AP EAPCET 04-Jul-2022 Shift 1 Paper

Let initially the cone be in equilibrium i.e. floating From the diagram, it is clear that Buoyant force $(F_b) =$ weight of the cone. $(F_w)$ $\rho_w \times \left(\frac{1}{3} \pi r^2 x\right) \times g = \rho_c \times \left(\frac{1}{3} \pi R^2 H\right) \times g$ $\Rightarrow 2 r^2 x = R^2 H$ $\left(\because \rho_c = \frac{\rho_w}{2}\right)$ $\Rightarrow x = \frac{R^2 H}{2 r^2}$ ........(i) From the diagram, $\tan \theta = \frac{r}{x} = \frac{R}{H}$ $\Rightarrow x = \frac{r H}{R}$ .....(ii) From Eqs. (i) and (ii) $\frac{R^2 H}{2 r^2} = \frac{r H}{R}$ $\Rightarrow \frac{R^3}{2} = r^3 \Rightarrow r = \left(\frac{1}{2}\right)^{1/3} R$ Now, the cone. is depressed by small distance $\delta(\ll)$ and released it executes simple harmonic motion. In simple harmonic motion, the buoyancy force acts as restoring force which is given as $F_b = \text{density} \times \text{volume} \times \text{gravity}$ Since, the displacement is very small, the excess submerged shape of cone. can be considered as a cylinder whose volume is given by $\pi r^2 \delta$ . $\therefore F_b = \rho_w \pi r^2 \delta g = \rho_w \pi \left[ \left(\frac{1}{2}\right)^{1/3} R \right]^2 \delta g$ Thus, force is equal to restoring force, hence $F_b = k \delta = m \omega^2 \delta$ $\Rightarrow \frac{6 \left(\frac{1}{4}\right)^{1/3} g}{H} = \omega^2$ or