The ratio of the weight of a man in a stationary lift and when it is moving downward with uniform acceleration a is 3: 2 The value of a is (g = acceleration due to gravity of the earth)

Question

The ratio of the weight of a man in a stationary lift and when it is moving downward with uniform acceleration a is 3: 2 The value of a is (g = acceleration due to gravity of the earth) (A) 3/2 g (B) g/3 (C) 2/3 g (D) g

Tardigrade Admin · Accepted Answer

According to the question (W1/W2)=(m g/m(g-a))=(3/2) or (g/g-a)=(3/2) ⇒ a=(g/3) m / s2

Q. The ratio of the weight of a man in a stationary lift and when it is moving downward with uniform acceleration $a$ is $3 : 2$ The value of $a$ is
( $g =$ acceleration due to gravity of the earth)

3/2 g

g/3

2/3 g

g

According to the question
$\frac{W _{1}}{W _{2}} = \frac{m g}{m ( g - a )} = \frac{3}{2}$
or $\frac{g}{g - a} = \frac{3}{2}$
$\Rightarrow a = \frac{g}{3} m / s^{2}$

Q. The ratio of the weight of a man in a stationary lift and when it is moving downward with uniform acceleration a is 3:2 The value of a is (g= acceleration due to gravity of the earth)

3/2 g

g/3

2/3 g

g

According to the question W2​W1​​=m(g−a)mg​=23​ or g−ag​=23​ ⇒a=3g​m/s2

Q. The ratio of the weight of a man in a stationary lift and when it is moving downward with uniform acceleration $a$ is $3 : 2$ The value of $a$ is
( $g =$ acceleration due to gravity of the earth)

According to the question
$\frac{W _{1}}{W _{2}} = \frac{m g}{m ( g - a )} = \frac{3}{2}$
or $\frac{g}{g - a} = \frac{3}{2}$
$\Rightarrow a = \frac{g}{3} m / s^{2}$