MCQ Questions for Class 11 Engineering Physics Thermal Properties Of Matter Quiz 7

A closed cubicle box made of perfectly insulating material has walls of thickness

$8$ cm and the only way for heat to enter and leave the box is through the metal plugs

$A$ and

$B$ each of cross section

$12cm^2$ and length

$8$ cm fixed in the opposite walls of the box as shown in figure. Outer surface of

$A$ is kept at

$100^oC$ while outer surface of

$B$ is at

$4^oC$ . The thermal conductivity of the material of the plugs is

$0.5cal/s/cm/\space^oC$ . A source of energy generating

$36$ cal/s is enclosed inside the box. The equillibrium temperature of the inner surface of the box (assuming that it is same at all points on the inner surface) is

0%
$38^oC$
0%
$57^oC$
0%
$76^oC$
0%
$85^oC$

Explanation

The principle of conservation of energy is used to solve this question.
Heat from

$A$ to source +

$36$ = Heat from source to

$B$ .

$\dfrac{KA}{L}(100-T)+36=\dfrac{KA}{L}(T-4)$

$T$ =Temp of source

$K$ =0.5 cal/s/cm/degC

$L=8 cm$

$A=12 cm^2$
Substituting the above values, we get

$T= 76\ ^oC$

A sample of an ideal gas occupies a volume of 'V' at a pressure 'P' and absolute temperature 'T'. The mass of each molecule is 'm'. The equation for density is

0%
mkT
0%
P/kT
0%
P/(kTV)
0%
Pm/kT

Explanation

We know that PV = nRT =(m/M)RT
Where M = Molecular weight
Now

$P\times \left(\displaystyle \frac{m}{d}\right)$ =

$\left(\displaystyle\ \frac{m}{M} \right)RT$
where d = density of gas

$\displaystyle\ \frac{p}{d}$ =

$\displaystyle\ \frac{RT}{M}$ =

$\displaystyle\ \frac{kN_{A}T}{M}$
Where

$R$ =

$kN_{A}$ , k is Boltzmann constant
But

$\displaystyle\ \frac{ M}{N_{A}}$ = m = mass of each molecule so
So,

$d$ =

$\displaystyle\ \frac{P\times m}{kT}$

An electric heater kept in vacuum is heated continuously by passing an electric current. Its temperature

0%
will go rising with the time.
0%
will stop rising after sometime as it will lose heat to the surroundings by conduction.
0%
will rise for sometime and thereafter will start falling.
0%
will become constant after some time because of loss heat due to radiation.

A liquid used as a coolant in automobile engine is

0%
alcohol
0%
engine oil
0%
water
0%
benzene

The air density at Mount Everest is less than that at the sea level. It is found by mountaineers that for one trip lasting a few hours, the extra oxygen needed by them corresponds to 30,000cc at sea level (pressure 1 atmosphere, temperature 27

$^o$ C). Assume that the temperature around Mount Everest is -

$73^o$ C and that the oxygen cylinder has a capacity of 5.2 litters. The pressure at which oxygen be filled (at site) in the cylinder is

0%
3.86 atm
0%
5.00 atm
0%
5.77 atm
0%
1 atm

Explanation

Since moles of gas must remain constant,

$\dfrac { { P }_{ 1 }{ V }_{ 1 } }{ { T }_{ 1 } } =\dfrac { { P }_{ 2 }{ V }_{ 2 } }{ { T }_{ 2 } }$

$\dfrac { 1(30000) }{ 300 } =\dfrac { P(5200) }{ 143 }$

$P=3.86atm$

On which one of the factors does the nature of the thermal radiation depend inside an enclosure?

