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

A Centigrade and Fahrenheit thermometers are dipped in boiling water. The water temperature is lowered unit the Fahrenheit thermometre registers a temperature of

$140^o$ . The fall of temperature as registered by the Centigrade thermometre is :

0%
$80^o$ C
0%
$40^o$ C
0%
$50^o$ C
0%
$90^o$ C

A sample of gas expands from volume

$V _ { 1 }$ to

$V _ { 2 } .$ The amount of work done by the gas is greatest when the expansion is

0%
adiabatic
0%
isobaric
0%
isothermal
0%
Equal in all above cases

Explanation

$\begin{array}{l} \therefore \, \, w=\int { pdv } \\ Ans.\, \, (B) \end{array}$
1380421_1532617_ans_7449d13bacdc4e649fd7c00a8b61b00a.png

End

$A$ of a copper rod is. maintained in steam chamber and other end is maintained at

$0^{\circ}C$ . Assume

$x = 0$ at

$A$ . The

$T - x$ graph for the rod in steady state is

0%
0%
0%
0%
None of these

Heat current is maximum in which of the following (rods are of identical dimension)

0%
Copper
0%
Copper steel
0%
Steel Copper
0%
Steel

Explanation

Thermal resistance of

$Cu$ is lesser than the thermal resistance of

$steel$ . Hence ,only in option (a), thermal resistance is minimum so heat current is maximum .

An ideal gas is initially at

$P_{l}, V_{1}$ is expanded to

$P_{2}, V_{2}$ and then compressed adiabatically to the same volume

$V_{l}$ and pressure

$P_{3}$ .If W is the net work done by the gas in complete process which of the

0%
$W>0 ; P_{3}>P_{1}$
0%
$W<0 ; P_{3}>P_{1}$
0%
$W<0;{P_3}$
0%
$P_{2}>P_{1}$ ; $W<0$

Explanation

Hence, option

$(D)$ is correct answer.
1380775_1500553_ans_e7eac230606548cabd9bb4a5018acc59.jpg

A small quantity, mass m, of water at a temperature

$\theta$ in (

$^oC$ ) is poured on to a large mass M of ice which is at its melting point. If c is the specific heat capacity of water and L the latent heat of fusion of ice, then the mass of ice melted is given by :

0%
$\dfrac{ML}{mc\theta}$
0%
$\dfrac{mc\theta}{ML}$
0%
$\dfrac{Mc\theta}{L}$
0%
$\dfrac{mc\theta}{L}$

A fixed mass of gas is taken through a process

$A\rightarrow B\rightarrow C\rightarrow A$ . Here

$A\rightarrow B$ is isobaric

$B\rightarrow C$ is adiabatic and

$C\rightarrow A$ is isothermal

The pressure at

$C$ is given by

$(\gamma=1.5)$

0%
$\frac{10^5}{64}\,N/m^2$
0%
$\frac{10^5}{32}\,N/m^2$
0%
zero
0%
${10}^5\,N/m^2$

$C_p$ and

$C_v$ are molar specific heats of an ideal gas at constant pressure and volume respectively. If

$\gamma$ is ratio of two specific heats and

$R$ is universal gas constant then

$C_p$ is equal to?

0%
$\dfrac{R\gamma}{\gamma -1}$
0%
$\gamma R$
0%
$\dfrac{1+\gamma}{1-\gamma}$
0%
$\dfrac{R}{\gamma -1}$

Explanation

Given:

$\dfrac{C_p}{C_v} = \gamma$

We know,

$C_p - C_v = R$ .........................

$(1)$

Here,

$C_p$ and

$C_v$ are molar specific heat capacities of an ideal gas at constant pressure and volume and

$R$ is the universal gas constant.

Dividing both sides by

$C_p$ in equation

$(1)$ .

$1 - \dfrac{1}{\gamma} = \dfrac{R}{C_p}$

$\dfrac{\gamma - 1}{\gamma} = \dfrac{R}{C_p}$

$\therefore C_p = \dfrac{γ R}{\gamma-1}$

$2$ mole ideal He gas and

$3$ mole ideal

$H_2$ gas at constant volume find out

$C_v$ of mixture

0%
$\dfrac{21 R}{10}$
0%
$\dfrac{11 R}{10}$
0%
$\dfrac{21 R}{5}$
0%
$\dfrac{11 R}{5}$

Explanation

The given mixture of two moles of monoatomic gas mixed with three mole of diatomic gas can be given as

$C=\dfrac{n_1C_{v_1}+n_2C_{v_2}}{n_1+n_2}$

$=\dfrac{2\times \dfrac{3}{2}R+3\times \dfrac{5}{2}R}{2+3}$

$=\dfrac{3R+15\dfrac{R}{2}}{5}=\left(\dfrac{21 R}{10}\right)$

What happens in the process of thermionic emission?

