Question

If in case $A$, elongation in wire of length $L$ is $l$, then for same wire elongation in case $B$ will be:

• A. $4l$
• B. $2l$
• C. $l$
• D. $\frac{l}{2}$

Answer 1

The correct option is C $l$

Guven, when weight 'W is hanged by wire of length 'L' Elongation in it is 'l

$w=\mathrm{yA}\left(\frac{1}{L}\right)\Rightarrow l=\frac{wL}{y_A}$

Now, in case 2

$\begin{array}{c}\text{Net elongation}=\Delta L_1+\Delta L_2\\ W=YA\left(\frac{\Delta L_1}{L/2}\right)\\ W=YA\left(\frac{\Delta L_2}{4/2}\right)\end{array}$

$\begin{array}{c}\text{Adding beth:}2W=\frac{2yA}{L}(\Delta L_1+\Delta L_2)\\ \Delta L_1+\Delta L_2=\frac{WL}{YA}=l\\ \text{Hence, net econgation in second case is}L\\ \text{Option}c\text{is correct option:}\end{array}$