Question

If the ratio of diameters, lengths and Young's modulus of steel and copper wires shown in the figure are $p,q$ and $s$ respectively, then the corresponding ratio of increase in their lengths would be :

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

Answer 1

Answer: (B)

As $\gamma =\frac{FL}{A\Delta L}=\frac{4FL}{\pi D^2\Delta L}$
$\Delta L=\frac{4FL}{\pi D^2\gamma }$
$\therefore \frac{\Delta L_s}{\Delta L_c}=\frac{F_s{L}_s{D_c}^2{Y}_c}{F_c{L}_c{D_s}^2{Y}_s}$
where subscripts 5 and C refer to copper and steel respectively.
Here,
$F_s=(5m+2m)g=7mg$
$F_c=5mg$
$\frac{L_s}{L_c}=q,\frac{D_s}{D_c}=p,\frac{Y_s}{Y_c}=s$
$\therefore \frac{\Delta L_s}{\Delta L_c}=\left(\frac{7mg}{5mg}\right)(q)\left(\frac{1}{p}{)}^2\right(\frac{1}{s})=\frac{7q}{5p^{2}s}$