Question

A wire of length $L$ is hanging from a fixed support. The length changes to $L_{1}$ and $L_{2}$ when masses $M_{1}$ and $M_{2}$ are suspended respectively from its free end. Then $L$ is equal to

• A. $\frac{L_{1}+L_{2}}{2}$
• B. $\sqrt{L_{1} L_{2}}$
• C. $\frac{L_{1} M_{2}+L_{2} M_{1}}{M_{1}+M_{2}}$
• D. $\frac{L_{1} M_{2}-L_{2} M_{1}}{M_{2}-M_{1}}$

Answer 1

Answer: (D)

Let L be the original length of the wire

When the mass $M_{1}$ is suspended from the wire, change in length of wire is $\Delta L_{1}=L_{1}-L$
When the mass $M_{2}$ is suspended from it, change in length of wire is
$\Delta L_{2}=L_{2}-L$
From figure (b), $T_{1}=M_{1} g\,\,\,\,\,\,\dots(i)$
From figure (c), $T_{2}=M_{2} g\,\,\,\,\,\,\, \dots(ii)$
As Young's modulus, $ Y=\frac{T_{1} L}{A \Delta L_{1}}=\frac{T_{2} L}{A \Delta L_{2}}$
$\therefore \frac{T_{1}}{\Delta L_{1}}=\frac{T_{2}}{\Delta L_{2}} \quad \text { or } \quad \frac{T_{1}}{L_{1}-L}=\frac{T_{2}}{L_{2}-L}$
or $ \frac{M_{1} g}{L_{1}-L}=\frac{M_{2} g}{L_{2}-L}$ (Using (i) and (ii))
or $ M_{1}\left(L_{2}-L\right)=M_{2}\left(L_{1}-L\right)$ or $ M_{1} L_{2}-M_{1} L=M_{2} L_{1}-M_{2} L$
or $ L\left(M_{2}-M_{1}\right)=L_{1} M_{2}-L_{2} M_{1}$
or $ L=\frac{L_{1} M_{2}-L_{2} M_{1}}{M_{2}-M_{1}}$