Heat and Thermodynamics — Page 17 | JEE Physics

Q161

A lead bullet penetrates into a solid object and melts. Assuming that 40% of its kinetic energy is used to heat it, the initial speed of bullet is : (Given : initial temperature of the bullet = 127

^\circ

C, Melting point of the bullet = 327

^\circ

C, Latent heat of fusion of lead = 2.5

\times

104 J kg

-

1, Specific heat capacity of lead = 125 J/kg K)

A 125 ms

-

1

B 500 ms

-

1

C 250 ms

-

1

D 600 ms

-

1

Correct Answer

Option B

Solution

{2 \over 5} \times {1 \over 2}m{v^2} = mL + ms\Delta T

\Rightarrow {{{v^2}} \over 5} = 2.5 \times {10^4} + 125 + 200

\Rightarrow {{{v^2}} \over 5} = 5 \times {10^4}

\Rightarrow v = 500

m/s

Q162

A mixture of hydrogen and oxygen has volume 2000 cm3, temperature 300 K, pressure 100 kPa and mass 0.76 g. The ratio of number of moles of hydrogen to number of moles of oxygen in the mixture will be: [Take gas constant R = 8.3 JK

-

1mol

-

1]

A

{1 \over 3}

B

{3 \over 1}

C

{1 \over 16}

D

{16 \over 1}

Correct Answer

Option B

Solution

{P_1}V = {n_1}RT

{P_2}V = {n_2}RT

$\Rightarrow$ (100 kPa) V = (n1 + n2)RT

\Rightarrow {n_1} + {n_2} = {{(100\,kPa)(2000\,c{m^3})} \over {8.3 \times 300}}

..... (1) Also, n1 $\times$ 2 + n2 $\times$ 32 = 0.76 ...... (2) Solving (1) and (2), n1 = 0.06 n2 = 0.02

\Rightarrow {{{n_1}} \over {{n_2}}} = 3

Q163

The efficiency of a Carnot's engine, working between steam point and ice point, will be :

A 26.81%

B 37.81%

C 47.81%

D 57.81%

Correct Answer

Option A

Solution

\eta = 1 - {{{T_2}} \over {{T_1}}} = 1 - {{273} \over {373}}

= 0.26809

\approx 26.81\%

Q164

A thermally insulated vessel contains an ideal gas of molecular mass M and ratio of specific heats 1.4. Vessel is moving with speed v and is suddenly brought to rest. Assuming no heat is lost to the surrounding and vessel temperature of the gas increases by : (R = universal gas constant)

A

{{M{v^2}} \over {7R}}

B

{{M{v^2}} \over {5R}}

C 2

{{M{v^2}} \over {7R}}

D 7

{{M{v^2}} \over {5R}}

Correct Answer

Option B

Solution

Let there be n moles of gas Eloss = Egain

{1 \over 2}(nM){v^2} = n{C_v}\Delta T

{1 \over 2}M{v^2} = {C_v}\Delta T

here,

\gamma = 1.4 = {7 \over 5}

i.e. diatomic gas $\therefore$

{C_v} = {{5R} \over 2}

Now,

{1 \over 2}M{v^2} = {{5R} \over 2}\Delta T

\Delta T = {{M{v^2}} \over {5R}}

Q165

A flask contains hydrogen and oxygen in the ratio of

2: 1

by mass at temperature

27^{\circ} \mathrm{C}

. The ratio of average kinetic energy per molecule of hydrogen and oxygen respectively is:

A 1 : 1

B 4 : 1

C 1 : 4

D 2 : 1

Correct Answer

Option A

Solution

K.E. per molecule

= \left( {{f \over 2}KT} \right)

{{\mathrm{average{{(K.E)}_{hydrogen}}}} \over {\mathrm{average{{(K.E)}_{oxygen}}}}} = {{{f_{\mathrm{hydrogen}}}} \over {{f_{\mathrm{oxygen}}}}} = 1

Q166

A solid metallic cube having total surface area 24 m2 is uniformly heated. If its temperature is increased by 10

^\circ

C, calculate the increase in volume of the cube. (Given

\alpha

= 5.0

\times

10

-

4

^\circ

C

-

1).

