No heat is added or removed in an adiabatic process

The change in internal energy in a constant pressure process from temperature $$T_1$$ to $$T_2$$ is equal to $$NC_V(T_2-T_1)$$, where $$C_V$$ is the molar specific heat at constant volume and n is the number of moles of the gas

The change in internal energy of the gas and the work done by the gas are equal in magnitude in an adiabatic process

The internal energy does not change in an isothermal process

first at the bottom due to increase in pressure

first at the bootom due to decress in pressure

first at the top due to iuncrease in pressure

first at the top due to increase in pressure

MCQ Questions for Class 11 Engineering Physics Thermal Properties Of Matter Quiz 9

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

$=11\times { 10 }^{ -6 }_{ }\quad ^{ o }{ { C }^{ -1 } }$ )

0%
${ 8 }^{ o }C$
0%
${ 9 }^{ o }C$
0%
${ 4.5 }^{ o }C$
0%
${ 10 }^{ o }C$

Explanation

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.

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:

0%
$64$ times
0%
$16$ times
0%
$32$ times
0%
$4$ times

Explanation

From Stefan's law, the energy radiated by sun is given by:-

$P=\sigma eAT^4$

$Ist$ case:-

$P_1= \sigma e\times 4\pi R^2\times T^4$

$2nd$ 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

$(A)$ .

Clear nights are colder than cloudy nights because of

0%
conduction
0%
condensation
0%
radiation
0%
insolation

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

0%
Sea and land breeze
0%
Trade wind
0%
Boiling of water
0%
Warming of glass of bulb due to filament

Explanation

$\textbf{Explanation:}$

$\bullet$ 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.

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:

0%
$0.75 \times 10^{-2} J$
0%
$1.50 \times 10^{-2} J$
0%
$0.25 \times 10^{-2} J$
0%
$0.50 \times 10^{-2} J$

Explanation

$K_i = \dfrac{1}{2} \times \dfrac{(500 gm) (10 cm)^2}{2} \times (3 rad/s)^2$

$= \dfrac{9}{8} \times 10^{-2} J$
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 =

$\left( \dfrac{9}{8} - \dfrac{3}{8} \right) \times 10^{-2} = 0.75 \times 10^{-2} J$

An ideal gas is initially at

$P_1$ ,

$V_1$ is expanded to

$P_2, V_2$ and then compressed adiabatically to the same volume

$V_1$ and pressure

$P_3$ . If W is the net work done by the gas in the complete process which of the following is true.

0%
$W > 0; P_3 > P_1$
0%
$W < 0; P_3 > P_1$
0%
$W > 0; P_3 < P_1$
0%
$W < 0; P_3 < P_1$

Explanation

Since work is done by the gas as area under the graph is

$-ve$ ( anticlockwise) ie energy is lost by the gas and

$w<0$ when

$P_3>P_1$ .

$300$ grams of water at

$25^o$ C is added to

$100$ grams of ice at

$0^o$ C. The final temperature of the mixture is ________

$^oC$ .

0%
2
0%
0
0%
3
0%
4

Explanation

$\textbf{Given}:$ m = 300g at

$25^{o}C$ for water,

$m = 100g$ at

$0^{o}C$ for ice

$\textbf{Solution}:$

The heat required for 100g of ice at

$0^{o}C$ to change into Water at

$0^{o}C=mL=100\times 80\times 4.2=33,600J$ …(i)

The heat released by 300g of water at

$25^{o}C$ to change its temperature to

$0^{0}C=mc \Delta T=300\times 4.2\times 25=31,500J$ ….(ii)

Since the energy in eq.(ii) is less than of eq(i) therefore the final temperature will be

$0^{o}C$

$\textbf{Hence B is the correct option}$

We have half a bucket (6l) of water at 20C. If we want water at 40C, how much steam at 100C should be added to it?

0%
200 g
0%
2000/9 g
0%
2 kg
0%
200/3 g

Explanation

6000 x 1 x (40 - 20) = m x 540 + m x 1 x (100 - 40)
12 x

$10^4$ = m x 600
m = 2 x

$10^2$ = 200 gm

Temperature of hot end and cold end of a rod, which is in steady state, are

$100^{\circ}C$ and

$40^{\circ}C$ respectively. The area of cross section of rod and its thermal conductivity are uniform. The temperature of rod at its mid point is :

(Assume no heat loss through lateral surface).

