MCQ Questions for Class 11 Engineering Physics Thermal Properties Of Matter Quiz 8

State whether given statement is True or False
Radiant heat can easily pass through air

0%
True
0%
False

State reason for unsolidification of water inside lakes in cold countries?

0%
anomalous expansion of water
0%
thermal conductivity of water
0%
High specific heat capacity of water
0%
Data insufficient

Explanation

$Answer:-$ A

Firstly, let me give you a little basic knowledge about a property of liquids, then i'll proceed to the answer.

As the temperature of a liquid/solid increases, its volume increases. Most liquids behave this way. But, water has a unique quality, above (almost) 4°C water behaves normally i.e its volume increases as the temperature increases. However, between 0°C and almost 4°C its volume decreases with rise in temperature( or increases with temperature drop, thus making it less dense)-this is anomalous behaviour of water.

Say the water at the surface is at 12°C and its cooling down. Now, as the temperature decreases, its volume decreases and density increases. As the surface water reaches about 4°C, it becomes more dense and "sinks" to the bottom allowing the less dense water to surface up. As temperature falls below 4°C, surface water becomes lighter( less dense) compared to the water below and floats until it freezes. Once frozen, the ice insulates the water at the bottom from cooling so its temperature is maintained between 0 and almost 4°C, keeping it from freezing.

When water cools below

$4^\circ{}$ C, its volume

0%
increases.
0%
decreases.
0%
remain constant.
0%
Data insufficient

Explanation

Answer is A

When water is cools below

$4^o$ C, its density decreases, hence volume increases.

If the pressure of the gas contained in a vessel is increased by

$0.4$ %, when heated through

$\displaystyle { 1 }^{ \circ }C$ . What is the initial temperature of the gas?

0%
$\displaystyle 250K$
0%
$\displaystyle { 250 }^{ \circ }C$
0%
$\displaystyle 2500K$
0%
$\displaystyle { 25 }^{ \circ }C$

Explanation

For an Ideal Gas,

$PV=nRT$

$\implies V\Delta P =nR\Delta T$

$\implies \dfrac{\Delta P}{P}=\dfrac{\Delta T}{T}=0.004$

$\implies T=\dfrac{\Delta T}{0.004}=\dfrac{1K}{0.004}=250K$

We consider the radiation emitted by the human body. Which of the following statements if true?

0%
The radiation is emitted during the summers and absorbed during the winters
0%
The radiation emitted lies in the ultraviolet region and hence is not visible
0%
The radiation emitted is in the infra-red region
0%
The radiation is emitted only during the day

Explanation

The heat radiation emitted by the human body is the infrared radiation. Their wavelength is of the order of

$7.9\times {10}^{-7}m$ to

${10}^{-3}m$ which is of course the range of infrared region.
Hence, human body emits radiation in infrared region.

Identify the statement in which the mode of heat transfer is convection ?

0%
The heat of the sun warming our planet
0%
The heat from an electric stove warming a frying pan
0%
Ice cubes cooling a drink
0%
A microwave oven cooking a meal
0%
An overhead fan cooling a room

Heat travels through vacuum by:

0%
convection
0%
radiation
0%
conduction
0%
of all these

$\displaystyle { 10 }^{ \circ }C$ , the value of the density of a fixed mass of an ideal gas divided by its pressure is

$'x'$ , at

$\displaystyle { 110 }^{ \circ }C$ this ratio is:

0%
$\displaystyle \frac { 10 }{ 110 } x$
0%
$\displaystyle \frac { 383 }{ 283 } x$
0%
$\displaystyle \frac { 110 }{ 10 } x$
0%
$\displaystyle \frac { 283 }{ 383 } x$

Explanation

Let the ratio of density and pressure be

$r$ .

From Ideal Gas Equation,

$\rho =\dfrac { PM }{ RT }$

$r=\dfrac { \rho }{ P } =\dfrac { M }{ RT }$

$\therefore \dfrac { { r }_{ 2 } }{ { r }_{ 1 } } =\dfrac { { T }_{ 1 } }{ { T }_{ 2 } } =\dfrac { 273+10 }{ 273+110 } =\dfrac { 283 }{ 383 }$

${ r }_{ 2 }=\dfrac { 283 }{ 383 } x$

Describe the anomalous expansion of water?

