Two identical solid cones of base radius R and height 2R float in water as shown in figure.
Cone A has height $h_{1}$ immersed in Water while Cone B has height $h_{2}$ lying outside the water. Relative density of each cone is $\frac{1}{3} .$

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Value of $\frac{h_{1}}{h_{2}}$ will be 1

Value of $\frac{h_{1}}{h_{2}}$ will be $\frac{1}{2^{\frac{1}{3}}}$

If cone A is further suppressed by small distance x and released, it will execute SHM.
The time period of oscillation would be ${(3)}^{- \frac{4}{3}} [2 π \sqrt{\frac{2 R}{g}}]$

If cone A is further suppressed by small distance x and released, it will execute SHM.
The time period of oscillation would ${(3)}^{- \frac{4}{3}} [2 π \sqrt{\frac{R}{g}}]$

answer is A.

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Detailed Solution

For cone A

$\frac{V_{i}}{V} = \frac{1}{3} \Rightarrow V_{i} = \frac{1}{3} V \Rightarrow \frac{1}{3} \times π \times \frac{h_{1}^{2}}{4} \times h_{1} = \frac{1}{3} \times \frac{1}{3} \times π R^{2} \times 2 R \frac{h_{1}^{3}}{4} = \frac{2}{3} R^{3} \Rightarrow h_{1}^{3} = \frac{8}{3} R^{3} \Rightarrow h_{1}^{} = \frac{2}{3 \sqrt{3}} R$

For cone B

$\frac{V - V_{i}}{V} = \frac{2}{3} \Rightarrow \frac{1}{3} \times π \times \frac{h_{2}^{2}}{4} \times h_{2} = \frac{2}{3} \times \frac{1}{3} π R^{2} \times 2 R h_{2} = 2 \sqrt[3]{\frac{2}{3} R} \Rightarrow \frac{h_{1}}{h_{2}} = \frac{1}{2^{(\frac{1}{3})}} = {(2)}^{- \frac{1}{3}}$

Time period:

$ρ π (\frac{h_{1}^{2}}{4}) x g = σ (\frac{1}{3} π R^{2}) (2 R) a ω = \frac{3}{2} \sqrt{\frac{g h_{1}^{2}}{2 R^{3}}} T = 3^{- \frac{2}{3}} [2 π \sqrt{\frac{2 R}{g}}]$