c
(c)Apparent weight \( = V(\rho - \sigma )g = \frac{m}{\rho }(\rho - \sigma )g\)
where \(m = \) mass of the body,
\(\rho = \) density of the body
\(\sigma = \) density of water
If two bodies are in equilibrium then their apparent weight must be equal.
\(\therefore \) \(\frac{{{m_1}}}{{{\rho _1}}}({\rho _1} - \sigma ) = \frac{{{m_2}}}{{{\rho _2}}}({\rho _2} - \sigma )\)
==> \(\frac{{36}}{9}(9 - 1) = \frac{{48}}{{{\rho _2}}}({\rho _2} - 1)\)
By solving we get \({\rho _2} = 3\).