A lift is moving down with an acceleration of $5 m s^{- 2}$ . The percentage change in weight of person in the lift is (g $= 10 m s^{- 2}$ ) :

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

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

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

Open in App

Solution

The correct option is C 50
w $=$ mg $=$ 10 m N
When lift is moving downwards apparent weight
$W = m (g - a)$
$= m (10 - 5) = 5 m N$
% change in weight
$= \frac{w^{1} - w}{w} \times 100$
$= \frac{5 m - 10 m}{10 m} \times 100$
$= - \frac{1}{2} \times 100 = - 50$ %

Suggest Corrections

Similar questions

Q. A lift is moving down with a retardation of

5 m s^{- 2}

. Calculate the percentage change in the weight of the person in the lift (

g = 10 m s^{- 2}

Q. If lift starts moving down with a retardation of

5 m s^{- 2}

, the percentage change in weight of a person in the lift is (g = 10 m

s^{- 2}

Q. When a lift is moving upward with acceleration of 5

m / s^{2}

then percentage change in the weight of the person in the lift is x%.If the lift is moving down with acceleration 5

m / s^{2}

then percentage change in the weight of same person is y%.Then ratio.

\frac{x}{y} i s (g = 10 m / s^{2})

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. In a lift moving up with an acceleration of

5 m s^{- 2}

, a ball is dropped from a height of

1.25

m. The time taken by the ball to reach the floor of the lift is .....(nearly)

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

Join BYJU'S Learning Program

A lift is moving down with an acceleration of 5ms−2. The percentage change in weight of person in the lift is (g =10ms−2) :

The correct option is C 50w = mg = 10 m NWhen lift is moving downwards apparent weightW=m(g−a)=m(10−5)=5mN% change in weight=w1−ww×100=5m−10m10m×100=−12×100=−50%

A lift is moving down with an acceleration of $5 m s^{- 2}$ . The percentage change in weight of person in the lift is (g $= 10 m s^{- 2}$ ) :

The correct option is C 50
w $=$ mg $=$ 10 m N
When lift is moving downwards apparent weight
$W = m (g - a)$
$= m (10 - 5) = 5 m N$
% change in weight
$= \frac{w^{1} - w}{w} \times 100$
$= \frac{5 m - 10 m}{10 m} \times 100$
$= - \frac{1}{2} \times 100 = - 50$ %