0%
When cooled below 4 degrees Celsius, it starts expanding until it becomes ice at 0 degrees Celsius
0%
When cooled below 4 degrees Celsius, it starts contacting until it becomes ice at 0 degrees Celsius
0%
When cooled below 4 degrees Celsius, it's volume remains same until it becomes ice at 0 degrees Celsius
0%
None of the above

Explanation

The anomalous expansion of water is the property of water where by it expands instead of contracting when temperature goes from

$4^oC$ to

$0^oC$

In which of the following phenomena, the heat waves travel along straight lines with the speed of light?

0%
Thermal conduction
0%
Forced convection
0%
Natural convection
0%
Thermal radiation

Choose the correct statement from the following

0%
Woollen clothes keep the body warm in winter as air is the bad conductor of heat
0%
Shining surfaces are good reflectors of heat
0%
Heat radiations travel with speed of light
0%
The SI unit of coefficient of cubical expansion is $K^{-1}$

Explanation

Air and wool are bad conductors of heat. Heat from the body does not flow outside to the atmosphere. Therefore woollen clothes keep the body warm in winter.

Shining surfaces are good reflectors of heat and so the roofs of factories ares are painted white.

Thermal radiation is electromagnetic radiation, i.e. photons, and so does travel at the same speed as light.

The coefficient of cubical expansion is defined as the ratio of the increase in volume to its original volume for every degree increase in temperature. its Si unit is

$K^{-1}$ .

The temperature of a 3 kg block of aluminium increased by 3 degrees. How much heat would be needed to raise the temperature of a 9 kg block of aluminum by 3 degrees?

0%
9 times as much heat as the 3 kg block
0%
3 times as much heat as the 3 kg block
0%
the same amount of heat as the 3 kg block
0%
one-third as much heat as the 3 kg block
0%
one-ninth as much heat as the 3 kg block

Explanation

Given :

$\Delta T = 3^o C$

Let the specific heat of aluminium be

$S$ .

The initial mass to be heated

$m_i = 3kg$

$\therefore$

$Q_i = m_i S \Delta T = 3 \times S \times 3 =9S J$

Now the mass of aluminium to be heated

$m_f = 9 kg$

$\therefore$ Heat required

$Q_f = 9 \times S \times 3 = 27 S = 3 Q_i\ J$

Thus option B is correct.

A gas in a flexible container initially has volume of 320 L. If we decrease the pressure of the gas from 4.0 atm to 1.0 atm and decrease the temperature form 273 degrees Celsius to 0 degrees Celsius, what is the new volume of the gas?

0%
0 L
0%
320 L
0%
160 L
0%
640 L
0%
80 L

Explanation

Given :

$V_1 = 320$ L ,

$P_1 = 4.0$ atm,

$P_2 = 1.0$ atm

Initial temperature,

$T_1 = 273 ^oC =546$ K

Final temperature,

$T_2 = 0 ^oC =273$ K

Using

$\dfrac{P_2V_2}{T_2} = \dfrac{P_1 V_1}{T_1}$

$\therefore$

$\dfrac{1.0 \times V_2}{273} = \dfrac{4.0 \times 320}{546}$

$\implies V_2 = 640$ L

A closed container of volume V contains an ideal gas at pressure P and Kelvin Temperature T. The temperature of the gas is changed to 4T due to passing heat to the container. Choose correct pairs of pressure and volume after change of the temperature.

0%
$P\;and \;V$
0%
$\displaystyle P\;and \frac{1}{2}V$
0%
$4\;P\;and \;4\;V$
0%
$4\;P\;and \;V$
0%
$\displaystyle \frac{1}{4}\;P\;and\;V$

Explanation

As the container is closed, thus the volume of the gas remains constant i.e

$V$

Using ideal gas equation,

$PV = nRT$

$\implies$

$\dfrac{P}{T} = \dfrac{nR}{V} =constant$

$\therefore$

$\dfrac{P}{T} =\dfrac{P'}{T'}$ where

$T' = 4T$

$\dfrac{P}{T} =\dfrac{P'}{4T}$

$\implies P' = 4P$

Hence the final pressure and volume of the gas is

$4P$ and

$V$ respectively.

