If I1 is the moment of inertia of a thin rod about an axis perpendicular to its length and passing through its centre of mass and I2 is the moment of inertia of the ring formed by the same rod about an axis tangent to the ring and perpendicular to the plane of the ring, then the ratio I1/I2 is

I1=mL212 for rod for ring I2 = mR2 + mR2 = 2mR2 But 2πR=L⇒R=L2πI2=2mL24π2I1I2=mL2126×4π22mL2=π26

If I1 is the moment of inertia of a thin rod about an axis perpendicular to its length and passing through its centre of mass and I2 is the moment of inertia of the ring formed by the same rod about an axis tangent to the ring and perpendicular to the plane of the ring, then the ratio I1/I2 is

I1=mL212 for rod for ring I2 = mR2 + mR2 = 2mR2 But 2πR=L⇒R=L2πI2=2mL24π2I1I2=mL2126×4π22mL2=π26

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If I₁ is the moment of inertia of a thin rod about an axis perpendicular to its length and passing through its centre of mass and I₂ is the moment of inertia of the ring formed by the same rod about an axis tangent to the ring and perpendicular to the plane of the ring, then the ratio I₁/I₂ is

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$\frac{π^{2}}{3}$

$\frac{π^{2}}{6}$

$\frac{π^{2}}{2}$

$\frac{π^{2}}{5}$

answer is B.

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Detailed Solution

$I_{1} = \frac{{mL}^{2}}{12}$ for rod
for ring I₂ = mR² + mR² = 2mR²
But $2 πR = L \Rightarrow R = \frac{L}{2 π}$
$\begin{array}{l} I_{2} = 2 m \frac{L^{2}}{4 π^{2}} \\ \frac{I_{1}}{I_{2}} = \frac{m L^{2}}{12_{6}} \times \frac{4 π^{2}}{2 {mL}^{2}} = \frac{π^{2}}{6} \end{array}$