Three blocks A,B and C are arranged as shown | vedclass

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(D) For the centre of mass of the system to remain unchanged,the net displacement of the centre of mass must be zero.Let the masses be $m_A = 10 \ kg$

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$1 \ m$,towards block $C$

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(D) For the centre of mass of the system to remain unchanged,the net displacement of the centre of mass must be zero.Let the masses be $m_A = 10 \ kg$,$m_B = 25 \ kg$,and $m_C = 15 \ kg$.Let the displacements be $\Delta x_A$,$\Delta x_B$,and $\Delta x_C$.Taking the direction towards the right as positive:Block $A$ is moved towards $B$ (right) by $2 \ m$,so $\Delta x_A = +2 \ m$.Block $C$ is moved towards $B$ (left) by $3 \ m$,so $\Delta x_C = -3 \ m$.Let block $B$ be moved by $\Delta x_B$.The condition for the centre of mass to remain unchanged is:$m_A \Delta x_A + m_B \Delta x_B + m_C \Delta x_C = 0$Substituting the values:$(10 \ kg)(2 \ m) + (25 \ kg)(\Delta x_B) + (15 \ kg)(-3 \ m) = 0$$20 + 25 \Delta x_B - 45 = 0$$25 \Delta x_B - 25 = 0$$25 \Delta x_B = 25$$\Delta x_B = 1 \ m$Since the result is positive,block $B$ must be moved by $1 \ m$ in the positive direction,which is towards block $C$.

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