Heat can be transferred to the inside surface of the walls of the container by which of the above?

0%
I only
0%
I and II only
0%
I and III only
0%
II only
0%
I, II, and III only

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)?

0%
$78$
0%
$66$
0%
$62$
0%
$58$

Explanation

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

State whether the following statements are True or False.
Liquids expand on heating.

0%
True
0%
False

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

$T_2$ :

0%
$\cfrac { T }{ 2 }$
0%
$2T$
0%
$\sqrt { 2 } T$
0%
$\cfrac { T }{ \sqrt { 2 } }$

Explanation

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

$T_2$

$T_2=\sqrt{\dfrac{2v}{v}}=\sqrt{2}T_1$

The temperature of the water of a pond is

${0}^{o}C$ while that of the surrounding atmosphere is

$-{20}^{o}C$ . If the density of ice is

$\rho$ , coefficient of thermal conductivity is

$k$ and latent heat of melting is

$L$ then the thickness

$Z$ of ice layer formed increases as a function of time

$t$ as :

0%
${Z}^{2}=\cfrac{60k}{\rho L}t$
0%
$Z=\sqrt { \cfrac { 40k }{ \rho L } t }$
0%
${Z}^{2}=\cfrac { 40k }{ \rho L } \sqrt { t }$
0%
${Z}^{2}=\cfrac{40k}{\rho L}t$

Explanation

According to the question, the above diagram can be drawn.

So,

$H=\dfrac{dQ}{dT}=\dfrac{kA(0-(-20))}{x}$

$\implies \dfrac{dm}{dt}L=\dfrac{kA(20)}{x}$

$\implies \rho A\dfrac{dx}{dt}L=\dfrac{kA(20)}{x}$

On integrating both sides, we get:

$\int _{ 0 }^{ z }{xdx } =\dfrac{20k}{\rho L}\int_{0}^{t}{dt}$

$\implies z^2=\dfrac{40k}{\rho L}t$

Lake water never completely freezes because

0%
Principle of conduction prevents it
0%
Principle of convection prevents it
0%
both of the above
0%
none of the above

Explanation

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.

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,

0%
Differences in densities produced between the bottom and the top of water , in turn produce the driving force which continuously circulates the water.
0%
A continuous conductive channel is produced between the different layers of water in the pipe which circulates the heat.
0%
both the above explains correctly
0%
all of the above

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.

0%
$\displaystyle m_2=2m_1$
0%
$\displaystyle m_1 =2m_2$
0%
$\displaystyle m_1=\frac{m_2}{4}$
0%
$\displaystyle m_1m_2=1$

Explanation

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

$\therefore \displaystyle\frac{p_1}{p_2}=\frac{m_1}{m_2}\Rightarrow \frac{\Delta p}{2\Delta p}=\frac{m_1}{m_2}$

$\therefore m_2=2m_1$ .

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.

0%
$4:1$
0%
$1:4$
0%
$8:1$
0%
$1:8$

Explanation

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.

$H=\displaystyle\frac{Q}{t}\propto \frac{1}{R}$

$\therefore \displaystyle\frac{H_1}{H_2}=\frac{R_2}{R_1}=\frac{1}{8}=1:8$ .

The thermal conductivity of a material in CGS system is 0.In steady state the rate of flow of heat

$10 \,cal/s-cm^2$ , then the thermal gradient will be

0%
$15^o C/cm$
0%
$25^o C/cm$
0%
$50^o C/cm$
0%
$75^o C/cm$

Explanation

$\dfrac{\Delta Q}{\Delta t}=\dfrac{KA\Delta T}{\Delta x}$
Thermal gradient

$\dfrac{\Delta T}{\Delta x}$ is given by

$\dfrac{\left ( \dfrac{\Delta Q}{\Delta t} \right )}{KA}=\dfrac {10}{0.4}= 25^o C/ cm$

Two metal rods, rod-1 and rod-2 of equal length are welded together end to end.Under steady state conditions, when the free end of the end of the rod-1 is kept at

$100^0C$ and the free end of the rod -2 is kept at

$0^0C$ , the temperature at the welded junction is approximately.

