A U-sphere is made up of two layers: The first is lead (density $11,340 \mathrm{~kg} / \mathrm{m}^{3}$ ) and is of radius $\mathrm{R}$. The second is tin, $\left(7,310 \mathrm{~kg} / \mathrm{m}^{3}\right)$ and is concentric with the first, from radius $R$ to radius $2 \mathrm{R}$. The sphere is placed in a pool of mercury (density 13,600 kg/m3). How much of the volume of the sphere is below the surface?
A) $57.0 \%$
B) $64.2 \%$
C) $57.5 \%$
D) $68.5 \%$

Question

Quizplus · Accepted Answer

To determine how much of the U-sphere's volume is submerged when floating in mercury, we apply the principle of buoyancy, which states that the volume of fluid displaced (in this case, mercury) must equal the weight of the object. The weight of the object is the sum of the weights of the lead and tin parts.First, calculate the volume of the lead sphere: $V_{\text{lead}} = \frac{4}{3}\pi R^3$.Then, calculate the volume of the tin shell: $V_{\text{tin}} = \frac{4}{3}\pi (2R)^3 - \frac{4}{3}\pi R^3 = \frac{4}{3}\pi (8R^3 - R^3) = \frac{28}{3}\pi R^3$.The total weight of the sphere is the sum of the weights of the lead and tin parts, which is the product of their volumes and densities:- Weight of lead: $W_{\text{lead}} = \rho_{\text{lead}} V_{\text{lead}} = 11,340 \times \frac{4}{3}\pi R^3$.- Weight of tin: $W_{\text{tin}} = \rho_{\text{tin}} V_{\text{tin}} = 7,310 \times \frac{28}{3}\pi R^3$.The total weight of the sphere is: $W_{\text{total}} = W_{\text{lead}} + W_{\text{tin}}$.The volume of mercury displaced by the sphere when it floats equals the total weight of the sphere divided by the density of mercury:$$V_{\text{displaced}} = \frac{W_{\text{total}}}{\rho_{\text{Hg}}}$$To find the percentage of the sphere's volume that is submerged, we divide the volume of mercury displaced by the total volume of the sphere and multiply by 100:$$\text{Percentage submerged} = \left(\frac{V_{\text{displaced}}}{V_{\text{total}}}\right) \times 100$$The total volume of the sphere ($V_{\text{total}}$) is the sum of the volumes of the lead and tin parts, which we already calculated.Substituting the given densities and simplifying, we find that the percentage of the sphere's volume submerged is approximately 57.5%, which matches option C. This calculation involves algebraic manipulation and substitution of the given values, leading to the conclusion that about 57.5% of the sphere's volume is below the surface when it floats in mercury.

A U-Sphere Is Made Up of Two Layers: the First 11,340 kg/m311,340 \mathrm{~kg} / \mathrm{m}^{3}11,340 kg/m3

A U-Sphere Is Made Up of Two Layers: the First $11,340 \mathrm{~kg} / \mathrm{m}^{3}$