Heat and Thermodynamics — Page 3 | JEE Physics

Q21

The relation between root mean square speed (vrms) and most probable sped (vp) for the molar mass M of oxygen gas molecule at the temperature of 300 K will be :

A

{v_{rms}} = \sqrt {{2 \over 3}} {v_p}

B

{v_{rms}} = \sqrt {{3 \over 2}} {v_p}

C

{v_{rms}} = {v_p}

D

{v_{rms}} = \sqrt {{1 \over 3}} {v_p}

Correct Answer

Option B

Solution

{v_{rms}} = \sqrt {{{3RT} \over M}}

{v_p} = \sqrt {{{2RT} \over M}}

\Rightarrow {v_{rms}} = \sqrt {{3 \over 2}} {v_p}

Q22

A Carnot engine takes 5000 kcal of heat from a reservoir at 727

^\circ

C and gives heat to a sink at 127

^\circ

C. The work done by the engine is

A 3

\times

106 J

B Zero

C 12.6

\times

106 J

D 8.4

\times

106 J

Correct Answer

Option C

Solution

Efficiency

\eta = 1 - {{{T_L}} \over {{T_H}}}

= 1 - {{400} \over {1000}} = 0.6

\Rightarrow 0.6 = {W \over Q}

\Rightarrow W = 0.6Q = 3000

kcal

=12.6 \times {10^6}

J

Q23

Two moles of helium are mixed with n moles of hydrogen. If

{{Cp} \over {Cv}} = {3 \over 2}

for the mixture, then the value of n is :

A 1

B 3

C 2

D 3 / 2

Correct Answer

Option C

Solution

{{{C_p}} \over {{C_v}}} = {{{f_{mix}} + 2} \over {{f_{mix}}}} = {3 \over 2}

$\Rightarrow$

\,\,\,

fmix = 4 As, fmix =

{{{n_1}{f_1} + {n_2}{f_2}} \over {{n_1} + {n_2}}}

$\Rightarrow$

\,\,\,

4 =

{{2 \times 3 + n \times 5} \over {2 + n}}

$\Rightarrow$

\,\,\,

n = 2 moles.

Q24

If the temperature of the sun were to increase from

T

to

2T

and its radius from

R

to

2R

, then the ratio of the radiant energy received on earth to what it was previously will be

A

32

B

16

C

4

D

64

Correct Answer

Option D

Solution

E = \sigma A{T^4};\,\,A \propto {R^2}

$\therefore$

E \propto {R^2}{T^4}

$\therefore$

{{{E_2}} \over {{E_1}}} = {{R_2^2T_2^4} \over {R_1^2T_1^4}}

\Rightarrow {{{E_2}} \over {{E_1}}}

= {{{{\left( {2R} \right)}^2}{{\left( {2T} \right)}^4}} \over {{R^2}{T^4}}}

= 64

Q25

A heat source at T = 103 K is connected to another heat reservoir at T = 102 K by a copper slab which is 1 mthick. Given that the thermal conductivity of copper is 0.1 WK–1m–1, the energy flux through it in the steady state is -

A 200 Wm

-

2

B 65 Wm

-

2

C 120 Wm

-

2

D 90 Wm

-

2

Correct Answer

Option D

Solution

\left( {{{dQ} \over {dt}}} \right) = {{kA\Delta T} \over \ell }

$\Rightarrow$

{1 \over A}\left( {{{dQ} \over {dt}}} \right) = {{\left( {0.1} \right)\left( {900} \right)} \over 1} = 90W/{m^2}

Q26

One mole of ideal monatomic gas

\left( {\gamma = 5/3} \right)

is mixed with one mole of diatomic gas

\left( {\gamma = 7/5} \right)

. What is

\gamma

for the mixture?

\gamma

Denotes the ratio of specific heat at constant pressure, to that at constant volume

A

35/23

B

23/15

C

3/2

D

4/3

Correct Answer

Option C

Solution

{{{n_1} + {n_2}} \over {\gamma - 1}} = {{{n_1}} \over {{\gamma _1} - 1}} + {{{n_2}} \over {{\gamma _2} - 1}}

\Rightarrow {{1 + 1} \over {\gamma - 1}}

= {1 \over {{5 \over 3} - 1}} + {1 \over {{7 \over 5} - 1}}

\Rightarrow \gamma = {3 \over 2}

Q27

Time taken by a

836

W

heater to heat one litre of water from

10{}^ \circ C

to

40{}^ \circ C

is

A

150

s

B

100

s

C

50

s

D

200

s

Correct Answer

Option A

Solution

\Delta Q = mC \times \Delta T

= 1 \times 4180 \times \left( {40 - 10} \right) = 80 \times 30

( $\therefore$

\Delta Q =

heat supplied in time

t

for heating

1L

water from

{10^ \circ }C

to

{40^ \circ }C

) also

\Delta Q = 836 \times t \Rightarrow t = {{4180 \times 30} \over {836}} = 150\,s

Q28

When 100 g of a liquid A at 100oC is added to 50 g of a liquid B at temperature 75oC, the temperature of the mixture becomes 90oC. The temperature of the mixture, if 100 g of liquid A at 100oC is added to 50 g of liquid B at 50oC, will be :

A 60oC

B 70oC

C 85oC

D 80oC

Correct Answer

Option D

Solution

100 $\times$ SA $\times$ [100 $-$ 90] = 50 $\times$ SB $\times$ (90 $-$ 75) 2SA = 1.5 SB SA =

{3 \over 4}

SB Now, 100 $\times$ SA $\times$ [100 $-$ T] = 50 $\times$ SB (T $-$ 50) 2 $\times$

\left( {{3 \over 4}} \right)

(100 $-$ T) = (T $-$ 50) 300 $-$ 3T = 2T $-$ 100 400 = 5T T = 80

Q29

Which statement is incorrect?

A all reversible cycles have same efficiency

B reversible cycle has more efficiency than an irreversible one

C Cannot cycle is a reversible one

D Cannot cycle has the maximum efficiency in all cycles.

Correct Answer

Option A

Solution

All reversible engines working for the same temperature of source and sink have same efficiencies.

If the temperatures are different, the efficiency is different.

Q30

Infrared radiation is detected by

A spectrometer

B pyrometer

C nanometer

D photometer

Correct Answer

Option B

Solution

Pyrometer is used to detect infra-red radiation.