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JEE Advanced 2025 · Paper 1, Physics question 8

A person sitting inside an elevator performs a weighing experiment with an object of mass 50 kg. Suppose that the variation of the height yy (in m) of the elevator, from the ground, with time tt (in s) is given by y=8[1+sin⁡(2πtT)]y = 8\left[1 + \sin\left(\dfrac{2\pi t}{T}\right)\right], where T=40πT = 40\pi s. Taking acceleration due to gravity, g=10g = 10 m/s2^2, the maximum variation of the object's weight (in N) as observed in the experiment is _____.

Answer and worked solution

Official answer: 2

Acceleration.
a=y¨=−8(2πT)2sin⁡2πtT,2πT=120 s−1.a=\ddot y=-8\left(\frac{2\pi}{T}\right)^2\sin\frac{2\pi t}{T},\qquad \frac{2\pi}{T}=\frac1{20}\ \text{s}^{-1}.
So amax=8400=0.02 m/s2a_{max}=\frac{8}{400}=0.02\ \text{m/s}^2.

Apparent weight. m(g+a)m(g+a) ranges from m(g−0.02)m(g-0.02) to m(g+0.02)m(g+0.02). The maximum variation is
2mamax=2×50×0.02=2 N.2ma_{max}=2\times50\times0.02=2\ \text{N}.