question_answer 16)

The cylinderical tube of a spray pump has a cross-section of \[=1\cdot 49\times {{10}^{-4}}m\] one of which has 40 fine holes each of diameter 1.0 mm. If the liquid flow inside the tube is 1.5 m per minute, what is the speed of ejection of the liquid through the holes ?

Answer:

Area of cross-section of tube, \[=1\cdot 3\times {{10}^{-4}}m\] No. of holes = 40 ; Diameter of each hole, \[\left( \text{1pa=1N}{{\text{m}}^{\text{-2}}} \right)\text{;g=10m/}{{\text{s}}^{\text{2}}}\text{.}\] \[A=0\cdot 10\times 0\cdot 10={{10}^{-2}}{{m}^{2}};\] radius of hole, \[F=mg=100\times 10N\] \[=\frac{\vartriangle L}{L}=\frac{\left( F/A \right)}{G}\] Area of cross-section of each hole \[\vartriangle L=\frac{FL}{AG}=\frac{\left( 100\times 10 \right)\times 0\cdot 10}{{{10}^{-2}}\times \left( 25\times {{10}^{9}} \right)}=4\times {{10}^{-7}}m.\]Total area of cross-section of 40 holes, \[2\cdot 0\times {{10}^{11}}pa.\] Speed of liquid inside the tube, \[A=\pi \left( r_{2}^{2}-r_{1}^{2} \right)=\frac{22}{7}\left[ {{\left( 0\cdot 60 \right)}^{2}}-{{\left( 0\cdot 30 \right)}^{2}} \right]\] If \[=\frac{22}{7}\times 0\cdot 27{{m}^{2}}\]is the velocity of ejection of the liquid through the holes, then \[=\frac{F/A}{Y}=\frac{F}{AY}\] \[\frac{50,000\times 9\cdot 8}{4\times \left( \frac{22}{7}\times 0\cdot 27 \right)\times 2\cdot 0\times {{10}^{11}}}=7\cdot 21\times {{10}^{-7}}\]