[Thermal conductivity of rod-1 =

$92 J s^{-1}m^{-1}K^{-1}$ , and rod-2 =

$16 J s^{-1}m^{-1}K^{-1}$ ]

0%
$85^0C$
0%
$50^0C$
0%
$75^0C$
0%
$95^0C$

Explanation

Heat Current =

$i = \dfrac{KA \Delta T}{L}$

$K_1(T_1-T) = K_2(T-T_2)$

$92(100-T)=16(T-0)$

$\implies T = 85^o C$

A body of specific heat

$0.2$ kcal

$/kg^o$ C is heated through

$100^o$ C. The percentage increase in its mass is?

0%
$9\%$
0%
$9.3\times 10^{-11}\%$
0%
$10\%$
0%
None of these

Explanation

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

$\displaystyle =\frac{\Delta m}{M}\times 100=\frac{20\times 4.2\times 10^3}{(3\times 10^8)^2}\times 100$

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

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

0%
White
0%
Rough
0%
Black
0%
Yellow

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.

0%
If $P_1 > P_2$ then $T_2 > T_1$
0%
If $V_2 > V_1$ then $T_2 < T_1$
0%
If $V_2 > V_1$ then $T_2 > T_1$
0%
If $P_1 > P_2$ then $V_2 > V_2$

Explanation

we know

$PV=nRT$

also

$PV^2=C$

$\dfrac{T^2}{P}=C_1$

$\dfrac{T_1^2}{P_1}=\dfrac{T_2^2}{P_2}$

Also

$T_1V_1=T_2V_2$

So answer B is correct.

'Thermos Flask' was invented by

0%
Dewar
0%
daimler
0%
Urey
0%
None of these

The temperature (T) of one mole of an ideal gas varies with its volume (V) as

$T = -\alpha { V }^{ 3 }+\ \beta { V }^{ 2 }$ , where

$\alpha$ and

$\beta$ are positive constants. The maximum pressure of gas during this process is

0%
$\dfrac { \alpha \beta }{ 2R }$
0%
$\dfrac { { \beta }^{ 2 }R }{ 4\alpha }$
0%
$\dfrac { (\alpha +\beta )R }{ { 2\beta }^{ 2 } }$
0%
$\dfrac { { \alpha }^{ 2 }R }{ { 2\beta } }$

Explanation

$T=-\alpha V^3+\beta V^2\\ \cfrac{PV}{R}=-\alpha V^2+\beta V^2\\ P=RV(-\alpha V+\beta)$

For maximum pressure

$\cfrac{dP}{dV}=0\\ \Rightarrow -2\alpha V+\beta =0\\ \Rightarrow V=\cfrac{\beta }{2\alpha}\\ P=R[-\alpha(\cfrac{\beta }{2\alpha})^2+\cfrac{\beta }{2\alpha}\times \beta]\\P=R[\cfrac{-\beta^2}{4\alpha}+\cfrac{\beta^2}{2\alpha}]\\P=\cfrac{R\beta^2}{4\alpha}$

A building at a temperature T(in k) is heated by an ideal heat pump which uses the amosphere at

$T_0(k)$ . What is the equilibrium temperature of the building ?

0%
$T_o+\dfrac{W}{2\alpha}+ \sqrt{T_0\dfrac{W}{\alpha}+(\dfrac{W}{2\alpha})^2}$
0%
$T_0-\dfrac{W}{2\alpha}_\sqrt{T_0\dfrac{W}{2\alpha}+(\dfrac{W}{2\alpha})^2}$
0%
$T_o+\dfrac{W}{2\alpha}- \sqrt{T_0\dfrac{W}{\alpha}+(\dfrac{W}{2\alpha})^2}$
0%
$T_0-\dfrac{W}{2\alpha}_\sqrt{T_0\dfrac{W}{2\alpha}+(\dfrac{W}{\alpha})^2}$

The radii of two spheres made of same metal are

$r$ and

$2r$ . These are heated to the same temperature and placed in the same surrounding. The ratio of rates of decrease of their temperature will be:

0%
$1 : 1$
0%
$4 : 1$
0%
$1 : 4$
0%
$2 : 1$

Explanation

Same metal

$\implies$ same density

Rate of cooling

$\dfrac { -dT }{ dt } =\dfrac { eA\sigma }{ mc } \left( { T }^{ 4 }-{ T }_{ 0 }^{ 4 } \right)$ formula used.

