Question 59 of 130

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:

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

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Answer: (A) $\dfrac{5a}{12},\dfrac{5b}{12}$

If the total mass of object mass $M$ then the mass removed part will be $\dfrac{M}{4}$ .

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

JEE Main Physics practice question on Center of Mass, Momentum & Collision (Centre of mass of continuous mass distribution), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.