The ratio of the weight of a man in a stationary lift and in a lift accelerating downwards with a uniform acceleration ' a ' is 3: 2 . The acceleration of the lift is

Question

The ratio of the weight of a man in a stationary lift and in a lift accelerating downwards with a uniform acceleration ' a ' is 3: 2 . The acceleration of the lift is (A) g / 3 (B) g / 2 (C) g (D) 2 g.

Tardigrade Admin · Accepted Answer

M g-N=M a or N=M(g-a) Given (M g/N)=(3/2) or (M g/M(g-a))=(3/2) ⇒ 2 M g=3 M g-3 M a ⇒ a=(g/3)

Q. The ratio of the weight of a man in a stationary lift and in a lift accelerating downwards with a uniform acceleration ' $a$ ' is $3 : 2$ . The acceleration of the lift is

$g /3$

$g /2$

$g$

$2 g$ .

$M g - N = M a$
or $N = M (g - a)$
Given $\frac{M g}{N} = \frac{3}{2}$ or $\frac{M g}{M ( g - a )} = \frac{3}{2}$
$\Rightarrow 2 M g = 3 M g - 3 M a \Rightarrow a = \frac{g}{3}$

Q. The ratio of the weight of a man in a stationary lift and in a lift accelerating downwards with a uniform acceleration ' a ' is 3:2 . The acceleration of the lift is

g/3

g/2

g

2g.

Mg−N=Ma or N=M(g−a) Given NMg​=23​ or M(g−a)Mg​=23​ ⇒2Mg=3Mg−3Ma⇒a=3g​

Q. The ratio of the weight of a man in a stationary lift and in a lift accelerating downwards with a uniform acceleration ' $a$ ' is $3 : 2$ . The acceleration of the lift is

$g /3$

$g /2$

$g$

$2 g$ .

$M g - N = M a$
or $N = M (g - a)$
Given $\frac{M g}{N} = \frac{3}{2}$ or $\frac{M g}{M ( g - a )} = \frac{3}{2}$
$\Rightarrow 2 M g = 3 M g - 3 M a \Rightarrow a = \frac{g}{3}$