Question

$a,A$ are areas, ${\rho }_m$ is the density of mercury for a venturimeter the speed of fluid at wide neck is

• A. $v_1=\sqrt{\frac{2{\rho }_{m}gh}{\rho }}{\left[\frac{A}{a}-1\right]}^{-\frac{1}{2}}$
• B. $v_1=\sqrt{\frac{2{\rho }_{m}gh}{\rho }}{\left[\frac{a}{A}+1\right]}^{-\frac{1}{2}}$
• C. $v_1=\sqrt{\frac{2\rho gh}{{\rho }_m}}{\left[\frac{A^2}{a^2}-1\right]}^{-\frac{1}{2}}$
• D. $v_1=\sqrt{\frac{2{\rho }_{m}gh}{\rho }}{\left[\frac{A^2}{a^2}-1\right]}^{-\frac{1}{2}}$

Answer 1

Answer: (D)

Here $A{v}_1=a{v}_2$ or $v_2=\left(\frac{A}{a}\right)v_1$
Using Bernoulli's theorem,
$P_1+\frac{1}{2}\rho v_1^2=P_2+\frac{1}{2}\rho v_2^2$
$\Rightarrow P_1-P_2=\frac{1}{2}\rho \left(v_2^2-v_1^2\right)$
or ${\rho }_m\cdot g\cdot h=\frac{1}{2}\rho v_1^2\left[{\left(\frac{A}{a}\right)}^2-1\right]$
Hence, $v_1=\sqrt{\frac{2{\rho }_m\cdot gh}{\rho }}{\left[{\left(\frac{A}{a}\right)}^2-1\right]}^{-\frac{1}{2}}$