Thermal Properties of Matter Test

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Thermal Properties of Matter Test - 43

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Question 1
1 / -0

Lake water never completely freezes because

A
Principle of conduction prevents it

B
Principle of convection prevents it

C
both of the above

D
none of the above

Solution

Lake water never completely freezes because principle of convection prevents it.
Warm water gets more dense when it gets colder and therefore sinks. This fact may lead to fact that ice should form on bottom of lake first. But water colder than $$4^oC$$ begins expanding and become less dense as it gets closer. As a result, close to freezing, colder water floats on top and warmer water sinks to bottom. Eventually, the coldest water, which has floated to top of lake in wintery condition freezes to form ice layer. When the water freezes to ice, the ice becomes significantly less dense than water can continues to float on lake's surface. And then lake never completely freezes. So, in this manner; principle of convection prevents it.
Question 2
1 / -0

In experiment of heat transfer by convection, water inside the pipe completely gets heated (with the application of heat source). This can be explained best by which of the following statements,

A
Differences in densities produced between the bottom and the top of water , in turn produce the driving force which continuously circulates the water.

B
A continuous conductive channel is produced between the different layers of water in the pipe which circulates the heat.

C
both the above explains correctly

D
all of the above

Solution

When the water in below part of pipe is heated, its density decreases. Hence less density water moves up (shown in red) and cooler water with more density comes down (shown in blue). It is nothing but convection as convection is mode of heat transfer in which heat is transferred in fluid (liquid or gas) by molecular motion.
Question 3
1 / -0

Two identical containers P and Q with frictionless piston contain the same ideal gas at the same temperature and the same volume. The mass of gas in P is $$m_1$$ and that in Q is $$m_2$$. The gas in each cylinder is now allowed to expand isothermally to the same final volume $$1.5$$V. The changes pressure in P and Q are found to be $$\Delta$$p and $$2\Delta$$p respectively then.

A
$$\displaystyle m_2=2m_1$$

B
$$\displaystyle m_1 =2m_2$$

C
$$\displaystyle m_1=\frac{m_2}{4}$$

D
$$\displaystyle m_1m_2=1$$

Solution

From ideal gas equation, we have
$$pV=mRT$$
So $$\displaystyle\frac{p}{m}=\frac{RT}{V}$$
As both gases expand isothermally to the same volume
So, $$RT/V=$$ constant
$$\therefore \displaystyle\frac{p}{m}=$$ constant
$$\therefore \displaystyle\frac{p_1}{p_2}=\frac{m_1}{m_2}\Rightarrow \frac{\Delta p}{2\Delta p}=\frac{m_1}{m_2}$$
$$\therefore m_2=2m_1$$.
Question 4
1 / -0

Two conductors of the same material have their diameters in the ratio $$1:2$$ and their lengths in the ratio $$2:1$$. If the temperature difference between their ends is the same, then the ratio of amounts of heat conducted per second through them will be.

A
$$4:1$$

B
$$1:4$$

C
$$8:1$$

D
$$1:8$$

Solution

The resistance of conductor
$$R=\displaystyle\frac{\rho l}{A}=\frac{\rho l}{\pi r^2}$$
or $$\displaystyle R\propto \frac{1}{r^2}$$
$$\therefore \displaystyle\frac{R_1}{R_2}=\frac{l_1}{l_2}\times \frac{r^2_2}{r^2_1}$$
$$=\displaystyle\frac{2}{1}\times \left(\displaystyle\frac{2}{1}\right)^2=\frac{8}{1}$$
Thermal potentials between the ends of the rods are same. So, heat conducted per second.
$$H=\displaystyle\frac{Q}{t}\propto \frac{1}{R}$$
$$\therefore \displaystyle\frac{H_1}{H_2}=\frac{R_2}{R_1}=\frac{1}{8}=1:8$$.
Question 5
1 / -0

In a mercury thermometer, the ice point is marked as $$10^o$$ and the steam point is marked as $$130^o$$. At $$40^oC$$ temperature, what will this thermometer read (in degrees)?

A
$$78$$

B
$$66$$

C
$$62$$

D
$$58$$

Solution

$$\cfrac { steam\; point-ice\; point }{ given\; temp.-ice\; point } =constant\\ \cfrac { 100-0 }{ 40-0 } =\cfrac { 130-10 }{ x-10 } \\ \Rightarrow \cfrac { 10 }{ 4 } =\cfrac { 120 }{ x-10 } \\ \Rightarrow x-10=48\Rightarrow x=58℃$$
Question 6
1 / -0

_____ surface absorbs more heat that any other surface under identical conditions.

A
White

B
Rough

C
Black

D
Yellow

Solution

The black colour means the absence of any other colour. Our visible spectrum ranges from 400-700 nm. Now if thing an object appears red to us it means that it reflects the wavelength of red light and absorbs all the other wavelength light. A black object absorbs all the colour light and reflects none thats why it absorbs more heat.
Question 7
1 / -0

A body of specific heat $$0.2$$ kcal$$/kg^o$$C is heated through $$100^o$$C. The percentage increase in its mass is?

