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  4. A boy is standing on a weighing machine inside a lift. When the lift goes upwards with acceleration g/4, the machine shows the reading 50 kg. wt. When the lift goes downward with acceleration g/4, the reading of the machine in kg. wt. would be

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Q. A boy is standing on a weighing machine inside a lift. When the lift goes upwards with acceleration $g/4$, the machine shows the reading $50 \,kg$. wt. When the lift goes downward with acceleration $g/4$, the reading of the machine in kg. wt. would be

KEAMKEAM 2020

Solution:

Reading of the weighing balance
$W_{1}=m\left(1+\frac{a}{g}\right) kgwt$.
when it is upwards
Given:
$a=\frac{g}{4} ; \omega_{1}=50\, kgw t$
$\Rightarrow m=\left(1+\frac{1}{4}\right)=50\, kgwt$
or $ \frac{5}{4} M=50\, kgw t$
Reading of the weighing balance when lift is accelerating downwards is
$\omega_{2}=M\left(1-\frac{a}{g}\right)=\frac{3}{4}\, m \,kgwt =30 \,kgw t$