Question

Solid sphere $A$ is rotating about an axis $PQ$ If the radius of the sphere is $5 cm$ then its radius of gyration about $PQ$ will be $\sqrt{x} cm$ The value of $x$ is __

Answer 1

Correct Answer: 110

The radius of gyration k for a solid sphere about an axis through its center is given by:

\( k = \sqrt{\frac{2}{5}R^2} \)

where \( R \) is the radius. Here, \( R = 5 \text{ cm} \).

First, calculate \( k \) about the center:

\( k = \sqrt{\frac{2}{5}(5)^2} = \sqrt{\frac{2}{5} \times 25} = \sqrt{10} \text{ cm} \)

According to the parallel axis theorem, the radius of gyration about an axis at a distance \( d \) from the center is:

\( k_{PQ} = \sqrt{k^2 + d^2} \)

Given \( d = 10 \text{ cm} \), substitute to find \( k_{PQ} \):

\( k_{PQ} = \sqrt{10 + 100} = \sqrt{110} \text{ cm} \)

The value of \( x \) where \( \sqrt{x} = \sqrt{110} \) is:

x = 110

This confirms it fits within the given range of 110, 110.