QJEEM21Ph9b8d: JEE Main 2021 Physics - Class 11 JEEM21Ph9b8d Solution: A person standing on a spring balance inside a stationary lift…

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.