0%
$70^{\circ}C$
0%
$60^{\circ}C$
0%
$80^{\circ}C$
0%
None of these

Explanation

We know,

$\dfrac{dQ}{dt}=\dfrac{k\times A\times (T_2-T_1)}{l}$

Suppose the temperature at midpoint is

$T^0 C$ ,

Now,

$\dfrac{dQ}{dt}=\dfrac{kA(T-40)}{l/2}=\dfrac{kA(100-T)}{l/2}$

$100-T=T-40\\ 2T=140 \\ T=70^0 C$

Hence option

$\textbf A$ is correct answer.

For a ideal gas.

0%
The change in internal energy in a constant pressure process from temperature $T_1$ to $T_2$ is equal to $NC_V(T_2-T_1)$ , where $C_V$ is the molar specific heat at constant volume and n is the number of moles of the gas
0%
The change in internal energy of the gas and the work done by the gas are equal in magnitude in an adiabatic process
0%
The internal energy does not change in an isothermal process
0%
No heat is added or removed in an adiabatic process

Explanation

Change in internal energy of a gas

$=\dfrac{f}{2}nR\triangle T$

In an isothermal process,

$\triangle T=0$

$\Rightarrow$ Change in internal energy

$=0$

So, option (C) is correct.

Now,

$C_v=\dfrac{f}{2}R$

$\Rightarrow \triangle u=n\left(C_v\right) \triangle T=nC_v\left(T_2-T_1\right)$

So, option (A) is also correct.

First law of thermodynamics :

$dQ=dv+dw$

In an adiabatic process,

$dQ=0$ [ option (D) is also correct. ]

$\Rightarrow dv=-dw$

$\Rightarrow \left| dv \right| =\left| dw \right|$

So, option (B) is also correct.

Hence, this question has multiple answers.

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.

0%
4
0%
13
0%
22
0%
25

Explanation

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

$(100+m)\times 66=546\ m$

$6600+66\ m=546\ m$

$600=486\ m$

$\boxed {m=13.75\ g\ of\ steam}\simeq 13\ gm$

$(B)$

A long metallic bar is carrying heat from one of its ends to the other end under steady-state. The variation of temperature

$\theta$ along the length x of the bar from its hot end is best described by which of the following figures?

Explanation

$KA\cfrac { d\theta }{ dx }$ will be same at all

$x$

$\Rightarrow \theta \quad vsx$ will be straight line

Water of volume

$2L$ in a closed container is heated with a coil of

$1kW$ . While water is heated, the container loses energy at a rate of

$60J/s$ . In how much time will the temperature of water rise from

${27}^{o}C$ to

${77}^{o}C$ (Specific heat of water is

$4.2kJ/kg$ and that of the container is negligible)

0%
$8min$ $20s$
0%
$6min$ $2s$
0%
$7$ min
0%
$14$ min

Explanation

$\textbf{Given}:$ Volume of water = 2L

$\textbf{Solution}:$

From question heat gained by water in 1 second = 1 KW - 160 J/s = 1000 J/s - 160 J/s = 840 J/s

Total heat required to raise the temperature of water(volume 2L) from

$27^o C$ to

$77^o C$

$m_{water} \times \text{specific heat} \times \Delta \theta$

$= 2 \times 10^3 \times 4.2 \times 50 [ Q_{mass} = density \times volume]$

Thus

$840 \times t = 2 \times 10^3 \times 4.2 \times 50$

$t = 2 \times 10^3 \times 4.2 \times \dfrac{50}{840}$

t = 500 seconds

t = 8 min 20 sec

$\textbf{Hence A is the correct option}$

A cylinder of mass

$1$ kg is given heat of

$20000$ J at atmospheric pressure. If initially temperature of cylinder is

$20^o$ C, find final temperature of the cylinder.

0%
$T_{final}=90^oC$ .
0%
$T_{final}=70^oC$ .
0%
$T_{final}=80^oC$ .
0%
$T_{final}=60^oC$ .

Two monatomic ideal gases

$1$ and

$2$ of molecular masses

$m_1$ and

$m_2$ respectively are enclosed in separate containers kept at the same temperature. The ratio of the speed of sound in gas

$1$ to that in gas

$2$ is given by.

