Question

With what acceleration ${}^{\prime }{a}^{\prime }$ shown, the elevator descends so that the block of mass $M$ exerts a force of $\frac{Mg}{10}$ on the weighing machine? [$g$ = acceleration due to gravity]

• A. $0.3g$
• B. $0.1g$
• C. $0.9g$
• D. $0.6g$

Answer 1

The correct option is C $0.9g$

as lift is moving down pseudo force acts up
$\Rightarrow N=m(g-a)$
Given $N=\frac{mg}{10}$
$\frac{mg}{10}=m(g-a)$
$g=10g-10a$
$10a=9g$
$a=0.9g$