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JEE Main 2019 (8 Apr, Shift 1) · Elastic Potential Energy

Elastic Potential Energy question from JEE Main 2019 (8 Apr, Shift 1)

Q13JEE Main 2019 · 8 APR · SHIFT 1Medium

Elastic Potential Energy

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 0.02kg 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: (b)

  1. Energy stored in the cord becomes kinetic energy: 12mv2=12×0.02×400=4 J\dfrac{1}{2}mv^2 = \dfrac{1}{2} \times 0.02 \times 400 = 4\text{ J}.
  2. Stored energy =YAx22L= \dfrac{YAx^2}{2L}, with A=π(3×10−3)2=2.83×10−5 m2A = \pi (3 \times 10^{-3})^2 = 2.83 \times 10^{-5}\text{ m}^2.
  3. Y=2×4×0.422.83×10−5×(0.2)2≈3×106 N m−2Y = \dfrac{2 \times 4 \times 0.42}{2.83 \times 10^{-5} \times (0.2)^2} \approx 3 \times 10^6\text{ N m}^{-2}
  4. This is of order 10610^6. So the answer is (b) 106 N m−210^6\text{ N m}^{-2}.
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