }{ k(55-{ \theta }_{ s }) }$$
$$\ { 2 } =\dfrac { (75-{ \theta }_{ s }) }{ (55-{ \theta }_{ s }) }$$
$$2(55-{ \theta }_{ s })=(75-{ \theta }_{ s })$$
$$110-{ 2\theta }_{ s }=75-{ \theta }_{ s }$$
$$110-75=2{ \theta }_{ s }-{ \theta }_{ s }$$
$${ \theta }_{ s }=35$$
Hence, Temperature of surroundings is 35 degree celcious.Hence, Option (D) is correct option.
Question 6
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Two closed containers of equal volume filled with air at pressure $$P_{0}$$ and temperature $$T_{0}$$. Both are connected by narrow tube. If one of the container is maintained at temperature $$T_{0}$$ and another at temperature T, then new pressure in the containers will be

A
$$\dfrac{2P_{0}T}{T+T_{0}}$$

B
$$\dfrac{P_{0}T}{T+T_{0}}$$

C
$$\dfrac{P_{0}T}{2(T+T_{0}})$$

D
$$\dfrac{T+T_{0}}{P_{0}}$$

Solution
Question 7
1 / -0

Compare rate of radiation of metal sphere at $$627^0C$$ annd $$ 327^0C$$.

A
0.7 : 1

B
3 : 1

C
63 : 1

D
5.063 : 1

Solution

According to stafan's Law
$$E=k{ T }^{ 4 }$$
Then,
$$\dfrac { { E }_{ 1 } }{ { E }_{ 2 } } ={ \left( \dfrac { { T }_{ 1 } }{ { T }_{ 2 } } \right) }^{ 4 }$$
$$\dfrac { { E }_{ 1 } }{ { E }_{ 2 } } ={ \left( \dfrac { 627+273 }{ 327+273 } \right) }^{ 4 }$$
$$\dfrac { { E }_{ 1 } }{ { E }_{ 2 } } ={ \left( \dfrac { 900 }{ 600 } \right) }^{ 4 }$$
$$\dfrac { { E }_{ 1 } }{ { E }_{ 2 } } ={ \left( \dfrac { 3 }{ 2 } \right) }^{ 4 }$$
$$\dfrac { { E }_{ 1 } }{ { E }_{ 2 } } ={ \left( \dfrac { 1.5 }{ 1 } \right) }^{ 4 }$$
$$\dfrac { { E }_{ 1 } }{ { E }_{ 2 } } =\dfrac { { 1.5 }^{ 4 } }{ 1 }$$
$$\dfrac { { E }_{ 1 } }{ { E }_{ 2 } } =\dfrac { 5.0625 }{ 1 }$$
Hence, option (D) is correct Option.
Question 8
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In the given figure, if the heat flowing through the $$5 \Omega$$ resistance is 10 calorie then how much heat is flowing through the $$4\Omega $$ resistance ?

A
$$1$$ calorie

B
$$2$$ calorie

C
$$3$$ calorie

D
$$4$$ calorie

Solution
Question 9
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Two rods with the same dimensions have thermal conductivity in the ratio $$1:\ 2$$. They are arranged between heat reservoirs with the same temperature difference, in two different configurations, in two different configurations. $$A\ and\ B$$. The rates of heat flow in $$A\ and\ B$$ are $$I_{A}\ and\ I_{B}$$ respectively. The ratio $$\dfrac{I_{A}}{I_{B}}$$ is equal to

A
$$1:\ 2$$

B
$$1:\ 3$$

C
$$2:\ 5$$

D
$$2:\ 9$$

Solution
Question 10
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Two bodies $$A$$ and $$B$$ having equal surface areas are maintained at temperatures $$10 ^ { \circ } \mathrm { C }$$ and $$20 ^ { \circ } \mathrm { C }$$. 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
$$1 : 16$$

Solution