1

### JEE Main 2019 (Online) 9th January Morning Slot

Temperature difference of 120oC is maintained between ends of a uniform rod AB of length 2L. Another bent rod PQ, of same cross-section as AB and length ${{3L} \over 2},$ is connected across AB (see figure). In steady state, temperature difference between P and Q will be close to : A
45oC
B
75oC
C
60oC
D
35oC

## Explanation

We know,

Resistance, R = ${{\rho L} \over A}$

$\therefore$   R $\propto$ L

So, Resistance is directly proportional to lengt5h of the rod. Equivalent resistance between A and B is,

Req = ${R \over 2} + {{R \times {{3R} \over 2}} \over {R + {{3R} \over 2}}} + {R \over 2}$

= R + ${{3{R^2}} \over 2} \times {2 \over {5R}}$

= R + ${{3R} \over 5}$

= ${{8R} \over 5}$

Thermal current between point A and B is,

I = ${{\Delta {T_{AB}}} \over {{{\mathop{\rm R}\nolimits} _{eq}}}}$

= ${{120 - 0} \over {{{8R} \over 5}}}$

= ${{120 \times 5} \over {8R}}$

Resistance between P and Q = ${{3R} \over 5}$

$\therefore$   $\Delta$TPQ = ${\rm I} \times {{3R} \over 5}$

= ${{120 \times 5} \over {8R}} \times {{3R} \over 5}$

= 45oC
2

### JEE Main 2019 (Online) 9th January Morning Slot

A gas can be taken from A to B via two different processes ACB and ADB. When path ACB is used 60 J of heat flows into the system and 30 J of work is done by the system. If path ADB is used work done by the system is 10 J. The heat Flow into the system in path ADB is :
A
40 J
B
80 J
C
100 J
D
20 J

## Explanation

Using law of thermodynamics in path ACB,

$\Delta$QACB = $\Delta$UACB + $\Delta$WACB

$\Rightarrow$   60 = $\Delta$UACB + 30

$\Rightarrow$   $\Delta$UACB = 30 J

As value of $\Delta$U is path independent,

$\therefore$   $\Delta$UACB = $\Delta$UADB = 30 J

$\therefore$   In path ADB,

$\Delta$QADB = $\Delta$UADB + $\Delta$WADB

$\Rightarrow$$\Delta$QADB = 30 + 10 = 40 J
3

### JEE Main 2019 (Online) 9th January Evening Slot

Two Carnot engines A and B are operated in series. The first one, A, receives heat at T1 (= 600 K) and rejects to a reservoir at temperature T2. The second engine B receives heat rejected by the first engine and, in tum, rejects to a heat reservoir at T3 (=400 K). Calculate the temperature T2 if the work outputs of the two engines are equal :
A
600 K
B
400 K
C
300 K
D
500 K

## Explanation Here, Q1 = W1 + Q2

and Q2 = W2 + Q3

Given that,

W1 = W2

$\therefore$  Q1 $-$ Q2 = Q2 $-$ Q3

$\Rightarrow$  nCv(T1 $-$ T2) = nCv(T2 $-$ T3)

$\Rightarrow$  T1 $-$ T2 = T2 $-$ T3

$\Rightarrow$   T2 = ${{{T_1} + {T_3}} \over 2}$

= ${{600 + 400} \over 2}$

= 500 K
4

### JEE Main 2019 (Online) 9th January Evening Slot

A 15 g mass of nitrogen gas is enclosed in a vessel at a temperature 27oC. Amount of heat transferred to the gas, so that rms velocity of molecules is doubled, is about : [Take R = 8.3 J/K mole]
A
0.9 kJ
B
6 kJ
C
10 kJ
D
14 kJ

## Explanation

We know,

Vrms $\propto$ $\sqrt T$

So, to make Vrms double we have to make temperature 4 times.

$\therefore$   Final temperature = 300 $\times$ 4 = 1200 K

As N2 gas present in the closed vessel

So it is a isochoric process.

$\therefore$   Q = nCv $\Delta$ T

= ${{15} \over {28}} \times \left( {{5 \over 2}R} \right)\left( {1200 - 300} \right)$

= 10000 J

= 10 kJ