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Complete molecules are emitted from the surface of a hot liquid
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Single atoms are emitted from the surface of a hot liquid
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Electrons are emitted from the surface of a hot metal
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Protons are emitted from the surface of a hot metal

Two plastic cups are placed one inside the other. A small spacer keeps the two cups separated. How water is poured into the inner cup and a lid is put on top, as shown.
Which statement is correct?

0%
The bench is heated by conduction from the bottom of the outer cups
0%
The lid reduced the energy gain by convection
0%
There is no thermal conduction through the sides of either cup
0%
Thermal radiation is prevented by the small air gap

r.m.s. speed of ideal gas at

$127^oC$ is

$200m/s$ , the r.m.s. speed of same ideal gas at temperature

$227^oC$ is:

0%
$100\sqrt{5}$
0%
$200\sqrt{5}$
0%
$100\sqrt{15}$
0%
$100\sqrt{10}$

Explanation

$v = \sqrt{\dfrac{3RT}{M}}$

$\dfrac{v_1}{v_2} = \sqrt{\dfrac{T_1}{T_2}} = \dfrac{200}{v_2} = \sqrt{\dfrac{400}{500}}$

$v_2 = 100\sqrt{5}$

Two moles of an ideal gas with

$\dfrac{C_P}{C_V}=\dfrac{5}{3}$ are mixed with

$3$ moles of another ideal gas with

$\dfrac{C_P}{C_V}=\dfrac{4}{3}$ . The value of

$\dfrac{C_P}{C_V}$ for the mixture is :

0%
$1.47$
0%
$1.45$
0%
$1.42$
0%
$1.50$

Explanation

Let

$n_1,n_2$ be the number of moles of gases

$1$ and

$2$ .

We have

$\dfrac{C_{p_1}}{C_{v_1}}=\dfrac{5}{3}$ and We know

$C_{p_1}=C_{v_1}+R$

$\therefore \dfrac{5C_{v_1}}{3}=C_{v_1}+R,\dfrac{2C_{v_1}}{3}R_1,C_{v_1}=\dfrac{3R}{2}$

$A$ and

$C_{p_1}=C_{ v_ 1 }+R=\dfrac{5R}{2}$

Also,

$\dfrac{C_{ p_ 2}}{C_{ v_ 2 }}=\dfrac{4}{3}$ and

$C_{p_2}=C_{v_2}+R$

$\therefore \dfrac{4C_{ v_ 2 }}{3}=C_{ v_ 2 }+R, \dfrac{C_{ v_ 2 }}{3}=R,C_{ v_ 2 }=3R$

$C_{ p_ 2}=C_{ v_ 2 }+ R=4R$

Now

$C_{ v_ {mix} }=\dfrac{n_1C_{ v_ 1 }+n_2C_{ v_ 2 }}{n_1+n_2}=\dfrac{2\times \dfrac{3R}{2}+3\times 3R}{2+3}$

$=\dfrac{3R+9R}{5}=\dfrac{12R}{5}$

$C_{ p_ {mix} }=\dfrac{n_1C_{ p_ 1 }+n_2C_{ p_ 2 }}{n_1+n_2}=\dfrac{2\times \dfrac{5R}{2}+3\times 4R}{2+3}$

$=\dfrac{17R}{5}$

$l_{mix}=\dfrac{C_{ p_ {mix} }}{C_{ v_ {mix} }}=\dfrac{17}{12}=1.4166=1.42$

Option (C) is correct.

Pressure of an ideal gas is increased by keeping temperature constant. The kinetic energy of molecules.

0%
Decreases
0%
Increases
0%
Remains same
0%
Increases or decreases depending on the nature of gas

In which of the following phenomena do heat waves travel along a straight line with the speed of light ?

