Question

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Water flows through a circular pipe whose internal diameter is 2 cm at the rate of 6 m/sec into a cylindrical tank, the radius of whose base is 60 cm. Find the rise in the level in water in 30 minutes

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Answer 1

EXPLANATION.

Water flows through a circular pipe.

Internal diameter = 2 cm.

Rate = 6m/sec into a circular tank.

The radius of whose base = 60 cm.

To find the rise in the level in water in 30 minutes.

As we know that,

Diameter = 2 x Radius.

Radius = Diameter/2.

Radius = 2/2 = 1 cm = 1/100 m.

Volume of cylinder = πr²h.

Volume of water flows through a circular pipe in 1 seconds = πr²h.

π x (1/100)² x 6.

The raise in the water level in 30 minutes = π x (1/100)² x 6 x 30 x 60.

Radius whose base = 60 cm = 60/100 m.

Volume = πr²h.

⇒ π x (60/100)² x h.

⇒ π x (60/100)² x h = π x (1/100)² x 6 x 30 x 60.

⇒ 60/100 x 60/100 x h = 1/100 x 1/100 x 6 x 30 x 60.

⇒ 60 x 60 x h = 6 x 30 x 60.

⇒ 60 x h = 6 x 30.

⇒ 10 x h = 30.

⇒ h = 3m.

Answer 2

Answer:-

$Internal \: diameter \: of \: the \: pipe = 2m.$

$So, its \: radius = 1cm = \frac{1}{100} m$

$Water \: that \: flows \: out \: through \: the \\ pipe \: in \: 6ms {}^{ - 1} \\ so \: volume \: of \: water \: that \: flows \: out \: \\ through \: the \: pipe \: in \: \\ 1sec \: = \pi \times \frac{1 {}^{2} }{100} \times 6m {}^{3}$

$∴ In \: 30 \: minutes,volume \: of \: water \: \\ flow = \pi \frac{1}{100 \times 100} \times 6 \times 30 \times 60m {}^{3}$

This must be equal to the volume of water that rises in the cylindrical tank after 30 minutes and height up to which it rises say h.

$Radius \: of \: tank = 60cm = \frac{60}{100} m$

$Volume \: = \pi( \frac{60}{100} ) {}^{2} h$

$=\pi( \frac{60}{100} ) {}^{2} h = \\ \pi\times \frac{1}{100 \times 100} \times 6\times 30\times 60$

$=> \frac{60 \times 60}{100 \times 100} h = \frac{60 \times 30 \times 60}{100 \times 100}$

$= > h = \frac{3 \times 36}{36} = 3m$

$\fbox\pink{So,required height will be 3m}$