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.

Question

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. (A) (g/19) (B) (9 g/11) (C) zero (D) g

Tardigrade Admin · Accepted Answer

N=m(g + a) Nâ²=m(g - a) Nâ²=(10/100)× N m(g - a)=(m (g + a)/10) 10g-10a=g+a 9g=11a ⇒ a=(9 g/11)

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.

$\frac{g}{19}$

$\frac{9 g}{11}$

zero

$g$

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

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

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.

19g​

119g​

zero

g

N=m(g+a) N′=m(g−a) N′=10010​×N m(g−a)=10m(g+a)​ 10g−10a=g+a 9g=11a ⇒a=119g​

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.

$\frac{g}{19}$

$\frac{9 g}{11}$

$g$

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

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