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

Step 1: Recall how much a wire stretches.
A loaded wire stretches by an amount given by Hooke's law: \[ \Delta L = \frac{FL}{AY} \] For the same wire, only the pulling force $F$ changes, so the stretch is proportional to the force.
Step 2: Find the first force.
At first a sphere of radius $R$ hangs in air, so the force is just its weight: \[ F_1 = \frac{4}{3}\pi R^{3}\rho g \] Call its volume $V$.
Step 3: See how doubling the radius changes volume.
When the radius doubles to $2R$, the volume becomes $2^{3} = 8$ times bigger: \[ V_2 = 8V \]
Step 4: Find the new force in the liquid.
Now the sphere is in a liquid of density $0.6\rho$. The liquid pushes up with a buoyant force, so the net pull is weight minus upthrust: \[ F_2 = 8V\rho g - 8V(0.6\rho)g = 8V\rho g(1 - 0.6) \]
Step 5: Simplify the new force.
\[ F_2 = 8V\rho g(0.4) = 3.2\,V\rho g = 3.2\,F_1 \] So the stretch becomes $3.2$ times the original.
Step 6: Find the percentage increase.
\[ \frac{\Delta L_2 - \Delta L_1}{\Delta L_1}\times 100 = (3.2 - 1)\times 100 \] \[ \boxed{= 220\%} \]