For the arrangement shown in the figure, the pulleys are light and smooth and the strings are ideal. When the system is released from rest, the $10 - k g$ and $5 - k g$ blocks are found to remain at rest. This is possible:

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If $m_{1} = 2.5 kg, m_{2} = 5 kg$

For all values of $m_{1}$ and $m_{2}$

If the arithmetic mean of $m_{1}, m_{2} = 5 kg$

If the harmonic mean of $m_{1}, m_{2} = 2.5 kg$

answer is D.

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Detailed Solution

Tension in the thread over the topmost pulley is $10 g$ . Tension in the thread over the middle pulley is $5 g .$ Tension in the thread over the bottom-most pulley is $2.5 g .$

The question does not state whether the two blocks

$m_{1}, m_{2}$ are at rest. So, let their acceleration be a (see figure). Thus, we have

$2.5 g - m_{1} g = m_{1} a$ .[1]

and $m_{2} g - 2.5 g = m_{2} a$ ...[2]

Eliminating a from the above equations, we have

$\begin{array}{l} \frac{2.5 - m_{1}}{m_{2} - 2.5} = \frac{m_{1}}{m_{2}} \\ \Rightarrow 2.5 (m_{1} + m_{2}) = 2 m_{1} m_{2} \\ \Rightarrow \frac{1}{m_{1}} + \frac{1}{m_{2}} = \frac{2}{2.5} \\ \Rightarrow HM of m_{1}, m_{2} equals 2.5 \end{array}$