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  4. 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. Find the acceleration of lift.

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Q. 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. Find the acceleration of lift.

NTA AbhyasNTA Abhyas 2022

Solution:

$N=m\left(g + a\right)$
$N^{′}=m\left(g - a\right)$
$N^{′}=\frac{10}{100}\times N$

Solution
$m\left(g - a\right)=\frac{m \left(g + a\right)}{10}$
$10g-10a=g+a$
$9g=11a$
$\Rightarrow a=\frac{9 g}{11}$