Apparent Weight in Lift PYQ | JEE Main Physics 2025

 

❓ Question

A person is standing in a lift.

The lift accelerates with magnitude

$$a=5\ \text{m/s}^2.$$

If the lift moves

1. Upward, and
2. Downward,

find the percentage change in the person's apparent weight in each case.

Also find the ratio of these percentage changes.

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✍️ Solution

The apparent weight of a person is equal to the normal reaction $N$.

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Case 1: Lift Accelerating Upward

Applying Newton's Second Law,

$$N-mg=ma.$$

Hence,

$$N_{\text{up}}=m(g+a).$$

The percentage change in weight is

$$\frac{N_{\text{up}}-mg}{mg}\times100.$$

Substituting $N_{\text{up}}$,

$$\frac{m(g+a)-mg}{mg}\times100 = \frac{a}{g}\times100.$$

Taking

$$g=10\ \text{m/s}^2,\qquad a=5\ \text{m/s}^2,$$
$$\frac{5}{10}\times100 = 50\%.$$

Thus,

$$\boxed{\text{Apparent weight increases by }50\%.}$$

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Case 2: Lift Accelerating Downward

Applying Newton's Second Law,

$$mg-N=ma.$$

Hence,

$$N_{\text{down}}=m(g-a).$$

The percentage change in weight is

$$\frac{N_{\text{down}}-mg}{mg}\times100.$$

Substituting $N_{\text{down}}$,

$$\frac{m(g-a)-mg}{mg}\times100 = -\frac{a}{g}\times100.$$

With

$$a=5\ \text{m/s}^2,\qquad g=10\ \text{m/s}^2,$$
$$-\frac{5}{10}\times100 = -50\%.$$

Thus,

$$\boxed{\text{Apparent weight decreases by }50\%.}$$

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Ratio of Percentage Changes

Magnitude of increase

$$=50\%.$$

Magnitude of decrease

$$=50\%.$$

Therefore,

$$\boxed{1:1.}$$

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✅ Final Answer

• Upward acceleration: $\boxed{50\%}$ increase
• Downward acceleration: $\boxed{50\%}$ decrease
• Ratio of percentage changes: $\boxed{1:1}$