Question

If the ratio of diameters, lengths and Young’s module 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

Answer: (B)

$\frac{5q}{3p^{2}r}$

Answer 2

:

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, $F_{S} = (2m + 3m)g = 5mg$

$F_{B} = 3mg$

$\frac{L_{S}}{L_{B}} = q,\frac{D_{S}}{D_{B}} = p,\frac{Y_{S}}{Y_{B}} = r$

$\therefore\frac{\Delta L_{S}}{\Delta L_{B}} = \left( \frac{5mg}{3mg} \right)(q)\left( \frac{1}{p} \right)^{2}\left( \frac{1}{r} \right) = \frac{5q}{3p^{2}r}$