The weight of a man in a lift moving upwards with an acceleration ' $a$ ' is $620 \mathrm{~N}$. When the…

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
  1. $620 \mathrm{~N}$
  2. $680 \mathrm{~N}$
  3. $380 \mathrm{~N}$
  4. $480 \mathrm{~N}$

Solution

$\begin{aligned} & \mathrm{m}(\mathrm{g}+\mathrm{a})=620 \mathrm{~N} \\ & \mathrm{~m}(\mathrm{~g}-\mathrm{a})=340 \mathrm{n} \\ & \Rightarrow \frac{\mathrm{g}+\mathrm{a}}{\mathrm{g}-\mathrm{a}}=\frac{620}{340} \\ & \mathrm{a}=\frac{7}{24} \mathrm{~g} \end{aligned}$ Hence, from eq.(i) actual weight $\mathrm{mg}=480 \mathrm{~N}$

Asked in: MHT CET 2021 (20 Sep Shift 2)

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