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Water is flowing at the rate of $2.52 kmph$ through a cylindrical pipe into a cylindrical tank, the radius of whose base is $40 cm$ . If the increase in the level of water in the tank, in half an hour is $3.15 m$ , find the internal diameter of the pipe.

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4 cm

5 cm

9 cm

1 cm

answer is A.

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Detailed Solution

Given that, flow rate of water is 2.52kmph, base radius of cylindrical tank is 50cm.
Water level increase is 3.15m in half an hour.
Volume of cylinder,

V = π r 2 h

, where r is radius, h is height.
Let the internal diameter of the pipe be

r

m and water flows in

1 h o u r = 2.52 km

.
The water flows in

12 h o u r

= 2.52 2

⇒ 1.26 k m

⇒ 1260 m

The volume of water flows in

12

hour

= π r 2 h

⇒ π r 2 × 1260

The volume of the water in a cylindrical tank

= π × 401002 × 3.15

Volume of water flow is equal to Volume of increased water,

⇒ π r 2 × 1260 = π 252 × 3.15

⇒ 1260 r 2 = 25 × 25 × 3.15

⇒ r 2 = 425 × 315100 × 11260 = 12500

⇒ r = 150 m

⇒ r = 2 cm

The diameter of a pipe.

D = 2 R

⇒ D = 2 × 2 c m

⇒ D = 4 cm

Thus, the internal diameter of pipe is 4cm.
Therefore, option (1) is correct.

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