0%
$\sqrt{\dfrac{m_1}{m_2}}$
0%
$\sqrt{\dfrac{m_2}{m_1}}$
0%
$\dfrac{m_1}{m_2}$
0%
$\dfrac{m_2}{m_1}$

Explanation

$V_{rm}=\sqrt {\dfrac {rRT}{m}}$

i.e

$vr\dfrac {1}{\sqrt m}$

which given

$v-1\times \dfrac {1}{\sqrt {m_1}}$ and

$v_2\times \dfrac {1}{\sqrt {m_1}}$

$\dfrac {v_1}{v_2}=\sqrt {\dfrac {m_2}{m_1}}$

A double pane window used for insulating a room thermally from outside, consists of two glass sheets each of area

$1m^2$ and thickness

$0.01$ m separated by a

$0.05$ m thick segment air space. In the steady state the room glass interface and the glass outdoor interface are at constant temperature of

$27^o$ C and

$0^o$ C respectively. Calculate the rate of heat flow through the window pane. Also find the temperature of other interfaces. Given thermal conductivities of glass and air are as

$0.8$ and

$0.08$

$Wm^{-1}K^{-1}$ respectively.

0%
$41.54$ W, $46.48^oC, 0.52^oC$ .
0%
$41.54$ W, $26.48^oC, 0.52^oC$ .
0%
$41.54$ W, $36.48^oC, 0.52^oC$ .
0%
$41.54$ W, $66.48^oC, 0.52^oC$ .

Explanation

$\dfrac {dQ}{dt}=K\dfrac {A\Delta Q}{L}=\dfrac {\Delta Q}{R}$

Required

$=\displaystyle \sum \dfrac {L}{KA}=\dfrac {1}{A} \left [\dfrac {0.01}{0.8}\times 2+\dfrac {0.05}{0.08}\right]$

$A=1\ m^2$ , Required

$=\dfrac {1}{40}+\dfrac {5}{8}=\dfrac {26}{40}$

$\dfrac {dQ}{dt}=\dfrac {\Delta Q}{R}= \dfrac {(27-0\times 40)}{26}=41.5\ W$

Now

$41.5=0.8\times 1^3 \dfrac {27-Q_2}{0.01}$ or

$Q_2=26.48^oC$

$41.5=\dfrac {0.8\times 1^3 (Q_2 -0)}{0.01}, Q_1=0.52^oC$

$\dfrac {dQ}{dt}=41.5\ W$

$Q_2=26.48^oC$

$Q_1=0.52^oC$

$(B)$

One kg of a diatomic gas is at a pressure of

$8\times {10}^{4}N/{m}^{2}$ . The density of gas is

$4kg/{m}^{2}$ . What is the energy of the gas due to its thermal motion?

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

Out of three thermometers one is kept in contact with the mans skin, second in contact with cotton vest and polyester shift and third in between shirt and woolen coat. The reading of the thermometers are

$35^{o}C$ ,

$32^{o}C$ ,

$25^{o}C$ . The ratio of

$K$ for cotton and polyester will be

0%
$7\ :\ 3$
0%
$3\ :\ 7$
0%
$10\ :\ 3$
0%
$3\ :\ 10$

Explanation

Based on heat balance, the distribution of temperature is described by a second order differential equation.

$\dfrac {d^2T}{dx^2} = -h' (T_o-T)$

The negative sign indicates a temperature decrease with time.

$T_o$ is surrounding temperature.

Temperature decreases towards right so, the ratio of the temperature

$k$ for cotton and polyester is

$k=\dfrac {h'(35-32)}{h'(32-25)}=\dfrac {3}{7}$

So,

$k=\dfrac {3}{7}$ .

Hence, Option

$B$ is the correct answer.

Rate of heat flow through cylindrical rod is

${Q}_{1}$ . Temperatures of ends of rod are

${T}_{1}$ and

${T}_{2}$ . If all the linear dimensions of the rod become double and temperature difference remains same, its rate of heat flow is

${Q}_{2}$ , then

0%
${Q}_{1}=2{Q}_{2}$
0%
${Q}_{2}=2{Q}_{1}$
0%
${Q}_{2}=4{Q}_{1}$
0%
${Q}_{1}=4{Q}_{2}$

Explanation

$H = kA \dfrac {(Q_{1} - Q_{2})}{L}T$

Rate of neat flow

$Q = \dfrac {H}{T} = \dfrac {KA(Q_{1} - Q_{2})}{t}$

$Q_{1} \times \dfrac {A}{L}$ Dimension of

$Area = A = (L^{2})$ Dimension of distance

$= (L)$

$QQL$

$\Rightarrow \dfrac {Q_{1}}{Q_{2}} = \dfrac {L}{L_{2}} = 2$

$\Rightarrow Q = 2Q_{2}$ .

