1

### JEE Main 2019 (Online) 10th January Morning Slot

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

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

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

### JEE Main 2019 (Online) 10th January Morning Slot

Three Carnot engines operate in series between a heat source at a temperature T1 and a heat sink at temperature T4 (see figure). There are two other reservoirs at temperature T2 and T3, as shown, with T1 > T2 > T3 > T4. The three engines are equally efficient if - A
T2 = (T13T4)1/4;  T3 = (T1T43)1/4
B
T2 = (T1T4)1/2;  T3 = (T12T4)1/3
C
T2 = (T1T42)1/3;  T3 = (T12T4)1/3
D
T2 = (T12T4)1/3;  T3 = (T1T42)1/3

## Explanation

${t_1} = 1 - {{{T_2}} \over {{T_1}}} = 1 - {{{T_2}} \over {{T_2}}} = 1 - {{{T_4}} \over {{T_3}}}$

$\Rightarrow \,\,\,{{{T_2}} \over {{T_1}}} = {{{T_3}} \over {{T_4}}} = {{{T_4}} \over {{T_3}}}$

$\Rightarrow \,\,\,{T_2} = \sqrt {{T_1}{T_3}} = \sqrt {{T_1}\sqrt {{T_2}{T_4}} }$

${T_3} = \sqrt {{T_2}{T_4}}$

$T_2^{3/4} = \sqrt {T_1^{1/2}} T_4^{1/4}$

${T_2} = T_1^{2/3}T_4^{1/3}$
3

### JEE Main 2019 (Online) 10th January Evening Slot

Half mole of an ideal monoatomic gas is heated at constant pressure of 1 atm from 20oC to 90oC. Work done by gas is close to – (Gas constant R = 8.31 J/mol.K)
A
581 J
B
73 J
C
146 J
D
291 J

## Explanation

WD = P$\Delta$V = nR$\Delta$T = ${1 \over 2} \times 8.31 \times 70$
4

### JEE Main 2019 (Online) 10th January Evening Slot

Two kg of a monoatomic gas is at a pressure of 4 $\times$ 104 N/m2. The density of the gas is 8 kg/m3. What is the order of energy of the gas due to its thermal motion ?
A
104 J
B
103 J
C
105 J
D
106 J

## Explanation

Thermal energy of N molecule

= N$\left( {{3 \over 2}kT} \right)$

= ${N \over {{N_A}}}{3 \over 2}$RT

= ${3 \over 2}$(nRT)

= ${3 \over 2}$PV

= ${3 \over 2}$P$\left( {{m \over 8}} \right)$

= ${3 \over 2}$ $\times$ 4 $\times$ 104 $\times$ ${2 \over 8}$

= 1.5 $\times$ 104

order will 104