0%
size of the enclosure
0%
temperature
0%
nature of the walls
0%
color of the walls

A vessel contains air at a temperature of

$15^{0}C$ and 60% R.H. What will be the R.H if it is heated to

$20^{0}C$ ? (S.V.P at

$15^{0}C$ is 12.67 & at

$20^{0}C$ is 17.36mm of Hg respectively)

0%
262%
0%
26.2%
0%
44.5%
0%
46.2%

Explanation

By definition,

$\displaystyle\ \frac{V.P }{S.V.P }\times100$ =

$\displaystyle\ \frac{60}{100}$ (both VP and SVP are in room temperature)

$= \displaystyle\ \frac{(V.P)_{15}}{12.67}$

$(V.P)_{15} = 7.6 mm$

$of$

$Hg$
Now since unsaturated vapour obeys gas equation & mass of water remains constant so

$P$ =

$\left (\displaystyle\ \frac{nRT}{V} \right)$

$\Rightarrow$

$P\ \alpha \ T$

$\displaystyle\ \frac{(V.P)_{15}}{(V.P)_20}$

$= \displaystyle\ \frac{273+15}{273+20}$

$\Rightarrow$

$(RH)_{20}$ = 7.73 mm of Hg
So

$(R.H)_{20} =\displaystyle\ \frac{(V.P)_{20}}{(S.V.P)_{20}}\times 100$

$= 44.5$ %

Final volume of the system will be nearly -

0%
$6.2m^{3}$
0%
$2.8m^{3}$
0%
$4.5m^{3}$
0%
$1.4m^{3}$

Explanation

$PV = nRT$
Initial volume

$= \displaystyle\ \frac{ 1.45\times10^{3}}{29} \times \displaystyle\ \frac{8.3\times300}{1.5\times10^{5}}m^{3}$

$\displaystyle\ \frac{V_{i}}{P_{i^{2}}}$ =

$\displaystyle\ \frac{V_{f}}{P_{f}^{2}}$

$\Rightarrow$

$V_{f}$ =

$\left (\displaystyle\ \frac{3.5}{1.5}\right)^{2} V_{i}$ =

$4.5 m^{3}$

$T = \displaystyle\ \frac{P_{f}V_{f}}{nR} = 3800K$

$\triangle U = nC_{V}\triangle T = n \displaystyle\ \frac{5R}{2} \triangle T = 3.6\times10^{6}J$

A vessel has 6g of hydrogen at pressure P and temperature 500K. A small hole is made in it so that hydrogen leaks out. How much hydrogen leaks out if the final pressure is

$P/2$ and temperature falls to 300K?

0%
2g
0%
3g
0%
4g
0%
1g

Explanation

$PV$ =

$\displaystyle\ \frac{m}{M}RT$
Initally,

$PV$ =

$\displaystyle\ \frac{6}{M}R\times500$
Finally,

$\displaystyle\ \frac{P}{2}V$ =

$\displaystyle\ \frac{(6-x)}{M}R\times300$ ( if x g gas leaks out)
Hence,

$2 = \displaystyle\ \frac{6}{6-x}\times\displaystyle\ \frac{5}{3}$

$\therefore x = 1$ gram

A graph is plotted with PV/T on y-axis and mass of the gas along x-axis for different gases. The graph is

0%
a straight line parallel to x-axis for all the gases
0%
a straight line passing through origin with a slope having a constant value for all the gases
0%
a straight line passing through origin with a slope having different values for different gases
0%
a straight line parallel to y-axis for all the gases

Explanation

$\displaystyle\ \frac{PV}{T}$ = nR =

$\left( \displaystyle\ \frac{m}{M}\right)R$
Or,

$\displaystyle\ \frac{PV}{T} = \left (\displaystyle\ \frac{R}{M}\right)m$
i.e.,

$\displaystyle\ \frac{PV}{T}$ versus on graph is straight line passing through origin with slope R/M. i.e., the slope depends on molecular mass of the gas M and is different for different gases

If pressure of a gas contained in a closed vessel is increased by 0.4% when heated by

$1^{0}C$ , the initial temperature must be

0%
250K
0%
$250^{0}C$
0%
2500K
0%
$25^{0}C$

Explanation

We know that

$\displaystyle\ \frac{P_{1}}{T_{1}}$ = constant for constant volume.

