Answer
The height h of the liquid will be $h=\frac{225}{49}\ in$.
Work Step by Step
The key to solving this problem is to realize that the volume of the liquid must be the same in both cylinders.
The volume of a cylinder is given by the expression: $V=\pi r^2h$, where V is the volume, r is the radius and h is the height of the cylinder. Since we are given the diameter of both cylinders, we must remember that the diameter is twice the radius, $D=2r$, so:
$V=\pi (\frac{D}2)^2h$
$V=\frac{\pi D^2h}4$
Calculate the volume of the first cylinder, $D_1=2\frac12\ in=\frac52\ in$ and $h_1=9\ in$:
$V_1=\frac{\pi D_1^2h_1}4$
$V_1=\frac{\pi\cdot(\frac52)^2\cdot 9}4$
$V_1=\frac{\pi\cdot\frac{25}4\cdot 9}4$
$V_1=\frac{225\pi}{16}$
As discussed before, $V_1=V2$, and for the second cylinder, $D_2=3\frac12\ in=\frac72$, so:
$V_2=\frac{\pi D_2^2h_2}4$
$\frac{225\pi}{16}=\frac{\pi \cdot (\frac72)^2\cdot h_2}4$
$\frac{225\pi}{16}=\frac{\pi \cdot \frac{49}4\cdot h_2}4$
$\frac{225\pi}{16}=\frac{49\pi h_2}{16}$
Cancel out $\pi/16$ on both sides of the equation:
$225=49h_2$
$h_2=\frac{225}{49}\ in\approx 4\frac 35\ in$
The height h of the liquid will be $h=\frac{225}{49}\ in$.
