JEE Advanced 2025 Paper 1 · Q08 · Friction & NLM
A person inside an elevator weighs an object of mass $50$ kg. The elevator's height varies as $y = 8\bigl[1 + \sin(2\pi t/T)\bigr]$ m, with $T = 40\pi$ s. Take $g = 10$ m/s². Find the maximum variation in the object's weight (in N) observed during the experiment.
Reveal answer + step-by-step solution
Correct answer:2
Solution
Acceleration $a = d^2 y/dt^2 = -8\,(2\pi/T)^2 \sin(2\pi t/T)$. With $T = 40\pi$, $\omega = 2\pi/T = 1/20$. $a_{\max} = 8\,\omega^2 = 8/400 = 0.02$ m/s². Apparent weight $= m(g + a)$. Variation from $g = m\cdot a$, so amplitude of weight variation $= m\cdot a_{\max} = 50 \times 0.02 = 1$ N. Maximum VARIATION (peak-to-peak swing in some readings, or peak from mean) — official answer is $2$, indicating peak-to-peak: weight oscillates between $m(g - a_{\max})$ and $m(g + a_{\max})$, spanning $2 m a_{\max} = 2 \times 50 \times 0.02 = 2$ N. Answer: 2 N.
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