T is same for both spheres

${ T }_{ 0 }$ is same

L is same

$\sigma$

e is same

Here rate of cooling proportional

$\propto \dfrac { A }{ m }$

$\propto \dfrac { 4{ \pi r }^{ 2 } }{ D\times 4{ \pi r }^{ 3 } }$

Density

$\propto\dfrac { 1 }{ Dr }$ , D=Density =Mass/volume

${ R }_{ 1 }$ = rate of cooling of radius

${ r }_{ 1 }$

${ R }_{ 2 }$ = rate of cooling of radius

${ r }_{ 2 }$

$\dfrac { { R }_{ 1 } }{ { R }_{ 2 } } =\dfrac { { D }_{ 2 }{ \quad r }_{ 2 } }{ { D }_{ 1 }\quad { r }_{ 2 } } \quad \left[ { D }_{ 1 }={ D }_{ 2 }\quad same\quad density \right]$

$=2:1$

Vijay performed an experiment to compare the heat conductivity of four different materials P, Q, R and S. Thumbtacks were fixed with wax on each material and then the materials were fixed to a trough containing hot water as shown in the diagram. Vijay tabulated his results as shown here.

Material	Time taken for thumbtacks to drop(minutes)
P	$5$
Q	$6$
R	$2$
S	$4$

What can be inferred from his experiment?

0%
Material R is a better conductor of heat than material S.
0%
Material S is a better conductor of heat than material R.
0%
Material Q is a better conductor of heat than material P.
0%
Materials P, Q, R and S are equally good conductors of heat.

Explanation

Material R is better conductor of heat because it took only

$2$ minutes for the wax attached to it to melt as compared to material S which took

$4$ minutes.

Thermal radiation can be detected by.

0%
A thermometer
0%
A radiometer
0%
A bolometer
0%
All of the above

Which one of the following statements is not true regarding thermal radiations?

0%
All bodies emit thermal radiation at all temperatures
0%
They are electromagnetic in nature
0%
They travel in straight line with a velocity of $3\times 10^8$ m/s in free space
0%
They cannot be reflected by mirrors

A cube, a thin circular plate, and a sphere are made of the same metal and having the same mass and same volume. They are initially heated to the same temperature. When left in air at the room temperature, which of these bodies will cool fastest and slowest respectively?

0%
The cube; the sphere
0%
The circular plate; the cube
0%
The circular plate; the sphere
0%
The sphere; the circular plate

If the pressure and the volume of certain quantity of ideal gas are halved, then its temperature

0%
Is doubled
0%
Becomes one-fourth
0%
Remains constant.
0%
Become four times

Explanation

According to ideal gas equation

$\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}$ or

$T_2=\dfrac{P_2V_2}{P_1V_1}$
Here,

$P_1=P, V_1=V, T_1=T, P_2=\dfrac{P}{2}, V_2=\dfrac{V}{2}, T_2=?$

$\therefore T_2=\dfrac{T\left(\dfrac{P}{2}\right)\left(\dfrac{V}{2}\right)}{PV}$ ,

$T_2=\dfrac{T}{4}$ .

Two identical glass bulbs are interconnected by a thin glass tube at

$0^{\circ}C$ . A gas is filled at N.T.P. in these bulb is placed in ice and another bulb is placed in hot bath, then the pressure of the gas becomes

$1.5\ times$ . The temperature of hot bath will be

0%
$100^{\circ}$
0%
$182^{\circ}C$
0%
$256^{\circ}C$
0%
$546^{\circ}C$

Explanation

Quantity of gas in these bulbs is constant i.e,

Initial no. of moles in both bulb

$=$ final no. of moles

$\Rightarrow n_1+n_2={ n }_{ 1 }^{ ' }+{ n }_{ 2 }^{ ' }$

$\Rightarrow \dfrac{PV}{R(273)}+\dfrac{PV}{R(273)}=\dfrac{1.5PV}{R(273)}+\dfrac{1.5PV}{R(T)}$

