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(D) Let the maximum tension in the rope be $T = 60 ~kg-wt = 60g ~N$. The two boys have masses $m_1 = 20 ~kg$ and $m_2 = 30 ~kg$. Let their acceleratio

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$3$ : $2$

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(D) Let the maximum tension in the rope be $T = 60 ~kg-wt = 60g ~N$. The two boys have masses $m_1 = 20 ~kg$ and $m_2 = 30 ~kg$. Let their accelerations be $a_1$ and $a_2$ respectively.The total force exerted on the rope is the sum of the forces exerted by each boy: $T = m_1(g + a_1) + m_2(g + a_2)$.Substituting the given values: $60g = 20(g + a_1) + 30(g + a_2)$.$60g = 20g + 20a_1 + 30g + 30a_2$.$60g = 50g + 20a_1 + 30a_2$.$10g = 20a_1 + 30a_2$.Dividing by $10$, we get $g = 2a_1 + 3a_2$.To find the ratio of maximum accelerations, we consider the individual limits. If boy $1$ climbs with maximum acceleration $a_1$, then $a_2 = 0$, so $g = 2a_1 \Rightarrow a_1 = g/2$. If boy $2$ climbs with maximum acceleration $a_2$, then $a_1 = 0$, so $g = 3a_2 \Rightarrow a_2 = g/3$.The ratio of their maximum possible accelerations is $a_1 : a_2 = (g/2) : (g/3) = 3 : 2$.

The maximum tension a rope can withstand is $60 ~kg-wt$. The ratio of maximum acceleration with which two boys of masses $20 ~kg$ and $30 ~kg$ can climb up the rope at the same time is

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