The disc of mass M with uniform surface mass density $\sigma$ is shown in the figure. The centre of mass of the quarter disc (the shaded area) is at the position $\frac{\mathrm{x}}{3} \frac{\mathrm{a}}{\pi}, \frac{\mathrm{x}}{3} \frac{\mathrm{a}}{\pi}$ where $x$ is--------. (Round off to the Nearest integer) [$a$ is an area as shown in the figure]

JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Conservation of Momentum

The disc of mass M with uniform surface mass density

σ

is shown in the figure. The centre of mass of the quarter disc (the shaded area) is at the position

\frac{x}{3} \frac{a}{π}, \frac{x}{3} \frac{a}{π}

where

x

is--------. (Round off to the Nearest integer) [

a

is an area as shown in the figure]

Answer: 4

The shaded lamina is a quarter of the disc, of radius

a

, lying in the first quadrant with its corner at the origin. With

d A = r d r d θ

and the abscissa of the element

r cos θ

\overset{x}{ˉ} = \frac{\int _{0}^{a} \int _{0}^{π /2} ( r cos θ ) r d r d θ}{\int _{0}^{a} \int _{0}^{π /2} r d r d θ} = \frac{\frac{a ^{3}}{3}}{\frac{a ^{2}}{2} \cdot \frac{π}{2}} = \frac{4 a}{3 π}

The lamina is symmetric about the line

y = x

, so

\overset{y}{ˉ} = \overset{x}{ˉ} = \frac{4 a}{3 π}

. Comparing with the quoted form

\frac{x}{3} \frac{a}{π}

gives

x = 4

JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Integer answer