Question

The radius of gyration of a solid sphere of mass \(5 \, \text{kg}\) about \(XY\) is \(5 \, \text{m}\) as shown in figure.
The radius of the sphere is \(\frac{5x}{\sqrt{7}} \, \text{m}\), then the value of \(x\) is:

• A. \(5\)
• B. \(\sqrt{2}\)
• C. \(\sqrt{3}\)
• D. \(\sqrt{5}\)

Answer 1

The Correct Option is D

Step 1: Moment of inertia about XY axis using parallel axis theorem

$$ I_{xy} = I_{CM} + MR^2 $$ $$ I_{xy} = \frac{2}{5}MR^2 + MR^2 $$ $$ I_{xy} = \frac{2}{5}MR^2 + \frac{5}{5}MR^2 = \frac{7}{5}MR^2 $$

Given $M = 5$, so:

$$ I_{xy} = \frac{7}{5} \times 5 \times R^2 = 7R^2 \quad \text{...(1)} $$

Step 2: Moment of inertia using radius of gyration $K$

$$ I_{xy} = MK^2 $$

Given $M = 5$ and $K = 5$, we get:

$$ I_{xy} = 5 \times 5^2 = 5 \times 25 = 125 \quad \text{...(2)} $$

Step 3: Equate equations (1) and (2)

$$ 7R^2 = 5 \times 5^2 $$ $$ 7R^2 = 125 $$ $$ R^2 = \frac{125}{7} $$ $$ R = \sqrt{\frac{125}{7}} = \sqrt{\frac{25 \times 5}{7}} = 5\sqrt{\frac{5}{7}} $$

Step 4: Compare with given form

The problem states that:

$$ R = \frac{5x}{\sqrt{7}} \quad \text{(Given)} $$

Equating both expressions for $R$:

$$ 5\sqrt{\frac{5}{7}} = \frac{5x}{\sqrt{7}} $$ $$ 5 \times \frac{\sqrt{5}}{\sqrt{7}} = \frac{5x}{\sqrt{7}} $$

Cancel $\frac{5}{\sqrt{7}}$ from both sides:

$$ \sqrt{5} = x $$ $$ \therefore x = \sqrt{5} $$