1039385_1022930_ans_53ec0da6f70a4d40841b07df73f91974.jpg

In an experiment, 1.35 mol of oxygen (O2) are heated at constant pressure starting at 11.0ºC. How much heat must be added to the gas to double its volume?

0%

$1.12\times 10^4 J$
0%

$1.40\times 10^4$
0%
$$2.12\times 10^4$$J
0%
$$3.12×10^4 J$$

Explanation

Since oxygen

$(O_2)$ is a diatomic gas

$\therefore C_v=5/2R$

So,

$C_p=C_v+R\\ \quad=5/2R+R\\ \quad =7/2R$

At constant pressure

If we double the volume the temperature will be doubled.

$T_i=11.0°C=(11.0+273)K=284K$

So,

$T_p=2(284)K=568K$
To obtain heat,

$Q=nC_p(T_p-T_i)$

$Q=(1.35mol)(7/2R)(508K-284K)\\Q=1.12\times10^4J$

Three metal rods made of copper ,aluminium and brass, each 20 cm long and 4 cm in diameter,are placed end to end with aluminium between the other two.The free ends of copper and brass are maintained at 100 and

$0^0$ C respectively. Assume that the thermal conductivity of copper is twice that of aluminium and four times that of brass. The equilibrium temperatures of the copper aluminium and aluminium brass junctions are respectively.

0%
$68^0$ C and $75^0$
0%
$75^0$ C and $68^0$
0%
$57^0$ C and $86^0$
0%
$86^0$ C and $57^0$

By the ideal gas law, the pressure of

$0.60$ moles

${NH}_{3}$ gas in a

$3.00\ L$ vessel at

${25}^{o}C$ is, given that

$R=0.082\ L$ atm

${mol}^{-1}{k}^{-1}$ :

0%
$48.9\ atm$
0%
$4.89\ atm$
0%
$0.489\ atm$
0%
$489\ atm$

Explanation

Pressure is given as

$P=\dfrac{nRT}{V}=\dfrac{0.6\times .082\times (25+273)}{3}=4.89atm$

Option B is correct.

Time taken by a

$836$ W heater to heat one litre of water from

$10^o$ C to

$40^o$ C is?

0%
$50$ s
0%
$100$ s
0%
$150$ s
0%
$200$ s

Explanation

According to the given question we should find the time taken

It is given that :

$\Rightarrow P = 836$ W

$\Rightarrow$ mass of water

$= 1L = 1000$ g

$\Rightarrow$ temperature rise

$=\left({\theta}_{1}-{\theta}_{2}\right) =\left(40-10\right) ={30}{\circ}$

$\Rightarrow$ specific heat of water is

$C = 1$ cal

${g}^{-1}{c}^{-1}$

Now,heat produced by the heater in time

$t$ is

$= P \times T = 836 t$ joules ( we should convert to calories )

$=\dfrac{836 t}{4.2}$ ......

$(1)$ calories---------------------(1)

Now, heat taken by the water

$= mc\left({\theta}_{1}-{\theta}_{2}\right)= 1000 \times 1 \times 30=30000$ -----------

$(2)$

we know that ,heat produced by the heater in time

$t =$ heat taken by the water implies,

$\dfrac{836 t}{4.2} = 30000$

$\Rightarrow t = \dfrac{\left(30000 \times 4.2\right)}{836}$

$t =\dfrac{126000}{836}$

$t = 150.7$ seconds .

$\therefore$ the time taken is

$150.7$ seconds.

A glass flask is filled up to a mark with

$50\ cc$ of mercury at

${18}^{o}C$ . If the flask and content are heated to

${38}^{o}C$ , how much mercury will be above the mark? (

$\alpha$ for glass is

$9\times { { 10 }^{ -6 } }/{ C }$ and coefficient of real expansion of mercury is

$180\times { { 10 }^{ -6 } }/{ C })$

0%
$0.85\ cc$
0%
$0.46\ cc$
0%
$0.153\ cc$
0%
$0.05\ cc$

Explanation

Volume expansion coefficient of glass

$=3\times linear =3\alpha$

Change in temperature is

$38-18=20^0 C$

Increment of volume of flask is

$\Delta V=50cc\times 3\alpha \times 20^0 C=0.027cc$

Expansion in mercury

$50cc \times \gamma \Delta t=50\times 180\times 10^{-6}\times 20=0.180cc$

Apparent increment=

$0.180-0.027=0.153cc$

Option C is correct.

