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A boy stands on a weighing machine inside a lift. When the lift is going down with acceleration $g / 4$ , the machine shows a reading $30 k g f$ . When the lift goes upwards with acceleration $g / 4$ , the reading would be

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$18 k g f$

$50 k g f$

$67.5 k g f$

$37 . 5 k g f$

answer is C.

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Detailed Solution

While going down, machine reading is $30 k g f$ .

$m g - N = m (\frac{g}{4}) m g - 30 g = (\frac{m g}{4}) \frac{3 m g}{4} = 30 g \Rightarrow m = \frac{30 \times 4}{3} = 40 k g$

While going up the reaction would be

$N^{'} - m g = m \frac{g}{4} N^{'} = 5 \frac{m g}{4} = \frac{5}{4} (40 k g) g N^{'} = 50 k g f$

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