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Q.

A large tank is completely filled with water. A small hole is made at a depth of 10 m below the free surface of water. The range of water issuing out of the hole is R on the ground. How much extra pressure must be applied on the free surface of water so that the range now becomes 2R? 
(1 atm = 105 Pa and g = 10 m/s2)

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a

1 atm

b

4 atm 

c

3 atm

d

2 atm

answer is D.

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Detailed Solution

R=V2Hg=2gh×2Hg so for the range to be doubled, ‘h’ should  be made 4 times i.e.,40 m water column.
So, increase in pressure = 30 m water  column= 3 atm

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