Q.1 the frame is made by seven meter wire, each has conductivity K. length of each segment is

$\ell$ lThe temperature of points E,F and A are maintained at

$160^o$ c,

$200^oC$ and

$100^oC$ respectively.(assume heat flows through the wire only and there is no loss of heat from the side of the wire).Then the tempera-ture at C is ( approx).

0%
$135^oC$
0%
$124^oC$
0%
$120^oC$
0%
$112^oC$

Temperature of a room is maintained at

$20^{\circ}$ by a heater of power

$2\ kW$ . The room has a glass window of area

$1\ m^{2}$ , thickness

$0.2\ cm$ and thermal conductivity

$0.2\ cal/m/^{\circ}C/s$ . Temperature of outside is about.

0%
$-63.6^{\circ}C$
0%
$15^{\circ}C$
0%
$20^{\circ}C$
0%
$12^{\circ}C$

Explanation

If temperature of outside be

$\theta$ then putting

$k=0.2, A=1m^2 , l=.002meter$

and

$H=2kW=2000J/s=2000\times 4.18 cal/s =8360cal/s$

in the formula

$H=KA\dfrac{20-\theta }{l}$ , we get

$20-\theta =83.6^0 C$

$\theta =-63.6^0 C$

Two diagonally opposite comers of a square made of four thin rods of the same material, same dimensions are at temperature

$40^o C$ and

$10^o C$ . If only heat conduction takes place, then the temperature difference between the other two corners will be:

0%
$0^{o}C$
0%
$10^{o}C$
0%
$25^{o}C$
0%
$15^{o}C$

Explanation

energy flow rate is

$\dfrac{dQ}{dt}=\dfrac{KA(T_{1}-T_{2})}{L}$

let T be the temperature of the other diagonal of the square.

The rate of flow of heat from one rode is equal to another rod.

$\dfrac{KA(40^0-T)}{L}=\dfrac{KA(T-10^0)}{L}$

$2T=30$

$T=15^0$

hence D is the correct option.

Three perfect gasses at absolute temperatures

$T_1$ ,

$T_2$ and

$T_3$ are mixed. If number of molecules of the gasses are

$n_1$ ,

$n_2$ and

$n_3$ respectively then temperature of mixture will be (assume no loss of energy)

0%
$\dfrac{ T_{1}+ T_{2}+T_{3}}{3}$
0%
$\dfrac{ { n }_{ 1 }^{ 2 } T_{1}+ { n }_{ 2 }^{ 2 } T_{2}+ { n }_{ 3 }^{ 2 } T_{3}}{{ n }_{ 1 }+{ n }_{ 2 }+{ n }_{ 3 }}$
0%
$\dfrac{ { n }_{ 1 } T_{1}+ { n }_{ 2 }T_{2}+ { n }_{ 3 }T_{3}}{{ n }_{ 1 }+{ n }_{ 2 }+{ n }_{ 3 }}$
0%
$\dfrac{ T_{1}+ T_{2}+T_{3}}{{ n }_{ 1 }+{ n }_{ 2 }+{ n }_{ 3 }}$

Explanation

For perfect gas,

Kinetic Energy of n molecule, $K.E=n\left( \dfrac{1}{2}{{K}_{B}}T \right)$

Where, ${{K}_{B}}$ is Boltzmann constant

If there is no loss of energy.

Total kinetic energy of mixture is sum of each gas kinetic energy.

${{n}_{total}}K.{{E}_{total}}={{n}_{1}}K.{{E}_{1}}+{{n}_{2}}K.{{E}_{2}}+{{n}_{3}}K.{{E}_{3}}$

$\left( {{n}_{1}}+{{n}_{2}}+{{n}_{3}} \right)\left( \dfrac{1}{2}{{K}_{B}}T \right)={{n}_{1}}\left( \dfrac{1}{2}{{K}_{B}}{{T}_{1}} \right)+{{n}_{2}}\left( \dfrac{1}{2}{{K}_{B}}{{T}_{2}} \right)+{{n}_{3}}\left( \dfrac{1}{2}{{K}_{B}}{{T}_{3}} \right)$

$T=\dfrac{{{n}_{1}}{{T}_{1}}+{{n}_{2}}{{T}_{2}}+{{n}_{3}}{{T}_{3}}}{{{n}_{1}}+{{n}_{2}}+{{n}_{3}}}$

In case of heat flow across the composite rod shown above, temperature

$\theta$ is

0%
$80^{\circ}C$
0%
$-80^{\circ}C$
0%
$0^{\circ}C$
0%
$-20^{\circ}C$

Explanation

Rods have been joined in series so the heat rate should be same in both the rods.

