In the diagram shown, the difference in the two tubes of the...

$6{v}_{A} = 10{v}_{B} . . . . . . \left(1\right)$

$\dfrac{1}{2}ρ{v}_{A}^{2} - ρgh = \dfrac{1}{2}ρ{v}_{B}^{2}$ .........(2)

$⇒ {v}_{A} = \dfrac{1}{0 .8}m{s}^{- 1}$

$\dfrac{dV}{dt} = A{v}_{A}$

$= 6 \times {10}^{- 2}{cm}^{2} \times \dfrac{100}{0 .8}cm{s}^{- 1} = 7 . 5cc{s}^{- 1}$

OR

By the concept of the venturi meter tube

rate of flow$= Q = {A}_{1}{A}_{2}\sqrt{\dfrac{2g( {h}_{1} - {h}_{2} )}{( {A}_{2}^{2} - {A}_{1}^{2} )}}$

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

$= 7 .5 cc{s}^{- 1}$