JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Rotational Mechanics

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Let the moment of inertia of a hollow cylinder of length 30 cm (inner radius 10 cm and outer radius 20 cm ), about its axis be I . The radius of a thin cylinder of the same mass such that its moment of inertia about its axis is also I, is:

(A)12 cm
(B)16 cm✓ Correct
(C)14 cm
(D)18 cm

Answer: (B)

Solution

For a hollow, thick-walled cylinder the moment of inertia about its axis is

I = \frac{m}{2} (R_{out}^{2} + R_{in}^{2})

. With

R_{out} = 20 cm

and

R_{in} = 10 cm

I = \frac{m}{2} (400 + 100) = 250 m

cm^{2}

units. A thin-walled cylinder of the same mass and radius

r_{0}

has

I = m r_{0}^{2}

. Setting the two equal,

m r_{0}^{2} = 250 m

, so

r_{0} = 250 \approx 15.8 cm

, which rounds to

16 cm

JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Single correct option

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