A) \[\frac{4}{3}\]
B) \[\frac{3}{2}\]
C) 3
D) 5
Correct Answer: C
Solution :
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 weights must be equal. \[\therefore \] \[\frac{{{m}_{1}}}{{{\rho }_{1}}}({{\rho }_{1}}-\sigma )=\frac{{{m}_{2}}}{{{\rho }_{2}}}({{\rho }_{2}}-1)\] \[\Rightarrow \] \[\frac{36}{9}(9-1)=\frac{48}{{{\rho }_{2}}}({{\rho }_{2}}-1)\] By solving, we get\[{{\rho }_{2}}=3\]
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