Question

One end of steel wire is fixed to ceiling of an elevator moving up with an acceleration of $2{\text{m/s}}^2$ and a load of $10\text{kg}$ hangs from the other end. Area of cross-section of the wire is $2{\text{cm}}^2$. The longitudinal strain in the wire is

$[Y=2\times {10}^{11}{\text{N/m}}^2]$

• A. $5\times {10}^{-6}$
• B. $3\times {10}^{-6}$
• C. $7\times {10}^{-6}$
• D. $8\times {10}^{-6}$

Answer 1

The correct option is B $3\times {10}^{-6}$

From F.B.D of block in accelerating frame of elevator,

$F_{\text{pseudo}}=m{a}_0$

Tension in the wire,
$T=m(g+a_0)=10(10+2)=120\text{N}$
So, stress in the wire,
$\sigma =\frac{T}{A}=\frac{120}{2\times {10}^{-4}}$
$\Rightarrow \sigma =6\times {10}^5{\text{N/m}}^2$
From Hooke's law, strain in the wire,
$ϵ=\frac{\sigma }{Y}$
$\Rightarrow ϵ=\frac{6\times {10}^5}{2\times {10}^{11}}=3\times {10}^{-6}$.