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### JEE Main 2021 (Online) 27th August Evening Shift

Numerical
A heat engine operates between a cold reservoir at temperature T2 = 400 K and a hot reservoir at temperature T1. It takes 300 J of heat from the hot reservoir and delivers 240 J of heat to the cold reservoir in a cycle. The minimum temperature of the hot reservoir has to be ______________ K.

## Explanation

Qin = 300 J ; Qout = 240 J

Work done = Qin $$-$$ Qout = 300 $$-$$ 240 = 60 J

Efficiency = $${W \over {{Q_{in}}}} = {{60} \over {300}} = {1 \over 5}$$

efficiency = $$1 - {{{T_2}} \over {{T_1}}}$$

$${1 \over 5} = 1 - {{400} \over {{T_1}}} \Rightarrow {{400} \over {{T_1}}} = {4 \over 5}$$

T1 = 500 k
2

### JEE Main 2021 (Online) 27th August Morning Shift

Numerical
A rod CD of thermal resistance 10.0 KW$$-$$1 is joined at the middle of an identical rod AB as shown in figure. The end A, B and D are maintained at 200$$^\circ$$C, 100$$^\circ$$C and 125$$^\circ$$C respectively. The heat current in CD is P watt. The value of P is ................. . ## Explanation Rods are identical so

RAB = RCD = 10 Kw$$-$$1

C is mid-point of AB, so

RAC = RCB = 5 Kw$$-$$1

at point C

$${{200 - T} \over 5} = {{T - 125} \over {10}} + {{T - 100} \over 5}$$

2(200 $$-$$ T) = T $$-$$ 125 + 2(T $$-$$ 100)

400 $$-$$ 2T = T $$-$$ 125 + 2T $$-$$ 200

$$T = {{725} \over 5}$$ = 145$$^\circ$$C

$${I_h} = {{145 - 125} \over {10}}w = {{20} \over {10}}w$$

Ih = 2w
3

### JEE Main 2021 (Online) 25th July Evening Shift

Numerical
A system consists of two types of gas molecules A and B having same number density 2 $$\times$$ 1025/m3. The diameter of A and B are 10 $$\mathop A\limits^o$$ and 5 $$\mathop A\limits^o$$ respectively. They suffer collision at room temperature. The ratio of average distance covered by the molecule A to that of B between two successive collision is ____________ $$\times$$ 10$$-$$2

## Explanation

$$\because$$ mean free path

$$\lambda = {1 \over {\sqrt 2 \pi {d^2}n}}$$

$${{{\lambda _1}} \over {{\lambda _2}}} = {{d_2^2{n_2}} \over {d_1^2{n_1}}}$$

$$= {\left( {{5 \over {10}}} \right)^2} = 0.25 = 25 \times {10^{ - 2}}$$
4

### JEE Main 2021 (Online) 22th July Evening Shift

Numerical
In 5 minutes, a body cools from 75$$^\circ$$C to 65$$^\circ$$C at room temperature of 25$$^\circ$$C. The temperature of body at the end of next 5 minutes is _________$$^\circ$$ C.

## Explanation

$${{75 - 65} \over 5} = k\left( {{{75 + 65} \over 2} - 25} \right)$$

$$\Rightarrow k = {2 \over {45}}$$

$${{65 - T} \over 5} = k\left( {{{65 + T} \over 2} - 25} \right)$$

$$\Rightarrow T = 57^\circ$$ C

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