Physics - Fluid Question with Solution | TestHub

$6v_{\mathrm{A}}=10v_{\mathrm{B}} ......\left(1\right)$

$\frac{1}{2}\rho v_{\mathrm{A}}^2-\rho gh=\frac{1}{2}\rho v_{\mathrm{B}}^2$  .........(2)

$\Rightarrow v_{\mathrm{A}}=\frac{1}{0.8} \mathrm{m} {\mathrm{s}}^{-1}$ 

 $\frac{dV}{dt}=A{v}_{\mathrm{A}}$

$=6\times {10}^{-2} {cm}^2\times \frac{100}{0.8} \mathrm{cm} {\mathrm{s}}^{-1}=7.5 \mathrm{cc} {\mathrm{s}}^{-1}$

OR

By the concept of the venturi meter tube

rate of flow$=Q=A_1{A}_2\sqrt{\frac{2g(h_1-h_2)}{(A_2^2-A_1^2)}}$  

$=6\times 10\times {10}^{-4}\sqrt{\frac{2\times 100\times 5}{(-36+100)\times {10}^{-2}}}$

$=7.5 \mathrm{cc} {\mathrm{s}}^{-1}$