JEE MAIN - Physics (2021 - 26th February Morning Shift - No. 21)

A person standing on a spring balance inside a stationary lift measures 60 kg. The weight of that person if the lift descends with uniform downward acceleration of 1.8 m/s2 will be ______________ N. [g = 10 m/s2]
Answer
492

Explanation

The apparent weight $W_{\text{app}}$ of a person in an elevator moving with acceleration is given by:

$$W_{\text{app}} = m(g - a)$$

where:

  • $W_{\text{app}}$ is the apparent weight,
  • $m$ is the mass of the person,
  • $g$ is the acceleration due to gravity,
  • $a$ is the acceleration of the elevator.

Given that the person's mass is 60 kg, the acceleration due to gravity is 10 m/s², and the acceleration of the lift is 1.8 m/s², we can substitute these values into the formula:

$$W_{\text{app}} = 60 \, \text{kg} \times (10 \, \text{m/s}^2 - 1.8 \, \text{m/s}^2) = 60 \, \text{kg} \times 8.2 \, \text{m/s}^2 = 492 \, \text{N}$$

So, the apparent weight of the person when the lift descends with a uniform downward acceleration of 1.8 m/s² will be 492 N.

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