The heat rate in the thick rod is

$H_1=(3K)(2A)\dfrac{(80-40)}{2L}=120KA/L$

For the thin rod the heat rate will be

$H_2=KA\dfrac{40-\theta}{L}$

Both

$H_1$ and

$H_2$ should be equal so

$40-\theta=120$ so

$\theta =-80^0C$

Option B is correct.

A vessel has

$6g$ of oxegen at pressure

$P$ and temperature

$400\ K$ . A small hole is made in it so that oxygen leaks out. How much oxygen leaks out if the final pressure is

$P/2$ and temperature is

$300\ K$

0%
$3g$
0%
$2g$
0%
$4g$
0%
$5g$

Explanation

From ideas gas equation

$PV=nRT$

$PV=\dfrac{m}{{{M}_{o}}}RT$

$m=\dfrac{PV{{M}_{o}}}{RT}\$

In first event

${{m}_{1}}=\dfrac{{{P}_{1}}V{{M}_{o}}}{R{{T}_{1}}}$

$6\ gram=\dfrac{PV{{M}_{o}}}{R\times 400}\$

$\dfrac{{{\operatorname{PVM}}_{o}}}{R}=400\times 6\ .\ .....\ (1)$

In the second event when hole is made:

${{m}_{2}}=\dfrac{{{P}_{2}}V{{M}_{o}}}{R{{T}_{2}}}$

${{m}_{2}}=\dfrac{\dfrac{P}{2}V{{M}_{o}}}{R\times 300}\ =\dfrac{1}{300\times 2}\dfrac{PV{{M}_{o}}}{R}$

${{m}_{2}}=\dfrac{1}{300\times 2}\times \left( 400\times 6 \right)=4\ gram$

Leak out oxygen is $\left( 6gram-4\,gram \right)\ =\ 2\,gram$

If a thermometer reads freezing point of water as

$20^{\circ}C$ and boiling point at

$150^{\circ}C$ how much thermometer read when the actual temperature is

$60^{\circ}C$ .

0%
$98^{\circ}C$
0%
$110^{\circ}C$
0%
$40^{\circ}C$
0%
$60^{\circ}C$

Explanation

We let the unknown degree scale be called

${x}^{\circ}$

Now we know that the freezing point

$={20}^{\circ}x$

The boiling point

$={150}^{\circ}x$

This also means that

${20}^{\circ}x= {0}^{\circ}$ C(

$\because$ on a Celcius Scale, the freezing point is

${0}^{\circ}$ )

And

${150}^{\circ}x={100}^{\circ}$ C (On a Celcius Scale, boiling point is

${100}^{\circ}$ )

Also

${(150-20)}^{\circ}x= {100}^{\circ}$ C (

$\because$ on a Celcius Scale

${100}^{\circ}-{0}^{\circ}={100}^{\circ}$ )

So, if

${100}^{\circ}$ C

$={130}^{\circ}x$

Then

${1}^{\circ}$ C

$={1.3}^{\circ}x$

To go from

$C$ degree to

$x$ , we need to use:

$x=1.3C+20$

And

$C=\dfrac{(x-20)}{1.3}$

So if the temperature given is

${60}^{\circ}$ C, then we calculate

$x$ as:

$C=\dfrac{(x-20)}{1.3}$

$\Rightarrow 60=\dfrac{(X-20)}{1.3}$

$\Rightarrow 60\times 1.3=x-20$

$\Rightarrow 78=x-20$

$\Rightarrow x= {98}^{\circ}$

${60}^{\circ}$ C will be equal to

${98}^{\circ}$

When

$500 kg$ of water is heated from

$20^oC$ to

$100^oC$ , then the increase in mass of water will be

0%
$3.2\times { 10 }^{ -9 }$
0%
$1.87\times { 10 }^{ -9 }$
0%
$0.96\times { 10 }^{ -9 }$
0%
$2.8\times { 10 }^{ -9 }$

