JEE MAIN - Physics (2023 - 13th April Morning Shift - No. 4)
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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
$$27 \mathrm{~Pa}$$
$$175 \mathrm{~Pa}$$
$$135 \mathrm{~Pa}$$
$$36 \mathrm{~Pa}$$
Explanation
From continuity theorem $\mathrm{A}_1 \mathrm{~V}_1=\mathrm{A}_2 \mathrm{~V}_2$
$$ \begin{aligned} & 1.5 \times \mathrm{V}_1=25 \times 10^{-2} \times 60 \\\\ & \mathrm{~V}_1=\frac{25 \times 60 \times 10^{-2} \times 10}{1.5} \\\\ & \mathrm{~V}_1=10 \mathrm{~cm} / \mathrm{s} \end{aligned} $$
By Bernoulli's theorem
$$ \begin{aligned} & P_1+\frac{1}{2} \times 1000 \times(0.1)^2=P_2+\frac{1}{2} \times 1000 \times(0.6)^2 \\\\ & P_1+5=P_2+\frac{1}{2} \times 1000 \times 36 \times 10^{-2} \\\\ & P_1+5=P_2+180 \\\\ & P_1-P_2=175 \mathrm{~Pa} \end{aligned} $$
$$ \begin{aligned} & 1.5 \times \mathrm{V}_1=25 \times 10^{-2} \times 60 \\\\ & \mathrm{~V}_1=\frac{25 \times 60 \times 10^{-2} \times 10}{1.5} \\\\ & \mathrm{~V}_1=10 \mathrm{~cm} / \mathrm{s} \end{aligned} $$
By Bernoulli's theorem
$$ \begin{aligned} & P_1+\frac{1}{2} \times 1000 \times(0.1)^2=P_2+\frac{1}{2} \times 1000 \times(0.6)^2 \\\\ & P_1+5=P_2+\frac{1}{2} \times 1000 \times 36 \times 10^{-2} \\\\ & P_1+5=P_2+180 \\\\ & P_1-P_2=175 \mathrm{~Pa} \end{aligned} $$
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