First slide

Elastic Modulie

Question

The Young's modulus of a rope of 10m length and having diameter of 2 cm is 20 x 10¹¹ dyne cm^-2. If the elongation produced in the rope is 1 cm, the force applied on the rope is

Easy

A

6.28 x 10⁵ N
B

6.28 x 10⁴ N
C

6.28 x 10⁴ dyne
D

6.28 x 10⁵ dyne

Solution

Young's modulus of a rope, $Y = \frac{F L}{A Δ l}$
Given, $L = 10 m, A = π r^{2} = π (1)^{2} = π$
$Y = 20 \times 10^{11} dyne {cm}^{- 2}, Δ l = 1 cm$
$\Rightarrow F = \frac{Y \cdot A \cdot Δ l}{L} \Rightarrow F = \frac{20 \times 10^{11} \times π \times 1}{10 \times 10^{2}}$
$F = 6.28 \times 10^{9} dyne$
$or F = 6.28 \times 10^{4} N$

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