a pipe of radius r contains a vertical column of water. a hole in the pipe of radius r ejects water…

Explanation:

Step1: Apply the principle of continuity

The principle of continuity is $A_1v_1 = A_2v_2$, where $A$ is the cross - sectional area and $v$ is the speed of the fluid. The cross - sectional area of a circle is $A=\pi r^{2}$. Let the cross - sectional area of the pipe be $A_1=\pi R^{2}$ and the speed of water in the pipe be $v_1$. The cross - sectional area of the hole is $A_2 = \pi r^{2}$ and the speed of water coming out of the hole is $v_2=v_0$.

Step2: Substitute the areas into the continuity equation

We have $A_1v_1=A_2v_2$, so $\pi R^{2}v_1=\pi r^{2}v_0$.

Step3: Solve for $v_1$

Dividing both sides of the equation $\pi R^{2}v_1=\pi r^{2}v_0$ by $\pi R^{2}$, we get $v_1 = v_0\frac{r^{2}}{R^{2}}$.

Answer:

C. $v_0r^{2}/R^{2}$