JEE Advanced · Physics · 5. Laws of Motion
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 _______ .
- A 1
- B 2
- C 3
- D 4
Answer & Solution
Correct Answer
(B) 2
Step-by-step Solution
Detailed explanation
\(y=8+8 \sin \frac{2 \pi t}{T}\)
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
\(\begin{aligned} & \Delta \mathrm{W}=2 \mathrm{~m} \times\left(\frac{2 \pi}{\mathrm{~T}}\right)^2 \times \mathrm{A} \mathrm{N} \\ & \Delta \mathrm{W}=2 \times 50 \times\left(\frac{2 \pi}{40 \pi}\right)^2 \times 8 \mathrm{~N} \\ & \Delta \mathrm{~W}=2 \times 50 \times \frac{1}{400} \times 8 \mathrm{~N} \\ & \Delta \mathrm{~W}=\frac{800}{400} \mathrm{~N}=2 \mathrm{~N}\end{aligned}\)
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
\(\begin{aligned} & \Delta \mathrm{W}=2 \mathrm{~m} \times\left(\frac{2 \pi}{\mathrm{~T}}\right)^2 \times \mathrm{A} \mathrm{N} \\ & \Delta \mathrm{W}=2 \times 50 \times\left(\frac{2 \pi}{40 \pi}\right)^2 \times 8 \mathrm{~N} \\ & \Delta \mathrm{~W}=2 \times 50 \times \frac{1}{400} \times 8 \mathrm{~N} \\ & \Delta \mathrm{~W}=\frac{800}{400} \mathrm{~N}=2 \mathrm{~N}\end{aligned}\)
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