Question

4. Water is flowing through a horizontal pipe of
non-uniform cross-section. The speed of water is
30 cm s at a place where pressure is 10 cm (of
water). Calculate the speed of water at the other place
where the pressure is half of that at the first place.​

Answer 1

Correct Question:

Water is flowing through a horizontal pipe of non-uniform cross-section. the speed of water is 30 cm/s at a place where pressure is 10 cm (of water). calculate the speed of water at the other place where the pressure is half of that at the first place?

Answer:

• Speed of the Water (v₂) = 1.04 m/s.

Given:

1. Velocity at one End (v₁) = 30 cm/s.
2. Height where pressure (h₁) = 10 cm.
3. Height half the Pressure (h₂) =5 cm

Note:-

• We took h₂ as 5 cm as Question states to find the Velocity where the pressure is half of that at the first place.

Explanation:

$\rule{300}{1.5}$

This is an Application of Bernoullis theorem.

From Bernoullis theorem we Know,

$\large\star \: {\boxed{\bold{P + \dfrac{1}{2}\rho v^2 + \rho gh = Constant}}}$

$\bold{Here}\begin{cases} \rho \: \text{ denotes Density} \\ \text{v denotes Velocity} \\ \text{P denotes Pressure} \\ \text{h denotes Height} \\ \text{g denotes Acceleration due to gravity}\end{cases}$

$\large{\boxed{\tt P + \dfrac{1}{2}\rho v^2 + \rho gh = Constant}}$

But Question says that the Pipe is Horizontal. Therefore the Height (h) will be Zero.

Therefore, Equation becomes,

$\large{\tt \hookrightarrow \bold{P + \dfrac{1}{2}\rho v^2 = Constant}}$

We Know, P = hρg

$\large{\tt \hookrightarrow h \rho g+ \dfrac{1}{2}\rho v^2 = Constant}$

$\large{\tt \hookrightarrow \rho( h g+ \dfrac{1}{2} v^2) = Constant}$

$\large{\tt \hookrightarrow hg+ \dfrac{1}{2} v^2 = Constant}$

Now,

$\large{\tt \hookrightarrow h_1g+ \dfrac{1}{2} v_1^2 = h_2g+ \dfrac{1}{2} v_2^2}$

Substituting the values,

$\large{\tt \hookrightarrow 10 \times 10^3 + \dfrac{1}{2} \times (30)^2 = 5 \times 10^3+ \dfrac{1}{2} v_2^2}$

∵ [g = 10 m/s² = 1000 cm/s²]

$\large{\tt \hookrightarrow 10 \times 10^3+ \dfrac{1}{2} \times 900 = 5 \times 10^3+ \dfrac{1}{2} v_2^2}$

$\large{\tt \hookrightarrow 10 \times 10^3 + \dfrac{1}{\cancel{2}} \times \cancel{900} = 5 \times 10^3+ \dfrac{1}{2} v_2^2}$

$\large{\tt \hookrightarrow 10 \times 10^3 + 450 = 5 \times 10^3 + \dfrac{1}{2} v_2^2}$

$\large{\tt \hookrightarrow 10 \times 10^3 - 5 \times 10^3 + 450= \dfrac{1}{2} v_2^2}$

$\large{\tt \hookrightarrow 5 \times 10^3 + 450 = \dfrac{v_2^2}{2}}$

$\large{\tt \hookrightarrow (5 \times 10^3 + 450) \times 2 = v_2^2}$

$\large{\tt \hookrightarrow v_2^2 = 10 \times 10^3 + 900}$

$\large{\tt \hookrightarrow v_2^2 = 10000 + 900}$

$\large{\tt \hookrightarrow v_2^2 = 10900}$

$\large{\tt \hookrightarrow v_2 = \sqrt{10900}}$

$\large{\tt \hookrightarrow v_2 = 104.4 \: cm/s}$

We know, [1cm = 10⁻²m]

$\large{\tt \hookrightarrow v_2 = 104.4 \times 10^{-2}}$

$\huge{\boxed{\boxed{\tt v_2 = 1.04 \: m/s}}}$

∴ Velocity will be 1.04 m/s.

$\rule{300}{1.5}$

Answer 2

Correct Question:

Given:

→ Water is flowing through a horizontal pipe of non-uniform cross-section.

The speed of water is 30 cm/s at a place where pressure is 10 cm of water.

Find:

→ Calculate the speed of water at the other place where the pressure is half of that at the first place?

According to given question:

→ 31.62 m/s = (v₂ Speed of Water).

→ 30 cm/s = (v₁ Velocity at one End).

→ 10 cm = (h₁ height of pressure).

→ 5 cm = (h₂ half height of pressure).

Calculations:

→ P + 1/2 pv² = constant

→ P = hpg

→ hpg + 1/2 pv² = constant

→ p(hg + 1/2 v²) = constant

→ hg + 1/2 v² = constant

→ h₁ g + 1/2 v²₁ = h = g + 1/2 v²₂

Adding values to the question:

→ 10 × 10 + 1/2 × (3)² = 5 × 10 + 1/2 v²₂

→ 100 + 1/2 × 900 = 50 + 1/2 v²₂

→ 100 + 1/2 × 900 = 50 + 1/2 v²₂

→ 100 + 450 = 50 + 1/2 v²₂

→ 550 = 50 + 1/2 v²₂

→ 550 = 50 + 1/2 v²₂/2

→ 500 = v²₂/2

→ v²₂ = 500 × 2

→ v²₂ = 1000

→ v₂ = √1000

→ v₂ = 31.62 m/s

Therefore, 31.62 m/s is the velocity.

#AnswerWithQuality

#BAL