Question

Find the ratio of the weights, as measured by a spring balance, of a 1 kg block of iron and a 1 kg block of wood. Density of iron = 7800 kg $m^{-3}$, density of wood = 800 kg $m^{-}3$ and density of air = 1.293 kg $m^{-3}$.

Answer 1

Net weight of iron block,

$w_1=mg-V_1{\rho }_{air}\times g$

= $[m-\left(\frac{m}{{\rho }_1}\right){\rho }_{air}]g$

= $[1-\left(\frac{1}{7800}\right)\times 1.293]\times \left(9.8\right)$

Again, Net weight of wood

= $W_w=mg-V_w\rho airg$

= $[m-\left(\frac{m}{{\rho }_w}\right){\rho }_{air}]g$

= $(1-\frac{1}{800}\times 1.293)9.8$

$\therefore \frac{W_1}{w_w}=\frac{9.8\left(\frac{7800-1.293}{7800}\right)}{9.8\left(\frac{800-1.293}{800}\right)}$

= $\frac{7800-1.293}{800-1.293}\times \frac{8}{78}=1.0015$