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} $
BETA
