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Two capillary tubes of the same length but different radii r1 and r2 are fitted in parallel to the bottom of a vessel. The pressure head is P . What should be the radius of a single tube that can replace the two tubes so that the rate of flow is same as beforeA. `r_1+r_2`B. `(r_1r_2)/(r_1+r_2)`C. `(r_1+r_2)/(2)`D. None of these |
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Answer» Correct Answer - D When the tubes are fitted in parallel, `V=V_1 + V_2` `rArr (pipr^4)/(8etal) = (pr_1^4)/(8etal) + (pipr_2^4)/(8) rArr r^4 = r_1^4 + r_2^4` `therefore r=(r_1^4 + r_2^4)^(1//4)` |
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