$\Rightarrow \dfrac{2}{273}=\dfrac{1.5}{273}+\dfrac{1.5}{T}$

$\Rightarrow T=819K$

$=546°C$

Hence, the answer is

$546°C.$

When a thermometer is taken from the melting ice to a warm liquid, the mercury level rises to

$\\ \\ \left(\dfrac { 2}{ 5 }\right)^{ th }$ of the distance between the lower and the upper fixed points. The temperature of liquid in K is

0%
$217.15$
0%
$313.15$
0%
$220$
0%
$330$

Explanation

Given:

$T_b-T_f=100$

Temperature of liquid,

$T=\dfrac{2}{5}$ distance between lower and upper fixed point

Temperature of liquid,

$T=\dfrac{2}{5}\times 100=40^0C$

On kelvin scale,

$T_k=273.15+40$

$T_k=313.15K$

The correct option is B.

A cubical box of side

$0.20\ m$ is constructed from

$0.012\ m$ thick concrete panels of thermal conductivity

$K = 0.8\ W\ m^{-1} K^{-1}$ . A

$100\ W$ heater is sealed inside the box and switched on. The temperature of air outside the box is

$20^{\circ}C$ . After a long period of time, the air temperature inside the box will be close to

0%
$20^{\circ}C$
0%
$22^{\circ}C$
0%
$24^{\circ}C$
0%
$26^{\circ}C$

Explanation

Let the air temperature inside the box be

$x^oC$

So,

$x-20=\cfrac { 100\times 0.012 }{ 6\times 0.2\times 0.2\times 0.8 } \quad \quad \quad \left\{ H=\cfrac { \triangle T }{ \cfrac { l }{ kA } } \Rightarrow \triangle T=\cfrac { Hl }{ kA } \right\} \\ \Rightarrow x-20=6.25\\ \Rightarrow x=26.25℃\\ \Rightarrow x\approx 26℃$

Heat from the sun is received by the earth through

0%
Radiation
0%
Convection
0%
Conduction
0%
None of the above

two bars of same length and same cross-sectional area but of different thermal conductivities

${ K }_{ 1 }$ and

${ K }_{ 2 }$ are joined end to end as shown in the figure. One end of the compound bar is at temperature

${ T }_{ 1 }$ and the opposite end at temperature

${ t }_{ 2 }$ (where

${ T }_{ 1 }>{ T }_{ 2 }$ ). the temperature of the junction is

0%
$\dfrac { { K }_{ 1 }{ T }_{ 1 }+{ K }_{ 2 }{ T }_{ 2 } }{ { K }_{ 1 }+{ K }_{ 2 } }$
0%
$\dfrac { { K }_{ 1 }{ T }_{ 2 }+{ K }_{ 2 }{ T }_{ 1 } }{ { K }_{ 1 }+{ K }_{ 2 } }$
0%
$\dfrac { { K }_{ 1 }\left( { T }_{ 1 }+{ T }_{ 2 } \right) }{ { K }_{ 2 } }$
0%
$\dfrac { { K }_{ 2 }\left( { T }_{ 1 }+{ T }_{ 2 } \right) }{ { K }_{ 1 } }$

Explanation

Let L and A be length and area of cross-section of each bar respectively.

$\therefore$ Heat current through the bar 1 is

${ H }_{ 1 }=\dfrac { { K }_{ 2 }A\left( { T }_{ 1 }-{ T }_{ 0 } \right) }{ L }$

Here

${ T }_{ 0 }$ is junction temperature.

Heat current through the bar 2 is

${ H }_{ 1 }=\dfrac { { K }_{ 2 }A\left( { T }_{ 0 }-{ T }_{ 2 } \right) }{ L }$

at steady state,

${ H }_{ 1 }={ H }_{ 2 }$

$\therefore \quad \dfrac { { K }_{ 1 }A\left( { { T }_{ 1 } }-{ T }_{ 0 } \right) }{ L } =\dfrac { { K }_{ 2 }A\left( { T }_{ 0 }-{ T }_{ 2 } \right) }{ L }$

${ K }_{ 1 }\left( { T }_{ 1 }-{ T }_{ 0 } \right) ={ K }_{ 2 }\left( { T }_{ 0 }-{ T }_{ 2 } \right)$

