Correct Option: (B) Let $r_c$ be the internal radius of the cylindrical vessel. Let $r_s$ be the radius of the sphere. Let $r_{co}$ be the radius of one cone and $h_{co}$ be its height. Given values: Internal radius of cylindrical vessel, $r_c = 5 \, \text{cm}$. Rise in water level due to sphere, $\Delta h_s = 2 \, \text{cm}$. Rise in water level due to two cones, $\Delta h_{co} = 1.5 \, \text{cm}$. For each cone, $h_{co} = 3r_{co}$. **Step 1: Determine the volume of the sphere.** When the sphere is submerged, the volume of water displaced is equal to the volume of the sphere. This volume of water displaced causes a rise in the water level in the cylindrical vessel. The volume of water displaced is the volume of the cylindrical column of water that rose: $V_{\text{water rise due to sphere}} = \pi r_c^2 \Delta h_s$ Substituting the given values: $V_{\text{water rise due to sphere}} = \pi (5)^2 (2) = \pi (25)(2) = 50\pi \, \text{cm}^3$. The volume of a sphere is given by the formula: $V_s = \frac{4}{3} \pi r_s^3$ Equating the volume of the sphere to the volume of water displaced: $\frac{4}{3} \pi r_s^3 = 50\pi$ Divide both sides by $\pi$: $\frac{4}{3} r_s^3 = 50$ $r_s^3 = \frac{50 \times 3}{4}$ $r_s^3 = \frac{150}{4}$ $r_s^3 = 37.5$ \quad (Equation 1) **Step 2: Determine the volume of one cone.** When two identical cones are submerged, the total volume of water displaced is equal to the total volume of the two cones. This displacement causes a rise in the water level in the cylindrical vessel. The volume of water displaced by the two cones is the volume of the cylindrical column of water that rose: $V_{\text{water rise due to cones}} = \pi r_c^2 \Delta h_{co}$ Substituting the given values: $V_{\text{water rise due to cones}} = \pi (5)^2 (1.5) = \pi (25)(1.5) = 37.5\pi \, \text{cm}^3$. The volume of one cone is given by the formula: $V_{co} = \frac{1}{3} \pi r_{co}^2 h_{co}$ We are given that for each cone, $h_{co} = 3r_{co}$. Substitute this into the cone volume formula: $V_{co} = \frac{1}{3} \pi r_{co}^2 (3r_{co})$ $V_{co} = \pi r_{co}^3$ The total volume of two identical cones is $2 V_{co} = 2 \pi r_{co}^3$. Equating the total volume of the two cones to the volume of water displaced: $2 \pi r_{co}^3 = 37.5\pi$ Divide both sides by $\pi$: $2 r_{co}^3 = 37.5$ $r_{co}^3 = \frac{37.5}{2}$ $r_{co}^3 = 18.75$ \quad (Equation 2) **Step 3: Calculate the ratio of the radius of the sphere to the radius of one of the cones.** We need to find the ratio $\frac{r_s}{r_{co}}$. We have $r_s^3 = 37.5$ (from Equation 1) and $r_{co}^3 = 18.75$ (from Equation 2). Divide Equation 1 by Equation 2: $\frac{r_s^3}{r_{co}^3} = \frac{37.5}{18.75}$ Simplify the right side: $37.5 = 2 \times 18.75$, so $\frac{37.5}{18.75} = 2$. Therefore, $\left(\frac{r_s}{r_{co}}\right)^3 = 2$ To find the ratio $\frac{r_s}{r_{co}}$, take the cube root of both sides: $\frac{r_s}{r_{co}} = \sqrt[3]{2}$ Thus, the ratio of the radius of the sphere to the radius of one of the cones is $\sqrt[3]{2}$.