Question

If the ratio of diameters, lengths and Young’s moduli of steel and brass wires shown in the figure are p, q and r respectively. Then the corresponding ratio of increase in their lengths would be

• A. $\frac{3q}{5p^{2}r}$
• B. $\frac{5q}{3p^{2}r}$
• C. $\frac{3q}{5pr}$
• D. $\frac{5q}{3pr}$

Answer 1

The correct option is B $\frac{5q}{3p^{2}r}$


As Young's modulus, $Y=\frac{FL}{A\Delta L}=\frac{4FL}{\pi D^2\Delta L}$
where the symbols have their usual meanings.
$\therefore \Delta L=\frac{4FL}{\pi D^{2}Y}$
$\therefore \frac{\Delta L_S}{\Delta L_B}=\frac{F_S}{F_B}\frac{L_S}{L_B}\frac{D_B^2}{D_S^2}\frac{Y_B}{Y_S}$
Where subscripts S and B refer to steel and brass respectively.
Here, $F_S=(2m+3m)g=5mg$
$F_B=3mg$
$\frac{L_S}{L_B}=q,\frac{D_S}{D_B}=p,\frac{Y_S}{Y_B}=r$
$\therefore \frac{\Delta L_S}{\Delta L_B}=\left(\frac{5mg}{3mg}\right)\left(q\right){\left(\frac{1}{p}\right)}^2\left(\frac{1}{r}\right)=\frac{5q}{3p^{2}r}$