Question

A bucket, full of water is revolved in a vertical circle of radius 2 m. What should be the maximum time-period of revolution so that the water doesn’t fall out of the bucket?(a) 1 sec (b) 2 sec(c) 3 sec (d) 4 sec

Answer 1

Question : -

A bucket full of water is revolved in a vertical circle of radius 2 m. What should be the maximum time - period of revolution, so that, the water doesn't fall out of the bucket?

Options : -

(a) 1 second

(b) 2 seconds

(c) 3 seconds

(d) 4 seconds

Answer : -

Given that,

Radius of the vertical circle (R) = 2 m

Assumption : -

Let the velocity of the vertical circle be V m / s.

Acceleration due to gravity (g) = 10 m/s^2

Here, the centrifugal force is acting on the bucket which keep it's rotationing. But the centrifugal force acting upon the bucket should be greater than the gravitational force so that the water inside the bucket wouldn't fall down.

As we know,

Centrifugal force (Fc) = $\frac{mV^2} {R}$

Where, m is the mass of the object, V is the velocity of the bucket and r is the radius of the vertical circle

And force (f) = mg

Where, m is the mass of the object and g is the acceleration due to gravity.

As centrifugal force is a force, then,

Fc = f

That is,

$\frac{mV^2}{R}$ = mg

=> $\frac{\cancel{m}V^2}{R} = \cancel{m}g$

=> $\frac{V^2}{R} = g$

=> V^2 = R $\times$ g

=> V = $\sqrt{Rg}$

=> V = $\sqrt{2 \times 10}$

=> V = $\sqrt{20}$

=> V = $\sqrt{4} \sqrt{5}$

=> V = $2 \sqrt{5}$

So, the velocity of the bucket will be $2 \sqrt{5}$ m/s

As we know,

The circular velocity =

=> V = $\frac{2 \pi r}{T}$

Where, V is the circular velocity, 2πr is the circumference of the circle and t is the time period of revolution.

So, T = $\frac{2 \pi r}{V}$

=> T = $\frac{2 \times 22/7 \times 2}{2 \sqrt{5}}$

=> T = $\frac{\cancel{2} \times 22/7 \times 2}{\cancel{2} \sqrt{5}}$

=> T = $\frac{22/7 \times 2}{\sqrt{5}}$

=> T = $\frac{44/7}{\sqrt{5}}$

=> T = $\frac{6.2\:(approx)}{\sqrt{5}}$

=> T = 2.8 seconds (approximately)

By rounding 2.8 we'll get 3.

Therefore, T = 3 seconds.

Therefore, option (c) 3 seconds is correct.

Answer 2

Answer:

C) 3 sec

Explanation:

Radius of the vertical circle (R) = 2m  (Given)

Let the velocity of the vertical circle be = V m / s.  

Acceleration due to gravity (g) = 10 m/s²

Minimum angular velocity  ωmin =  √g/R

Therefore,

Tmax = 2π/ ω

min

T = 2π√R/g

T = 2π√2/10

T = 2√2

T = 3

Thus, the maximum time-period of revolution so that the water doesn’t fall out of the bucket is 3 seconds.