You visited us 1 times! Enjoying our articles? Unlock Full Access!

Variation of G Due to the Rotation of Earth

If lift start...

Question

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 $s^{- 2}$ ):

100

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
Initially, the weight of the person is $w = m g = 10 m$
When lift is moving down, the effective acceleration will be $g^{'} = g + 5 = 10 + 5 = 15 m / s^{2}$
Now weight of person becomes, $w^{'} = m g^{'} = 15 m$
% change in weight $= \frac{w^{'} - w}{w} \times 100 = \frac{15 m - 10 m}{10 m} \times 100 = 50$

Suggest Corrections

Similar questions

5 m s^{- 2}

g = 10 m s^{- 2}

5 m s^{- 2}

= 10 m s^{- 2}

m / s^{2}

m / s^{2}

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

60 kg

2 {m/s}^{2}

2 {m/s}^{2}

60 k g

940 k g

m / s^{2}

g = 10 m / s^{2}

Join BYJU'S Learning Program