0%
Thermal conduction
0%
Thermal convection
0%
Thermal radiation
0%
Both, thermal conduction and radiation

Water does not freeze at the bottom of the lakes in winter because :

0%
ice is a good conductor of hear
0%
ice reflects heat and light
0%
of anomalous expansion of water between $4^oC$ to $0^oC$
0%
nothing can be said

Explanation

lets look at the freezing process of water in lakes :

The top layer of water which is exposed to surrounding emits heat energy to the surrounding due to which temperature of water lower , if we divide the process in two parts:

$T>4^oC$ : when the temperature of top layer lowers its density increases so momentarily its density is more than densities of bottom layers therefore the top layer of water sinks down

$T\leq 4^oC$ : now when top layers emits heat its temperature lower but density decreases so it does not sinks down and remains at the top and when freezing process will start at

$0^oc$ it will be from top to bottom.

The ice formed on the surface is less dense than water and floats at the water surface. This layer acts as an insulation layer which prevents contact between the water bulk and the freezing surrounding, thus preventing further freezing of water

1533974_1702958_ans_2e5d3522bbc24398b14bbf81be666586.JPG

Let

$C_v$ and

$C_p$ denote the molar heat capacities of an ideal gas at constant volume and constant pressure respectively.Which of the following is a universal constant?

0%
$\dfrac{C_p}{C_v}$
0%
$C_p C_v$
0%
$C_p - C_v$
0%
$C_p + C_v$

A steel scale gives correct reading at

$t_1 ^oC$ . When temperature changes to

$t_2 ^oC$ , then :

0%
if $t_2> t_1$ , reading is greater that true value
0%
if $t_2> t_1$ , reading is lesser that true value
0%
the reading is always equal to true value
0%
the reading is always less than true value

A physicist says "a body contains

$10\ joule$ heat" but a physics learner says "this statement is correct only when the body is in liquid state". Mark correct option or options :

0%
physicist statement is correct
0%
physics learner's statement is correct
0%
both statements are correct
0%
both statements are wrong

A metal rod of length

$L$ and cross-sectional area

$A$ converts a large tank of water at temperature

$\theta_{0}$ and a small vessel containing mass

$m$ of water at initial temperature of

$\theta_{1}(< \theta_{0})$ . If the thermal conductivity of rod is

$K$ , then the time taken for the temperature of water in smaller vessel to become

$\theta_{2}(\theta_{1} < \theta_{2} < \theta_{0})$ is:
(Given : Specific heat capacity of water is

$s$ and all other heat capacities are neglected).

0%
$T = \dfrac {ms}{KA} \log \dfrac {(\theta_{0} - \theta_{1})}{(\theta_{0} - \theta_{2})}$
0%
$T = \dfrac {Lms}{KA}\log \left (\dfrac {\theta_{0} - \theta_{1}}{\theta_{0} - \theta_{2}}\right )$
0%
$T = \dfrac {2mLs}{KA} \left (\log \dfrac {(\theta_{0} - \theta_{1})}{\theta_{0} - \theta_{2}}\right )$
0%
None of the above

The quantity

$\dfrac{PV}{kT}$ represents

0%
mass of the gas
0%
kinetic energy of the gas
0%
number of moles of the gas
0%
number of molecules in the gas

Explanation

The ideal gas equation is given as:

$PV=nRT$

The Boltzmann constant

$k$ is represented as:

$k=\dfrac{R}{N}$

Now, substitute the value of the Boltzmann constant :

$PV=nkNT$

$\therefore \dfrac{PV}{kT}=nN$

It is the number of molecules of the particular sample. SO, option

$D$ is correct.

The temperature of a body is increased from

$27^oC$ to

$127^oC$ . The radiation emitted by it increases by a factor of :

0%
$(256/81)$
0%
$(15/9)$
0%
$(4/3)$
0%
$(12/27)$

Three rods of identical cross-sectional area and made from the same material form the sides of an equilateral triangle. The point

$A$ and

$B$ are maintained at

$T$ and

$2T$ respectively. In steady state temperature of point

$C$ is

$T_c$ . ( Assuming only heat conduction takes place.) The value of

$T_c$ is

0%
$T/2$
0%
$T$
0%
$\dfrac{2}{3}T$
0%
$\dfrac{3}{2}T$

Explanation

Let

$H_1$ heat flow from

$2T$ to

$T_c$

and,

$H_2$ heat flow from

$T_c$ to

$T$

In steady state,

$H_1 =H_2$

$\dfrac {(2T-T_c)KA}{l}$

$=$

$\dfrac{T_c- T}KA{l}$

$2T-T_c = T_c- T$

$3T= 2T_c$

$T_c= \dfrac {3 T}{2}$

1634498_1738956_ans_32810a9fa7a84796a2c0d7c3a0d58edb.jpeg

The net rate of heat loss by a hot body depends upon:

