Question

38. A pipe of length 1m and mass 1 kg is suspended vertically from one end of a massless string passing over a smooth fixed Pulley A ball of mass 2 kg is attached to the other end and the arrangement is released from the rest from the position shown time taken by the ball to cross the pipe?
$a) \: \sqrt{ \frac{3}{2g} }$
$b) \: \sqrt{ \frac{3}{g} }$
$c) \: \sqrt{ \frac{3}{4g} }$
$d) \: \sqrt{ \frac{3}{8g} }$
No spam answers please.​

Answer 1

Answer:

Firstly , draw FREE BODY DIAGRAM. Here 'T' is tension in the string.

From the free body diagram of the pipe.

$T- 1g = 1a$

From the free body diagram of ball.

$2g - T= 2a$

On using elimination method ,

$T - 1g = 1a \\ 2g -T = 2a$

we get ,

$g = 3a \:$

$\frac{g}{ 3} = a \: ....(1)$

we know from Kinematics that ,

$s = ut + \frac{1}{2} a {t}^{2}$

Here , u = 0.

so,

$s = \frac{1}{2} a {t}^{2}$

We know that the pipe has length of 1 m. When the pipe goes up , the ball also comes down. Hence relative displacement will be 1/2 metres

Substituting values ,

$\frac{1}{2} = \frac{1}{2} \times \frac{g}{3} \times {t}^{2}$

$\frac{3}{g } = {t}^{2}$

$\sqrt{ \frac{3}{g} } = t$

Hence root 3/g is your answer.

Hope it helps.

...