Try "smart" numbers. The logic of above answers is the same, but it was easier for me to see, might be for others.

If a problem doesn't forbid emptying contents of one container into another when different dimensions are at issue but NOT different units, almost always, it helps to assume that one empties into the other. (I started trying to account for -3 inches per second. Bad idea.)

So the volume of water in the first tank, with height of 3 is EMPTIED in 3 seconds.

1. Numbers chosen

Let base area of Tank 1 = 3

Let height of Tank 1 = 3

Base area of Tank 1 is three times that of Tank 2=>

Tank 2 base area= 3/1 = 1

BASE AREA of both is (π*\(r^2\)) - No need to calculate it in volume formula; its value is assigned

2. Volume of water in Tank 1

--> V =(π*\(r^2\))*h

Base area = 3

Height of water is 3

V=3*3

Volume of water equals 9

3. Find height of Tank 2, into which the water goes. Volume is 9. Same formula, but solve for height (to which water will rise)

9 = (π*\(r^2\))*h

Base area = 1,so height of water:

9=(1)(h), h=9

Answer B: 9 inches per second

_________________

In the depths of winter, I finally learned

that within me there lay an invincible summer.

-- Albert Camus, "Return to Tipasa"