0%
temperature of body
0%
temperature of surroundings
0%
material of body
0%
nature of the surface

Three rods of the same cross-sectional and made of the same material from the sides of a triangle

$ABC$ as shown. The points

$A$ and

$B$ are maintained at temperature

$T$ and

$\sqrt 2 T$ respectively in the steady state. Assuming that only heat conduction takes plane, the temperature at point

$C$ is:

0%
$\left[\dfrac {2\sqrt 2 +\sqrt 3}{2+\sqrt 3}\right]T$
0%
$\left[\dfrac {3 }{1+\sqrt 2}\right]T$
0%
$\left[\dfrac {2}{\sqrt 3}\right]T$
0%
$\left[\dfrac {\sqrt 5}{2}\right]T$

Three identical rods of same material are joined to from an equilateral triangle. The temperature of end

$A$ and

$B$ is maintained constant as

$\sqrt {3}T$ and

$T$ . The ratio of

$T_C / T_B$ will be: (Assuming no loss of heat from surfaces)

0%
$\dfrac {1+\sqrt 3}{2}$
0%
$\dfrac {1+\sqrt 3}{2}$
0%
$\dfrac {1+\sqrt 2}{2}$
0%
$\dfrac {1-\sqrt 2}{2}$

The temperature of end

$A$ of a rod us maintained at

$0^oC$ . The temperature of end

$B$ is changing slowly such that the rod may be considered in steady state at all time and is given by

$T_B$ =

$\alpha T$ ; where

$\alpha$ is positive constant and

$t$ is time. Temperature of point

$C$ , at a distance

$x$ from end

$A$ , at any time is ____.

0%
$\dfrac{\alpha xt}{L}$
0%
$\dfrac{\alpha xt^2}{2L}$
0%
$\dfrac{\alpha x^2t}{2L}$
0%
$\dfrac{\alpha (L-x)}{L}t$

Explanation

According to temperature gradient,

$\dfrac{T_A - T_x}{x}$ =

$\dfrac{T_A - T_B}{L}$

$\Rightarrow$

$\dfrac{0 - T_x}{x}$ =

$\dfrac{0 - \alpha t}{L}$

$\Rightarrow$

$T_x$ =

$\dfrac{ \alpha xt}{L}$

1634225_1738942_ans_2066200b412248aaa039d51a6d87eae0.png

If for hydrogen

$C_{p}-C_{v}=m$ and for nitrogen

$C_{p}-C_{v}=$

$n,$ where

$C_{p}$ and

$C_{v}$ refer to specific heats per unit mass respectively at constant pressure and constant volume, the relation between

$m$ and

$n$ is (molecular weight of hydrogen

$=2$ and molecular weight of nitrogen

$=14$ )

0%
$n=14m$
0%
$n=7m$
0%
$m=7n$
0%
$m=14n$

Explanation

$C_{p}-C_{v}=\mathrm{m},$ for hydrogen

$\left(M_{1}=2\right)$

$C_{p}-C_{v}=n,$ for nitrogen

$\left(M_{2}=14\right)$

For hydrogen,

$\quad C_{p}-C_{v}=\dfrac{1}{M_{1}} \dfrac{d Q}{d T}=m$

For nitrogen,

$\quad C_{p}-C_{v}=\dfrac{1}{M_{2}} \dfrac{d Q}{d T}=n$

$\therefore \quad \dfrac{m}{n}=\dfrac{\dfrac{1}{M_{1}} \dfrac{d Q}{d T}}{\dfrac{1}{M_{2}} \dfrac{d Q}{d T}}=\dfrac{M_{2}}{M_{1}}=\dfrac{14}{2}=7$

$\therefore m=7n$

An ideal gas undergoes the cyclic process shown in a graph below

0%
$T_1 = T_2$
0%
$T_1 > T_2$
0%
$V_aV_c = V_bV_d$
0%
$V_aV_b = V_cV_d$

Explanation

$\text{For adiabatic process 'bc'}$

$T_1V{^{\gamma-1}_B}=T_2V{^{\gamma-1}_C}......(i)$

$\text{For adiabatic process 'da'}$

$T_2V{^{\gamma-1}_d}=T_1V{^{\gamma-1}_a}......(ii)$

Multiplying (i) and (ii)