Let the initial pressure be

$P_1$ and initial temperature be

$T_1$ .

Thus final pressure

$P_2 =1.004 P_1$ and final temperature

$T_2 = T_1 + 1$ .

Using

$\dfrac{P_1}{T_1} = \dfrac{P_2}{T_2}$
So,

$\displaystyle\ \frac{P_{1}}{T_{1}}$ =

$\displaystyle\ \frac{1.004 P_{1}}{T_{1}+1}$

$T_1 + 1 = 1.004 T_1$

$0.004 T_1 = 1$

$\Rightarrow T_1 =250K$

An ideal gas is found to obey an additional law

$VP^{2}$ = constant. The gas is initially at temperature T and volume V. When it expands to a volume 2V, the temperature becomes :

0%
$\displaystyle\ \frac{T}{\sqrt{2}}$
0%
$2T$
0%
$\displaystyle\ \frac{2T}{\sqrt{2}}$
0%
$4T$

Explanation

We are given an ideal gas which is following an additional law,

$PV^2=$ a constant ....................(1)

As the gas is ideal, the gas law is

$PV=nRT$

Taking the value of

$P$ from the above equation, we get

$P=\dfrac{nRT}{V}$

Putting this value in equation (1), we get

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

$n$ and

$R$ are constant, above equation becomes,

$TV=$ a constant .....................(2)

Now, the gas is expanding from initial volume

$V_1$ to final volume

$V_2$ by

$2$ units, it becomes

$V_2=2V_1$

From equation (2), we get

$T_1 V_1 = T_2 V_2$

$T_1 V_1 = T_2 (2V_1)$

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

So the correct answer is

$\dfrac{T}{2}$

The density of water at

$4^{\circ}C$ is

$1000\:kg\:m^{-3}$ and at

$104^{\circ}C$ it is

$958.4\:kg\:m^{-3}$ . The cubic expansivity of water between these temperature is

0%
$4.16\times 10^{-5}(K)^{-1}$
0%
$2.08\times 10^{-5}(^{\circ}C)^{-1}$
0%
$4.16\times 10^{-4}(^{\circ}C)^{-1}$
0%
$2.08\times 10^{-4}(K)^{-1}$

Explanation

$\rho =\rho _{0}(1-\gamma \Delta T)$

$\therefore \displaystyle \gamma =\frac{\rho _{0}-\rho }{\rho _{0}\Delta T}=\frac{1000-958.4}{(100)(1000)}$

$=4.16\times 10^{-4}(^{\circ}C)^{-1}$

Two containers A and B contain liquid at 50C. Two metal balls one at 50C and the other at 30C are dropped in the containers A and B respectively. In which container will the transfer of heat take place?

0%
$B$
0%
$A$
0%
$A$ & $B$
0%
None of these

A glass of cold water whose particles had a low average kinetic energy was placed on a table. The average kinetic energy in the cold water increased, while the average kinetic energy of the part of the table under the glass decreased. What do you think happened?

0%
Thermal energy was transferred from the table to the glass of cold water.
0%
Thermal energy was transferred from the glass of cold water to the table.
0%
Entropy was transferred from glass of water to table
0%
None of these.

In the diagram shown, will the convection current be clockwise or anticlockwise on the left part of the figure?

0%
Anticlockwise
0%
Clockwise
0%
Will keep on changing
0%
Can't be said

When the specific heat of a solid is measured by the method of mixture, the heat is lost to surroundings from the calorimeter by the process of

0%
radiation
0%
conduction
0%
convection
0%
both conduction and convection

The amount of heat required to raise the temperature of 100g of copper (sp. heat = 390 J/kg

$^o$ C) from

$40^o$ to

$90^o$ is

0%
1950 J
0%
2350 J
0%
2550 J
0%
None of these

Thermal energy is always transferred from .............. to ............. temperatures

