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

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

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$1$

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(A) According to Torricelli's law,the velocity of efflux $v$ at a depth $h$ is given by $v = \sqrt{2gh}$.Since the depth $h$ is the same for both tanks,the velocity of efflux for both tanks is the same: $v_A = v_B = v = \sqrt{2gh}$.The mass flux (mass flow rate) is given by $\dot{m} = A \cdot v \cdot \rho$,where $A$ is the area of the hole,$v$ is the velocity,and $\rho$ is the density.Given that the mass flux is equal for both tanks: $\dot{m}_A = \dot{m}_B$.$A_A \cdot v_A \cdot \rho_A = A_B \cdot v_B \cdot \rho_B$.Since $v_A = v_B$,we can cancel them from both sides:$A_A \cdot \rho_A = A_B \cdot \rho_B$.Given that the area of the hole in $B$ is twice that of $A$,$A_B = 2 A_A$.Substituting this into the equation:$A_A \cdot \rho_A = (2 A_A) \cdot \rho_B$.Dividing both sides by $A_A \cdot \rho_B$:$\frac{\rho_A}{\rho_B} = 2$.Wait,re-evaluating the provided options and the logic: If $A_B = 2 A_A$,then $\rho_A = 2 \rho_B$,so $\rho_A / \rho_B = 2$. However,looking at the provided solution logic in the prompt,it assumes $v_2 = v_1/2$ which is incorrect for Torricelli's law. Given the standard physics,the ratio is $2$. Since $2$ is not an option,$I$ will re-verify the prompt's provided solution logic. The prompt's solution claims $A_1 v_1 = A_2 v_2$ which is the continuity equation for a pipe,not for efflux from a tank. For efflux,$v = \sqrt{2gh}$. Thus $\rho_A / \rho_B = A_B / A_A = 2$. If the intended answer is $1$,the area must be equal. Given the constraints,$I$ will provide the mathematically correct derivation.

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