Water flows through a horizontal pipe at a speed 'V'. Internal diameter of the pipe is 'd'. If the water is…

Water flows through a horizontal pipe at a speed 'V'. Internal diameter of the pipe is 'd'. If the water is emerging at a speed ' $V_{1}$ ' then the diameter of the nozzle is
  1. $\frac{\mathrm{V}}{\mathrm{V}_{1}}$
  2. $\mathrm{d} \sqrt{\frac{\mathrm{V}_{1}}{\mathrm{~V}}}$
  3. $\mathrm{d} \sqrt{\frac{\mathrm{V}}{\mathrm{V}_{1}}}$
  4. $\frac{\mathrm{d} V_{1}}{\mathrm{~V}}$

Solution

By equation of continuity $A_{1} V_{1}=A_{2} V_{2}$ $\begin{array}{l} \therefore \frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}=\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}=\frac{\mathrm{d}_{1}^{2}}{\mathrm{~d}_{2}^{2}} \\ \therefore \frac{\mathrm{d}_{2}^{2}}{\mathrm{~d}_{1}^{2}}=\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}} \\ \therefore \frac{\mathrm{d}_{2}}{\mathrm{~d}_{1}}=\sqrt{\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}} \\ \therefore \mathrm{d}_{2}=\mathrm{d}_{1} \sqrt{\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}}=\mathrm{d} \sqrt{\frac{\mathrm{v}}{\mathrm{v}_{1}}} \end{array}$

Asked in: MHT CET 2020 (13 Oct Shift 1)

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