Question

One end of a steel wire is fixed to the ceiling of a moving elevator. A load of $20\text{kg}$ hangs from the other end. If the elevator is moving up with an acceleration $1{\text{m/s}}^2$ and Area of cross- section of wire is $2{\text{cm}}^2$ then find the longitudnal strain in the wire.
[Take $g=10{\text{ms}}^{-2},Y=2\times {10}^{11}{\text{Nm}}^{-2}$ ]

• A. $5.5\times {10}^{-4}$
• B. $5.5\times {10}^{-6}$
• C. $3\times {10}^{-6}$
• D. $2\times {10}^{-4}$

Answer 1

The correct option is B $5.5\times {10}^{-6}$


FBD of block from the frame of reference of elevator
$T=mg+ma=m(g+a)$
From the data given in the question ,
$T=20(10+1)=220\text{N}$
From the definition of modulus of elasticity, we know that
$Y=\frac{F\times l}{A\times \Delta l}\Rightarrow \frac{T\times l}{A\times \Delta l}$
$\Rightarrow$ Longitudinal Strain $\frac{\Delta l}{l}=\frac{T}{AY}\Rightarrow \frac{220}{2\times {10}^{-4}\times 2\times {10}^{11}}\Rightarrow 5.5\times {10}^{-6}$
Thus, option (b) is the correct answer.