Question

A uniform metal wire is suspended from a rigid ceiling and a solid sphere is attached to the second end of the wire. If the radius of the sphere is doubled and then immersed in a liquid whose density is 60% of the density of the material of the sphere, then the percentage increase in the elongation of the wire is:

• A. 420%
• B. 120%
• C. 220%
• D. 320%

Hint:

Doubling the radius scales the volume (and therefore the raw weight) by a factor of \( 2^3 = 8 \). Keeping track of this cubic scaling relation is crucial for getting the correct answer.

Answer 1

The Correct Option is C

Concept: The elongation \( \Delta L \) of a wire under load is given by Hooke’s Law: \[ \Delta L = \frac{F \cdot L}{A \cdot Y} \] The stretching force \( F \) is equal to the effective weight of the suspended sphere, accounting for any buoyant upthrust forces when immersed in a liquid.

Step 1: Finding the initial elongation context.
Initially, the sphere of radius \( R \) hangs in air, so the load force is simply its weight: \[ F_1 = V \cdot \rho \cdot g = \frac{4}{3}\pi R^3 \cdot \rho \cdot g \]

Step 2: Finding the new elongation context after transformation.
The radius of the sphere is doubled to \( 2R \), which changes its volume to \( V_2 = \frac{4}{3}\pi (2R)^3 = 8V \). The sphere is then immersed in a liquid with density \( \rho_L = 0.6\rho \). The net effective downward force is: \[ F_2 = \text{Weight} - \text{Buoyant Upthrust} = V_2 \cdot \rho \cdot g - V_2 \cdot \rho_L \cdot g \] \[ F_2 = 8V\rho g - 8V(0.6\rho)g = 8V\rho g(1 - 0.6) = 8V\rho g(0.4) = 3.2 (V\rho g) = 3.2 F_1 \]

Step 3: Calculating the percentage increase.
Since elongation is directly proportional to force, \( \Delta L_2 = 3.2 \Delta L_1 \). \[ \text{Percentage Increase} = \frac{\Delta L_2 - \Delta L_1}{\Delta L_1} \times 100\% = \frac{3.2 - 1}{1} \times 100\% = 2.2 \times 100\% = 220\% \] This matches option (C) perfectly.