Tanks A and B open at the top contain two dif | vedclass

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(D) According to Torricelli's law,the velocity of efflux $v$ at a depth $h$ is given by $v = \sqrt{2gh}$. Since both tanks have holes at the same dept

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(D) According to Torricelli's law,the velocity of efflux $v$ at a depth $h$ is given by $v = \sqrt{2gh}$. Since both tanks have holes at the same depth $h$,the velocity of efflux for both liquids is the same: $v_1 = v_2 = \sqrt{2gh}$.The mass flux (mass per unit time) is given by $\dot{m} = \rho A v$,where $\rho$ is the density,$A$ is the area of the hole,and $v$ is the velocity of efflux.For tank $A$: $\dot{m}_A = \rho_A A_A v_A$.For tank $B$: $\dot{m}_B = \rho_B A_B v_B$.Given that the mass flux is equal,$\dot{m}_A = \dot{m}_B$,and $A_B = 2A_A$:$\rho_A A_A v = \rho_B (2A_A) v$.Canceling $A_A$ and $v$ from both sides:$\rho_A = 2 \rho_B$.Therefore,the ratio of the densities $\frac{\rho_A}{\rho_B} = \frac{2}{1} = 2$. Wait,re-evaluating the provided options: If the mass flux is equal,$\rho_A A_A v = \rho_B A_B v$. Since $A_B = 2A_A$,then $\rho_A A_A = \rho_B (2A_A)$,which implies $\rho_A = 2\rho_B$,so $\frac{\rho_A}{\rho_B} = 2$. Given the options provided,there might be a misunderstanding of the question's intent or a typo in the options. However,following the logic strictly,the ratio is $2$. If the question implies $\frac{\rho_B}{\rho_A}$,the answer is $\frac{1}{2}$.

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