Four blocks are connected as shown in the fig | vedclass

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(C) Let the mass of each block be $m$. The total mass of the system is $M = m_1 + m_2 + m_3 + m_4 = 4m$.The force $T_1$ is applied to the system,so th

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$2 : 1$

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(C) Let the mass of each block be $m$. The total mass of the system is $M = m_1 + m_2 + m_3 + m_4 = 4m$.The force $T_1$ is applied to the system,so the acceleration $a$ of the system is given by $a = T_1 / (4m)$.Now,consider the block $m_1$. The only force acting on it is the tension $T_4$. Therefore,$T_4 = m_1 a = m(T_1 / 4m) = T_1 / 4$.Next,consider the blocks $m_1$ and $m_2$ together. The force acting on this system is $T_3$. Therefore,$T_3 = (m_1 + m_2) a = 2m(T_1 / 4m) = T_1 / 2$.Finally,the ratio $T_3 / T_4$ is $(T_1 / 2) / (T_1 / 4) = 2 / 1 = 2 : 1$.

Four blocks are connected as shown in the figure on a horizontal frictionless surface. If $m_1 = m_2 = m_3 = m_4 = m$,then the ratio $T_3 / T_4$ will be:

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