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The range of water flowing out of a small hole made at a depth `10 m` below water surface in a large tank is `R`. Find the extra pressure (in atm) applied on the water surface so that range becomes `2R`. Take `1atm=10^(5)Pa`.

A

9atm

B

4atm

C

5atm

D

3atm

Text Solution

Verified by Experts

The correct Answer is:
4
`P=(F)/(A)=(2Arhov^(2))/(A)=2rhov^(2)`
`R=vsqrt((2h)/(g)),Rpropv,Ppropv^(2),PpropR^(2)`
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