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JEE Mains · Physics · STD 11- 8. mechanical properties of solids

A stone of mass \(20\, {g}\) is projected from a rubber catapult of length \(0.1\, {m}\) and area of cross section \(10^{-6} \,{m}^{2}\) stretched by an amount \(0.04\, {m}\). The velocity of the projected stone is \(....\,m\,/s.\) (Young's modulus of rubber \(=0.5 \times 10^{9}\, {N} / {m}^{2}\) )

  1. A \(10\)
  2. B \(15\)
  3. C \(25\)
  4. D \(20\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(20\)

Step-by-step Solution

Detailed explanation

\(\frac{1}{2} \cdot k \cdot x ^{2}=\frac{1}{2} \cdot \frac{ YA }{ L }\cdot x ^{2}\) By energy conservation \(\frac{1}{2} \cdot \frac{ YA }{ L } \cdot x ^{2}=\frac{1}{2} mv ^{2}\) \(\frac{0.5 \times 10^{9} \times 10^{-6} \times(0.04)^{2}}{0.1}=\frac{20}{1000} v ^{2}\)…
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