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 \(=0.5 \times 10^{9}\, {N} / {m}^{2}\) )
- A \(10\)
- B \(15\)
- C \(25\)
- D \(20\)
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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