Question

If lift starts moving down with a retardation of $5m{s}^{-2}$, the percentage change in weight of a person in the lift is (g = 10 m $s^{-2}$):

• A. 100
• B. 25
• C. 50
• D. 75

Answer 1

The correct option is C 50

Initially, the weight of the person is $w=mg=10m$

When lift is moving down, the effective acceleration will be $g^{\prime }=g+5=10+5=15m/s^2$

Now weight of person becomes, $w^{\prime }=m{g}^{\prime }=15m$

% change in weight $=\frac{w^{\prime }-w}{w}\times 100=\frac{15m-10m}{10m}\times 100=50$