A) \[\frac{V\rho g}{8}\]
B) \[\frac{V\rho g}{3}\]
C) \[\frac{V\rho g}{2}\]
D) \[\frac{3V\rho g}{4}\]
Correct Answer: A
Solution :
[a] V= volume of cone \[=\frac{1}{3}\pi {{R}^{2}}h\] Volume of protruding part \[{{V}_{out}}=\frac{1}{3}r{{\left( \frac{R}{2} \right)}^{2}}\frac{h}{2}=\frac{V}{8}\] Volume of cone inside water \[{{V}_{in}}V-\frac{V}{8}=\frac{7V}{8}\] Imagine the cone to have that part of it which protrudes through the hole removed and the space under the container filled with water. The buoyancy force would then be \[{{F}_{0}}=\rho g{{V}_{in}}=\frac{7V\rho g}{8}\] As there is no water beneath the hole, a contribution \[\pi {{\left( \frac{R}{2} \right)}^{2}}\rho gh\]is missing. \[\therefore \] Actual buoyancy force is \[F={{F}_{0}}-\pi \frac{{{R}^{2}}}{4}\rho gh\] \[=\frac{7}{8}V\rho g-\frac{3V}{4}\rho g=\frac{V\rho g}{8}\]
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