JEE Main 20198 AprilEvening Shift

Exam: JEE Main
Year: 2019
Date: 8 April
Shift: Evening Shift
Subject: Physics
Chapter: Center of Mass, Momentum and Collision

JEE Main 2019 · 8 April · Evening Shift · Physics · Center of Mass, Momentum and Collision

A uniform rectangular thin sheet ABCD of mass M has length a and...

A uniform rectangular thin sheet $ABCD$ of mass $M$ has length $a$ and breadth $b$ , as shown in the figure. If the shaded portion $HBGO$ is cut-off, the coordinates of the centre of mass of the remaining portion will be:

Options

A $\dfrac{5a}{12}, \dfrac{5b}{12}$
B $\dfrac{5a}{3}, \dfrac{5b}{3}$
C $\dfrac{2a}{3}, \dfrac{2b}{3}$
D $\dfrac{3a}{4}, \dfrac{3b}{4}$

Correct answer

A. $\dfrac{5a}{12}, \dfrac{5b}{12}$

Step-by-step solution

If the total mass of object mass $M$ then the mass removed part will be $\dfrac{M}{4}$ .
A uniform rectangular thin sheet ABCD of mass M has length a and breadth b , as shown in the figure. If the shaded

${X}_{CM} = \dfrac{M \cdot \dfrac{a}{2} - \dfrac{M}{4}\left(\dfrac{3a}{4}\right)}{M - \dfrac{M}{4}} = \dfrac{\dfrac{Ma}{2} - \dfrac{3Ma}{16}}{\dfrac{3M}{4}} = \dfrac{5}{12}a$
${Y}_{CM} = \dfrac{M \cdot \dfrac{b}{2} - \dfrac{M}{4}\left(\dfrac{3b}{4}\right)}{M - \dfrac{M}{4}} = \dfrac{\dfrac{Mb}{2} - \dfrac{3Mb}{16}}{\dfrac{3M}{4}} = \dfrac{5}{12}b$

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