${ K }_{ 1 }{ T }_{ 1 }-{ K }_{ 1 }{ T }_{ 0 }={ K }_{ 2 }{ T }_{ 0 }-{ K }_{ 2 }{ T }_{ 2 }$

${ K }_{ 1 }{ T }_{ 0 }-{ K }_{ 2 }{ T }_{ 0 }={ K }_{ 1 }{ T }_{ 1 }-{ K }_{ 2 }{ T }_{ 2 }$

${ T }_{ 0 }\left( { K }_{ 1 }+{ K }_{ 2 } \right) ={ K }_{ 1 }{ T }_{ 1 }+{ K }_{ 2 }{ T }_{ 2 }$

${ T }_{ 0 }=\dfrac { { K }_{ 1 }{ T }_{ 1 }+{ K }_{ 2 }{ T }_{ 2 } }{ \left( { K }_{ 1 }+{ K }_{ 2 } \right) }$

1028042_942008_ans_838c5638d2004884aa0b7241763cf71d.png

One Kg of a diatomic gas is at a pressure of

$8\times 10^4$ N/

$m^{2}$ . The density of the gas is

$4$ kg/

$m^{3}$ . The energy of the gas due to its thermal motion will be

0%
$3\times 10^4$ J
0%
$5\times 10^4$ J
0%
$6\times 10^4$ J
0%
$7\times 10^4$ J

Explanation

For a diatomic molecular gas, internal energy per molecule is

$U=\dfrac{5}{2}NK_BT$

$PV=NK_BT$
From equation is (i) and (ii)

$U=\dfrac{5}{2}$ PV
Here,

$P=8\times 10^4N$

$m^{-2}, \rho =4kg$

$m^{-3}$ ,

$m=1$ kg

$\therefore U=\dfrac{5}{2}\times 8\times 10^4\times \dfrac{1}{4}=5\times 10^4$ J

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?

Explanation

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.

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

$80\,cal \,g^{-1}$ )

0%
$20\,cal\,s^{-1}\,m^{-1}\,^oC^{-1}$
0%
$10\,cal\,s^{-1}\,m^{-1}\,^oC^{-1}$
0%
$40\,cal\,s^{-1}\,m^{-1}\,^oC^{-1}$
0%
$30\,cal\,s^{-1}\,m^{-1}\,^oC^{-1}$

Explanation

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

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

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

$\Rightarrow30\,cal\,s^{-1}m^{-1} \,^oC^{-1}$

Which one of the following is not an assumption of kinetic theory of gases?

0%
The volume occupied by the molecules of the gas is negligible
0%
The force of attraction between the molecules is negligible
0%
The collision between the molecules are elastic
0%
All molecules have same speed

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:

0%
$12^oC$
0%
$18^oC$
0%
$6^oC$
0%
$24^oC$

Explanation

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

$A=T_A-T=12^oC$

Two blocks with heat capacities

${ C }_{ 1 }$ and

${ C }_{ 2 }$ are connected by a rod of length l, cross-sectional are A and heat conductivity K. Initial temperature difference between the two blocks is

${ T }_{ 0 }$ . Assuming the entire system to be isolated from surroundings, heat capacity of the rod to be negligible. The temperature difference between the blocks as a function of time is.

0%
${ T }_{ 0 }exp\left( \frac { -KA\left( { C }_{ 1 }+{ C }_{ 2 } \right) t }{ { C }_{ 1 }{ C }_{ 2 } } \right)$
0%
${ T }_{ 0 }exp\left( \frac { -KA\left( { C }_{ 1 }+{ C }_{ 2 } \right) }{ { C }_{ 1 }{ C }_{ 2 } } \right)$
0%
${ T }_{ 0 }exp\left( \frac { KA\left( { C }_{ 1 }+{ C }_{ 2 } \right) t }{ { C }_{ 1 }{ C }_{ 2 } } \right)$
0%
${ T }_{ 0 }exp\left( \frac { -KA\left( { C }_{ 1 }+{ C }_{ 2 } \right) { t }^{ 2 } }{ { C }_{ 1 }{ C }_{ 2 } } \right)$

Explanation

Let T be the temperature difference between two blocks at time t.

