wires

Length of wire in $P = 2\pi r$

Length of wire in $Q = 2\pi (nr)$

Since the wires are of equal length:

$$ 2\pi r_P = 2\pi r_Q \implies r_P = r_Q $$

But here, the loops are of radii $r$ and $nr$, so the length of the wire for $P$ is $2\pi r$, and for $Q$ it is $2\pi n r$.

Let mass per unit length = $\lambda$.

So,

$$ m_P = \lambda \cdot 2\pi r $$

$$ m_Q = \lambda \cdot 2\pi n r $$

Step 2: Write the moments of inertia

The moment of inertia of a thin ring (loop) about its axis:

$$ I = m r^2 $$

So,

$$ I_P = m_P r^2 = (\lambda \cdot 2\pi r) \cdot r^2 = 2\pi\lambda r^3 $$

$$ I_Q = m_Q (nr)^2 = (\lambda \cdot 2\pi n r) \cdot (n^2 r^2) = 2\pi\lambda n r \cdot n^2 r^2 = 2\pi\lambda n^3 r^3 $$

Step 3: Use the condition $I_Q = 4I_P$

From above,

$$ I_Q = 2\pi\lambda n^3 r^3 $$

$$ I_P = 2\pi\lambda r^3 $$

Given,

$$ I_Q = 4 I_P $$

So,

$$ 2\pi\lambda n^3 r^3 = 4 \cdot 2\pi\lambda r^3 $$

$$ n^3 = 4 $$

$$ n = 4^{1/3} = (2^2)^{1/3} = 2^{2/3} $$

Final Answer:

$$ \boxed{(2)^{2/3}} $$

So, the correct option is Option B.

Rotational Motion

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