Question

A wire elongates by $\mathrm{l}\mathrm{mm}$ when a load $\mathrm{W}$ is hanged from it. If the wire goes over a pulley and two weighs $\mathrm{W}$ each is hung at the two ends, the elongation of the wire will be (in $\mathrm{mm}$)

• A. $\frac{\mathrm{l}}{2}$
• B. $\mathrm{l}$
• C. $2I$
• D. Zero

Answer 1

The correct option is B

$\mathrm{l}$

Step 1: Given the information and formula used

Length is given as $\mathrm{l}\mathrm{mm}$ and load is $\mathrm{W}$.

According to Hooke's law

Modules of elasticity, $\mathrm{E}=\frac{\mathrm{W}}{\mathrm{A}}\times \frac{\mathrm{L}}{\mathrm{l}}$

$\mathrm{L}$ - original length of the wire

$\mathrm{A}$ - cross-sectional area of the wire

Step 2: Determine the elongation

Elongation $∆=\frac{\mathrm{WL}}{\mathrm{E}}$

On either side of the wire, tension is W and length is $\frac{\mathrm{l}}{2}$

$$\begin{gathered} ∆\mathrm{l}=\frac{\left({\displaystyle \frac{\mathrm{WL}}{2}}\right)}{\mathrm{AE}} \\ =\frac{\mathrm{WL}}{2\mathrm{AE}} \\ =\frac{\mathrm{l}}{2} \end{gathered}$$

Total elongation in the wire $$\begin{gathered} =\frac{\mathrm{l}}{2}+\frac{\mathrm{l}}{2} \\ =\mathrm{l} \end{gathered}$$

Therefore, the total elongation in the wire is calculated as $\mathbf{l}$.

Hence, option (B) is correct.