Question

Young's modulus of rubber is ${10}^4$ N$/m^2$ and area of cross-section is $2$ $c{m}^2$. If force of $2\times {10}^5$ dyne is applied along its length, then its final length becomes.

• A. $3L$
• B. $4L$
• C. $2L$
• D. None of these

Answer 1

Answer: (C)

$2L$

Given, $Y={10}^{4}N/m^2,A=2c{m}^2=2\times {10}^{-4}{m}^2$
$F=2\times {10}^5$ dyne $=2$N
$\therefore$ Initial length $I=\frac{FL}{AY}$
$=\frac{2\times L}{2\times {10}^{-4}\times {10}^4}=L$
$\therefore$ Final length $=$ Initial length $+$ Increment
$=2L$.