Answer :

Let the length of pipe filling the tank be x m.

Given the diameter of the pipe is 14cm

⇒ radius = 7cm

= 0.07m

Now the pipe is in shape of the cylinder, so the volume of the pipe, V_{P} becomes,

V_{P} = πr^{2}h

Substituting the corresponding values, we get

V_{P} = π(0.07)^{2}(x)

V_{P} = (0.0049) πx m^{3}………..(i)

Now the water tank is rectangular, so it will be in the form of the cuboid.

So, given the length of the tank, l = 50 m long

and given the breadth of the tank, b = 44 m wide

We need to find the level of water in the tank will rise by 21cm, hence this will be the height, i.e., h = 21cm = 0.21m

So, the volume of the tank, V_{T}, will be

V_{T} = lbh

Substituting the corresponding values, we get

V_{T} = 50 × 44 × (0.21)

⇒ V_{T} = 462m^{3}…………. (ii)

Now as the pipe is filling the water, so

The volume of the pipe should be equal to the volume of the tank, i.e.,

V_{P} = V_{T}

Substituting values from equation(i) and (ii), we get

(0.0049) πx = 462

⇒ x = 30000m

= 30km

This is the length of the pipe.

And it is given water in the pipe is flowing at the rate of 15km/hr, means

15km travels in the pipe in = 1hour

⇒ 1km travels in the pipe in

So, in 30km long pipe water travels in

= 2 hr

**So, in 2 hours the level of water in the tank will rise by 21cm.**

**OR**

As the solid sphere is melted and recast into three small spherical balls, so

The volume of the big sphere = volume of small balls

⇒ The volume of big sphere = number of small balls × volume of small balls ……(i)

We know, the volume of the sphere

And let the radius big sphere, R = 3cm

And given diameter of small sphere = 0.6cm, so the radius of each small sphere,

= 0.3 cm

And let the number of balls be n

Substituting this in equation (i), we get

Canceling the like terms, we get

R^{3} = n × r^{3}

Substituting corresponding values, we get

(3)^{3} = n × (0.3)^{3}

⇒ 27 = n × (0.027)

⇒ n = 1000

**Hence 1000 number of balls can be made from the big sphere.**

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