Answer is: option3

1.5 ft per hr

Solution:

We are given the volume of water in a spherical tank as:

$$V=\frac{\pi h^2}{3}(18-h)$$

where $V$ is the volume of water and $h$ is the depth of the water. The rate of change of volume is given as:

$$\frac{dV}{dt}=30\pi \text{cubic feet per hour}$$

We need to find $\frac{dh}{dt}$ when $h=2$ feet.

Expanding the given equation:

$$V=\frac{\pi }{3}(18h^2-h^3)$$

Differentiate both sides with respect to $t$:

$$\frac{dV}{dt}=\frac{\pi }{3}(36h\frac{dh}{dt}-3h^2\frac{dh}{dt})$$

Factor out $\frac{dh}{dt}$:

$$\frac{dV}{dt}=\frac{\pi }{3}(36h-3h^2)\frac{dh}{dt}$$ $$\frac{dV}{dt}=\frac{\pi }{3}(36h-3h^2)\frac{dh}{dt}$$

Solve for $\frac{dh}{dt}$:

$$\frac{dh}{dt}=\frac{3}{\pi (36h-3h^2)}\frac{dV}{dt}$$

Substituting $\frac{dV}{dt}=30\pi$:

$$\frac{dh}{dt}=\frac{3\times 30\pi }{\pi (36h-3h^2)}$$

Cancel $\pi$:

$$\frac{dh}{dt}=\frac{90}{36h-3h^2}$$

Evaluate at $h=2$:

$$\frac{dh}{dt}=\frac{90}{36(2)-3(2^2)}$$ $$=\frac{90}{72-12}$$ $$=\frac{90}{60}$$ $$=1.5\text{ ft per hour}$$

Final Answer: 1.5 ft per hr.