A
$$9\%$$

B
$$9.3\times 10^{-11}\%$$

C
$$10\%$$

D
None of these

Solution

From Einstein's mass energy relation
$$E=\Delta mc^2$$ .............(i)
where $$\Delta$$m is mass lost, c is speed of light. Also heat given by body is
$$E=mC\Delta \theta$$ ...........(ii)
where, C is specific heat.
Equating Eqs. (i) and (ii) we get
$$\Delta m=\displaystyle\frac{MC\Delta \theta}{C^2}=\frac{M\times 0.2\times 100\times 4.2\times 10^3}{(3\times 10^8)^2}$$
$$\displaystyle\frac{\Delta m}{M}=\frac{20\times 4.2\times 10^3}{(3\times 10^8)^2}$$
$$\%$$ increase in mass
$$\displaystyle =\frac{\Delta m}{M}\times 100=\frac{20\times 4.2\times 10^3}{(3\times 10^8)^2}\times 100$$
$$=9.3\times 10^{-11}\%$$.
Question 8
1 / -0

During an experiment an ideal gas is found to obey an additional law $$V{p}^{2}=$$ constant. The gas is initially at temperature $$T$$ and volume $$V$$, when it expands to volume $$2V$$, the resulting temperature is $$T_2$$:

A
$$\cfrac { T }{ 2 } $$

B
$$2T$$

C
$$\sqrt { 2 } T $$

D
$$\cfrac { T }{ \sqrt { 2 } } $$

Solution

Ideal gas: $$VP^2=$$ constant
Again $$PV =nRT$$ from equation of state,
Hence, $$VP\times P=$$ constant i.e. $$nRT\times P=$$ constant
Again, $$P=\dfrac{nRT}{V}$$

$$\therefore$$ $$\dfrac{(nRT)^2}{V}=$$ constant

$$\dfrac{T^2}{V}=$$ constant

Thus volume $$V$$ when expanded to $$2V$$, temperature $$T_2$$

$$T_2=\sqrt{\dfrac{2v}{v}}=\sqrt{2}T_1$$
Question 9
1 / -0

An ideal gas follows a process described by $$PV^2=C$$ from $$(P_1, V_1, T_1)$$ to $$(P_2, V_2, T_2)$$ (C is a constant). Then.

A
If $$P_1 > P_2$$ then $$T_2 > T_1$$

B
If $$V_2 > V_1$$ then $$T_2 < T_1$$

C
If $$V_2 > V_1$$ then $$T_2 > T_1$$

D
If $$P_1 > P_2$$ then $$V_2 > V_2$$

Solution

we know $$PV=nRT$$
also $$PV^2=C$$
So $$\dfrac{T^2}{P}=C_1$$
$$\dfrac{T_1^2}{P_1}=\dfrac{T_2^2}{P_2}$$
Also$$T_1V_1=T_2V_2$$
So answer B is correct.
Question 10
1 / -0

The temperature of water at the surface of a deep lake is $${2}^{0}$$ C. The temperature expected at the bottom is

A
$${0}^{0}$$ C

B
$${2}^{0}$$ C

C
$${4}^{0}$$ C

D
$${6}^{0}$$ C

Solution

It is supposed that at the deepest point density is maximum. But density is maximum at $$4^oC$$. Hence, temperature at bottom is $$4^oC$$

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Re-Attempt Weekly Quiz Competition

Thermal Properties of Matter Test - 43

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Thermal Properties of Matter Test - 43

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SHARING IS CARING

Question 1

Principle of conduction prevents it

Principle of convection prevents it

both of the above

none of the above

Solution

Question 2

Differences in densities produced between the bottom and the top of water , in turn produce the driving force which continuously circulates the water.

A continuous conductive channel is produced between the different layers of water in the pipe which circulates the heat.

both the above explains correctly

all of the above

Solution

Question 3

$$\displaystyle m_2=2m_1$$

$$\displaystyle m_1 =2m_2$$

$$\displaystyle m_1=\frac{m_2}{4}$$

$$\displaystyle m_1m_2=1$$

Solution

Question 4

$$4:1$$

$$1:4$$

$$8:1$$

$$1:8$$

Solution

Question 5

$$78$$

$$66$$

$$62$$

$$58$$

Solution

Question 6

White

Rough

Black

Yellow

Solution

Question 7

$$9\%$$

$$9.3\times 10^{-11}\%$$

$$10\%$$

None of these

Solution

Question 8

$$\cfrac { T }{ 2 } $$

$$2T$$

$$\sqrt { 2 } T $$

$$\cfrac { T }{ \sqrt { 2 } } $$

Solution

Question 9

If $$P_1 > P_2$$ then $$T_2 > T_1$$

If $$V_2 > V_1$$ then $$T_2 < T_1$$

If $$V_2 > V_1$$ then $$T_2 > T_1$$

If $$P_1 > P_2$$ then $$V_2 > V_2$$

Solution

Question 10

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

$${2}^{0}$$ C

$${4}^{0}$$ C

$${6}^{0}$$ C

Solution

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