From a uniform wire, two circular loops are made (`i`) `P` of radius `r` and (`ii`) `Q` of radius `nr`. If the moment of inertia of `Q` about an axis passing through its center and perpendicular to tis plane is `8` times that of `P` about a similar axis, the value of `n` is (diameter of the wire is very much smaller than `r` or `nr`)

Options

A. `8`
B. `6`
C. `4`
D. `2`

Explanation

To solve the problem, we need to find the value of \( n \) given that the moment of inertia of loop \( Q \) is 8 times that of loop \( P \). Let's break this down step by step.

### Step 1: Understand the Moment of Inertia Formula
The moment of inertia \( I \) of a circular loop about an axis passing through its center and perpendicular to its plane is given by the formula:
\[ I = m r^2 \]
where \( m \) is the mass of the loop and \( r \) is its radius.
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