 ### JEE Mains Previous Years Questions with Solutions

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1

### JEE Main 2021 (Online) 31st August Evening Shift

A mixture of hydrogen and oxygen has volume 500 cm3, temperature 300 K, pressure 400 kPa and mass 0.76 g. The ratio of masses of oxygen to hydrogen will be :-
A
3 : 8
B
3 : 16
C
16 : 3
D
8 : 3

## Explanation

PV = nRT

400 $$\times$$ 103 $$\times$$ 500 $$\times$$ 10$$-$$6 = n$$\left( {{{25} \over 3}} \right)$$ (300)

n = $${{2 \over {25}}}$$

n = n1 + n2

$${{2 \over {25}}}$$ = $${{{M_1}} \over 2} + {{{M_2}} \over {32}}$$

Also, M1 + M2 = 0.76 gm

$${{{M_2}} \over {{M_1}}} = {{16} \over 3}$$
2

### JEE Main 2021 (Online) 31st August Evening Shift

Two thin metallic spherical shells of radii r1 and r2 (r1 > r2) are placed with their centres coinciding. A material of thermal conductivity K is filled in the space between the shells. The inner shell is maintained at temperature $$\theta$$1 and the outer shell at temperature $$\theta$$2($$\theta$$1 < $$\theta$$2). The rate at which heat flows radially through the material is :-
A
$${{4\pi K{r_1}{r_2}({\theta _2} - {\theta _1})} \over {{r_2} - {r_1}}}$$
B
$${{\pi {r_1}{r_2}({\theta _2} - {\theta _1})} \over {{r_2} - {r_1}}}$$
C
$${{K({\theta _2} - {\theta _1})} \over {{r_2} - {r_1}}}$$
D
$${{K({\theta _2} - {\theta _1})({r_2} - {r_1})} \over {4\pi {r_1}{r_2}}}$$

## Explanation Thermal resistance of spherical sheet of thickness dr and radius r is

$$dR = {{dr} \over {K(4\pi {r^2})}}$$

$$R = \int\limits_{{r_1}}^{{r_2}} {{{dr} \over {K(4\pi {r^2})}}}$$

$$R = {1 \over {4\pi K}}\left( {{1 \over {{r_1}}} - {1 \over {{r_2}}}} \right) = {1 \over {4\pi K}}\left( {{{{r_2} - {r_1}} \over {{r_1}{r_2}}}} \right)$$

Thermal current (i) $$= {{{\theta _2} - {\theta _1}} \over R}$$

$$i = {{4\pi K{r_1}{r_2}} \over {{r_2} - {r_1}}}({\theta _2} - {\theta _1})$$
3

### JEE Main 2021 (Online) 31st August Morning Shift

For an ideal gas the instantaneous change in pressure 'p' with volume 'v' is given by the equation $${{dp} \over {dv}} = - ap$$. If p = p0 at v =0 is the given boundary condition, then the maximum temperature one mole of gas can attain is : (Here R is the gas constant)
A
$${{{p_0}} \over {aeR}}$$
B
$${{a{p_0}} \over {eR}}$$
C
infinity
D
0$$^\circ$$C

## Explanation

$$\int\limits_{{p_0}}^p {{{dp} \over P} = - a\int\limits_0^v {dv} }$$

$$\ln \left( {{p \over {{p_0}}}} \right) = - av$$

$$p = {p_0}{e^{ - av}}$$

For temperature maximum p-v product should be maximum

$$T = {{pv} \over {nR}} = {{{p_0}v{e^{ - av}}} \over R}$$

$${{dT} \over {dv}} = 0 \Rightarrow {{{p_0}} \over R}\{ {e^{ - av}} + v{e^{ - av}}( - a)\}$$ = 0

$${{{p_0}{e^{ - av}}} \over R}\{ 1 - av\} = 0$$

$$v = {1 \over a},\infty$$

$$T = {{{p_0}1} \over {Rae}} = {{{p_0}} \over {Rae}}$$

at v = $$\infty$$

T = 0

Option (a)
4

### JEE Main 2021 (Online) 31st August Morning Shift

A reversible engine has an efficiency of $${1 \over 4}$$. If the temperature of the sink is reduced by 58$$^\circ$$C, its efficiency becomes double. Calculate the temperature of the sink :
A
174$$^\circ$$C
B
280$$^\circ$$C
C
180.4$$^\circ$$C
D
382$$^\circ$$C

## Explanation

T2 = sink temperature

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

$${1 \over 4} = 1 - {{{T_2}} \over {{T_1}}}$$

$${{{T_2}} \over {{T_1}}} = {3 \over 4}$$ .... (i)

$${1 \over 2} = 1 - {{{T_2} - 58} \over {{T_1}}}$$

$${{{T_2}} \over {{T_1}}} = {{58} \over {{T_1}}} = {1 \over 2}$$

$${3 \over 4} = {{58} \over {{T_1}}} + {1 \over 2}$$

$${1 \over 4} = {{58} \over {{T_1}}} \Rightarrow {T_1} = 232$$

$${T_2} = {3 \over 4} \times 232$$

$${T_2} = 174$$ K

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