Q194 / 200 · MHT CET - Medical 2026 (25 April Shift 2) · Physics
The ratio of the weight of the man in stationary lift and in a lift accelerating upwards with uniform acceleration is $2:3$. The acceleration of lift is
- A$\dfrac{g}{3}$
- B$g$
- Correct answerC$\dfrac{g}{2}$
- D$\dfrac{5g}{2}$
Solution
The weight of the man in a stationary lift is $W_1 = mg$.
The apparent weight of the man in a lift accelerating upwards with uniform acceleration $a$ is $W_2 = m(g + a)$.
Given the ratio of the weights is $2:3$, we have:
$\dfrac{W_1}{W_2} = \dfrac{2}{3}$
$\dfrac{mg}{m(g + a)} = \dfrac{2}{3}$
$\dfrac{g}{g + a} = \dfrac{2}{3}$
$3g = 2(g + a)$
$3g = 2g + 2a$
$2a = g$
$a = \dfrac{g}{2}$
Answer: $\dfrac{g}{2}$
The apparent weight of the man in a lift accelerating upwards with uniform acceleration $a$ is $W_2 = m(g + a)$.
Given the ratio of the weights is $2:3$, we have:
$\dfrac{W_1}{W_2} = \dfrac{2}{3}$
$\dfrac{mg}{m(g + a)} = \dfrac{2}{3}$
$\dfrac{g}{g + a} = \dfrac{2}{3}$
$3g = 2(g + a)$
$3g = 2g + 2a$
$2a = g$
$a = \dfrac{g}{2}$
Answer: $\dfrac{g}{2}$
Hence, the correct answer is option: 'C'.