JEE MAIN - Physics (2021 - 27th July Morning Shift - No. 24)
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)
(Young's modulus of rubber = 0.5 $$\times$$ 109 N/m2)
Answer
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
$${1 \over 2}.{{YA} \over L}.{x^2} = {1 \over 2}m{v^2}$$
$${{0.5 \times {{10}^9} \times {{10}^{ - 6}} \times {{(0.04)}^2}} \over {0.1}} = {{20} \over {1000}}{v^2}$$
$$\therefore$$ $${v^2} = 400$$
$$v = 20$$ m/s
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