TS EAMCET 6-Aug-2021 Shift 2 Question Paper

Radius of each circular opening, $r_{1}=1\,\text{mm}$ $=10^{-3}\,\text{m}$ $\therefore$ Cross-sectional area of each opening, $A_{1}=\pi r_{1}^{2}=\pi \times 10^{-6}\,\text{m}^{2}$ Radius of the pipe, $r_{2}=2\,\text{cm}=0.02\,\text{m}$ $\therefore$ Cross-sectional area of pipe, $A_{2} \;\; = \pi r_{2}^{2}$ $\;\; = \pi \times (0.02)^{2}\,\text{m}^{2}$ The speed of water in the pipe, $v_{2} \;\; =25\,\text{cm}/\text{s}$ $\;\; =0.25\,\text{m}/\text{s}$ Let the speed of water when it exits from the circular openings is $v_{1}\,\text{m}/\text{s}$. Number of circular openings, $n=25$ As, the flow rate is always constant. $\therefore \;\; 25 A_{1} v_{1} \;\; = A_{2} v_{2}$ $\Rightarrow \;\; v_{1} \;\; = \frac{A_{2}}{25 A_{1}} v_{2}$ $\;\; = \frac{\pi \times (0.0)^{2}}{\pi \times 10^{-6}} \times \frac{0.25}{25} =4\,\text{m}/\text{s}$