A block attached between two springs is in equilibrium. If the upper spring is cut, then the acceleration of the block just after cut is ${6 m/s}^{2}$ downwards. Now, if instead of upper spring, the lower spring is cut, then the magnitude of acceleration of the block, just after the cut will be (Take g = ${10 m/s}^{2}$ and identical springs)

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${16 m/s}^{2}$

${4 m/s}^{2}$

${10 m/s}^{2}$

None of these

answer is B.

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Detailed Solution

$For equillibrium, F_{U} + F_{L} = mg . .. (i)$

${When F}_{U} = 0, mg - F_{L} = m \times 6$

$\Rightarrow F_{L} = 10 m - 6 m = 4 m$

$Again, when F_{L} = 0, mg - F_{U} = ma$

$\Rightarrow a = \frac{mg - F_{U}}{m} = \frac{mg - (mg - F_{L})}{m} (∵ F_{U} = mg - F_{L})$

$= \frac{F_{L}}{m} = \frac{4 m}{m}$

$\Rightarrow {a = 4 m/s}^{2}$

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