Question

The Young's modulus of a rope of $10m$ length and having diameter of $2cm$ is $200\times 10^{11}dyne/c{m}^2$ . If the elongation produced in the rope is $1cm,$ 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 $Y=\frac{FL}{A\Delta l}$
Given, $L=10m,A=\pi r^2=\pi (1{)}^2=\pi ;$
$Y=20\times 10^{11}$ dyne $/c{m}^2,\Delta l=1cm$,
$F=\frac{Y\cdot A\cdot \Delta l}{L}$
$F=\frac{20\times 10^{11}\times 1\times 1}{10\times 10^2}$
$F=6.28\times 10^9$ dyne
$F=6.28\times 10^{4}N$