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Water flows through a horizontal pipe of varying cross-seetions. The pressure of water equals 0.1 m of mercury at a place where the velocity of flow is 0.4 ms^(-1). What will be the pressure at another place, where the velocity of flow is 0.5ms^(-1). |
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Answer» <P> Solution :Given `P_(1)=0.1"M of HG"=0.1 xx 13600 xx 9.8``v_(1)=0.4ms^(-1) =1332Nm^(-2) P_(2)=? V_(2)=0.5 ms^(-1)` Applying `(P_(1)-P_(2))=1/2 rho(v_(2)^(2)-v_(1)^(2))` we get , `=1/2 xx 10^(3) (0.5^(2)-0.4^(2))` `13328-P_(2)=500(0.9) (0.1)` `therefore P_(2)=13328-45=13283 Nm^(-2)` `or P_(2)=(13283)/(13600 xx 9.8)=0.0996"m of mercury"` |
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