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In the figure...

Question

In the figure, the blocks $A, B$ and $C$ of mass $m$ each have acceleration $a_{1}, a_{2}$ and $a_{3}$ respectively. $F_{1}$ and $F_{2}$ are external forces of magnitudes $2 m g$ each. What is the relation between the acceleration of the masses $A$ , $B$ and $C$ ?

A

a_{1} = a_{2} = a_{3}

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B

a_{1} > a_{2} > a_{3}

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C

a_{1} = a_{3} > a_{2}

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D

a_{1} > a_{2}, a_{2} = a_{3}

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Solution

The correct option is C $a_{1} = a_{3} > a_{2}$
From the FBDs of all three systems we have,

For $A$
$a_{1} = \frac{2 m g - m g}{m} = g$ .......(i)

For $B$
$2 m g - T = 2 m a_{2}$
$T - m g = m a_{2}$
$a_{2} = g / 3$

$a_{3} = \frac{2 m g - m g}{m} = g$
$∴ a_{1} = a_{3} > a_{2}$

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