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JEE Mains · Physics · STD 11- Mechanical properties of solids
  • JEE Mains 2019
  • Medium
  • MCQ

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

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Answer & Solution

Correct Answer

(B) \(10^6\, Nm^{-2}\)

Step-by-step Solution

Detailed explanation

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

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

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

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

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

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