$\Rightarrow T_1T_2(V_b V_d)^\gamma-1=T_1T_2(V_aV_c)^\gamma-1$

$\Rightarrow V_bV_d=V_aV_c$

Hence option 'B' and 'D' are correct

An immersion heater, in an insulated vessel of negligible heat capacity, brings 100 g of water to the boiling point from $16^{0}C$ in 7 min. Then

The heat of vaporization of alcohol is

0%
$854 J/kg$
0%
$854 \times 10^{3} J/kg$
0%
$204 cal/g$
0%
$204 cal/kg$

Explanation

$Q''_a=P\times t_{3}=84\times(60\times 5+6)=84\times 306J$

$Q''_a=mL=30L$

$\Rightarrow L=\dfrac{84\times 306}{30}=854J/kg$

$=845\times 10^{3}J/kg$

Consider two rods of same length and different specific heats (

$s_1$ and

$s_2$ ), conductivities

$K_1$ and

$K_2$ and areas of cross section (

$A_1$ and

$A_2$ ) and both having temperature

$T_1$ and

$T_2$ at their ends. If the rate of heat loss due to conduction is equal, then

0%
$K_1 A_1$ = $K_2 A_2$
0%
$K_2 A_1$ = $K_1 A_2$
0%
$\dfrac{K_1 A_1}{s_1}$ = $\dfrac{K_2 A_2}{s_2}$
0%
$\dfrac{K_2 A_1}{s_2}$ = $\dfrac{K_1 A_2}{s_1}$

At temperature TK, the pressure of 4.0 g argon in a bulb is P. the bulb is put in a bath having temperature higher by 50 K than the first one. 0.8 g of argon gas had to be removed to maintained original pressure .The temperature T is equal to :

0%
510 K
0%
200 K
0%
100 K
0%
73 K

Heat is added to an ideal gas and the gas expands. In such a process the temperature:

0%
must always increase
0%
will remain the same if the work done equals the hear added
0%
must always decrease
0%
will remain the same if change in internal energy equals the heat added

Three rods of identical cross-sectional area and made from the same metal form the sides of an isosceles triangle ABC right angles at B. The points A and B are maintained at temperatures T and

$\sqrt{2}T$ , respectively, in the steady state. Assuming that only heat conduction takes place, temperature of point C

0%
$\frac {3T} {\sqrt{2} + 1}$
0%
$\frac {T} {\sqrt{2} + 1}$
0%
$\frac {3T} {3(\sqrt{2} - 1)}$
0%
$\frac {T} {\sqrt{2} - 1}$

Explanation

$T_s > T_A$ , heat flows from B to A through both paths BA and BCA.

Rate of heat flow in BC = Rate of heat flow in CA

$\dfrac {KA(\sqrt{2}T - T_c)} {l} = \dfrac{KA(T_c - T)} {\sqrt{2}l}$

Solving this, we get

$T_c = \dfrac{3T} {\sqrt{2} +1}$

1644035_1750218_ans_fca3b392b94046ae91215f011fd84200.png

Write 'true' or 'false' for each statement:

Medium is necessary for the transfer of heat by radiation.

0%
True
0%
False

A marble tile would feel cold as compared to a wooden tile on a winter morning, because the marble tile

0%
is a better conductor of heat than the wooden tile.
0%
is polished while wooden tile is not polished.
0%
Reflects more heat than wooden tile.
0%
is a poor conductor of heat than the wooden tile.

Write 'true' or 'false' for each statement:

On touching a lump of ice, we feel cold because some heat passes from our body to the ice.

0%
True
0%
False

Choose the correct alternative:

In the process of convection, heat travels

0%
sideways
0%
downwards
0%
upwards
0%
in all direction

$S.I.$ unit of temperature is:

0%
calorie
0%
Joule
0%
Celsius
0%
Kelvin

Explanation

Through temperature is usually measured in Celsius,

$kelvin\ (K)$ is the

$S.I.$ unit.

Absolute scale of temperature is reproduced in the laboratory by making use of a

0%
Radiation pyrometer
0%
Platinum resistance thermometer
0%
Constant volume helium gas thermometer
0%
Constant pressure ideal gas thermometer

Wires A and B have identical lengths and have circular cross-sections. The radius of A is twice the radius of B i.e.

