MHT CET · Physics · Laws of Motion
The weight of a man in a lift moving upwards with an acceleration ' \(a\) ' is \(620 \mathrm{~N}\). When the lift moves downwards with the same acceleration, his weight is found to be \(340 \mathrm{~N}\). The real weight of the man is
- A \(620 \mathrm{~N}\)
- B \(680 \mathrm{~N}\)
- C \(380 \mathrm{~N}\)
- D \(480 \mathrm{~N}\)
Answer & Solution
Correct Answer
(D) \(480 \mathrm{~N}\)
Step-by-step Solution
Detailed explanation
\( \begin{aligned} & \mathrm{m}(\mathrm{g}+\mathrm{a})=620 \mathrm{~N} \\ & \mathrm{~m}(\mathrm{~g}-\mathrm{a})=340 \mathrm{n} \\ & \Rightarrow \frac{\mathrm{g}+\mathrm{a}}{\mathrm{g}-\mathrm{a}}=\frac{620}{340} \\ & \mathrm{a}=\frac{7}{24} \mathrm{~g} \end{aligned} \)
Hence, from eq.(i) actual weight
\(\mathrm{mg}=480 \mathrm{~N}\)
Hence, from eq.(i) actual weight
\(\mathrm{mg}=480 \mathrm{~N}\)
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