Question

A spiral spring loaded with a piece of metal extends by 10.5 cm in air, when the metal is fully submerged in water the spring extends by 6.8 cm .Assuming Hooke's law is obeyed calculate the different forces given the different extensions.

Answer 1

The Hooke's law says

$$kl_1=mg,\quad kl_2=mg-mg*\frac{\rho_w}{\rho}$$

Hence,

$$l_2=l_1(1-\frac{\rho_w}{\rho})$$

$$\rho=\rho_w(1-\frac{l_2}{l_1})=1000*(1-\frac{6.8}{10.5})=352\:\rm kg/m^3$$