Physics · System Of Particles

JEE Main 2025 — 7 April, Morning Shift — Question 54

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 $L=10 \mathrm{~cm}$ )

Answer: B

$m_{1}=(2 m) ; x_{1}=L, y_{1}=0$ $m_{2}=(3 m) ; x_{2}=0, y_{2}=\frac{3 L}{2}$

\text { So, } x_{\mathrm{cm}}=\frac{2 m(L)+0}{5 m}=\frac{2 L}{5}=4 \hat{i}(\mathrm{~cm})

y_{\mathrm{cm}}=\frac{(0)+\frac{9 m L}{2}}{5 m}=\frac{9 L}{10}=9 \hat{j}(\mathrm{~cm})

So $r_{\mathrm{cm}}=(4 \hat{i}+9 \hat{j}) \mathrm{cm}$

Answer key and solution verified before publishing.

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