JEE Main · 2021 · 24 Feb 2021 · Shift 2 · Conservation of Momentum

A circular hole of radius

(\frac{a}{2})

is cut of a circular disc of radius 'a' as shown in figure. The centroid of the remaining circular portion with respect to point 'O' will be:

Answer: (A)

Solution

Measure

x

from

O

, which lies on the rim of the complete disc, so the centre of that disc is at

x = a

. The hole of radius

\frac{a}{2}

touches the disc centre on one side and the rim on the other, so its centre is at

x = \frac{3 a}{2}

. With a uniform sheet the masses go as the areas,

M_{1} \propto a^{2}

for the full disc and

M_{2} \propto \frac{a ^{2}}{4}

for the piece removed. Treating the hole as negative mass,

x_{c m} = \frac{a ^{2} ( a ) - \frac{a ^{2}}{4} ( \frac{3 a}{2} )}{a ^{2} - \frac{a ^{2}}{4}} = \frac{a ^{3} - \frac{3 a ^{3}}{8}}{\frac{3 a ^{2}}{4}} = \frac{5 a}{6}

The centroid of the remaining portion lies at

\frac{5}{6} a

from

O

JEE Main · 2021 · 24 Feb 2021 · Shift 2 · Single correct option