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

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The Young's modulus of a rope of 10m length and having diameter of 2 cm is $20 \times 10^{11}$ dyne cm^-2.If the elongation produced in the rope is 1 cm, the force applied on the rope is

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$6.28 \times 10^{5} N$

$6 .28 \times 10^{4} N$

$6 .28 \times 10^{4} dyne$

$6 .28 \times 10^{5} dyne$

answer is B.

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Detailed Solution

Young's modulus of a rope, $Y = \frac{FL}{AΔl}$
Given, $L = 10 m, A = {πr}^{2} = π (1)^{2} = π$
$Y = 20 \times 10^{11} dyne {cm}^{- 2}, Δl = 1 cm \Rightarrow F = \frac{Y \cdot A \cdot Δl}{L} \Rightarrow F = \frac{20 \times 10^{11} \times π \times 1}{10 \times 10^{2}} \begin{aligned} F = 6 .28 \times 10^{9} dyne \\ or F = 6 .28 \times 10^{4} N \end{aligned}$