Ratio of radius of curvature of cylindrical emitters of same type is

$1:4$ and their temp. are in ration

$2:1$ . Then ration of amount of heat emitted by them is-(For Cylinder length = radius);-

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

$0.1\,{m^3}$ of water at

$80^\circ C$ is mixed with

$0.3\,{m^3}$ of water at

$60^\circ C$ . The final temperature of the mixture is :

0%
$65^\circ C$
0%
$70^\circ C$
0%
$60^\circ C$
0%
$75^\circ C$

Explanation

Given volume of water

${V}_{1}=0.1 {m}^{3}$ and initial temperature

${T}_{1}={80}^{0}C$ and volume

${V}_{2}=0.3 {m}^{3}$ and

${T}_{2}={60}^{0}C$

Thus

${m}_{1}{c}_{p}{dt}_{1}={m}_{2}{c}_{p}{dt}_{2}$

as we know

$m=\rho V$

$\rho (0.1){c}_{p}(80-{T}_{f})=\rho (0.3){c}_{p}({T}_{f}-60)$

$4{T}_{f}=260$

${T}_{f}={65}^{0}C$

Find the temperature of the junction shown in the figure for three rods; identical in dimensions:

0%
$40^0C$
0%
$60^0C$
0%
$30^0C$
0%
$80^0C$

A laboratory thermometer gave the reading of

$-1^oC$ and

$99^o C$ when inserted into melting ice and boiling water respectively, both at standard atmospheric pressure. What is the error when the same thermometer is used to measure the difference between two arbitrary temperatures?

0%
$-1^o C$
0%
$1^o C$
0%
$0^o C$
0%
$2^o C$

The figure shows a glass tube (linear coefficient of expansion is

$\sigma$ ) completely filled with a liquid of volume expansion coefficient

$\gamma$ . On heating, length of the liquid column does not change. Choose the correct relation between

$\gamma$ and

$\sigma$ .

0%
$\gamma = \alpha$
0%
$\gamma =2 \alpha$
0%
$\gamma = 3\alpha$
0%
$\gamma = \alpha/3$

Explanation

When length of liquid coloumn remains constant,

then the level of liquid moves down w.r.t container

thus

$\gamma=3\alpha$

Now,

$V=V_o(1+\gamma^{\Delta T})$

Since,

$V=Al_o=[A_o(1+2\alpha \Delta T)]10=V_o(1+2\alpha \Delta t)$

Hence,

$V_o(1+\gamma^{\Delta T})=V_o(1+2 \alpha \Delta T)$

$\Rightarrow \gamma= 2 \alpha$

Three identical iron rods are welded together to form the shape of

$Y$ . The top ends of the

$Y$ are maintained at

$0^{o}C$ and the bottom ends is maintained at

$60^{o}C$ . The temperature of the junction of the three rods is

0%
$10^{o}\ C$
0%
$20^{o}\ C$
0%
$25^{o}\ C$
0%
$40^{o}\ C$

A vessel contains 28 gm of

$N-2$ and 32 gm of

$O_2$ at temperature T = 1800 K and pressure 2 atm pressure it

$N_2$ dissociates

$30%$ and

$O_2$ dissociates

$50%$ if temperature remains constant.

0%
2 atm
0%
1 atm
0%
2.8 atm
0%
1.4 atm

An ideal gas has an initial volume

$V$ and pressure

$P$ . In doubling its volume the minimum work done will be in (of the given processes):

0%
Isobaric process
0%
Isothermal process
0%
Adiabatic process
0%
Same in all given processes

Which of the following statements is true for a thermometer?

0%
Coefficient of cubical expansion of liquid must be greater than that of bulb material
0%
Coefficient of cubical expansion of liquid may be equal to that of bulb material
0%
Coefficient of cubical expansion of liquid must be less than of bulb material
0%
None of the above

The velocities of sound in an ideal gas at temperature

$T_1$ and

$T_2 k$ are found to be

$V_1$ and

$V_2$ respectively.If the r.m.s. velocities of the molecules of the same gas at the same temperatures

$T_1$ and

$T_2$ are

$v_1$ and

$v_2$ respectively then

0%
$v_2=v_1(\dfrac{V_1}{V_2})$
0%
$v_2=v_1(\dfrac{V_2}{V_1})$
0%
$v_2=v_1\sqrt{\dfrac{V_2}{V_1}}$
0%
$v_2=v_1\sqrt{\dfrac{V_1}{V_2}}$

A certain perfect gas occupying

$1\ litre$ at

$80\ cm$ of Hg, suddenly expands to

$1190\ cc$ , while the pressure falls to

$60\ cm$ of Hg. Therefore, the gas is ?

