JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Elasticity & SHM

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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 0.02 kg 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:

(A) $1 0^{3} Nm^{- 2}$
(B) $1 0^{6} Nm^{- 2}$ ✓ Correct
(C) $1 0^{8} Nm^{- 2}$
(D) $1 0^{4} Nm^{- 2}$

Answer: (B)

Solution

The elastic energy stored in the stretched cord is released wholly as kinetic energy of the stone, so

\frac{1}{2} \frac{Y A ( Δ ℓ ) ^{2}}{ℓ} = \frac{1}{2} m v^{2}

. Here

ℓ = 0.42

Δ ℓ = 0.20

m, the radius

r = 3 \times 1 0^{- 3}

m gives

A = π r^{2} = 9 π \times 1 0^{- 6} m^{2}

, with

m = 0.02

kg and

v = 20

m/s. Solving,

Y = \frac{m v ^{2} ℓ}{A ( Δ ℓ ) ^{2}} = \frac{0.02 \times 400 \times 0.42}{9 π \times 1 0 ^{- 6} \times 0.04} \approx 3 \times 1 0^{6} N/m^{2}

, closest in order of magnitude to

1 0^{6} N/m^{2}

JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Single correct option

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JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Elasticity & SHM

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