A 2.4

\times

106 cm3

B 1.2

\times

105 cm3

C 6.0

\times

104 cm3

D 4.8

\times

105 cm3

Correct Answer

Option B

Solution

6 \times {l^2} = 24

\Rightarrow l = 2

m $\therefore$

{{\Delta V} \over V} = 3 \times {{\Delta l} \over l}

\Rightarrow \Delta V = 3 \times (\alpha \Delta T) \times V

= 3 \times 5 \times {10^{ - 4}} \times 10 \times (8)

= 120 \times {10^{ - 3}}

m3

= 120 \times {10^{ - 3}} \times {10^6}

cm3

= 1.2 \times {10^5}

cm3

Q167

A copper block of mass 5.0 kg is heated to a temperature of 500

^\circ

C and is placed on a large ice block. What is the maximum amount of ice that can melt? [Specific heat of copper : 0.39 J g

-

1

^\circ

C

-

1 and latent heat of fusion of water : 335 J g

-

1]

A 1.5 kg

B 5.8 kg

C 2.9 kg

D 3.8 kg

Correct Answer

Option C

Solution

mL = \Delta Q = ms\Delta T

\Rightarrow m = {{5 \times 0.39 \times {{10}^3} \times 500} \over {335}}

= 2.9

kg

Q168

A Carnot engine whose heat sinks at 27

^\circ

C, has an efficiency of 25%. By how many degrees should the temperature of the source be changed to increase the efficiency by 100% of the original efficiency?

A Increases by 18

^\circ

C

B Increases by 200

^\circ

C

C Increases by 120

^\circ

C

D Increases by 73

^\circ

C

Correct Answer

Option B

Solution

Initially :

{1 \over 4} = 1 - {{300} \over {{T_H}}}

\Rightarrow {T_H} = 400\,K

Finally : Efficiency becomes

{1 \over 2}

\Rightarrow {1 \over 2} = 1 - {{300} \over {T{'_H}}}

\Rightarrow T{'_H} = 600\,K

$\Rightarrow$ Temperature of the source increases by 200

^\circ

C.

Q169

The pressure of the gas in a constant volume gas thermometer is 100 cm of mercury when placed in melting ice at 1 atm. When the bulb is placed in a liquid, the pressure becomes 180 cm of mercury. Temperature of the liquid is : (Given 0

^\circ

C = 273 K)

A 300 K

B 400 K

C 600 K

D 491 K

Correct Answer

Option D

Solution

Here volume is constant. $\therefore$

{{{P_1}} \over {{T_1}}} = {{{P_2}} \over {{T_2}}}

\Rightarrow {{100} \over {273}} = {{180} \over {{T_2}}}

\Rightarrow {T_2} = {{180} \over {100}} \times 273

= 1.8 \times 273

= 491\,K

Q170

A sample of monoatomic gas is taken at initial pressure of 75 kPa. The volume of the gas is then compressed from 1200 cm3 to 150 cm3 adiabatically. In this process, the value of workdone on the gas will be :

A 79 J

B 405 J

C 4050 J

D 9590 J

Correct Answer

Option B

Solution

For monoatomic gas degree of freedom f = 3 and $\gamma$ =

{5 \over 3}

Here for gas, Initial pressure (P1) = 75 kPa Initial volume (V1) = 1200 cm3 Final volume (V2) = 150 cm3 Final pressure (P2) = ?

For adiabatic process,

{P_1}V_1^\gamma = {P_2}V_2^\gamma

\Rightarrow 75 \times {(1200)^\gamma } = {P_2}{(150)^\gamma }

\Rightarrow {P_2} = 75 \times {\left( {{{1200} \over {150}}} \right)^{{5 \over 3}}}

= 75 \times {(8)^{{5 \over 3}}}

= 75 \times 32

kPa = 2400 kPa Work done in adiabatic process,

W = {{{P_2}{V_2} - {P_1}{V_1}} \over {1 - \gamma }}

= {{2400 \times {{10}^3} \times 150 \times {{10}^{ - 6}} - 75 \times {{10}^3} \times 1200 \times {{10}^{ - 6}}} \over {1 - {5 \over 3}}}

= {{(2400 \times 150 - 75 \times 1200) \times {{10}^{ - 3}}} \over { - {2 \over 3}}}

= {{ - 810000 \times {{10}^{ - 3}}} \over 2}

= - 405

kJ