MHT CET · Physics · Laws of Motion
The ratio of weight of a man in a stationery lift and weight when the lift is moving downward with a uniform acceleration ' \(a\) ' is \(3: 2\). Then the value of ' \(a\) ' is
- A \(\frac{3}{2} \mathrm{~g}\)
- B \(\frac{\mathrm{g}}{3}\)
- C \(\frac{2}{3} \mathrm{~g}\)
- D g
Answer & Solution
Correct Answer
(B) \(\frac{\mathrm{g}}{3}\)
Step-by-step Solution
Detailed explanation
Weight of a man when the lift is stationary, \(\mathrm{W}_1=\mathrm{mg}\).
Weight when the lift is going down, \(\mathrm{W}_2=\mathrm{m}(\mathrm{g}-\mathrm{a})\)
\(\begin{array}{ll}
\therefore & \frac{W_1}{W_2}=\frac{m g}{m(g-a)}=\frac{3}{2} \\
\therefore & \frac{g}{g-a}=\frac{3}{2} \\
\therefore & 2 g=3 g-3 a \\
\therefore & g=3 a \\
\therefore & a=\frac{g}{3}
\end{array}\)
....(Given)
Weight when the lift is going down, \(\mathrm{W}_2=\mathrm{m}(\mathrm{g}-\mathrm{a})\)
\(\begin{array}{ll}
\therefore & \frac{W_1}{W_2}=\frac{m g}{m(g-a)}=\frac{3}{2} \\
\therefore & \frac{g}{g-a}=\frac{3}{2} \\
\therefore & 2 g=3 g-3 a \\
\therefore & g=3 a \\
\therefore & a=\frac{g}{3}
\end{array}\)
....(Given)
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