Water is flowing in a conical tube as shown in figure. Velocity of water at area ' $\mathrm{A}_2$ ' is $60…

Water is flowing in a conical tube as shown in figure. Velocity of water at area ' $\mathrm{A}_2$ ' is $60 \mathrm{~cm} / \mathrm{s}$. The value of ' $\mathrm{A}_1$ ' and ' $\mathrm{A}_2$ ' is $10 \mathrm{~cm}^2$ and $5 \mathrm{~cm}^2$ respectively. The pressure difference at both the cross-section is
  1. $230 \mathrm{~N} / \mathrm{m}^2$
  2. $200 \mathrm{~N} / \mathrm{m}^2$
  3. $135 \mathrm{~N} / \mathrm{m}^2$
  4. $105 \mathrm{~N} / \mathrm{m}^2$

Solution

$\begin{aligned} A_1 \times V_1 & =A_2 \times V_2 \\ 10 \times V_1 & =5 \times 60 \\ V_1 & =30 \mathrm{~cm} / \mathrm{s} \end{aligned}$
From Bernoulli's equation $\begin{aligned} \left(\mathrm{P}_{\mathrm{A}}-\mathrm{P}_{\mathrm{B}}\right) & =\frac{1}{2} \rho\left(\mathrm{~V}_2^2-\mathrm{V}_1^2\right) \\ & =\frac{1}{2} \times 10^3\left(60^2-30^2\right) \times 10^{-4} \\ & =\frac{1}{2} \times 10^{-1}(3600-900) \\ & =\frac{1}{2} \times 2700 \\ & =1350 \mathrm{dyne} / \mathrm{cm}^2 \\ & =135 \mathrm{~N} / \mathrm{m}^2 \end{aligned}$

Asked in: MHT CET 2024 (09 May Shift 1)

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