SATHEE: Laws Of Motion Question 423

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

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

D) $g m / \sec^{2}$

Show Answer

Answer:

Correct Answer: B

Solution:

[b]

$N = m (g + a)$

$N^{'} = m (g - a)$

$N^{'} = \frac{10}{100} \times N$

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

$10 g - 10 a = g + a$

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

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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:

Answer:

Solution:

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