# Water is flowing Let the length of pipe filling the tank be x m.

Given the diameter of the pipe is 14cm

= 0.07m

Now the pipe is in shape of the cylinder, so the volume of the pipe, VP becomes,

VP = πr2h

Substituting the corresponding values, we get

VP = π(0.07)2(x)

VP = (0.0049) πx m3………..(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, VT, will be

VT = lbh

Substituting the corresponding values, we get

VT = 50 × 44 × (0.21)

VT = 462m3…………. (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.,

VP = VT

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

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.

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