Thermal Properties of Matter Test

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Thermal Properties of Matter Test - 46

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Weekly Quiz Competition

Question 1
1 / -0

Select the correct option given below:
In order that the heat flows from one part of solid to another part,what is required. ?

A
uniform density

B
temperature gradient

C
density gradient

D
uniform temperature

Solution

As heat flows from higher temperature to lower temperature,so a temperature gradient is required.
Question 2
1 / -0

A steel metre scale is to be ruled so that the millimetre intervals are accurate within about $$5\times { 10 }^{ -5 }mm$$ at a certain temperature. The maximum temperature variation allowable during the ruling of the millimetre marks is ($$\alpha$$ for steel $$=11\times { 10 }^{ -6 }_{ }\quad ^{ o }{ { C }^{ -1 } }$$)

A
$${ 8 }^{ o }C$$

B
$$ { 9 }^{ o }C$$

C
$${ 4.5 }^{ o }C$$

D
$${ 10 }^{ o }C$$

Solution

Use formula of thermal expansion ,

$$\Delta L = L_{0} \alpha \Delta T$$

where $$\Delta L$$ is the change in length of object after changing temperature

$$L_{0}$$ is the initial length of object

$$\Delta T$$ is the temperature change

and $$\alpha$$ is the coefficient of linear expansion

Here, $$ \Delta L = 5 \times 10^{-5} mm = 5 \times 10^{-8} m$$

$$L_{0} = 1 mm = 10^{-3} m$$

$$\alpha = 11 \times 10^{-6} C^{-1}$$

Now, from formula,

$$5 \times 10^{-8} = 10^{-3} \times 11 \times 10^{-6} \times \Delta T\\$$

$$\Delta T = \dfrac{50}{11}\\$$

$$\Delta T = {4.5} \ { C^{-1}}$$

Option C is correct.
Question 3
1 / -0

If the temperature of the Sun were to increase from $$T$$ to $$2T$$ and its radius from $$R$$ to $$2R$$. The ratio of power radiated by it would become:

A
$$64$$ times

B
$$16$$ times

C
$$32$$ times

D
$$4$$ times

Solution

From Stefan's law, the energy radiated by sun is given by:-
$$P=\sigma eAT^4$$

In $$Ist$$ case:-
$$P_1= \sigma e\times 4\pi R^2\times T^4$$

In $$2nd$$ case:-
$$P_2=\sigma e\times 4\pi (2R)^2\times (2T)^4$$

Required ratio of power radiated $$=\dfrac{P_2}{P_1}= \dfrac{\sigma e\times 4\pi (2R)^2\times (2T)^4}{\sigma e\times 4\pi R^2\times T^4}= 64$$

So, correct answer is $$(A)$$.
Question 4
1 / -0

Clear nights are colder than cloudy nights because of

A
conduction

B
condensation

C
radiation

D
insolation

Solution

Cloud in effect reduce the radiation imbalance between the upward flux from the ground and downward flux from the atmosphere, relative to that on clear nights.
Question 5
1 / -0

In which of the following process convection does not take place primarily?

A
Sea and land breeze

B
Trade wind

C
Boiling of water

D
Warming of glass of bulb due to filament

Solution

$$\textbf{Explanation:}$$
$$\bullet$$In convection, transfer of heat took place by movement of gaseous and liquid molecules.
$$\bullet$$It generally happens in liquid and gas only where bulk portion is fluid. Sea and land breeze, trade wind and boiling water are example of such fluids. So, in those heat transfer is because of convection. But in bulb filament, resistance is responsible for transferring the heat which is example of conduction. So, in warming of glass bulb due to filament convection doesn’t take place.

Hence, option D is the correct answer.
Question 6
1 / -0

Select the correct option given below:
Which of the following graphs correctly shows variation of coefficient of volume expansion of copper as a function of temperature?

A

B

C

D

Solution

The volume expansion coefficient of the copper as a function of temperature is given by:
$$\alpha_v=\dfrac{dV}{dT}$$

So, $$\dfrac{\Delta V}{V}=\alpha_v\Delta T$$

taking log both sides:
$$ln\dfrac{V+\Delta V}{V}=\int\alpha_vTdT$$

So, $$\dfrac{\Delta V}{V}=exp\ {\int\alpha_vTdT}$$

So, initially it varies linearly and then after some time it starts increasing exponentially. So, option $$(C)$$ is incorrect.
Question 7
1 / -0

One end of a $$0.25m$$ long metal bar is in steam and the other is in contact with ice .If $$12g$$ of ice melts per minute, then the thermal conductivity of the metal is (Given cross section of the bar$$=5\times 10^{-4}m^{2}$$ and latent heat of ice is $$80\,cal \,g^{-1}$$)

A
$$20\,cal\,s^{-1}\,m^{-1}\,^oC^{-1}$$

B
$$10\,cal\,s^{-1}\,m^{-1}\,^oC^{-1}$$

C
$$40\,cal\,s^{-1}\,m^{-1}\,^oC^{-1}$$

D
$$30\,cal\,s^{-1}\,m^{-1}\,^oC^{-1}$$

Solution

Given: The length of the metal bar is $$x=0.25m$$
The change temperature across the ends of the rod is $$T_1-T_2=100-0=100^oC$$
The time taken to melt the ice is $$t=1min=60s$$
$$A=5\times 10^{-4}m^2$$ $$,L=80\,\,cal\,\,g^{-1}$$

