Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
A rubber cord catapult has cross-sectional area $25 \mathrm{~mm}^2$ and initial length of rubber cord is $10 \mathrm{~cm}$ It is stretched to ${ }^{5 \mathrm{~cm}}$ and then released to project a missile of mass $5 \mathrm{gm}$ Taking $Y_{\text {rubber }}=5 \times 10^8 \mathrm{~N} / \mathrm{m}^2$ velocity of projected missile is
PhysicsMechanical Properties of SolidsJEE Main
Options:
  • A $20 \mathrm{~ms}^{-1}$
  • B $100 \mathrm{~ms}^{-1}$
  • C $250 \mathrm{~ms}^{-1}$
  • D $200 \mathrm{~ms}^{-1}$
Solution:
1937 Upvotes Verified Answer
The correct answer is: $250 \mathrm{~ms}^{-1}$
Potential energy stored in the rubber cord catapult will be converted into kinetic energy of mass.
$\frac{1}{2} m v^2=\frac{1}{2} \frac{Y A l^2}{L} \Rightarrow v=\sqrt{\frac{Y A l^2}{m L}}$
$=\sqrt{\frac{5 \times 10^8 \times 25 \times 10^{-6} \times\left(5 \times 10^{-2}\right)^2}{5 \times 10^{-3} \times 10 \times 10^{-2}}}=250 \mathrm{~m} / \mathrm{s}$

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.