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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?

$(1 a t m = 10^{5} P a a n d g = 10 m / s^{2})$

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3 atm

2 atm

1 atm

4 atm

answer is D.

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Detailed Solution

$R = V \sqrt{\frac{2 H}{g}} = \sqrt{2 g h} \times \sqrt{\frac{2 H}{g}}$ ,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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