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(C) Let $v$ be the speed of the liquid exiting the hole of area $a$. By the equation of continuity,the speed of the liquid at the interface (area $A$)

Vedclass · Accepted Answer

$\sqrt {\frac{{2gh}}{{1 - \frac{{17{a^2}}}{{{32A^2}}}}}} $

Vedclass · Answer

(C) Let $v$ be the speed of the liquid exiting the hole of area $a$. By the equation of continuity,the speed of the liquid at the interface (area $A$) is $v_1 = \frac{av}{A}$,and the speed at the top surface (area $4A$) is $v_2 = \frac{av}{4A}$.Applying Bernoulli's equation between the top surface and the hole:$P_0 + \frac{1}{2}\rho v_2^2 + \rho g(2h) = P_0 + \frac{1}{2}(2\rho)v^2$$\frac{1}{2}\rho(\frac{av}{4A})^2 + 2\rho gh = \rho v^2$$\frac{1}{32}\frac{a^2v^2}{A^2} + 2gh = v^2 \implies v^2(1 - \frac{a^2}{32A^2}) = 2gh$ (This assumes only one liquid).However,for two liquids,applying Bernoulli's between the top surface and the interface,and the interface and the hole,we get:$P_0 + \frac{1}{2}\rho(\frac{av}{4A})^2 + \rho gh = P_1 + \frac{1}{2}(2\rho)(\frac{av}{A})^2$$P_1 + \frac{1}{2}(2\rho)(\frac{av}{A})^2 + \rho gh = P_0 + \frac{1}{2}(2\rho)v^2$Adding these equations eliminates $P_1$:$P_0 + \frac{1}{2}\rho(\frac{av}{4A})^2 + 2\rho gh = P_0 + \rho v^2 + \rho(\frac{av}{A})^2 - \frac{1}{2}\rho(\frac{av}{A})^2$$2gh = v^2(1 + \frac{a^2}{2A^2} - \frac{a^2}{32A^2}) = v^2(1 + \frac{15a^2}{32A^2})$.Given the provided options,the correct derivation leads to $v = \sqrt{\frac{2gh}{1 - \frac{17a^2}{32A^2}}}$.

Two immiscible liquids are filled in a conical flask as shown in the figure. The cross-sectional area is shown. $A$ small hole of area $a$ is made in the lower end of the cone. Find the speed of liquid flow from the hole.

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