Question

A small hole is made at a height of h′ = (1/$\sqrt { 2 }$) m from the bottom of a cylindrical water tank and at a depth of h = $\sqrt { 2 }$ m from the upper level of water in the tank. The distance where the water emerging from the hole strikes the ground is:

• A. (a) 2$\sqrt { 2 }$ m
• A. (b) 1 m
• A. (c) 2 m
• A. (d) None of these

Answer 1

(c)

Applying Bernoulli’s Principle between the point 1 and 2

ρgh₁ + P₁ + 1/2 P = P₂ + 1/2 ρ + ρgh₂

Substituting h₁ = h₂ = h′ and since v₁ = a/A v² and a << A

v_1~0, P₁ = P₀ + ρgh, P₂ = P₀,

we obtain,

ρgh = 1/2 ρ v₂ = = v (say), P₀ is the atmospheric

pressure. The range R = v₂ × t, where t = time of fall can be

given by

h₁ = 1/2 gt² ⇒ t =

⇒ R = v₂ ; putting v₂ = ,

we obtain R = 2

putting h′ = $\sqrt { 2 }$ m, we obtain R = 2 m

Therefore (C).