Question

Water is flowing through a cylindrical pipe of internal diameter 2 cm, into a cylindrical tank of base radius 40 cm, at the rate of 0.4 m per second. Determine the rise in level of water in the tank in half an hour.

Answer 1

Answer:

VOLUME OF WATER FLOWONG THROUGH PIPE IN 1 SEC

$\pi \: r {h }^{2} \\ = \pi \times (1) {}^{2} \times 0.4 \times 100cm {}^{3}$

Volume of water flowing in 30 min (30×60 )sec

$\pi \times (1) {}^{2} \times 0.4 \times 30 \times 60$

volume of water in cylindrical tank in 30 min

$\pi \: r {}^{2} h \: = \pi \times (40) { \times h}^{2}$

Now

$\pi \times( 40 ) {}^{2} \times h = \pi \times( 1) {}^{2} \times \\ 0.4 \times 100 \times 30 \times 60$

rise in water level

$h = \pi \times( 1) {}^{2} \times 0.4 \times 100 \times \\ 30 \times 60 + \div \pi \times 40 \times 40$

$= 45cm$

$thus \: level \: of \: water \: in \: the \: \\ tank \: is \: 45cm$

HOPE IT HELPS U ✌✌✌

Answer 2

Answer:

Internal Radiusof pipe=1cm

Rate=4cm/s

Quantity of liquid passed in 1 sec=πr²h

                                                      =22/7×1×1×4 cm³=88/7 cm³

Quantity of liquid passed in 1/2hr that is 30 min=(88/7×1800) cm³

Volume of tank filled in 1/2 hour=π ×(40)² × h

Volume of tank filled in 1/2 hour= quantity of water passed through pipe in 1/2 hrs

⇒22/7 × (40)² × h=88/7 × 1800

⇒h=23 cm(aapproximately)

Step-by-step explanation: