A boy is standing in a lift- The force on the floor of the lift due to the weight of the boy becomes zero-When the lift is not movingNone of theseWhen the lift moves down with an acceleration of 9-8-ms-2When the lift moves down with an acceleration 9-8-ms-2

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

Application of Third Law of Motion

A boy is stan...

Question

A boy is standing in a lift. The force on the floor of the lift due to the weight of the boy becomes zero:

When the lift moves down with an acceleration of

9.8 m s^{- 2}

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

When the lift moves down with an acceleration

9.8 m s^{- 2}

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

When the lift is not moving

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

None of these

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

Open in App

Solution

The correct option is B When the lift moves down with an acceleration $9.8 m s^{- 2}$
When the lift moves down the reaction will be in the upward direction. Therefore tension mg acts down and reaction mg acts in the opposite direction. The net force acting will be zero.

Suggest Corrections

Similar questions

Q. A boy of 50 kg is in a lift moving down with an acceleration 9.8

m s^{- 2}

. The apparent weight of the body is (g = 9.8

m s^{- 2}

)

A body weighing 50 kg(on earth) is standing on a floor of the lift. Find the weight when --

1) the lift moves up with uniform velocity.

2)lift moves down with an acceleration of -4.9m/s²

3)lift moves up with an acceleration of 4.9m/s²

4)lift begans to fall freely under gravity.

Q. A boy of 50 kg is standing in a lift moving down with an acceleration

9.8 m / s^{2}

. The apparent weight of the boy is:

Q. A boy is standing in a lift. The force on the floor of the lift due to the weight of the boy becomes zero:

A boy of mass 40 kg is standing in a lift,
which is moving downwards with an acceleration 9.8 $m / s^{2}$ . The apparent weight of
the boy is (Take $g = 9.8 m / s^{2}$ )

Join BYJU'S Learning Program

A boy is standing in a lift. The force on the floor of the lift due to the weight of the boy becomes zero:

The correct option is B When the lift moves down with an acceleration 9.8 ms−2When the lift moves down the reaction will be in the upward direction. Therefore tension mg acts down and reaction mg acts in the opposite direction. The net force acting will be zero.

The correct option is B When the lift moves down with an acceleration $9.8 m s^{- 2}$
When the lift moves down the reaction will be in the upward direction. Therefore tension mg acts down and reaction mg acts in the opposite direction. The net force acting will be zero.