Question

A horizontal pipeline carries water in a streamline flow. At a point along the pipe, where the cross-sectional area is $10{cm}^{2}$, the water velocity is $1{m}/{s}$ and pressure is 2000Pa. The pressure of water at another point where cross-sectional area is $5{cm}^{2}$, is $k \times {10}^{2}Pa$, the value of k is then
A. 700 Pa
B. 5
C. 600 Pa
D. 800 Pa

Answer 1

To solve this problem, first we have to find the velocity of water at another end of the pipeline. So, to find the velocity at the other end, use the continuity equation. Substitute the values in the equation and find the velocity of water at another end. Then, use the equation for Bernoulli’s theorem. Substitute the given values and density of water as $1000 {kg}/{{m}^{3}}$ in the equation and find the value of k.

Formula used:
${ A }_{ 1 }{ v }_{ 1 }={ A }_{ 2 }{ v }_{ 2 }$
${ P }_{ 1 }+\dfrac { 1 }{ 2 } \rho { v }_{ 1 }^{ 2 }={ P }_{ 2 }+\dfrac { 1 }{ 2 } \rho { v }_{ 2 }^{ 2 }$

Complete step-by-step solution:
Let ${A}_{1}$, ${v}_{1}$ and ${P}_{1}$ be the cross-sectional area, water velocity and pressure respectively at one end of the pipeline.

${A}_{2}$, ${v}_{2}$ and ${P}_{2}$ be the cross-sectional area, water velocity and pressure respectively at the other end of the pipeline.

Given:
${A}_{1}= 10{cm}^{2}=0.001 {m}^{2}$
${A}_{2}= 5{cm}^{2}=0.0005 {m}^{2}$
${v}_{1}= 1{m}/{s}$
${P}_{1}=2000Pa$
${P}_{2}= k \times {10}^{2}Pa$

According to continuity equation,
${ A }_{ 1 }{ v }_{ 1 }={ A }_{ 2 }{ v }_{ 2 }$ …(1)

Substituting values in above equation we get,
$0.001 \times 1=0.0005\times { v }_{ 2 }$
$\Rightarrow { v }_{ 2 }=\dfrac { 0.001 \times 1 }{ 0.0005 }$
$\Rightarrow {v}_{2}= 2{m}/{s}$

According to Bernoulli’s theorem,
${ P }_{ 1 }+\dfrac { 1 }{ 2 } \rho { v }_{ 1 }^{ 2 }={ P }_{ 2 }+\dfrac { 1 }{ 2 } \rho { v }_{ 2 }^{ 2 }$ …(2)

We know, the density of water is $1000 {kg}/{{m}^{3}}$.
$\Rightarrow \rho= 1000 {kg}/{{m}^{3}}$

Substituting values in the equation. (2) we get,
$2000+ \dfrac {1}{2} \times 1000 \times {1}^{2}= k \times {10}^{2}+ \dfrac {1}{2} \times 1000 \times {2}^{2}$
$\Rightarrow 2000+ \dfrac {1}{2} \times 1000 \times 1= k \times {10}^{2}+ \dfrac {1}{2} \times 1000 \times 4$
$\Rightarrow 2000+ 500= k \times {10}^{2}+ 2000$
$\Rightarrow 2500= k \times 100+ 2000$
$\Rightarrow 2500-2000= k \times 100$
$\Rightarrow k= \dfrac {500}{100}$
$\Rightarrow k= 5$

Thus, the value of k is 5.

So, the correct answer is option B i.e. 5.

Note:
To solve these types of problems, students must be familiar with the continuity equation and Bernoulli's theorem. Here, we had to convert to units of cross-sectional area. So, students must take care of the conversion units. Not converting the units of the quantities to their SI units may lead them to a wrong answer.