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A large tank is filled with water $(density = 10^{3} kg / m^{3})$ A small hole is made at a depth 10 m below water surface. The range of water issuing out of the hole is R on ground. What extra pressure (in atmosphere) must be applied on the water surface so that the range becomes 2R (take 1 atmosphrere $= 10^{5} Pa and g = 10 m / s^{2}).$

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answer is 3.

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Detailed Solution

$Δp = \frac{1}{2} {ρv}^{2} ⇒ v = \sqrt{\frac{2 (Δp)}{ρ}}$

$R = vt \Rightarrow R \propto \sqrt{Δp}$

Two make R two times $Δp$ should be 4 times. Initially

$ΔP = ρ_{ω} gh = 10^{3} \times 10 \times 10 = 10^{5} = 1 atm$

In second case, $ρ_{ω} gh + extra pressure = 4 atm$

$∴ extra pressure = 3 atm$

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