In the system shown in the figure at some instant acceleration of pulley was $10 m / s e c^{2}$ in upward direction. At that instant spring was in its natural length. If acceleration of block A and B are $a_{A} a n d a_{B}$ at that instant then : $(G i v e n m_{A} = 10 k g, m_{B} = 20 k g, g = 10 m / s e c^{2}, p u l l e y i s l i g h t a n d s m o o t h)$

a_{A} = \frac{10}{3} m / s e c^{2} u p w a r d a_{B} = \frac{10}{3} m / s e c^{2} d o w n w a r d

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a_{A} = a_{B} = 5 m / s e c^{2} u p w a r d

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a_{A} = a_{B} = 10 m / s e c^{2} d o w n w a r d

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None of these

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Solution

The correct option is C $a_{A} = a_{B} = 10 m / s e c^{2} d o w n w a r d$
Answer is C
Since the spring is in its natural length, then tension in the string is zero. This implies that blocks are in the free fall state and
$a_{A} = a_{B} = g w i t h d i r e c t i o n d o w n w a r d s$

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In the system shown in figure at some instant acceleration of pulley was $10 m / s^{2}$ in upward direction. At that instant spring was in its natural length. If acceleration of block A and B are $a_{A}$ and $a_{B}$ at that instant then $a_{A}$ and $a_{B}$ are respectively: (Given $m_{A}$ = 10 kg, $m_{B}$ = 20 kg, g = $10 m / s^{2}$ , pulley is light and smooth)