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Water from a tap emerges vertically down...

Water from a tap emerges vertically downwards with an initial spped of `1.0ms^-1`. The cross-sectional area of the tap is `10^-4m^2`. Assume that the pressure is constant throughout the stream of water, and that the flow is steady. The cross-sectional area of the stream 0.15 m below the tap is

A

`2xx10^(-5)m^2`

B

`3xx10^(-5)m^2`

C

`4xx10^(-5)m^2`

D

`5xx10^(-5)m^2`

Text Solution

Verified by Experts

The correct Answer is:
B
From the equation of continuity
`A_1V_1=A_2V_2`
`:." "V_2=(A_1V_1)/(A_2)`

for the freely falling flow of water,
`v_2^2=v_1^2+2gh`
`:." "V_2=sqrt(v_1^2+2gh)`
Using (2) in (1), we get
`sqrt(v_1^2+2gh)=(A_1V_1)/(A_2)`
`:." "A_2=(A_1V_1)/(sqrt(V_1^2+2gh))`
`=(10^(-4)xx1)/(sqrt(1^2+2xx10xx0.15))=(10^(-4))/2`
`=5xx10^(-5)m^2`
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