1.

A small hole of area of cross-section 2mm^(2) is present near of the bottom of a fully filled open tank of height 2m. Taking g= 10 m//s^(2), the rate of flow of water through the open hole would be nearly:

Answer»

`6.4 XX 10^(-6) m^(3)//s`
`12.6 xx 10^(-6) m^(3)//s`
`8.9 xx 10^(-6) m^(3)//s`
`2.23 xx 10^(-6) m^(3)//s`

Solution :Velocity of water COMING out of hole, `v= sqrt(2gh)`
Time rate of flow of water, = AV
`=A sqrt(2gh)`
`=(2 xx 10^(-6)) (2 xx 10 xx 2)^(½)`
`=4 sqrt10 xx 10^(-6)`
`=4 xx 3.16 xx 10^(-6)`
`=12.64 xx 10^(-6)`
`~~ 12.6 xx 10^(-6) m^(3)//s`


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