0%
Lower, higher
0%
Higher, lower
0%
Both $A$ and $B$
0%
Can't be determined

A ........... is a device used to measure heat

0%
Calorimeter
0%
Barometer
0%
Lactometer
0%
None of these

When you heat a bowl of water, we can observe that water starts circulating. This demonstrates:

0%
conduction
0%
convection
0%
radiation
0%
none

A gas behaves more closely like an ideal gas at

0%
low pressure and low temperature
0%
low pressure and high temperature
0%
at all pressure and temperature
0%
none of these

If T represents the absolute temperature of an ideal gas, the volume coefficient of thermal expansion at constant pressure is proportional to:

0%
T
0%
T $^2$
0%
1/T
0%
1/T $^2$

Explanation

Coefficient of thermal expansion at constant pressure is:

$\implies { \alpha }_{ V }=\dfrac { 1 }{ V } \dfrac { dV }{ dT }$

$PV=nRT$

$V=\dfrac { nRT }{ P }$

${ \alpha }_{ V }=\dfrac { P }{ nRT } \dfrac { dV }{ dT }$

${ \alpha }_{ V }\rightarrow \dfrac { 1 }{ T }$

If two conducting slabs of thickness

$d_1$ and

$d_2$ , and thermal conductivity

$K_1$ and

$K_2$ are placed together face to face as shown in figure in the steady state temperatures of outer surfaces are

$O_1$ , and

$O_2$ . The temperature of common surface is

0%
$\displaystyle{\frac{K_1O_1d_1 + K_2O_2d_2}{K_1d_1 + K_2d_2}}$
0%
$\displaystyle{\frac{K_1O_1 + K_2O_2}{K_1 + K_2}}$
0%
$\displaystyle{\frac{K_1O_1 + K_2O_2}{O_1 + O_2}}$
0%
$\displaystyle{\frac{K_1O_1d_2 + K_2O_2d_1}{K_1d_1 + K_2d_2}}$

Explanation

In steady state of temperature , the rate of flow of heat in both the slabs will be same .

Let

$\theta$ be the temperature of common surface ,

then ,

$Q/t=K_{1}A(O_{1}-\theta)/d_{1}=K_{2}A(\theta-O_{2})/d_{2}$ ,

$\dfrac{K_{1}}{d_{1}}.(O_{1}-\theta)=\dfrac{K_{2}}{d_{2}}.(\theta-O_{2})$ ,

$\dfrac{K_{1}O_{1}}{d_{1}}+\dfrac{K_{2}O_{2}}{d_{2}}=\theta(\dfrac{K_{1}}{d_{1}}+\dfrac{K_{2}}{d_{2}})$ ,

$\theta=\dfrac{K_{1}O_{1}d_{2}+K_{2}O_{2}d_{1}}{K_{1}d_{1}+K_{2}d_{2}}$

Two models of a windowpane are made. In one model, two identical glass panes of thickness 3 mm are separated with an air gap of 3 mm. This composite system is fixed in the window of a room. The other model consists of a single glass pane of thickness 6 mm, the temperature difference being the same as for the first model. The ratio of the heat flow for the double pane to that for the single pane is (

$K_{glass}\, =\, 2.5\, \times\, 10^{- 4}$ cal / m.

$^{\circ}C$ and

$K_{air}\, =\, 6.2\, \times\, 10^{- 6}$ cal/s.m.

$^{\circ} C$ )

0%
1/20
0%
1/70
0%
31/1312
0%
31/656

Explanation

Thermal Resistance

$R_A= 2R_g+R_{Air}$

Thermal Resistance

$R_B=\dfrac { 6mm }{ { K }_{ g }A }$

$RA=\dfrac { 2\times 3mm }{ { K }_{ g }A } +\dfrac { 3mm }{ { K }_{ air }A }$

The ratio of heat flow

$\dfrac { { i }_{ A } }{ { i }_{ B } } =\dfrac { \Delta T\times { R }_{ b } }{ { R }_{ A }\times \Delta T }$

$=\dfrac { 6 }{ { K }_{ g }A\times \left( \dfrac { 6 }{ { K }_{ g }A } +\dfrac { 3 }{ { K }_{ air }A } \right) }$

