 ### JEE Mains Previous Years Questions with Solutions

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1

### JEE Main 2021 (Online) 1st September Evening Shift

Numerical
A steel rod with y = 2.0 $$\times$$ 1011 Nm$$-$$2 and $$\alpha$$ = 10$$-$$5 $$^\circ$$C$$-$$1 of length 4 m and area of cross-section 10 cm2 is heated from 0$$^\circ$$C to 400$$^\circ$$C without being allowed to extend. The tension produced in the rod is x $$\times$$ 105 N where the value of x is ____________.

## Explanation

Given, the Young's modulus of the steel rod, Y = 2 $$\times$$ 1011 Pa

Thermal coefficient of the steel rod, $$\alpha$$ = 10$$-$$5$$^\circ$$C

The length of the steel rod, l = 4 m

The area of the cross-section, A = 10 cm2

The temperature difference, $$\Delta$$T = 400$$^\circ$$C

As we know that,

Thermal strain = $$\alpha$$ $$\Delta$$T

Using the Hooke's law

Young's modulus (Y) = $${{Thermal\,stress} \over {Thermal\,strain}} = {{F/A} \over {\alpha \,\Delta \,T}}$$

Thermal stress, $$F = YA\,\alpha \,\Delta \,T$$

Substitute the values in the above equation, we get

$$F = 2 \times {10^{11}} \times 10 \times {10^{ - 4}} \times {10^{ - 5}} \times (400)$$

$$= 8 \times {10^5}N$$

Comparing with, $$F = x \times {10^5}N$$

The value of the x = 8.
2

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

Numerical
When a rubber ball is taken to a depth of __________m in deep sea, its volume decreases by 0.5%. (The bulk modulus of rubber = 9.8 $$\times$$ 108 Nm$$-$$2, Density of sea water = 103 kgm$$-$$3, g = 9.8 m/s2)

## Explanation

$$B = - {{\Delta P} \over {\left( {{{\Delta V} \over V}} \right)}} = - {{\rho gh} \over {\left( {{{\Delta V} \over V}} \right)}}$$

$$- {{B{{\Delta V} \over V}} \over {\rho g}} = h$$

$${{9.8 \times {{10}^8} \times 0.5} \over {100 \times {{10}^3} \times 9.8}} = h$$

h = 500
3

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

Numerical
Wires W1 and W2 are made of same material having the breaking stress of 1.25 $$\times$$ 109 N/m2. W1 and W2 have cross-sectional area of 8 $$\times$$ 10$$-$$7 m2 and 4 $$\times$$ 10$$-$$7 m2, respectively. Masses of 20 kg and 10 kg hang from them as shown in the figure. The maximum mass that can be placed in the pan without breaking the wires is ____________ kg. (Use g = 10 m/s2) ## Explanation

B.S1 = $${{{T_{1\max }}} \over {8 \times {{10}^{ - 7}}}}$$ $$\Rightarrow$$ T1 max = 8 $$\times$$ 1.25 $$\times$$ 100 = 1000 N

B.S2 = $${{{T_{2\max }}} \over {4 \times {{10}^{ - 7}}}}$$ $$\Rightarrow$$ T2 max = 4 $$\times$$ 1.25 $$\times$$ 100 = 500 N

m = $${{500 - 100} \over {10}} = 40$$ kg
4

### JEE Main 2021 (Online) 26th August Morning Shift

Numerical
A soap bubble of radius 3 cm is formed inside the another soap bubble of radius 6 cm. The radius of an equivalent soap bubble which has the same excess pressure as inside the smaller bubble with respect to the atmospheric pressure is ................ cm.

## Explanation

Image

Excess pressure inside the smaller soap bubble

$$\Delta P = {{4S} \over {{r_1}}} + {{4S} \over {{r_2}}}$$ .... (i)

The excess pressure inside equivalent soap bubble

$$\Delta P = {{4S} \over {{R_{eq}}}}$$ ....... (ii)

From (i) & (ii)

$${{4S} \over {{R_{eq}}}} = {{4S} \over {{r_1}}} + {{4S} \over {{r_2}}}$$

$${1 \over {{R_{eq}}}} = {1 \over {{r_1}}} + {1 \over {{r_2}}}$$

$$= {1 \over 6} + {1 \over 3}$$

Req = 2 cm

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