Question

9.Water flows'along a horizontal pipe of cross sectional area 48cm2 which has a constriction of

cross sectional area 12cm2 at one place. If the speed of the water at the constriction is 36m/s,
calculate the speed in the wider section,
(3mk)
​

Answer 1

v₁ = 9 m/s

Step-by-step explanation:

Here, we can use the continuity equation, as follows:

`A_1v_1 = A_2v_2`

where,

A₁ = Cross-sectional area of the wider section = 48 cm²

A₂ = Cross-sectional area of the constriction = 12 cm²

v₁ = speed of flow in wider section = ?

v₂ = speed flow in constriction = 36 m/s

Therefore,

`(48\ cm^2)v_1 = (12\ cm^2)(36\ m/s)\\\\v_1 = ((12\ cm^2)(36\ m/s))/((48\ cm^2)) \\\\`

v₁ = 9 m/s