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A horizontal pipeline carries water in a streamline flow. At a point along the tube where the cross sectional area is `10^(-2)(m^2)`, the water velcity is `2m//s` and the pressure is 8000Pa. The pressure of water at another point where cross sectional area is `0.5xx(10)^(-2)(m^2)` isA. `4000Pa`B. `1000Pa`C. `2000Pa`D. `3000Pa` |
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Answer» Correct Answer - C Area at other point is half. So, speed will be double. Now, `p_(1)+(1)/(2)rhov_(1)^(2)=p_(2)+(1)/(2) rho v_(2)^(2)` `:. p_(2)=p_(1)+(1)/(2) rho v_(1)^(2)-v_(2)^(2)` `=(8000)+(1)/(2)xx1000(4-16)` `=2000Pa`. |
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