Q40 / 48 · JEE Advanced 2025 (Paper 1) · Physics
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 $y=8\left[1+\sin \left(\frac{2 \pi t}{T}\right)\right]$, where $T=40 \pi \mathrm{~s}$. Taking acceleration due to gravity, $g=10 \mathrm{~m} / \mathrm{s}^2$, the maximum variation of the object's weight (in N ) as observed in the experiment is _______ .
Correct answer: 2
Solution
$y=8+8 \sin \frac{2 \pi t}{T}$
With respect to elevator, variation in weight will be
$\begin{aligned} & \Delta \mathrm{W}=\mathrm{m}(\Delta \mathrm{a})_{\max } \\ & \Delta \mathrm{W}=\mathrm{m} \times 2 \omega^2 \mathrm{~A}\end{aligned}$
Here elevator is performing SHM
$\begin{aligned} & \Delta \mathrm{W}=2 \mathrm{~m} \times\left(\frac{2 \pi}{\mathrm{~T}}\right)^2 \times \mathrm{A} \mathrm{N} \\ & \Delta \mathrm{W}=2 \times 50 \times\left(\frac{2 \pi}{40 \pi}\right)^2 \times 8 \mathrm{~N} \\ & \Delta \mathrm{~W}=2 \times 50 \times \frac{1}{400} \times 8 \mathrm{~N} \\ & \Delta \mathrm{~W}=\frac{800}{400} \mathrm{~N}=2 \mathrm{~N}\end{aligned}$
Hence, the correct answer is 2.