$=\dfrac { 2 }{ { K }_{ g }\left( \dfrac { 2 }{ { K }_{ g } } +\dfrac { 1 }{ { K }_{ air } } \right) }$

$=\dfrac { 2\times { K }_{ air } }{ { { 2K }_{ air }+K }_{ g } }$

$=\dfrac { 2\times 6.2\times { 10 }^{ -6 } }{ 2\times 6.2\times { 10 }^{ -6 }+2.5\times { 10 }^{ -4 } }$

$=\dfrac { 2\times 6.2\times { 10 }^{ -2 } }{ 2\times 6.2\times { 10 }^{ -2 }+2.5 }$

$=\dfrac { 0.124 }{ 0.124+2.5 } =\dfrac { 0.124 }{ 2.624 }$

$=\dfrac { 124 }{ 2624 }=\dfrac { 31 }{ 656 }$

Three copper rods and three steel rods each of length

$l$ = 10 cm and area of cross - section

$1\, cm^2$ are connected as shown If ends A and E are maintained at temperatures

$125^{\circ}$ C and

$o^{\circ}$ C respectively, calculate the amount of heat flowing per second from the hot to cold function.

$[K_{cu}\, =\, 400\, W/m-K,\, K_{steel}\, =\, 50\, W/m-K]$

0%
2 watt
0%
9 watt
0%
4 watt
0%
10 watt

Explanation

$R_{steel}\, =\, \displaystyle \frac{L}{KA}\, =\,\frac{10^{-1}m}{50(w/m-^{\circ}C)\, \times\, 10^{-4}m^2}\, =\, \frac{1000}{50}\, ^{\circ} C/W$
Similarly

$R_cu\, =\,\displaystyle \frac{1000}{50}\, ^{\circ} C/W$
Junction C and 0 are identical in every respect and both will have same temperature. Consequently.the rod CD is in thermal equilibrium and no heat will flow through it. Hence it can be neglected in further analysis.Now rod BC and CE are in series their equivalent resistance is

$R_1 \, =\, R_s\, +\, R_{cu}$ similarly rods SO and DE are in series with same equivalent resistance

$R_1 \, =\, R_s\, +\, R_{cu}$ these two are in parallel giving an equivalent resistance of

$\displaystyle \frac{R_1}{2}\, =\, \frac{R_s\, +\,R_{cu}}{2}$
This resistance is connected in series with rod AB. Hence the net equivalent of the combination is

$R\, =\, R_{steel}\, +\, \displaystyle \frac{R_1}{2}\, =\, \frac{3R_{steel}\, +\, R_{cu}}{2}\, =\, 500 \left ( \frac{3}{50}\, +\, \frac{1}{400} \right ) ^{\circ}C/W$
Now

$i\, =\, \displaystyle \frac{T_H\, -\,T_c}{R}\, =\,\frac{125^{\circ}C}{{500\left ( \frac{3}{50}\, +\,\frac{1}{400} \right )}\,^{\circ}C/W}\, =\,4\, watt$

A manufacture marks the thermometer wrongly. At 0

$^o$ C it reads -10

$^o$ C, at 100

$^o$ C it reads 85

$^o$ C. Then the reading at 50

$^o$ C will be :

0%
40 $^o$ C
0%
32.5 $^o$ C
0%
37.5 $^o$ C
0%
42.5 $^o$ C

Explanation

From the interpolation we can get the value of x.

