If the system of blocks shown in the figure i | vedclass

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(D) Let the masses be $m_1 = 4 \ kg$ (left side), $m_2 = 3 \ kg$ (upper right), and $m_3 = 3 \ kg$ (lower right).The total mass on the right side is $

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$3: 4$

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(D) Let the masses be $m_1 = 4 \ kg$ (left side), $m_2 = 3 \ kg$ (upper right), and $m_3 = 3 \ kg$ (lower right).The total mass on the right side is $M_R = m_2 + m_3 = 3 \ kg + 3 \ kg = 6 \ kg$.The mass on the left side is $M_L = 4 \ kg$.Since $M_R > M_L$, the system accelerates towards the right with acceleration $a$.The acceleration $a$ is given by $a = \frac{(M_R - M_L)g}{M_R + M_L} = \frac{(6 - 4)g}{6 + 4} = \frac{2g}{10} = 0.2g$.For the lower $3 \ kg$ block, the equation of motion is $T_1 - m_3g = m_3a$.$T_1 = m_3(g + a) = 3(g + 0.2g) = 3(1.2g) = 3.6g$.For the entire system, the tension $T_2$ in the string passing over the pulley is $T_2 = \frac{2 M_L M_R g}{M_L + M_R} = \frac{2(4)(6)g}{4 + 6} = \frac{48g}{10} = 4.8g$.The ratio of tensions is $\frac{T_1}{T_2} = \frac{3.6g}{4.8g} = \frac{36}{48} = \frac{3}{4}$.

If the system of blocks shown in the figure is released from rest, the ratio of the tensions $T_1$ and $T_2$ is (Neglect the mass of the string shown in the figure).

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