Question

A conical vessel of height 2 meters carries water whose volume is one half of the volume of the cone find the speed at which the basis may be rotated to make the water just spell out

Answer 1

Answer:

A 90 degree conical vessel of height 2 m carries water whose volume is one half of the volume of the cone. Find the speed at which the vessel may be rotated to make the water just spill out.

Answer 2

The speed at which the basis may be rotated to make the water just spell out is 1.4rad/sec.

Given:-

Height of the conical vessel = 2m

To Find:-

The speed at which the basis may be rotated to make the water just spell out.

Solution:-

We can simply find out the speed at which the basis may be rotated to make the water just spell out by using these simple steps.

As

Height of the conical vessel (h) = 2m

Let the radius of conical vessel be r.

For cone, the vertical angle 45° with the horizontal,

So,

$\tan(45) = \frac{h}{r}$

$1 = \frac{h}{r}$

$h = r$

$h = r = 2m$

Now, According to the Volume of conical vessel,

$V = \frac{1}{3} \pi {r}^{2}h$

$V = \frac{1}{3} \pi {r}^{3}$

According to the question, volume of water = 1/2 of volume of conical vessel

$\frac{1}{3} \pi {r'}^{2}h' = \frac{1}{2} \times \frac{1}{3} \pi {r}^{3}$

on cancelling the common parts, we get

${r'}^{3} = \frac{1}{2} {r}^{3}$

or

${h'}^{3} = \frac{1}{2} {r}^{3}$

on taking cube roots on both the sides,

$h = \frac{r}{ \sqrt[3]{2} }$

$h = 0.79r$

$h = 0.79 \times 2$

$h = 1.58m$

Now, we all now the formula of speed of water flow is √2gh.

Here we will take the difference of heights as

∆h = 2-1.58 = 0.42m

So,

$v = \sqrt{2g∆h}$

$v = \sqrt{2 \times 10 \times 0.42}$

$v = \sqrt{2 \times 4.2}$

$v = \sqrt{8.4}$

$v = 2.8$

In question we have to find the rotatory speEd so equating normal speed with the rotating speed, we get

$v = wr$

$w = \frac{v}{r}$

$w = \frac{2.8}{2}$

$w = 1.4$

Hence, The speed at which the basis may be rotated to make the water just spell out is 1.4rad/sec.

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