Question

A large open tank containing water has two holes to its wall. A square hole of side 'a' is made at a depth 'y' and a circular hole of radius 'r' is made at a depth 16y from the surface of water. If equal amount of water comes out through both the holes per second, then the relation between r and a will be ______.

Options

• r = `"a"/(2sqrtpi)`

• r = `"a"/(2pi)`

• r = `"2a"/(pi)`

• r = `"2a"/(sqrtpi)`

Answer 1

A large open tank containing water has two holes to its wall. A square hole of side 'a' is made at a depth 'y' and a circular hole of radius 'r' is made at a depth 16y from the surface of water. If equal amount of water comes out through both the holes per second, then the relation between r and a will be `underline(r = "a"/(2sqrtpi))`.

Explanation:

Using principle of continuity,

A₁v₁ =A₂v₂   ...(i)

where, A₁ is the area of square hole, A₁ = a²,

A₂ is the area of circular hole, A₂ = πr²,

v₁ is the velocity at depth y, v₁ = `sqrt(2"gy")`

and v₂ is the velocity at depth 16y, v₂ = `sqrt(2"g"("16y"))`.

Substituting values in Eq. (i), we get

`"a"^2(sqrt(2"gy")) = pi"r"^2 (sqrt("2g"(16"y")))`

`"a"^2(sqrt(2"gy")) = 4pi"r"^2 (sqrt("2gy"))`

`"r"^2 = "a"^2/(4pi)`

∴ r = `"a"/(2sqrtpi)`