Question

The moment of inertia of a solid sphere of mass 'm' and radius 'R' about its diametric axis is 'I'. Its moment of inertia about a tangent in the plane is ______.

• A. 2.5I
• B. 3.0I
• C. 3.5I
• D. 4I

Hint:

Be careful to read whether the object is a solid sphere ($I = \frac{2}{5}mR^2$) or a hollow sphere ($I = \frac{2}{3}mR^2$). A hollow sphere tangent would yield $I_t = \frac{5}{3}mR^2$, which would equal $2.5 I$.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given the moment of inertia (M.I.) of a solid sphere about an axis passing through its center (diameter). We need to express its M.I. about an axis tangent to its surface in terms of this initial value '$I$'.

Step 2: Key Formula or Approach:
1. The standard formula for M.I. of a solid sphere about a diametric axis is $I = \frac{2}{5}mR^2$.
2. To find the M.I. about a tangent, we must use the Parallel Axis Theorem:
$I_{tangent} = I_{center} + md^2$, where $d$ is the perpendicular distance between the two parallel axes.

Step 3: Detailed Explanation:
The distance from the center of the sphere to a tangent line resting on its surface is exactly equal to its radius, so $d = R$.
Apply the Parallel Axis Theorem:
$$I_{tangent} = I_{center} + mR^2$$
Substitute the known standard value $I_{center} = \frac{2}{5}mR^2$:
$$I_{tangent} = \frac{2}{5}mR^2 + mR^2$$
Find a common denominator:
$$I_{tangent} = \frac{2}{5}mR^2 + \frac{5}{5}mR^2 = \frac{7}{5}mR^2$$
The question requires us to express this in terms of the original '$I$'.
Since $I = \frac{2}{5}mR^2$, we can write $mR^2 = \frac{5}{2}I$.
Substitute this back into the tangent equation:
$$I_{tangent} = \frac{7}{5} \left( \frac{5}{2}I \right)$$
The $5$ cancels out:
$$I_{tangent} = \frac{7}{2}I = 3.5 I$$

Step 4: Final Answer:
The moment of inertia about the tangent is 3.5I, matching option (c).