A) 0.5
B) 0.7
C) 0.6
D) 0.8
Correct Answer: C
Solution :
| [c] In 1st situation |
| \[{{V}_{b}}{{\rho }_{b}}g={{V}_{s}}{{\rho }_{w}}g\] |
| \[\frac{{{V}_{s}}}{{{V}_{b}}}=\frac{{{\rho }_{b}}}{{{\rho }_{w}}}=\frac{4}{5}\] (i) |
| here \[{{V}_{b}}\] is volume of block |
| \[{{V}_{s}}\] is submerged volume of block |
| \[{{\rho }_{b}}\] is density of block |
| \[{{\rho }_{w}}\] is density of water |
| & Let \[{{\rho }_{o}}\] is density of oil |
| finally in equilibrium condition |
| \[{{V}_{b}}{{\rho }_{b}}g=\frac{{{V}_{b}}}{2}{{\rho }_{o}}g+\frac{{{V}_{b}}}{2}{{\rho }_{w}}g\] |
| \[2{{\rho }_{b}}={{\rho }_{0}}+{{\rho }_{w}}\] |
| \[\Rightarrow \]\[\frac{{{\rho }_{o}}}{{{\rho }_{w}}}=\frac{3}{5}=0.6\] |
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