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A block of mass M placed inside a box descends vertically with acceleration 'a'. The block exerts a force equal to one-fourth of its weight on the floor of the box. The value of 'a' will be

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$g$

$\frac{g}{2}$

$\frac{3 g}{4}$

$\frac{g}{4}$

answer is C.

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Detailed Solution

mg – N = ma
$a = g - \frac{g}{4}$
$a = \frac{3 g}{4}$

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