Masses 8,kg, 2,kg, 4,kg, 2,kg are placed at t | vedclass

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(B) Let corner $A$ of the square $ABCD$ be at the origin $(0,0)$. The mass $8\,kg$ is placed at $A$. The diagonal of the square is $d = a\sqrt{2} = 80

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$30$

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(B) Let corner $A$ of the square $ABCD$ be at the origin $(0,0)$. The mass $8\,kg$ is placed at $A$. The diagonal of the square is $d = a\sqrt{2} = 80\,cm$,where $a$ is the side length.$a = \frac{80}{\sqrt{2}} = 40\sqrt{2}\,cm$.The masses are: $m_A = 8\,kg$,$m_B = 2\,kg$,$m_C = 4\,kg$,$m_D = 2\,kg$.The coordinates of the corners are: $A(0,0)$,$B(a,0)$,$C(a,a)$,$D(0,a)$.The coordinates of the centre of mass $(X_{CM}, Y_{CM})$ are given by:$X_{CM} = \frac{m_A x_A + m_B x_B + m_C x_C + m_D x_D}{m_A + m_B + m_C + m_D} = \frac{8(0) + 2(a) + 4(a) + 2(0)}{8 + 2 + 4 + 2} = \frac{6a}{16} = \frac{3a}{8}$.$Y_{CM} = \frac{m_A y_A + m_B y_B + m_C y_C + m_D y_D}{m_A + m_B + m_C + m_D} = \frac{8(0) + 2(0) + 4(a) + 2(a)}{8 + 2 + 4 + 2} = \frac{6a}{16} = \frac{3a}{8}$.Substituting $a = 40\sqrt{2}\,cm$:$X_{CM} = \frac{3(40\sqrt{2})}{8} = 15\sqrt{2}\,cm$.$Y_{CM} = \frac{3(40\sqrt{2})}{8} = 15\sqrt{2}\,cm$.The distance of the centre of mass from $A(0,0)$ is $d = \sqrt{X_{CM}^2 + Y_{CM}^2} = \sqrt{(15\sqrt{2})^2 + (15\sqrt{2})^2} = \sqrt{450 + 450} = \sqrt{900} = 30\,cm$.

Masses $8\,kg, 2\,kg, 4\,kg, 2\,kg$ are placed at the corners $A, B, C, D$ respectively of a square $ABCD$ of diagonal $80\,cm$. The distance of the centre of mass from $A$ will be ........ $cm$.

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