Heat transferred per second,

$\dfrac { dQ }{ dt } =\dfrac { KAT }{ l }$

$\dots$ (i)

Also,

$dT={ dT }_{ 1 }+{ dT }_{ 2 }$

$\dots$ (ii)

Heat lost by one block is equal to the heat gained by the

other,

${ C }_{ 1 }{ dT }_{ 1 }={ C }_{ 2 }{ dT }_{ 2 }$

$\dots$ (iii)

From equations, (ii) and (iii), we get

$dT=\left( \dfrac { { C }_{ 1 }+{ C }_{ 2 } }{ { C }_{ 2 } } \right) { dT }_{ 1 }$

$\dots$ (iv)

If a block loses heat,

$dQ=-{ C }_{ 1 }{ dT }_{ 1 }$

From equation (i),

$\frac { dQ }{ dt } =-{ C }_{ 1 }\frac { { dT }_{ 1 } }{ dt } =\dfrac { KAT }{ l }$ (v)

From equations (iv) and (v), we get

$\dfrac { -{ C }_{ 1 }{ C }_{ 2 } }{ { C }_{ 1 }+{ C }_{ 2 } } \dfrac { dT }{ dt } =\dfrac { KA }{ l } T$

$\dfrac { dT }{ T } =\dfrac { -KA\left( { C }_{ 1 }+C_{ 2 } \right) }{ { C }_{ 1 }{ C }_{ 2 } } dt$

$\int _{ { T }_{ 0 } }^{ T }{ \dfrac { dT }{ T } =\dfrac { -KA\left( { C }_{ 1 }+{ C }_{ 2 } \right) }{ { C }_{ 1 }{ C }_{ 2 } } \int _{ 0 }^{ t }{ dt } }$

$\dfrac { T }{ { T }_{ 0 } } =\dfrac { -KA\left( { C }_{ 1 }+{ C }_{ 2 } \right) }{ { C }_{ 1 }{ C }_{ 2 } } t$

$T={ T }_{ 0 }exp\left( \dfrac { -KA\left( { C }_{ 1 }+{ C }_{ 2 } \right)t }{ { C }_{ 1 }{ C }_{ 2 } } \right)$

1028175_943127_ans_4303e290bb8840b4abdfe179cca60416.png

A vessel contains a mixture consisting of

${m}_{1}=7kg$ of nitrogen

$\left( { M }_{ 1 }=28 \right)$ and

${m}_{2}=11g$ of carbon dioixide

$\left( { M }_{ 2 }=44 \right)$ at temeprature

$T=300K$ and pressure

${ P }_{ 0 }=1\quad atm$ . The density of the mixture is:

0%
$1.446g$ per litres
0%
$2.567g$ per litre
0%
$3.752g$ per litre
0%
$4.572g$ per litre

Explanation

Let V is the volume of the vessel.

Now, let $p_{1}$ and $p_{2}$ be the partial pressure, then using gas law:

$p_{1}V = \dfrac{m_1}{M_1}RT\\$

$p_{2}V = \dfrac{m_2}{M_2}RT,\ p_{0} = p_{1} + p_{2}\\$

$p_{0} = \left(\dfrac{m_1}{M_1} + \dfrac{m_2}{M_2}\right)\dfrac{RT}{V}\\$

$V = \left(\dfrac{m_1}{M_1} + \dfrac{m_2}{M_2}\right)\dfrac{RT}{p_{0}}\\$

$\because \rho_{mix}=\dfrac{(m_{1} + m_{2})}{V}\\$

$\rho_{mix}=\dfrac {(m_1 + m_2)M_1 M_2} {(m_1M_2 + m_2M_1)} \times \dfrac{p_0}{RT}\\$

Substituting values,

$\rho_{mix}=\dfrac {(7 + 11) \times 28 \times 44\times 10^{-3}} {(7 \times 44 + 11\times 28))} \times \dfrac{10^{5}}{8.3 \times 300}\\$

$= 1.446 \ per \ litre$

Option A is correct.

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

0%
uniform density
0%
temperature gradient
0%
density gradient
0%
uniform temperature

CBSE Questions for Class 11 Engineering Physics Thermal Properties Of Matter Quiz 8 - MCQExams.com

0:0:1

Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Thermal Properties Of Matter Quiz 8 - MCQExams.com

0:0:1

Practice Class 11 Engineering Physics Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.