Question

Three equal masses $3kg$ are connected by massless string of cross-sectional area $0.005{cm}^2$ and Young's modulus $2\times {10}^{11}N{m}^2$. In the absence of friction, the longitudinal strain in the wire:

• A. $A$ is ${10}^{-4}$
• B. $B$ is $2\times {10}^{-4}$
• C. Both a and b
• D. None of these

Answer 1

Answer: (C)

Both a and b

$3g-T_B=3a$
$T_A=3a$
$T_B-T_A$ are the equations defining the motion
Solving we get, $a=\frac{3g}{9}=\frac{g}{3},T_A=g;T_B=2g$
So, strain in $A=\frac{T_A}{aY}=\frac{g}{0.005\times {10}^{-4}\times Y}={10}^{-4}$
Strain in $B=\frac{T_B}{aY}=\frac{2g}{0.005\times {10}^{-4}\times Y}=2\times {10}^{-4}$
So choices (a) and (b) are correct and so choice (c) is correct answer