JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Rotational Mechanics

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The solid cylinder of length 80 cm and mass M has a radius of 20 cm . Calculate the density of the material used if the moment of inertia of the cylinder about an axis

C D

parallel to

A B

as shown in figure is

2.7 kg m^{2}

(A) $7.5 \times 1 0^{1} kg / m^{3}$
(B) $1.49 \times 1 0^{2} kg / m^{3}$ ✓ Correct
(C) $14.9 kg / m^{3}$
(D) $7.5 \times 1 0^{2} kg / m^{3}$

Answer: (B)

Solution

The axis

C D

is parallel to the cylinder's own axis

A B

and lies a perpendicular distance

\frac{L}{2}

from it, so the parallel-axis theorem gives

I_{C D} = \frac{1}{2} M R^{2} + M (\frac{L}{2})^{2}

With

R = 0.2

m and

L = 0.8

2.7 = M [\frac{( 0.2 ) ^{2}}{2} + (0.4)^{2}] = M (0.02 + 0.16) = 0.18 M

M = 15

kg. The volume of the cylinder is

π R^{2} L = π (0.2)^{2} (0.8) = 0.1005 m^{3}

, and the density is

ρ = \frac{15}{0.1005} = 1.49 \times 1 0^{2} kg m^{- 3}

JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Single correct option

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JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Rotational Mechanics

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