$r_A=2r_B$ . For a given temperature difference between the two ends, both wires conduct heat at the same rate. The relation between the thermal conductivities is given by :

0%
$K_A =4K_B$
0%
$K_A =2K_B$
0%
$K_A =K_B/2$
0%
$K_A =K_B/4$

Explanation

$\dfrac{Q}{t}=\dfrac{KA\Delta \theta}{l}\,\rightarrow \dfrac{K_A}{K_B}=\dfrac{A_B}{A_A}=\Bigg(\dfrac{r_B}{r_B}\Bigg)^2=\dfrac{1}{4}\,\implies K_A=\dfrac{K_B}{4}$

The length of the two rods made up of the same metal and having the same area of the cross-section are

$0.6$ m and

$0.8$ m respectively. The temperature between the ends of the first rod is

$90 ^\circ C$ and

$60 ^\circ C$ and that for the other rod is

$150$ and

$110 ^\circ C$ . For which rod the rate of conduction will be greater

0%
First
0%
Second
0%
Same for both
0%
None of the above

Explanation

$\dfrac{Q_1}{t}=\dfrac{KA(90-60)}{0.6}=50 \ KA$ and

$\dfrac{Q_2}{t}=\dfrac{KA(150-110)}{0.8}=50 \ KA$ .

Calories is the unit of

0%
Heat
0%
Work
0%
Temperature
0%
Food

Explanation

Heat is measured in calorie. (

$1$ cal

$=4.186\ J$ )

In cold countries, water pipes sometimes burst, because

0%
Pipe contracts
0%
Water expands on freezing
0%
When water freezes, pressure increases
0%
When water freezes, it takes heat from pipes

Explanation

In anomalous expansion, water contracts on heating and expands on cooling in the range

$0^\circ C$ to

$4^\circ C$ . Therefore water pipes sometimes burst, in cold countries.

Absolute zero (0 K) is that temperature at which

0%
Matter ceases to exist
0%
Ice melts and water freezes
0%
Volume and pressure of a gas becomes zero
0%
None of these

Explanation

We know that

$P = P_0 (1 + \gamma t)$ a

$V = V_0 (1 + \gamma t)$

and

$\gamma = (1 / 273) /^\circ C$ for

$t = -273^\circ C,$ we have

$P = 0$ and

$V = 0$

Hence, at absolute zero, the volume and pressure of the gas become zero.

Calorimeters are made of which of the following

0%
Glass
0%
Metal
0%
Wood
0%
Either (a) or (c)

Water is cooled from

$4^oC$ to

$0^oC$ it:

0%
Contracts
0%
Expands
0%
First contrast, then expands
0%
First expands then contrast

Explanation

Expansion of water occurs when water is cooled from

$4^oC$ to

$0^oC$ since density begins to decrease after attaining maximum value at

$4^oC$ .

The velocity of heat radiation in vacuum is

0%
Equal to that of light
0%
Less than that of light
0%
Greater than that of light
0%
Equal to that of sound

Explanation

The velocity of heat radiation in the vacuum is

$equal$ to that of light.

The coefficients of thermal conductivity of copper, mercury, and glass are respectively K, K, and K such that

$K > K > K$ . If the same quantity of heat is to flow per second per unit area of each and corresponding temperature gradients are X, X, and X, then

0%
$X_c=X_m=X_g$
0%
$X_c>X_m>X_g$
0%
$X_c < X_m < X_g$
0%
$X_m < X_c < X_g$

Explanation

The heat flowing is given by

$\dfrac{Q}{At}=K\dfrac{\Delta \theta}{l}\,\implies K\dfrac{\Delta \theta}{l}=constant\,\implies \dfrac{\Delta \theta}{l} \propto \dfrac{1}{K}$
Hence If

$K_c > K_m > K_g$ , then

$\Bigg(\dfrac{\Delta \theta }{l}\Bigg)_c < \Bigg(\dfrac{\Delta \theta }{l}\Bigg)_m < \Bigg(\dfrac{\Delta \theta }{l}\Bigg)_g\,\implies X_c < X_m < X_g$
because higher K implies lower value of the temperature gradient.

Four rods of identical cross-sectional area and made from the same the metal form the sides of square. The temperature of two diagonally opposite points and

$T$ and

$2T$ respective in the steady-state. Assuming that only heat conduction takes place, what will be the temperature difference between the other two points

0%
$\dfrac{\sqrt{2}+1}{2}T$
0%
$\dfrac{2}{\sqrt{2}+1}T$
0%
$0$
0%
None of these

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

0:0:2

Practice Class 11 Engineering Physics Quiz Questions and Answers

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