0%
polyatomic
0%
diatomic
0%
monoatomic
0%
data inadequate

Two new thermometer scales X and Y are plotted as shown. The LFP and UFP of Y-scale are

$0^{\circ}$ and

$90^{\circ}$ respectively. Then the corresponding values on X-scale are

0%
$-10^{\circ}$ , $100^{\circ}$
0%
$0^{\circ}$ , $80^{\circ}$
0%
$10^{\circ}$ , $90^{\circ}$
0%
$10^{\circ}$ , $100^{\circ}$

Explanation

$\textbf{Given}:$ LFP of the thermometer Y is

$0^o$ , UFP of the thermometer Y is

$100^0$

$\textbf{Solution}:$

Let us find the graph equation now, It is a straight line, so the equation will be of form y=mx+c.

The slope of the line is equal to m=

$tan 45^0$ , m = 1

The constant is the y coordinate when x is equal to 0.

Therefore, c=-10.

Now, the equation we get is,

y=x-10

Let us find the LFP of the thermometer X as Value of Y is given.,

y=mx+c

$\implies 0=x-10$

$\implies x=10^0$

$\therefore$ , the LFP of the thermometer X is 10 degrees.

Now, the UFP of the thermometer X can be founds similarly as,

y=x-10

$\implies 90=x-10$

$\implies x=100^0$

Hence, the UFP of the thermometer X is 100 degrees.

$\textbf{Hence D is the correct option}$

Equal temperature difference exists between the ends of two metallic rods

$1$ and

$2$ of equal length. Their thermal conductivities are

$K_1$ and

$K_2$ and cross sectional areas are respectively

$A_1$ and

$A_2$ . The condition for equal rate of heat transfer is:

0%
$K_1A_2=K_2A_1$
0%
$K_1A_1=K_2A_2$
0%
$K_1A_1^2=K_2A_2^2$
0%
$K_1^2A_2=K_2^2A_1$

Explanation

Equal temperature difference exists between the two metallic rods 1 and 2.

The heat transfer

$=\dfrac{KA \Delta T}{l}$

For equal rate of heat transfer,

$\dfrac{K_1 A_1 \Delta T}{l}=\dfrac{K_2A_2 \Delta T}{l}$

$K_1A_1=K_2A_2$

The correct option is B.

If water at

$0^0C$ , kept in a container with an open top, is placed in a large evacuated chamber,

0%
all the water will freeze
0%
all the water will vaporize
0%
part of the water will vaporize and the rest will freeze
0%
ice, water and water vapour will be formed and reach equilibrium at the triple point

In a waterfall the water falls from a height of 100 m. If the entire energy of water is converted into heat. The rise in temperature of water will be:

0%
$0.3^{o} C$
0%
$0.46^{o} C$
0%
$2.3^{o} C$
0%
$0.23^{o} C$

Glaciers melt:

0%
first at the bootom due to decress in pressure
0%
first at the top due to iuncrease in pressure
0%
first at the bottom due to increase in pressure
0%
first at the top due to increase in pressure

Two moles of an ideal gas X occupying a volume V excert a pressure P. The same pressure is excert by one mole of another gas Y occupying a volume 2V. (if the molecular weight of Y is 16 times the molecular weight of X), find the ratio of the 'rms' speed of the molecular of X and Y.

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

The temperature of a diatomic gas is T. The total kinetic energy of the gas is given as

$E = { K }_{ 1 }$ where

${ K }_{ 1 }$ is constant. Find out the total number of molecules of the gas in the sample. (K = Boltzaman's constant)

0%
$3K_{ 1 }K$
0%
$5K_{ 1 }/2K$
0%
$2K_{ 1 }/5K$
0%
None of these

CBSE Questions for Class 11 Engineering Physics Thermal Properties Of Matter Quiz 9 - MCQExams.com

0:0:2

Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Thermal Properties Of Matter Quiz 9 - MCQExams.com

0:0:2

Practice Class 11 Engineering Physics Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.