Question 61: Properties of Matter

Question

The figure shows a liquid of given density flowing steadily in horizontal tube of varying cross - section. Cross sectional areas at $$\mathrm{A}$$ is $$1.5 \mathrm{~cm}^{2}$$, and $$\mathrm{B}$$ is $$25 \mathrm{~mm}^{2}$$, if the speed of liquid at $$\mathrm{B}$$ is $$60 \mathrm{~cm} / \mathrm{s}$$ then $$\left(\mathrm{P}_{\mathrm{A}}-\mathrm{P}_{\mathrm{B}}\right)$$ is :
(Given $$\mathrm{P}_{\mathrm{A}}$$ and $$\mathrm{P}_{\mathrm{B}}$$ are liquid pressures at $$\mathrm{A}$$ and $$\mathrm{B}$$% points.
density $$\rho=1000 \mathrm{~kg} \mathrm{~m}^{-3}$$
$$\mathrm{A}$$ and $$\mathrm{B}$$ are on the axis of tube

Accepted Answer

From continuity theorem $\mathrm{A}_1 \mathrm{~V}_1=\mathrm{A}_2 \mathrm{~V}_2$
$$
\begin{aligned}
& 1.5 	imes \mathrm{V}_1=25 	imes 10^{-2} 	imes 60 \\
& \mathrm{~V}_1=\frac{25 	imes 60 	imes 10^{-2} 	imes 10}{1.5} \\
& \mathrm{~V}_1=10 \mathrm{~cm} / \mathrm{s}
\end{aligned}
$$
By Bernoulli's theorem
$$
\begin{aligned}
& P_1+\frac{1}{2} 	imes 1000 	imes(0.1)^2=P_2+\frac{1}{2} 	imes 1000 	imes(0.6)^2 \\
& P_1+5=P_2+\frac{1}{2} 	imes 1000 	imes 36 	imes 10^{-2} \\
& P_1+5=P_2+180 \\
& P_1-P_2=175 \mathrm{~Pa}
\end{aligned}
$$

Answer

$$27 \mathrm{~Pa}$$

Answer

$$175 \mathrm{~Pa}$$

Answer

$$135 \mathrm{~Pa}$$

Answer

$$36 \mathrm{~Pa}$$

Properties of Matter

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