JEE Advanced 2025 — Paper 1 — PHY-III solution

JEE Advanced2025Paper 1PHY-III

A person sitting inside an elevator performs a weighing experiment with an object of mass 50 kg. Suppose that the variation of the height $y$ (in m) of the elevator, from the ground, with time $t$ (in s) is given by $y = 8 [1 + sin (\frac{2 π t}{T})]$ , where $T = 40 π$ s. Taking acceleration due to gravity, $g = 10$ m/s $^{2}$ , the maximum variation of the object's weight (in N) as observed in the experiment is _______.

Your answer

Solution

The elevator executes SHM with amplitude $A = 8$ m and angular frequency $Ω = \frac{2 π}{T} = \frac{2 π}{40 π} = \frac{1}{20}$ rad/s.

Maximum acceleration of the elevator: $a_{m a x} = Ω^{2} A = (\frac{1}{20})^{2} \times 8 = \frac{8}{400} = 0.02$ m/s $^{2}$ .

As measured by an observer inside the elevator, the apparent weight changes by $ma$ . The weight swings between $m (g - a_{m a x})$ and $m (g + a_{m a x})$ , so the maximum variation is:

$Δ W = 2 m a_{m a x} = 2 \times 50 \times 0.02 = 2.00$ N

$∴ Δ W = 2.00$ N.

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