In a lift, the apparent weight becomes twice the actual weight when

moves down with a steady velocity of 19.6 m/s

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lift is moving down with acceleration equal to g

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lift is moving up with acceleration equal to g

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moves up with a steady velocity of 19.6m/s

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Solution

The correct option is C
lift is moving up with acceleration equal to g

When the lift is moving up with acceleration g, then we have

R = m (g+a) = m (g+g) = 2mg

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Similar questions

W_{1}

W_{2}

25 %

m

List I	List II
a) Lift is moving up with acceleration a	e) Apparent weight is greater than true weight
b) Lift is moving down with acceleration a	f) Apparent weight is zero
c) Lift is moving with uniformly velocity	g) Apparent weight is equal to true weight
d) Lift is freely falling	h) Apparent weight is less than true weight

900

300

(g = 10 m s^{- 2})

List - I	List-II
(a) Lift moves up with uniform acceleration a	(d) mg
(b) Lift moves down with uniform acceleration a	(e) m(g $+$ a)
(c) Lift moves down with uniform velocity	(f) m(g-a)

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