A person of mass 50 kg is standing in a lift. If the lift moves in upw

A person of mass 50 kg is standing in a lift. If the lift moves in upward direction with an acceleration of 1 m/s², then the weight of the person will be closest to :

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490 N

540 N

440 N

50 N

Answer is Right!

Answer is Wrong!

This question was previously asked in

UPSC CISF-AC-EXE – 2023

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The correct answer is B.

When a person is in a lift accelerating upwards, the apparent weight of the person is greater than their true weight. The forces acting on the person are the gravitational force (mg) downwards and the normal force (N) exerted by the lift floor upwards. According to Newton’s second law, the net force (N – mg) is equal to mass times acceleration (ma). Thus, the normal force N = mg + ma = m(g + a). This normal force is the apparent weight.

Given mass m = 50 kg and upward acceleration a = 1 m/s². Using the standard value of g ≈ 9.8 m/s²: Apparent weight N = 50 kg * (9.8 m/s² + 1 m/s²) = 50 kg * 10.8 m/s² = 540 N. If we use g ≈ 10 m/s², N = 50 kg * (10 m/s² + 1 m/s²) = 50 kg * 11 m/s² = 550 N. The option closest to 540 N (or 550 N) is 540 N.

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