Question

The weights of a hollow metallic sphere in air and when submerged in water are 264 gm-wt and 221 gm-wt respectively. If specific gravity of the metal is 8.8, then volume of the hollow portion is

• A. 11 cm\(^3\)
• B. 12 cm\(^3\)
• C. 13 cm\(^3\)
• D. 14 cm\(^3\)

Hint:

The buoyant force equals the weight of the displaced water. Use this principle to find the volume of a submerged object.

Answer 1

The Correct Option is B

Step 1: Understanding the problem.
The weight of the hollow metallic sphere in air is 264 gm-wt, and in water, it is 221 gm-wt. The difference in weight is due to the buoyant force, which is equal to the weight of the water displaced by the volume of the hollow portion of the sphere. The buoyant force is: \[ \text{Buoyant force} = 264 \, \text{gm-wt} - 221 \, \text{gm-wt} = 43 \, \text{gm-wt} \]
Step 2: Using Archimedes' principle.
The buoyant force is also
given by the equation:
\[ F_{\text{buoyant}} = \rho_{\text{water}} \cdot V_{\text{hollow}} \cdot g \] where \( \rho_{\text{water}} \) is the density of water, \( V_{\text{hollow}} \) is the volume of the hollow portion, and \( g \) is the acceleration due to gravity. Since the specific gravity of the metal is 8.8, the volume of the hollow portion can be calculated as: \[ V_{\text{hollow}} = \frac{43}{\text{density of water}} = 12 \, \text{cm}^3 \] Thus, the correct answer is
(B) 12 cm\(^3\)
.