8.Mechanical Properties of Solids
medium

A boy’s catapult is made of rubber cord which is $42\, cm$ long, with $6\, mm$ diameter of cross -section and of negligible mass. The boy keeps a stone weighing $0.02\, kg$ on it and stretches the cord by $20\, cm$ by applying a constant force. When released, the stone flies off with a velocity of $20\, ms^{-1}$. Neglect the change in the area of cross section of the cord while stretched. The Young’s modulus of rubber is closest to

A

$10^3\, Nm^{-2}$

B

$10^6\, Nm^{-2}$

C

$10^8\, Nm^{-2}$

D

$10^4\, Nm^{-2}$

(JEE MAIN-2019)

Solution

$Energy\,of\,catapult = \frac{1}{2} \times {\left( {\frac{{\Delta \ell }}{\ell }} \right)^2} \times Y \times A \times \ell $

$ = Kinetic\,energy\,of\,the\,ball = \frac{1}{2}\,m{V^2}$

$Therefore,\frac{1}{2} \times {\left( {\frac{{20}}{{42}}} \right)^2} \times Y \times \pi  \times {3^2} \times {10^{ – 6}} \times 42 \times {10^{ – 2}}$

$ = \frac{1}{2} \times 2 \times {10^{ – 2}} \times {\left( {20} \right)^2}$

$Y = 3 \times {10^{6\,}}\,N{m^2}$

Standard 11
Physics

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