Question

If the ratio of diameter, length and Young's modulus of steel and copper wire shown in the figure are $p,q$ and $s$ respectively, then the corresponding ratio of increase in their lengths would be
(Mass of blocks hanging are $5m$ and $2m$)

• A. $\frac{5q}{\left(7s{p}^2\right)}$
• B. $\frac{7q}{\left(5s{p}^2\right)}$
• C. $\frac{2q}{\left(5sp\right)}$
• D. $\frac{7q}{\left(5sp\right)}$

Answer 1

The correct option is B $\frac{7q}{\left(5s{p}^2\right)}$

Let
Force experienced by wire $=F$
Area of cross-section of wire $=A$
Original length of cross-section of wire $=L$
Change in length of wire $=\Delta L$
Young's modulus, $Y=\frac{\text{Stress}}{\text{Strain}}=\frac{F/A}{\Delta L/L}$
$\Rightarrow Y=\frac{FL}{A\Delta L}=\frac{4FL}{\pi D^2\Delta L}$
$\Rightarrow \Delta L=\frac{4FL}{\pi D^{2}Y}$
$\therefore \frac{\Delta L_s}{\Delta L_e}=\frac{F_s}{F_c}.\frac{L_s}{L_c}.\frac{D_c^2}{D_s^2}\frac{Y_c}{Y_s}$
Where subscripts $c,s$ refers to copper and steel respectively.
Here, $F_s=(5m+2m)g=7\text{mg}$
$F_c=5mg$
$\Rightarrow \frac{L_s}{L_c}=q,\frac{D_s}{D_c}=P,\frac{Y_s}{Y_c}=s$
$\therefore \frac{\Delta L_s}{\Delta L_c}=\left(\frac{7\text{mg}}{5\text{mg}}\right)\left(q\right){\left(\frac{1}{p}\right)}^2\left(\frac{1}{s}\right)=\frac{7q}{5p^{2}s}$
Option $\left(b\right)$ correct

$\begin{array}{c}\text{Why this question}?\\ \text{Important to understand how much weight is}\\ \text{attached to which string/rod}\end{array}$