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 weight must be equal. \[\therefore \] \[\frac{{{m}_{1}}}{{{\rho }_{1}}}({{\rho }_{1}}-\sigma )=\frac{{{m}_{2}}}{{{\rho }_{2}}}({{\rho }_{2}}-\sigma )\] Þ \[\frac{36}{9}(9-1)=\frac{48}{{{\rho }_{2}}}({{\rho }_{2}}-1)\] By solving we get \[{{\rho }_{2}}=3\].
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