Four particles of masses m_1 = 2m,m_2 = 4m,m_ | vedclass

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(D) Let the corners of the square be $A(0,0)$,$B(a,0)$,$C(a,a)$,and $D(0,a)$.Given masses are $m_1=2m$ at $(0,0)$,$m_2=4m$ at $(a,0)$,$m_3=m$ at $(a,a

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(D) Let the corners of the square be $A(0,0)$,$B(a,0)$,$C(a,a)$,and $D(0,a)$.Given masses are $m_1=2m$ at $(0,0)$,$m_2=4m$ at $(a,0)$,$m_3=m$ at $(a,a)$,and $m_4$ at $(0,a)$.The centre of mass $(X_{cm}, Y_{cm})$ is given by:$X_{cm} = \frac{m_1x_1 + m_2x_2 + m_3x_3 + m_4x_4}{m_1 + m_2 + m_3 + m_4} = \frac{2m(0) + 4m(a) + m(a) + m_4(0)}{2m + 4m + m + m_4} = \frac{5ma}{7m + m_4}$For the centre of mass to be at the centre of the square,$X_{cm} = \frac{a}{2}$.$\frac{5ma}{7m + m_4} = \frac{a}{2} \implies 10m = 7m + m_4 \implies m_4 = 3m$.Now check for $Y_{cm}$:$Y_{cm} = \frac{m_1y_1 + m_2y_2 + m_3y_3 + m_4y_4}{m_1 + m_2 + m_3 + m_4} = \frac{2m(0) + 4m(0) + m(a) + m_4(a)}{2m + 4m + m + m_4} = \frac{ma + m_4a}{7m + m_4}$For $Y_{cm} = \frac{a}{2}$:$\frac{a(m + m_4)}{7m + m_4} = \frac{a}{2} \implies 2m + 2m_4 = 7m + m_4 \implies m_4 = 5m$.Since the values of $m_4$ required for $X_{cm}$ and $Y_{cm}$ are different,it is impossible for the centre of mass to be at the centre of the square for any value of $m_4$. Thus,the correct option is $D$.

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