AP EAMCET · PHYSICS · Mechanical Properties of Solids
A rubber band catapult has initial length \(2 \mathrm{~cm}\) and cross-sectional area \(5 \mathrm{~mm}^2\). It is stretched to \(2 \mathrm{~cm}\) and then released to project a stone of mass of \(20 \mathrm{~g}\). The velocity of projected stone is
(Young's modulus of rubber \(=5 \times 10^8 \mathrm{Nm}^{-2}\) )
- A \(20 \mathrm{~ms}^{-1}\)
- B \(50 \mathrm{~ms}^{-1}\)
- C \(100 \mathrm{~ms}^{-1}\)
- D \(250 \mathrm{~ms}^{-1}\)
Answer & Solution
Correct Answer
(B) \(50 \mathrm{~ms}^{-1}\)
Step-by-step Solution
Detailed explanation
According to work-energy theorem, \(\mathrm{K} \mathrm{E}\) of stone \(=\) Elastic potential energy of rubber band. Now, for rubber band; Young's modulus \(=5 \times 10^8 \mathrm{~N} / \mathrm{m}^2\) Length, \(L=2 \times 10^{-2} \mathrm{~m}\) Change of length,…
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