The apparent weight of man inside a lift moving up with certain acceleration is $900$ N. When the lift is coming down with the same acceleration apparent weight is found to be $300$ N. The mass of the man is $(g = 10 m s^{- 2})$ :

45

No worries! We‘ve got your back. Try BYJU‘S free classes today!

60

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

75

No worries! We‘ve got your back. Try BYJU‘S free classes today!

80

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D $60$ kg
The weight of the man going up $= 900 N$
Weight while coming down $= 300 N$
So the weight of the men $= 600 N$
Now $w = m g$
$m = \frac{w}{g}$
$= \frac{600}{10} (t a k i n g . g = 10 \frac{m}{s^{2}})$
$= 60 k g$

Suggest Corrections

Similar questions

Q. A

60 k g

man is inside a lift which is moving up with an acceleration of

2.45 {m s}^{- 2}

. The apparent percentage change in his weight is,

Q. A man is inside a lift. When the lift is moving up with certain acceleration, the apparent change in weight is

25 %

. When the lift moves down with double the acceleration, the apparent change in weight is

Q. The apparent weight of a person inside a lift is

W_{1}

when lift moves up with certain acceleration and is

W_{2}

when lift moves down with same acceleration. The weight of person when lift moves up with constant speed is

A person of mass 60 kg is in a lift. The change in the apparent weight of the person, when the lift moves up with an acceleration of $2 m s^{- 2}$ and then comes down with an acceleration of $2 m s^{- 2}$ , is

Q. A man of mass

m

is on the floor of a lift. Match the following.

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

Join BYJU'S Learning Program

The apparent weight of man inside a lift moving up with certain acceleration is 900N. When the lift is coming down with the same acceleration apparent weight is found to be 300N. The mass of the man is(g=10ms−2):

The correct option is D 60kgThe weight of the man going up=900N Weight while coming down =300N So the weight of the men =600N Now w=mgm=wg =60010(taking.g=10ms2) =60kg

The apparent weight of man inside a lift moving up with certain acceleration is $900$ N. When the lift is coming down with the same acceleration apparent weight is found to be $300$ N. The mass of the man is $(g = 10 m s^{- 2})$ :

The correct option is D $60$ kg
The weight of the man going up $= 900 N$
Weight while coming down $= 300 N$
So the weight of the men $= 600 N$
Now $w = m g$
$m = \frac{w}{g}$
$= \frac{600}{10} (t a k i n g . g = 10 \frac{m}{s^{2}})$
$= 60 k g$