A man is inside a lift. When the lift is moving up with certain acceleration, the apparent change in weight is $25\%$. When the lift moves down with double the acceleration, the apparent change in weight is

The correct option is C $50\%$
As the weighing machine measures the normal reaction, we have


$W_{app}-W=ma$

Given that the apparent change in weight is 25%,

$\frac{25}{100}W=ma$

$\frac{mg}{4}=ma$

$a=\frac{g}{4}$

Now, when the lift moves downwards


$W-W_{app}=2a$

$mg-W_{app}=2a$

$W_{app}=m[g-2a]$

$W=m[g-2\left(\frac{g}{4}\right)]$

$W_{app}=\frac{mg}{2}$

Apparent change is thus
$W-W_{app}^{down}=(mg-\frac{mg}{2})=\frac{mg}{2}$

So, we have the percentage change as,

$\frac{W-W_{app}^{down}}{W}100\%=\frac{1}{2}100\%=50\%$