Question 36 of 130

A rod of length 5 L is bent right angle keeping one side length as 2 L.

The position of the centre of mass of the system: (Consider $\mathrm{L}=10 \mathrm{~cm}$)

A$2 \hat{i}+3 \hat{j}$B$3 \hat{i}+7 \hat{\mathrm{j}}$C$5 \hat{\mathrm{i}}+8 \hat{\mathrm{j}}$D$4 \hat{i}+9 \hat{j}$

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Answer: (D) $4 \hat{i}+9 \hat{j}$

$\mathrm{x}_{\mathrm{com}}=\frac{2 \mathrm{~m}(10)+3 \mathrm{~m}(0)}{5 \mathrm{~m}}=4 \mathrm{~cm}$
$\mathrm{y}_{\mathrm{com}}=\frac{2 \mathrm{~m}(0)+3 \mathrm{~m}(15)}{5 \mathrm{~m}}=9 \mathrm{~cm}$
$\overrightarrow{\mathrm{r}}_{\mathrm{com}}=4 \hat{\mathrm{i}}+9 \hat{\mathrm{j}}$

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