Water from a tap falls vertically with a velocity of 3 m.s^-1. The area of cross section of the mouth of the tap is 2.5 cm^2. If the flow of water is uniform and steady throughout, then determine the cross section area of the tube of flow of water at a depth of 0.8 m from the mouth of the tap. [g=10m.s^-2]

Water from a tap falls vertically with a velocity of 3 m.s^-1. The are

1.	Water from a tap falls vertically with a velocity of 3 m.s^-1. The area of cross section of the mouth of the tap is 2.5 cm^2. If the flow of water is uniform and steady throughout, then determine the cross section area of the tube of flow of water at a depth of 0.8 m from the mouth of the tap. [g=10m.s^-2]
Answer» Solution :The area of cross section of the mouth of the TAP `A_1=2.5 cm^2` and velocity of water FLOW there, `v_1=3m. S^-1`. LET the area of cross section of the tube of flow of water at a depth of 0.8 m below the tap be `A_2 ` and velocity of water of flow there be `v_2`. `therefore "" v_2^2=v_1^2+2gh` `=(3)^2+2xx10xx0.8` `=9+16=25` or, `""v_2=5m.s^-1` We know that , `A_1v_1=A_2v_2` `or,"" A_2=(A_1v_1)/(v_2)=(2.5xx3)/(5)=1.5 cm^2`.