A cylindrical bucket, open at the top, is 25.0 cm high and 10.0 cm in diameter. A circular hole with a cross-sectional area 1.50 cm² is cut in the center of the bottom of the bucket. Water flows into the bucket from a tube above it at the rate of 2.40 x 10^-4m³/s. How high will the water in the bucket rise?

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, $\dot{Q}=2.40\times {10}^{-4}{\text{\hspace{0.17em}m}}^{\text{3}}/\text{s}$.

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,

$p_1+\frac{1}{2}\rho v_1^2+\rho g{h}_1=p_2+\frac{1}{2}\rho v_2^2+\rho g{h}_2$.......(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, $\rho$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,

$\begin{array}{c}\dot{Q}=A_2{v}_2\\ 2.40\times {10}^{-4}{\text{\hspace{0.17em}m}}^{\text{3}}/\text{s}=A_2{v}_2\\ v_2=\frac{2.40\times {10}^{-4}{\text{\hspace{0.17em}m}}^{\text{3}}/\text{s}}{A_2}\end{array}$

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.

$\begin{array}{c}{v}_2=\frac{2.40\times {10}^{-4}{\text{\hspace{0.17em}m}}^{\text{3}}/\text{s}}{1.50\times {10}^{-4}{\text{\hspace{0.17em}m}}^{\text{2}}}\\ =1.6\text{\hspace{0.17em}m}/\text{s}\end{array}$

The area at point 1 (A₁ ) is greater than the value of A₂ i.e. $A_1>>A_2$, then $\frac{1}{2}\rho v_1^2<<\frac{1}{2}\rho v_2^2$so neglected the term $\frac{1}{2}\rho v_1^2$ . 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).

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

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.

$\begin{array}{c}{h}_1=\frac{{\left(1.6\text{\hspace{0.17em}m}/\text{s}\right)}^2}{2\times 9.81{\text{m}/\text{s}}^{\text{2}}}\\ =0.131\text{\hspace{0.17em}m}\end{array}$

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