$\dfrac { { 100 }^{ o }-{ 50 }^{ o } }{ { 85 }^{ o }-{ x }^{ o } } =\dfrac { { 50 }^{ o }-{ 0 }^{ o } }{ { x }^{ o }-{ 10 }^{ o } }$

${ x }^{ o }={ 37.5 }^{ o }$

Two identical beakers with negligible thermal expansion are filled with water to the same level at

$^o$ C. If one say A is heated while the other B is cooled, then :

0%
water level in A must rise
0%
water level in B must rise
0%
water level in A must fall
0%
water level in B must fall

Explanation

$Answer:-$ A,B

Thermal expansion is the tendency of matter to change in shape, area, and volume in response to a change in temperature, through heat transfer. Temperature is a monotonic function of the average molecular kinetic energy of a substance. When a substance is heated, the kinetic energy of its molecules increases.. The unusual behaviour of water between 0°C and 4°C i.e., contraction of water when heated from 0°C to 4°C is known as the anomalous expansion of water.

The density is the mass per unit volume of a substance and that mass of a given quantity of a substance does not change. So when volume changes density changes. Since water contracts form 0°C to 4°C,its volume decreases and the density increases to reach the maximum value at 4°C. When water is cooled from 100°C it contracts till it reaches 4°C; then it expands.

Since water contracts on melting and on heating from 0°C to 4°C

So in both beaker water must rise.

Absolute zero corresponds to :

0%
$- 273 K$
0%
$- 273^{\circ}C$
0%
$273^{\circ} F$
0%
None of these

Explanation

Hint: Absolute temperature is defined as 0K.

Step 1: $\textbf{Explanation:}$

We know that absolute temperature is $0K$ .

So now we will covert this temperature in Kelvin to degree Celsius, we know that ${{T}_{K}}=273+{{T}_{C}}$ (Where ${{T}_{K}}$ is temperature in Kelvin and ${{T}_{C}}$ is temperature in Celsius.)

$\Rightarrow 0=273+{{T}_{C}}$

$\Rightarrow {{T}_{C}}=-273{{C}^{\circ }}$

$\textbf{Thus, option (B) is correct.}$

A brass boiler has a base area of

$0.15 m^2$ and thickness 1.0 cm. It boils water at the rate of 6.0 kg/min when placed on a gas stove. The temperature (approximately) of the part of the flame in contact with the boiler is (Thermal conductivity of brass

$= 107.4Js^{-1}m^{-1} {\;} ^oC^{-1}$ , Heat of vaporization of water

$= 2256\times 10^3J kg^{-1})$ .

0%
2400 K
0%
$240^oC$
0%
$2400^oC$
0%
$24000^oC$

Explanation

Rate of supply of heat of water = Rate at which is conducted by boiler

$\Rightarrow L=\cfrac { dm }{ dt } =\cfrac { d\theta }{ dt } =\cfrac { kA(\theta -100) }{ t } \\ \Rightarrow \cfrac { 6\times 2256\times { 10 }^{ 3 } }{ 60 } =\cfrac { 107.4\times 0.15\times (\theta -100) }{ { 10 }^{ -2 } } \\ \Rightarrow 225600=\cfrac { 107.4\times 0.15\times (\theta -100) }{ { 10 }^{ -2 } } \\ \Rightarrow \theta -100=140.3\\ \Rightarrow \theta =240℃$

Which of the following curve represent variation of density of water with temperature best :

Explanation

$Answer:-$ D

This is due to anomalous expansion of water.

The anomalous expansion of water is an abnormal property of water whereby it expands instead of contracting when the temperature goes from $4^0$ C to $0^0$ C, and it becomes less dense. The density becomes less and less as it freezes because molecules of water normally form open crystal structures when in solid form.

So, maximum density of water is observed at

$4^0C$

The volume of a perfect gas at NTP is

0%
22.4 litres
0%
2.24 litres
0%
100 litres
0%
None of these

Explanation

At NTP

Pressure

$=1$ atm

Temperature

$=293k$

According to ideal gas equation

$PU=nRT$

$\Rightarrow V=\dfrac{1\;mol\;(0.08205)293}{1\;atm}$

$\Rightarrow (For\;n=1)\;V=22.4\;L$

Hence, the answer is

$22.4\;L.$

A thermometer uses 'density of water' as thermometric property. The actual reading in the thermometer is 'height of water' (h) which is inversely proportional to density of water (d). In a certain temperature range, density of water varies with temperature as shown. The graph is symmetric about the maximum
Two identical bodies ( of same mass and specific heat ) at different temperatures

$T_1\, and\, T_2$ show the same reading of height

$h_1\, =\,h_2$ the thermometer. The bodies are brought into contact and allowed to reach thermal equilibrium. The thermometer reading 'height of water for final equilibrium state

$h_1$ satisfies.

