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)

$6.28\times 10^{4}\,N$

Answer 2

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 $F=6.28 \times 10^{4} N$