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

$\bf{\Huge{\underline{\boxed{\sf{\green{ANSWER\::}}}}}}$

Given:

Water is flowing through a cylindrical pipe of internal diameter 2cm into a cylindrical tank of base radius 40cm, at rate of 0.4/ second.

To find:

The rise in level of water in the tank in half an hour.

$\bf{\Large{\underline{Explanation\::}}}}}$

We have,

• Internal diameter of cylindrical pipe= 2cm
• Radius of cylindrical tank,[R]= 40cm
• The length of water flowing in 1 second= 0.4m

We know that 1m= 100cm

So,

→ 0.4m = (0.4 × 100)cm

→ 40cm length of water flowing in 1 second.

Internal radius of cylindrical pipe,[r]= $\cancel{\frac{2}{2}}cm$

Internal radius of cylindrical pipe,[r]= 1cm.

According to the question:

→ Length of water flow in half an hour,[h]= 30minutes

→ Length of water flow in second=(30×40×60)cm

→ Length of water flow in second= 72000cm

Now,

We know that formula of the volume of cylinder: πr²h      [cubic units]

⇒ Volume of water in cylindrical tank = Volume of water flow from pipe

⇒ πR²H = πr²h

⇒ R²H= r²h

⇒ (40cm)² × H = (1cm)² × 72000cm

⇒ 1600cm² × H = 72000cm²

⇒ $H=\cancel{\frac{72000}{1600} }cm$

⇒ H= 45cm

Thus,

The rise in level of water in the tank in half an hour is 45cm.

Answer 2

${\bold{\boxed{\boxed{Answer = 45\:cm}}}}$

$\Large\underline{\underline{\sf \blue{Given}:}}$

• Internal diameter of cylindrical pipe=2cm
• Speed of water = 0.4 m/s
• Radius of cylindrical tank (R) = 40 cm

$\Large\underline{\underline{\sf \blue{To\:Find}:}}$

• Rise of level of water in the tank in 30 minutes.

$\Large\underline{\underline{\sf \blue{Solution }:}}$

$\implies$ Internal diameter of cylindrical pipe = 2 cm

$\implies$ So, radius of cylindrical pipe (r) = 2/2 = 1 cm

$\implies$ Area of cross-section of pipe = πr² = π(1)² = π cm²

$\implies$ Speed of the water = 0.4 m/s = 0.4 × 100 cm/s = 40 cm/s

$\implies$ Thus, Volume of water flown out in half an hour = (π×40×30×60) = 72,000π cm² .....(1)

$\implies$ Let level of water rise to the height of h cm

$\implies$ Thus, Volume of cylindrical tank = π(R²)h = π(40)² h = 1600πh cm² .....(2)

$\implies$ Now, Volume of cylindrical tank = Volume of the water flown out in half of an hour

$\hookrightarrow$ from equ(1) and (2) we get.

$\implies$ 1600πh = 72000π cm

$\implies$ h = $\sf\cancel\dfrac{72000\pi}{1600\pi}$

$\longrightarrow$ h = 45 cm.