0%
$h_f\, = \displaystyle \frac{h_1\, +\, h_2}{2}\, =\, h_1\, =\, h_2$
0%
$h_f\, <\, h_1\, =\, h_2$
0%
$h_f\, >\, h_1\, =\, h_2$
0%
$h_f$ may be greater or less than $h_1\, =\, h_2$ , depending on the specif heat of the bodies
0%
Information is not enough.

The boiling point of mercury is 367

$^o$ C. A mercury thermometer can be used to measure a temperature of 500

$^o$ C;

0%
by filling the space above mercury with oxygen at high pressure
0%
by filling the space above mercury with nitrogen at low pressure
0%
by filling the space above mercury with nitrogen at high pressure
0%
by keeping the space above mercury as vacuum

A quantity of air

$(\gamma = 1.4)$ at 27

$^o$ C is compressed suddenly, the temperature of the air system will

0%
fall
0%
rise
0%
remain unchanged
0%
first rise and then fall

The temperature to which a gas must be cooled before it can't be liquefied by pressure alone is called its

0%
Critical temperature
0%
Saturation point
0%
Boiling point
0%
Freezing point

Explanation

$Answer:-$ A

In this P-V curve the isotherm having

$T_c$ is that critical temperature and corresponding to that there is critical pressure

$P_c$ .

Critical temperature is the temperature of a gas in its critical state, above which it cannot be liquefied by pressure alone.

Critical pressure is the pressure of a gas or vapour in its critical state.

Which of the following curve represent variation of density of water with temperature best:

Real gases obey ideal gas laws more closely at:

0%
low pressure and low temperature
0%
low pressure and high temperature
0%
high pressure and low temperature
0%
high pressure and high temperature

A manufacture marks the thermometer wrongly.So At

$0^{\circ}C$ it reads

$10^{\circ}C$ , at

$100^{\circ}C$ it reads

$85^{\circ}C$ . Then the reading at

$50^{\circ}C$ will be :

0%
$40^{\circ}C$
0%
$32.5^{\circ}C$
0%
$37.5^{\circ}C$
0%
$42.5^{\circ}C$

Explanation

$0^o C$ there is a difference of

$10^o C$
At

$100^o C$ there is a difference of

$15^o C$
Hence, at

$50^o C$ there should be a difference of

$12.5^o C$ i.e. 50-12.5=

$37.5^o C$

Two gases

$A$ and

$B$ having the same temperature

$T$ , same pressure

$P$ and same volume

$V$ are mixed. If the mixture is at the same volume

$V$ , the pressure of the mixture is

0%
$P$
0%
${P}/{2}$
0%
$2P$
0%
$4P$

Explanation

From Ideal Gas Law,

$PV=nRT$

$\implies$ Moles of a gas=

$n=\dfrac{PV}{RT}$

When the mixture is created, the total number of moles is equal to the sum of that of individual gases.

Hence

$N=n_1+n_2$

$\implies \dfrac{P'V}{RT}=\dfrac{PV}{RT}+\dfrac{PV}{RT}$

$\implies P'=2P$

A temperature of

$15^o$ C on the Fahrenheit scale is ________________.

0%
$59^0$ F
0%
$27^0$ F
0%
$-27^0$ F
0%
$-59^0$ F

Explanation

Solution: There are two different units for the measurement of the temperature. the first one is Celsius and the other one is Fahrenheit.

The relation between the two units is given by,

$F=\dfrac{9}{5}C+32$ (1)

where C is the temperature in Celsius and F is the temperature in Fahrenheit.

