Two large,identical water tanks,1 and 2,kept | vedclass

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(B) For tank $1$,the velocity of efflux at height $y$ is $v_1 = \sqrt{2gy}$.By the equation of continuity,$A(-dy/dt) = a\sqrt{2gy}$,where $A$ is the c

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$3$

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(B) For tank $1$,the velocity of efflux at height $y$ is $v_1 = \sqrt{2gy}$.By the equation of continuity,$A(-dy/dt) = a\sqrt{2gy}$,where $A$ is the cross-sectional area of the tank and $a$ is the area of the hole.Integrating from $y=h$ to $y=0$:$dt = (A/a\sqrt{2g}) \cdot (-dy/\sqrt{y})$$t_1 = (A/a\sqrt{2g}) \int_0^h y^{-1/2} dy = (A/a\sqrt{2g}) \cdot 2\sqrt{h} = (A/a) \sqrt{2h/g}$.For tank $2$,the velocity of efflux at height $y$ is $v_2 = \sqrt{2g(y+H)}$.By the equation of continuity,$A(-dy/dt) = a\sqrt{2g(y+H)}$.Integrating from $y=h$ to $y=0$:$dt = (A/a\sqrt{2g}) \cdot (-dy/\sqrt{y+H})$$t_2 = (A/a\sqrt{2g}) \int_0^h (y+H)^{-1/2} dy = (A/a\sqrt{2g}) \cdot 2[\sqrt{y+H}]_0^h = (A/a\sqrt{2g}) \cdot 2(\sqrt{h+H} - \sqrt{H})$.Given $H = (16/9)h$,then $h+H = h + (16/9)h = (25/9)h$.$t_2 = (A/a\sqrt{2g}) \cdot 2(\sqrt{25h/9} - \sqrt{16h/9}) = (A/a\sqrt{2g}) \cdot 2(5/3\sqrt{h} - 4/3\sqrt{h}) = (A/a\sqrt{2g}) \cdot 2(1/3\sqrt{h}) = (A/3a) \sqrt{2h/g}$.Therefore,the ratio $t_1 / t_2 = [(A/a) \sqrt{2h/g}] / [(A/3a) \sqrt{2h/g}] = 3$.

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