Question

The pressure of water inside the closed pipe 3×10^5n/m^2 this pressure reducea to 2 ×10^5n/m^2 on opening the value of the pipe calculate the speed of water flowing through the pipe (density pf water=1000kg/m^3​

Answer 1

Answer:

Before opening the pipe,

P1= 3.5 x 10^(5) Nm^(-2),

upsilon 1 =0

After opening the pipe

P2= 3 x 10^(5) Nm^(-2)

, upsilon2 = ?

,rho= 10^(3) kg m^(-3)

Since the height of water does not change therefore the potential energy of the liquid will be same before and after opening the value .

Hence,

P1 + 1/2 rho upsilon(1)^(2) = P2+1/2 rho upsilon(2)^(2)

or

3.5 x 10^(5) + 1/2 x 10^(3) (0)^(2)

= 3 x 10^(5) +1/2 x 10^(3) x upsilon(2)^(2)

or

1/2 x 10^(3) x upsilon(2)^(2)

= (3.5 - 3) x 10^(5)

= 0.5 x 10^(5)

or

upsilon(2)^(2)

= (2 x 0.5 x 10^(5))/(10^3)

= 100 or upsilon(2)

= 10 ms^(-1).

Explanation: