Apparent weight in an elevator

The person experiences apparent weight, which is equal to the normal reaction $N$ exerted by the elevator floor.

If the actual weight is $mg$, then:

• When elevator accelerates upward with acceleration $a$: $N = m(g+a)$
• When elevator accelerates downward with acceleration $a$: $N = m(g-a)$

Condition for weight loss

A person feels weight loss when apparent weight becomes less than actual weight: $N < mg$

This happens when: $N = m(g-a)$ i.e. when the elevator has downward acceleration.

Check each option

A: When the elevator moves upward with constant acceleration $N = m(g+a) > mg$ This is weight gain, not weight loss.

B: When the elevator moves downward with constant acceleration $N = m(g-a) < mg$ This is weight loss.

C: When the elevator moves upward with uniform velocity Uniform velocity means acceleration $=0$. $N = mg$ No weight loss.

D: When the elevator moves downward with uniform velocity Again, acceleration $=0$. $N = mg$ No weight loss.

Final answer

The correct option is: $\boxed{\text{B}}$