On Substituting the value of

$C=15^{0}$ in equation (1) we get,

$F=\dfrac{9}{5}(15)+32$

$59^{0}$

So, temperature in Fahrenheit is

$59^{0}F$ .

Hence, the correct option is (A).

The volume of mole of a prefect gas at NTP is ______.

0%
22.4 litres
0%
2.24 litres
0%
100 litres
0%
none of these

Explanation

At NTP is normal temperature and pressure.

$\Rightarrow$ Pressure

$=1\;atm\;;T=293k\;;R=0.08205\dfrac{L\;atm}{mol}$

$\Rightarrow V=\dfrac{RT}{P}$

$\Rightarrow V=\dfrac{(0.08205)293}{1\;atm}$

$\Rightarrow V_{molar}=22.4\;L$

Hence, the answer is

$22.4\;L.$

At 4

$\displaystyle ^{\circ}C$ , the density of water is _____

$\displaystyle kg/m^{3}$

0%
1000
0%
1
0%
200
0%
0.1

Explanation

Graph showing variation of density of water with temperature in the

$0^oC$ to

$10^oC$ range, it is found that the density decreases when water at

$4^oC$ is heated further. The density also decreases when water at

$4^oC$ is cooled further.

Hence we can say water has maximum density at

$4^oC$ . The density of water is

$1g/cm^3$ or

$1000kg/m^3$ at

$4^oC$ .

Graph showing variation of density of water with temperature in the

$0^oC$ to

$10^oC$ range.

When an inflated tyre bursts, the air escaping out __________.

0%
will get heated up
0%
will be cooled
0%
will not undergo any change in its temperature
0%
will be liquefied

Explanation

Bursting of a tyre is an adiabatic expansive i.e,

$dQ=0$

$\&$

$dV=+ve$

So, According to first law of thermodynamic

$dQ=dV+PdV$

$\Rightarrow -PdV=dV=CvdT$

$P$ is

$+ve$

$dV$ is

$+V$ thus

$dV=-ve$

So,

$dT$ is

$-ve$ means decrease in temperature.

Hence, the answer is will be cooled.

The S.I unit of heat is

0%
Joule
0%
Calorie
0%
Newton
0%
None of these

State whether true or false.

When water is heated from 0

$^oC$ , it continuously expands.

0%
True
0%
False

Explanation

Water has a unique property of anomalous expansion. When water is heated from

$0^oC$ to

$4^oC$ it contracts continuously instead of expanding. That means its volume decreases upto

$4^oC$ . When water heated from

$4^oC$ its expansion becomes normal like the expansion of other liquids.

Two blocks of steel A and B, A being two times heavier than B, are at 40

$^o$ C. The ratio of heat content of A to B is:

0%
1
0%
4
0%
2
0%
$\displaystyle \frac{1}{2}$

Explanation

Let the mass of block B be

$m$ .

So, mass of block A is

$2m$

Both are made of steel, so, both has same specific heat capacity (let 'S')

So,

$\cfrac{Heat\;Capacity\;(A)}{Heat\;Capacity\;(B)}=\cfrac{2ms}{1ms}=2$

Alcohol expands more than mercury. True/False

0%
True
0%
False

Alcohol thermometers can be used for finding high temperatures

0%
True
0%
False

The figure shows that the level of water in the round bottom flask was initially at A. When heated the level fell to B and finally rises to C. Which of the following statements justify this observation

0%
That's how all liquids behave
0%
Anomalous expansion of water
0%
The dissolved gases come out decreasing the volume
0%
None of these

Explanation

Anomalous expansion of water is its unique property. When water is cooled from room temperature it first contracts in volume and becomes increasingly dense as do other liquids, but at

$4^oC$ water reaches its maximum density. On further cooling it starts expanding and becomes less dense. This will be continued till the temperature reaches

$0^oC$ , a point where it freezes into ice.

CBSE Questions for Class 11 Engineering Physics Thermal Properties Of Matter Quiz 7 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Thermal Properties Of Matter Quiz 7 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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