Question 93 of 210

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

A$g$B$\dfrac{3g}{4}$C$\dfrac{g}{2}$D$\dfrac{g}{4}$

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Answer: (B) $\dfrac{3g}{4}$

The forces acting on the block are shown below.
Here, $N$ is normal force and box is falling with acceleration $a$.

From figure, $Mg-N=Ma$
According to question, $N=\dfrac{Mg}{4}$
$\dfrac{Mg}{4}=M\left(g-a\right)$
$\Rightarrow a=\dfrac{3g}{4}$