Discussion :: Hydraulics and Fluid Mechanics
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A hemispherical tank of radius (R) containing liquid upto height (H1) has an orifice of cross-sectional area (a) at its bottom. The time required to lower the level of liquid from (H1) to (H2) will be
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A.
\(\frac { n } { C_d \times a\sqrt{2g} } [\frac {2}{3} R(H_1^\frac{3}{2} - H_2^\frac{3}{2}) - \frac{1}{5}(H_1^\frac{5}{2} - H_2^\frac{5}{2})]\) |
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B.
\(\frac { 2n } { C_d \times a\sqrt{2g} } [\frac {2}{3} R(H_2^\frac{3}{2} - H_1^\frac{3}{2}) - \frac{1}{5}(H_2^\frac{5}{2} - H_1^\frac{5}{2})]\) |
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C.
\(\frac { 2n } { C_d \times a\sqrt{2g} } [\frac {2}{3} R^2(H_1^\frac{3}{2} - H_2^\frac{3}{2}) - \frac{1}{5}(H_1^\frac{5}{2} - H_2^\frac{5}{2})]\) |
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D.
none of the above |
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Answer : Option C
Explanation :
No answer description available for this question.
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