Medium INTEGER +4 / -1 PYQ · JEE Mains 2021

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)

Answer (integer) 20

Solution

By energy conservation<br><br>${1 \over 2}.{{YA} \over L}.{x^2} = {1 \over 2}m{v^2}$<br><br>$${{0.5 \times {{10}^9} \times {{10}^{ - 6}} \times {{(0.04)}^2}} \over {0.1}} = {{20} \over {1000}}{v^2}$$<br><br>$\therefore$ ${v^2} = 400$<br><br>$v = 20$ m/s

About this question

Subject: Physics · Chapter: Properties of Solids and Liquids · Topic: Elasticity

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