Fig shows two pulley arrangements for lifting a mass m. In (a) the mass is lifted by attaching a mass $2 m$ while in (b) the mass is lifted by pulling the other end with a downward force $F = 2$ mg, if $f_{a}$ and $f_{b}$ are the accelerations of the two masses then:

f_{a} = f_{b}

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f_{a} = f_{b} / 2

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f_{a} = f_{b} / 3

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f_{a} = 2 f_{b}

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Solution

The correct option is C $f_{a} = f_{b} / 3$
From FBD of (A): $m f_{a} = T - m g . . . (1)$ and $2 m f_{a} = 2 m g - T . . . (2)$
$(1) + (2) \Rightarrow f_{a} = \frac{2 m - m}{2 m + m} g = g / 3$
From FBD of (B): $m f_{b} = F - m g = 2 m g - m g = m g \Rightarrow f_{b} = g$
thus, $f_{a} = \frac{f_{b}}{3}$

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