Question

Prove Toricelli's Equation of Efflux Velocity.

$\boxed{\sf \: v = \sqrt{2gh} } \: \: \: \: \: \bf...(show \: working)$
#Revision_Q24​

Answer 1

$\large \rm {\underbrace{\underline{✵Elucidation:-}}}$

$\sf \purple {\maltese{\underline{\underline{Torricelli's\: Law:}}}}$ The speed of efflux from an orifice of a tank is equal to that of the velocity acquired by a freely falling body from a height equal to that of the liquid above the orifice.

$\sf \pink {\maltese{\underline{\underline{Proof:}}}}$ Consider a liquid at rest in a container.Let a hole is made to the side of the container at a height $\bf {y_{1}}$ from the bottom.Let $\bf {y_{2}}$ be the height of the liquid and ρ be it's density.

$\sf \orange {\maltese{\underline{\underline{Using\: equation\: of\: continuity:}}}}$

$\cal \colon \implies {\boxed{\underline{v_{2}=\frac{A_{1}}{A_{2}}v_{1}}}}$

➻where $\bf {A_{1}}$ is the area of the surface of the liquid and $\bf {A_{2}}$ is the area of cross-section of hole.

➻At the surface of the liquid,it's velocity is approximated as zero.

➻Applying the "Bernoullis theorem" at (1) and (2) in the figure,

$\sf \mapsto {p_{1}+\frac{1}{2}ρv_{1}^{2}+ρgh_{1}=p_{2}+\frac{1}{2}ρv_{2}^{2}+ρgh_{2}}$

$\sf \mapsto {p_{a}+\frac{1}{2}ρv_{1}^{2}+ρgy_{1}=p+ρgy_{2}}$

$\sf \mapsto {\frac{1}{2}ρv_{1}^{2}=(p-p_{a})+ρg(y_{2}-y_{1})}$

$\large \sf \mapsto {\frac{1}{2}ρv_{1}^{2}=(p-p_{a})+ρgh}$

$\large \sf \mapsto {v_{1}^{2}=\frac{2[(p-p_{a})+ρgh]}{ρ}}$

$\large \sf \mapsto {v_{1}=\sqrt{\frac{2[(p-p_{a})+ρgh]}{ρ}}}$

➻If $\bf {p=p_{a}}$

$\large \sf \mapsto {v_{1}=\sqrt{\frac{2ρgh}{ρ}}}$

$\cal \implies \green {\boxed{v_{1}=\sqrt{2gh}}}$

➻(OR)

$\cal \implies \green {\boxed{\underline{v=\sqrt{2gh}}}}$

➻This is called Velocity of Efflux.

$\sf \blue {\maltese{\underline{\underline{Efflux\: velocity:}}}}$ The velocity with which the water comes out of the hole made to the container is called Velocity of efflux.The water which comes out of the hole follows a parabolic path.

!(: Hope it works :)!