JEE Advanced 2025 — Paper 1 — Question 6

$y=8+8 \sin \frac{2 \pi t}{T}$

With respect to elevator, variation in weight will be

$\Delta \mathrm{W}=\mathrm{m}(\Delta \mathrm{a})_{\text {max }}$

$\Delta \mathrm{W}=\mathrm{m} \times 2 \omega^{2} \mathrm{~A}$

Here elevator is performing SHM

$\Delta \mathrm{W}=2 \mathrm{~m} \times\left(\frac{2 \pi}{\mathrm{~T}}\right)^{2} \times \mathrm{A} \mathrm{N}$

$\Delta \mathrm{W}=2 \times 50 \times\left(\frac{2 \pi}{40 \pi}\right)^{2} \times 8 \mathrm{~N}$

$\Delta \mathrm{W}=2 \times 50 \times \frac{1}{400} \times 8 \mathrm{~N}$

$\Delta \mathrm{W}=\frac{800}{400} \mathrm{~N}=2 \mathrm{~N}$

Ans. is (B)