A tank of uniform cross sectional cross area (A) containing liqui
A. <span class="math-tex">\(\frac{{2A\sqrt {{H_1}} }}{{{C_d}a\sqrt {2g} }}\)</span>
B. <span class="math-tex">\(\frac{{2A{H_1}}}{{{C_d}a\sqrt {2g} }}\)</span>
C. <span class="math-tex">\(\frac{{2AH_1^{3/2}}}{{{C_d}a\sqrt {2g} }}\)</span>
D. <span class="math-tex">\(\frac{{2AH_1^2}}{{{C_d}a\sqrt {2g} }}\)</span>
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Right Answer is: A
SOLUTION
Concept:
Let h is the height of water at any instant in a container and 'dh' is a change in height in 'dt' time.
\( - Adh = {C_d}\sqrt {2gh} .a.dt\)
\(A\mathop \smallint \limits_{{H_1}}^{{H_2}} \frac{{ - dh}}{{\sqrt {2gh} }} = {c_d}\;a\mathop \smallint \limits_0^t dt\)
\( - A\left[ {\frac{1}{{\sqrt {2g} \;}}\frac{{{h^{\frac{{ - 1}}{2} + 1}}}}{{\frac{{ - 1}}{2} + 1}}} \right]_{{H_1}}^{{H_2}} = {C_d}.a.t\)
\(\frac{{ - A}}{{\sqrt {2g} }}.2\left[ {\sqrt {{H_2}} - \sqrt {{H_1}} } \right] = {C_d}.a.t\)
When tank is empty H2 = 0
\(\frac{{2A.\sqrt {{H_1}} }}{{\sqrt {2g} }} = {c_d}.a.t\)
\(t = \frac{{2A\sqrt {{H_1}} }}{{a.{c_d}\sqrt {2g} }}\)
