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1

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

The height of victoria falls is 63 m. What is the difference in temperature of water at the top and at the bottom of fall?

[Given 1 cal = 4.2 J and specific heat of water = 1 cal g$$-$$1 $$^\circ$$0C$$-$$1]
A
0.147$$^\circ$$ C
B
14.76$$^\circ$$ C
C
1.476$$^\circ$$ C
D
0.014$$^\circ$$ C

## Explanation

Change in P.E. = Heat energy

mgh = mS$$\Delta$$T

$$\Delta$$T = $${{gh} \over S}$$

= $${{10 \times 63} \over {4200J/kgC}}$$

= 0.147$$^\circ$$C
2

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

if the rms speed of oxygen molecules at 0$$^\circ$$C is 160 m/s, find the rms speed of hydrogen molecules at 0$$^\circ$$C.
A
640 m/s
B
40 m/s
C
80 m/s
D
332 m/s

## Explanation

$${V_{rms}} = \sqrt {{{3KT} \over M}}$$

$${{{{({V_{rms}})}_{{O_2}}}} \over {{{({V_{rms}})}_{{H_2}}}}} = \sqrt {{{{M_{{H_2}}}} \over {{M_{{O_2}}}}}} = \sqrt {{2 \over {32}}}$$

$${({V_{rms}})_{{H_2}}} = 4 \times {({V_{rms}})_{{O_2}}}$$

$$= 4 \times 160$$

$$= 640$$ m/s
3

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

An ideal gas is expanding such that PT3 = constant. The coefficient of volume expansion of the gas is :
A
$${1 \over T}$$
B
$${2 \over T}$$
C
$${4 \over T}$$
D
$${3 \over T}$$

## Explanation

PT3 = constant

$$\left( {{{nRT} \over v}} \right)$$T3 = constant

T4 V$$-$$1 = constant

T4 = kV

$$\Rightarrow 4{{\Delta T} \over T} = {{\Delta V} \over V}$$ ....... (1)

$$\Delta$$V = V$$\gamma$$$$\Delta$$T ........ (2)

comparing (1) and (2), we get

$$\gamma = {4 \over T}$$
4

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

A balloon carries a total load of 185 kg at normal pressure and temperature of 27$$^\circ$$C. What load will the balloon carry on rising to a height at which the barometric pressure is 45 cm of Hg and the temperature is $$-$$7$$^\circ$$C. Assuming the volume constant?
A
181.46 kg
B
214.15 kg
C
219.07 kg
D
123.54 kg

## Explanation

Pm = $$\rho$$RT

$$\therefore$$ $${{{P_1}} \over {{P_2}}} = {{{\rho _1}{T_1}} \over {{\rho _1}{T_2}}}$$

$${{{\rho _1}} \over {{\rho _2}}} \Rightarrow {{{P_1}{T_2}} \over {{P_2}{T_1}}} = \left( {{{76} \over {45}}} \right) \times {{266} \over {300}}$$

$${{{\rho _1}} \over {{\rho _2}}} \Rightarrow {{{M_1}} \over {{M_2}}} = {{76 \times 266} \over {45 \times 300}}$$

$$\therefore$$ $${M_2} \Rightarrow {{45 \times 300 \times 185} \over {76 \times 266}} = 123.54$$ kg

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