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QJEEM19Ph0548: JEE Main 2019 Physics - Class 11 JEEM19Ph0548 Solution: A boy's catapult is made of rubber cord which is 42 cm long,…

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:

Multiple Choice Options

  • A 104 Nm–2
  • B 106 Nm–2
  • C 108 Nm–2
  • D 103 Nm–2

Step-by-Step Verified Solution

When rubber cord is stretched, then it stores potential energy and when released, this potential energy is given to the stone as kinetic energy.

JEE Main 2019 (Online) 8th April Morning Slot Physics - Properties of Matter Question 218 English Explanation

So, potential energy of stretched cord = kinetic energy of stone

⇒12Y(ΔLL)2A . L=12mv2\Rightarrow {1 \over 2}Y{\left( {{{\Delta L} \over L}} \right)^2}A\,.\,L = {1 \over 2}m{v^2}

Here, Δ\Delta L = 20 cm = 0.2 m, L = 42 cm = 0.42 m, v = 20 ms −- 1, m = 0.02 kg, d = 6 mm = 6 ×\times 10 −- 3 m

∴\therefore A=πr2=π(d2)2=π(6×10−32)2A = \pi {r^2} = \pi {\left( {{d \over 2}} \right)^2} = \pi {\left( {{{6 \times {{10}^{ - 3}}} \over 2}} \right)^2}

=π(3×10−3)2=9π×10−6m2= \pi {(3 \times {10^{ - 3}})^2} = 9\pi \times {10^{ - 6}}{m^2}

On substituting values, we get

Y=mv2LA(ΔL)2=0.02×(20)2×0.429π×10−6×(0.2)2Y = {{m{v^2}L} \over {A{{(\Delta L)}^2}}} = {{0.02 \times {{(20)}^2} \times 0.42} \over {9\pi \times {{10}^{ - 6}} \times {{(0.2)}^2}}}

≈3.0×106Nm−2\approx 3.0 \times {10^6}N{m^{ - 2}}

So, the closest value of Young's modulus is 106 Nm −- 2.

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