Water is flowing at the rate of 15 km per hour through a pipe of diameter 14cm into a rectangular tank which is 50 m long and 44 m wide. Find the time in which the level of water in the tank will rise by 21cm.

OR

A solid sphere of radius 3 cm is melted and then recast into small spherical balls each of diameter 0.6cm. Find the number of balls.

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 cylinder, so volume of the pipe, V_P becomes,

V_P = πr²h

Substituting the corresponding values, we get

V_P = π(0.07)²(x)

V_P = (0.0049) πx m³………..(i)

Now the water tank is rectangular, so it will be in form of 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 = 462m³…………. (ii)

Now as the pipe is filling the water, so

Volume of the pipe should be equal to 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 pipe in = 1hour

⇒ 1km travels in 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

Volume of the big sphere = volume of small balls

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

We know, volume of sphere

And let the radius big sphere, R = 3cm

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

= 0.3 cm

And let number of balls be n

Substituting this in equation (i), we get

Cancelling the like terms, we get

R³ = n × r³

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.