Mechanical Properties of Solids

JEE Main level, numerical answer, medium

112 of 118

Question 112 of 118

A stone of mass 20g20g is projected from a rubber catapult of length 0.1m0.1m and area of cross section 10−6m2{10}^{-6}{m}^{2} stretched by an amount 0.04m.0.04m. The velocity of the projected stone is ms−1.m{s}^{-1}. (Young's modulus of rubber =0.5×109Nm−2=0.5\times {10}^{9}N{m}^{-2})

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Answer: 20

By energy conservation
12⋅YAL⋅x2=12mv2\dfrac{1}{2}\cdot \dfrac{YA}{L}\cdot {x}^{2}=\dfrac{1}{2}m{v}^{2}
0.5×109×10−6×(0.04)20.1=201000v2\dfrac{0.5\times {10}^{9}\times {10}^{-6}\times (0.04{)}^{2}}{0.1}=\dfrac{20}{1000}{v}^{2}
∴v2=400∴{v}^{2}=400
v=20ms−1v=20m{s}^{-1}

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