Question

Water contains in a tank flows through an orifice of diameter 2 cm on the a constant pressure difference of 10 cm of water column the rate of flow of water through the audio file is

Answer 1

$\huge\mathfrak {Answer:-}$

$Given:-$

$h = 10\:cm$

$Where,$

$h = 10 \times 10 ^{ - 2} m$

Radius of orifice = $\frac{2}{2} = 1 \: cm \: = 10^{ - 2} m$

The velocity of liquid flowing through the orifice is,

$v = \sqrt{2gh}$

$= \sqrt{2 \times 9.8 \times 10 \times 10 ^{ - 2} }$

$= \sqrt{1.96}$

$= 1.4 \: ms$

$Now,$

Rate of flow of liquid through orifice will be,

$Q \: = v \times \: area$

$= v \times \pi \: r ^{2}$

$= 1.4 \times 3.14 \times (10 ^{ - 2} ) ^{2}$

$So,$

$Q = 4.40 \times 10^{ - 4}$

$Therefore,$

The rate of flow of water through the orifice is $4.40 \times 10^{ - 4}$

$\huge{Be\:Brainly}$❤️

Answer 2

$\underline{\underline{\Huge\mathfrak{Answer ;}}}$

Dear ,

• Given:

h = 10 cm = 10×10-2 m

• The velocity of liquid flowing through the orifice is ;-

$= > \sqrt{2gh } = \sqrt{2 \times 9.8 \times 10 \times {10}^{ - 2} }$

$= > \sqrt{1.96} = 1.4m \: per \: s$

=> Radius of orifice =2/2 =1 cm = 10-2 m

Then rate of flow of liquid through orifice will be

• Q=v×area=v×πr2

=1.4×3.14×(10−2)2

⇒Q=4.40×10−4

__________________________

- Regards
@dmohit432