91Ó°ÊÓ

The given data can be listed below as,

• The height of cylinder bucket is, h = 25.0 cm.
• The diameter of cylinder bucket is, d₁ = 10cm .
• The cross-sectional area is, A₂ = 1.50 cm²
• The water flow rate is, $\stackrel{Ë™}{Q}=2.40×{10}^{−4}{\text{″¾}}^{\text{3}}/\text{s}$.

The water level in the bucket will rise in anticipation of the volume flow rate into the bucket, 2.40 x 10^-4 m³/s is equal to the volume flow rate out the hole in the bottom.

The relation of Bernoulli’s equation is expressed as,

$p_1+\frac{1}{2}\mathrm{ÒÏ}{v}_1^2+\mathrm{ÒÏ}g{h}_1=p_2+\frac{1}{2}\mathrm{ÒÏ}{v}_2^2+\mathrm{ÒÏ}g{h}_2$.......(i)

Here,p₁and p₂ are the pressure at point 1 and point 2, v₁ and v₂ are velocity at the point 1 and point 2, $\mathrm{ÒÏ}$is the density of water and g is the gravitational acceleration.

The volume flow rate from bucket is equal to the volume flow rate out of hole is expressed as,

$\begin{array}{c}\stackrel{Ë™}{Q}=A_2{v}_2\\ 2.40×{10}^{−4}{\text{″¾}}^{\text{3}}/\text{s}=A_2{v}_2\\ v_2=\frac{2.40×{10}^{−4}{\text{″¾}}^{\text{3}}/\text{s}}{A_2}\end{array}$

Here, A₂ is the area at point 2, and v₂ is the velocity at point 2.

Substitute 1.50 x 10^-4m² for A₂ in the above equation.

$\begin{array}{c}{v}_2=\frac{2.40×{10}^{-4}{\text{″¾}}^{\text{3}}/\text{s}}{1.50×{10}^{−4}{\text{″¾}}^{\text{2}}}\\ =1.6\text{″¾}/\text{s}\end{array}$

The area at point 1 (A₁ ) is greater than the value of A₂ i.e. $A_1>>A_2$, then $\frac{1}{2}\mathrm{ÒÏ}{v}_1^2<<\frac{1}{2}\mathrm{ÒÏ}{v}_2^2$so neglected the term $\frac{1}{2}\mathrm{ÒÏ}{v}_1^2$ . Now, the value of h₂ = 0 .because the height of point 2 from ground level is zero and $p_1=p_2=p_0$(atmospheric pressure), these values put in equation. (I).

$\begin{array}{c}{p}_0+\mathrm{ÒÏ}g{h}_1=p_0+\frac{1}{2}\mathrm{ÒÏ}{v}_2^2+0\\ \mathrm{ÒÏ}g{h}_1=\frac{1}{2}\mathrm{ÒÏ}{v}_2^2\\ h_1=\frac{v_2^2}{2g}\end{array}$

Here, h₁ is the height of water rise in the bucket.

Substitute 1.6 m/s for v₂ and 9.81m/s² for g in the above equation.

$\begin{array}{c}{h}_1=\frac{{\left(1.6\text{″¾}/\text{s}\right)}^2}{2×9.81\text{ }{\text{m}/\text{s}}^{\text{2}}}\\ =0.131\text{″¾}\end{array}$

Hence, the water rise in the bucket is 0.131m .