91Ó°ÊÓ

The expression for Bernoulli's equation for a flowing fluid, between two points $1$and $2$is written as,

$p_1+\frac{1}{2}\mathrm{ÒÏ}{v}_1^2+\mathrm{ÒÏ}g{y}_1=p_2+\frac{1}{2}\mathrm{ÒÏ}{v}_2^2+\mathrm{ÒÏ}g{y}_2$

Here, $p_1$is the pressure at point role="math" localid="1648179242044" $1$,$p_2$is the pressure at point $2$,$v_1$is the velocity at point$1$, $v_2$is the velocity at point $2$, is the$y_1$position of point $1$above the datum, $y_2$is the position of point $2$above the datum and $\mathrm{ÒÏ}$is the density of the flowing fluid.

Continuity equation between the above mentioned points is written as,

$\mathrm{ÒÏ}{a}_1{v}_1=\mathrm{ÒÏ}{a}_2{v}_2$

$a_1{v}_1=a_2{v}_2$

Here, $a_1$and $a_2$are the cross-sectional areas of the flow at the points $1$and $2$, respectively.

Understand that point $1$is the top of the bucket and the point $2$is at the hole. The height is measured from the top of the bucket. The pressure at points $1$and $2$is atmospheric pressure. So,

$p_1=p_2$

$=p_{\text{atm}}$

Here, localid="1648180078300" $p_{atm}$is atmospheric pressure.

Also,

$y_1=0\mathrm{cm}$

$y_2=-23\mathrm{cm}$

It is known that area of hole is much smaller to the area of water at the top of the bucket. Hence,

localid="1648180205607" $a_2<<a_1$

$\frac{a_2}{a_1}<<1.....1$

Apply continuity equation between points $1$and $2$

$a_1{v}_1=a_2{v}_2$

$\frac{a_2}{a_1}=\frac{v_1}{v_2}.............2$

Use equations $\left(1\right)$and $\left(2\right)$, to get:

$\frac{v_1}{v_2}≪1$

$v_1<<v_2$

Squaring both sides of the equation.

$v_1^2<<<<v_2^2$

Hence $v_1$can be neglected.

Substitute $p_{\text{atm}}$for $p_1$and$p_2$, $0cm$for $y_1$and $0\mathrm{m}/\mathrm{s}$for $v_1$.

$p_{atm=}+\frac{1}{2}\mathrm{ÒÏ}(0\mathrm{m}/\mathrm{s}{)}^2+\mathrm{ÒÏ}g(0\mathrm{cm})=p_{atm}+\frac{1}{2}\mathrm{ÒÏ}{v}_2^2+\mathrm{ÒÏ}g{y}_2$

$v_2^2=-2g{y}_2$

$v_2=\sqrt{-2g{y}_2}$

Substitute $-23\mathrm{cm}$for $y_2$and $9.8\mathrm{m}/{\mathrm{s}}^2$ for $\mathrm{g}$,

$v_2=\sqrt{-\left(2\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(-23\mathrm{cm}\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)\right)\right.}$

$v_2=2.12\mathrm{m}/\mathrm{s}$