Question

Water is flowing continuously from a tap having an internal diameter 0.008m. Water velocity as it leaves the tap is 0.4 ms-1. What will be the diameter of the water stream at a distance 0.2m below the tap?

Answer 1

Answer:

• Diameter of the stream is 3.6 × 10⁻³ meters.

${\huge\bold{\underline{Given}}}\begin{cases} \large\sf{Velocity (v_1) = 0.4 \: m/s} \\ \large\sf{Height (h) = 0.2 \: m} \\ \large\sf{Diameter (d_1) = 8 \times 10^{-3} \: m}\end{cases}$

Explanation:

$\rule{300}{1.5}$

Applying third kinematic equation,

$\large{\boxed{\bold{ v^2 = u^2 + 2as}}}$

But,

• u = v₁
• v = v₂
• a = g
• s = h

Substituting,

$\large{\tt \hookrightarrow v_2^2 = v_1^2 + 2gh}$

Substituting the values,

$\large{\tt \hookrightarrow v_2^2 = (0.4)^2 + 2 \times 10 \times 0.2}$

$\large{\tt \hookrightarrow v_2^2= 0.016 + 2 \times 2}$

$\large{\tt \hookrightarrow v_2^2 = 4 + 0.016}$

We can Neglect 0.016 as it is very small compared to 1.

[1 >> 0.016]

$\large{\tt \hookrightarrow v_2^2 = 4}$

$\large{\tt \hookrightarrow v_2 = \sqrt{4}}$

$\large{\boxed{\tt v_2 = 2 \: m/s}}$

$\rule{300}{1.5}$

$\rule{300}{1.5}$

From Equation of Continuity,

$\large{\boxed{\bold{A_1v_1 = A_2v_2}}}$

Substituting the values,

$\large{\tt \hookrightarrow \pi r_1^2 \times 0.4 = \pi r_2^2 \times 2}$

But we know the Relation r = d/2 (d = Diameter).

$\large{\tt \hookrightarrow \pi \left[\dfrac{d_1}{2}\right]^2 \times 0.4 = \pi \left[\dfrac{d_2}{2}\right]^2 \times 2}$

$\large{\tt \hookrightarrow \pi \dfrac{d_1^2}{4} \times 0.4 = \pi \dfrac{d_2^2}{4} \times 2}$

$\large{\tt \hookrightarrow \cancel{\pi} \dfrac{d_1^2}{\cancel{4}} \times 0.4 = \cancel{\pi} \dfrac{d_2^2}{\cancel{4}} \times 2}$

$\large{\tt \hookrightarrow d_1^2 \times 0.4 = d_2^2 \times 2}$

• d₁ = 8 × 10⁻³ meters.

Substituting,

$\large{\tt \hookrightarrow (8 \times 10^{-3})^2 \times 0.4 = d_2^2 \times 2}$

$\large{\tt \hookrightarrow 64 \times 10^{-6} \times 0.4 = d_2^2 \times 2}$

$\large{\tt \hookrightarrow d_2^2 = \dfrac{64 \times 10^{-6} \times 0.4}{2}}$

$\large{\tt \hookrightarrow d_2^2 = \dfrac{64 \times 10^{-6} \times \cancel{0.4}}{\cancel{2}}}$

$\large{\tt \hookrightarrow d_2^2 = 64 \times 10^{-6} \times 0.2}$

$\large{\tt \hookrightarrow d_2 = \sqrt{64 \times 10^{-6} \times 0.2}}$

$\large{\tt \hookrightarrow d_2 = 8 \times 10^{-3} \times \sqrt{0.2}}$

$\large{\tt \hookrightarrow d_2 = 8 \times 10^{-3} \times \dfrac{1}{2.236}}$

$\large{\tt \hookrightarrow d_2 = 3.57 \times 10^{-3} }$

Approximately,

$\huge{\boxed{\boxed{\tt d_2 \approx 3.6 \times 10^{-3}}}}$

So,the Diameter is 3.6 × 10⁻³ (or) 0.0036 meters.

$\rule{300}{1.5}$