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(D) Let $H = 0.1 \ m$ be the height and $R = 0.05 \ m$ be the base radius of the cone. The radius $r$ at any height $y$ from the vertex is given by $r

Vedclass · Accepted Answer

$1.9$

Vedclass · Answer

(D) Let $H = 0.1 \ m$ be the height and $R = 0.05 \ m$ be the base radius of the cone. The radius $r$ at any height $y$ from the vertex is given by $r = (R/H)y = (0.05/0.1)y = 0.5y$.Case $1$: The cone floats in a liquid of relative density $\sigma_1 = 0.8$ with depth $h_1 = H/3$. The volume of the submerged part is $V_1 = \frac{1}{3} \pi r_1^2 h_1$,where $r_1 = 0.5(H/3) = R/3$. Thus,$V_1 = \frac{1}{3} \pi (R/3)^2 (H/3) = \frac{1}{27} (\frac{1}{3} \pi R^2 H)$.By the principle of floatation,the weight of the cone $W = 	ext{Buoyant force} = \sigma_1 ho_w V_1 g$,where $ho_w$ is the density of water. So,$W = 0.8 ho_w g V_1$.Case $2$: Liquid of relative density $ho$ is filled inside up to $h_2 = H/3$. The weight of this liquid is $W_L = ho ho_w V_2 g$,where $V_2 = \frac{1}{3} \pi (R/3)^2 (H/3) = V_1$. The cone now floats with depth $h_3 = H/2$. The new buoyant force is $F_B = \sigma_1 ho_w V_3 g$,where $V_3 = \frac{1}{3} \pi (R/2)^2 (H/2) = \frac{1}{8} (\frac{1}{3} \pi R^2 H) = \frac{27}{8} V_1$.Force balance: $W + W_L = F_B$$0.8 ho_w g V_1 + ho ho_w g V_1 = 0.8 ho_w g (\frac{27}{8} V_1)$$0.8 + ho = 0.8 	imes \frac{27}{8} = 0.1 	imes 27 = 2.7$$ho = 2.7 - 0.8 = 1.9$.

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