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1

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)

Your Input ________

Answer

Correct Answer is 40

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
2

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.
Your Input ________

Answer

Correct Answer is 2

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
3

JEE Main 2021 (Online) 27th July Evening Shift

Numerical
The water is filled upto height of 12 m in a tank having vertical sidewalls. A hole is made in one of the walls at a depth 'h' below the water level. The value of 'h' for which the emerging steam of water strikes the ground at the maximum range is ________ m.
Your Input ________

Answer

Correct Answer is 6

Explanation


$$R = \sqrt {2gh} \times \sqrt {{{(12 - h) \times 2} \over g}} $$

$$\sqrt {4h(12 - h)} = R$$

For maximum R

$${{dR} \over {dh}} = 0$$

$$\Rightarrow$$ h = 6 m
4

JEE Main 2021 (Online) 27th July Morning Shift

Numerical
A stone of mass 20 g is projected from a rubber catapult of length 0.1 m and area of cross section 10$$-$$6 m2 stretched by an amount 0.04 m. The velocity of the projected stone is ______________ m/s.

(Young's modulus of rubber = 0.5 $$\times$$ 109 N/m2)
Your Input ________

Answer

Correct Answer is 20

Explanation

By energy conservation

$${1 \over 2}.{{YA} \over L}.{x^2} = {1 \over 2}m{v^2}$$

$${{0.5 \times {{10}^9} \times {{10}^{ - 6}} \times {{(0.04)}^2}} \over {0.1}} = {{20} \over {1000}}{v^2}$$

$$\therefore$$ $${v^2} = 400$$

$$v = 20$$ m/s

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