Laws Of Motion Question 423

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 decends with the same acceleration and reading is 10% of initial. Find the acceleration of lift?

Options:

A) $ \frac{g}{19}m/{{\sec }^{2}} $

B) $ \frac{9g}{11}m/{{\sec }^{2}} $

C) $ 0/{{\sec }^{2}} $

D) $ gm/{{\sec }^{2}} $

Show Answer

Answer:

Correct Answer: B

Solution:

[b]

$ N=m(g+a) $

$ N’=m(g-a) $

$ N’=\frac{10}{100}\times N $

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

$ 10g-10a=g+a $

$ 9g=11a\Rightarrow a=\frac{9g}{11} $



NCERT Chapter Video Solution

Dual Pane