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JEE Advanced 2025Paper 1
Exam
JEE Advanced
Year
2025
Paper
Paper 1
Subject
Physics
Chapter
Laws of Motion

JEE Advanced 2025 · Paper 1 · Physics · Laws of Motion

A person sitting inside an elevator performs a weighing experiment...

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(\frac{2 \pi t}{T}\right)\right], where T=40π sT=40 \pi \mathrm{~s}. Taking acceleration due to gravity, g=10 m/s2g=10 \mathrm{~m} / \mathrm{s}^2, the maximum variation of the object's weight (in N ) as observed in the experiment is _______ .

Correct answer

0

Step-by-step solution

y=8+8sin⁡2πtTy=8+8 \sin \frac{2 \pi t}{T} With respect to elevator, variation in weight will be ΔW=m(Δa)max⁡ΔW=m×2ω2 A\begin{aligned} & \Delta \mathrm{W}=\mathrm{m}(\Delta \mathrm{a})_{\max } \\ & \Delta \mathrm{W}=\mathrm{m} \times 2 \omega^2 \mathrm{~A}\end{aligned} Here elevator is performing SHM ΔW=2 m×(2π T)2×ANΔW=2×50×(2π40π)2×8 NΔ W=2×50×1400×8 NΔ W=800400 N=2 N\begin{aligned} & \Delta \mathrm{W}=2 \mathrm{~m} \times\left(\frac{2 \pi}{\mathrm{~T}}\right)^2 \times \mathrm{A} \mathrm{N} \\ & \Delta \mathrm{W}=2 \times 50 \times\left(\frac{2 \pi}{40 \pi}\right)^2 \times 8 \mathrm{~N} \\ & \Delta \mathrm{~W}=2 \times 50 \times \frac{1}{400} \times 8 \mathrm{~N} \\ & \Delta \mathrm{~W}=\frac{800}{400} \mathrm{~N}=2 \mathrm{~N}\end{aligned}
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