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. $\frac{3q}{5p^{}r}$
• D. $\frac{5q}{p^{}r}$

Answer 1

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

Given:

$\frac{F_S}{F_B}=\frac{5}{3}$ $\frac{D_B}{D_S}=p$ $\frac{L_S}{L_B}=q$ $\frac{Y_S}{Y_B}=r$

As Young's modulus $Y=\frac{FL}{A\Delta L}=\frac{4FL}{\pi D^2\Delta L}$
where the symbols have their usual meanings
$\therefore \Delta L=\frac{4FL}{\pi D^{2}Y}$
$\therefore \frac{\Delta L_S}{\Delta L_B}=\frac{F_S}{F_B}\frac{L_S}{L_B}\frac{D_B^2}{D_S^2}\frac{Y_B}{Y_S}$
$\therefore \frac{\Delta L_S}{\Delta L_B}=\left(\frac{5}{3}\right)\left(q\right){\left(\frac{1}{p}\right)}^2\left(\frac{1}{r}\right)=\frac{5q}{3p^{2}r}$