The amount of heat required by the ice is:
$$Q=mL=12\times 80=960\,\,cal$$

But,$$Q=\dfrac{KA(T_1-T_2)t}{x}$$ or

$$960=\dfrac{K\times5\times10^{-4}\times100\times60}{0.25}$$

$$\therefore K =\dfrac{960 \times 0.25}{5\times10^{-2}\times 60}$$

$$\Rightarrow30\,cal\,s^{-1}m^{-1} \,^oC^{-1}$$
Question 8
1 / -0

A disc of mass 500 g and radius 10 cm rotates about a fixed vertical axis passing through its centre, with an angular velocity 3 rad/s. A ring of same mass and radius is gently placed on it coaxially. The heat evolved after sufficient time is:

A
$$0.75 \times 10^{-2} J$$

B
$$1.50 \times 10^{-2} J$$

C
$$0.25 \times 10^{-2} J$$

D
$$0.50 \times 10^{-2} J$$

Solution

$$K_i = \dfrac{1}{2} \times \dfrac{(500 gm) (10 cm)^2}{2} \times (3 rad/s)^2$$
$$= \dfrac{9}{8} \times 10^{-2} J$$
Ans. momentum = $$ \dfrac{(500 gm) (10 cm)^2}{2} \times 3$$
$$\dfrac{3}{4} \times 10^{-2} J.s$$
$$K_f = \dfrac{ \left( \dfrac{3}{4} \times 10^{-2} \right)^2 }{2 \times \left[ \dfrac{(500 gm) \times (10 cm)^2}{2} + (500 gm) (10 cm)^2 \right] }$$
$$\dfrac{3}{8} \times 10^{-2} J$$
heat = $$\left( \dfrac{9}{8} - \dfrac{3}{8} \right) \times 10^{-2} = 0.75 \times 10^{-2} J$$
Question 9
1 / -0

A wall is made of equally thick layers A and B of different materials. Thermal conductivity of A is twice that that of B . In the steady state, the temperature difference across the wall is $$36^o C$$. The temperature difference across the layers A is:

A
$$12^oC$$

B
$$18^oC$$

C
$$6^oC$$

D
$$24^oC$$

Solution

Here, $$K_a=2K_B,T_A-T_B=36^oC$$
Let T is the temperature of the junction.
As $$\left(\dfrac{\triangle T}{\triangle t}\right)_A=\left(\dfrac{\triangle T}{\triangle t}\right)_B$$

$$\therefore \dfrac{K_AA(T_A-T)}{x}=\dfrac{K_BA(T-T_B)}{x}$$

$$2K_B(T_A-T)=K_B(T-T_B)$$
$$2(T_A-T)=T-T_B$$
Add $$(T_A-T)$$ on both sides,we get
$$3(T_A-T)=T_A-T+T-T_B$$
$$3(T_A-T)=T_A-T_B$$
$$T_A-T=\dfrac{T_A-T_B}{3}=\dfrac{36}{3}=12^o C$$
$$\therefore$$ temperature difference across the layer
$$A=T_A-T=12^oC$$
Question 10
1 / -0

The temperature of $$100$$g of water is to be raised form $$24^o$$C to $$90^o$$C by adding steam to it. Calculate the mass of the steam required for this purpose.

A
4

B
13

C
22

D
25

Solution

Let mass of steam required $$=m\ gm$$
each gram of steam of condensing release $$536$$ calories of heat steam condense at $$100^oC$$
cools finally to $$90^oC$$
Heat released by $$m\ gm$$ of steam on condensing $$=536\times m$$ calne
Final Temp of solution $$=m\times $$ specific heat of water $$\times$$ fall of temp
$$=m\times 1\times (100-90)$$
$$=m\times 1\times 10$$
$$=10\ m$$ clone
heat released $$=536\ m+10\ m$$
$$546\ m$$ calories of heat
Heat required to raised the temp of $$100\ gm$$ of water at $$24^oC+m\ gm$$ of condensed steam from $$24^oC-90^oC$$
$$=(100+m)\times 1\times (90-24)$$
$$=(100+m)\times 66$$ calories
heat gained = Heat lost
$$(100+m)\times 66=546\ m$$
$$6600+66\ m=546\ m$$
$$600=486\ m$$
$$\boxed {m=13.75\ g\ of\ steam}\simeq 13\ gm$$
$$(B)$$

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Thermal Properties of Matter Test - 46

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Thermal Properties of Matter Test - 46

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SHARING IS CARING

Question 1

uniform density

temperature gradient

density gradient

uniform temperature

Solution

Question 2

$${ 8 }^{ o }C$$

$$ { 9 }^{ o }C$$

$${ 4.5 }^{ o }C$$

$${ 10 }^{ o }C$$

Solution

Question 3

$$64$$ times

$$16$$ times

$$32$$ times

$$4$$ times

Solution

Question 4

conduction

condensation

radiation

insolation

Solution

Question 5

Sea and land breeze

Trade wind

Boiling of water

Warming of glass of bulb due to filament

Solution

Question 6

Solution

Question 7

$$20\,cal\,s^{-1}\,m^{-1}\,^oC^{-1}$$

$$10\,cal\,s^{-1}\,m^{-1}\,^oC^{-1}$$

$$40\,cal\,s^{-1}\,m^{-1}\,^oC^{-1}$$

$$30\,cal\,s^{-1}\,m^{-1}\,^oC^{-1}$$

Solution

Question 8

$$0.75 \times 10^{-2} J$$

$$1.50 \times 10^{-2} J$$

$$0.25 \times 10^{-2} J$$

$$0.50 \times 10^{-2} J$$

Solution

Question 9

$$12^oC$$

$$18^oC$$

$$6^oC$$

$$24^oC$$

Solution

Question 10

4

13

22

25

Solution

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