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

Answer 1

formulae,

$Y=\frac{FL}{A\Delta L}$

$=\frac{4FL}{\pi D^2\Delta L}$

$\Delta L=\frac{4FL}{\pi D^{2}Y}$

$\frac{\Delta L_S}{\Delta L_C}=\frac{F_S}{F_C}\frac{L_S}{L_C}\frac{D_C^2}{D_S^2}\frac{Y_C}{Y_S}$

as per given,

$F_S=(5m+2m)g=7mg$

$F_C=5mg$

$\frac{L_S}{L_G}=q$

$\frac{D_S}{D_G}=p$

$\frac{Y_S}{Y_G}=s$

$\frac{\Delta L_S}{\Delta L_G}=\left(\frac{7mg}{5mg}\right)\left(q\right){\left(\frac{1}{p}\right)}^2\left(\frac{1}{s}\right)$

$=\frac{7q}{5p^{2}s}$