Question

The Young's modulus of a rope of $ 10\,m$ length and having diameter of $2\,cm$ is $ 200\times 10^{11}\,dyne/cm^{2}$ . If the elongation produced in the rope is $ 1\,cm, $ the force applied on the rope is

• A. $ 6.28\times 10^{5}\,N $
• B. $ 6.28\times 10^{4}\,N $
• C. $ 6.28\times 10^{4}$ dyne
• D. $ 6.28\times 10^{5}$ dyne

Answer 1

Answer: (B)

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