Question

The length of wire, when $M_{1}$ is hung from it, is $l_{1}$ and is $l_{2}$ with both $M_{1}$ and $M_{2}$ hanging. The natural length of wire is

• A. $\frac{M_{1}}{M_{2}}\left(l_{1}-l_{2}\right)+l_{1}$
• B. $\frac{M_{2} l_{1}-M_{1} l_{2}}{M_{1}+M_{2}}$
• C. $\frac{l_{1}+l_{2}}{2}$
• D. $\sqrt{l_{1} l_{2}}$

Answer 1

Answer: (A)

Let the natural length of wire be $=l$
When only $M_{1}$ hanging
Using $\Delta l=\frac{F L}{A Y}$
$\left(l_{1}-l\right)=\frac{M_{1} g \cdot l}{A Y}$ ... (1)
When both $M_{1}, M_{2}$ hanging
$\left(l_{2}-l\right)=\frac{\left(M_{1}+M_{2}\right) g \cdot l}{A Y} \ldots$ (2)
Dividing (1) by (2)
$\frac{l_{1}-l}{l_{2}-l}=\frac{M_{1}}{M_{1}+M_{2}}$
Solving this we get
$l=\frac{M_{1}}{M_{2}}\left(l_{1}-l_{2}\right)+l_{1}$