Question

A man stands on a weighing machine kept inside a lift. Initially the lift is ascending with the acceleration 'a' due to which the reading is W. Now the lift descends with the same acceleration and reading is 10% of initial. The acceleration of the lift is:

• A. $\frac{9g}{19}m/se{c}^2$
• B. $\frac{9g}{11}m/se{c}^2$
• C. $0m/se{c}^2$
• D. $gm/se{c}^2$

Answer 1

Answer: (B)

$\frac{9g}{11}m/se{c}^2$

Taking the situation of ascending, we have
$N=mg+ma....\left(i\right)$
For descending, we have
$N^{\prime }=mg-ma....\left(ii\right)$
where, $N$ and $N^{\prime }$ are the weights during ascending and descending respectively.
Given : $N^{\prime }=\frac{N}{10}$
From (i) and (ii), we have
Or, $10(g-a)=g+a$
Therefore, $a=\frac{9g}{11}m{s}^{-2}$