Relating period, volume, surface area and the velocity of sound by dimensional analysis The question is:-
There is a dimensional relation between period T, volume V, surface area A and the velocity of sound C. Assume that period increases with volume and decrease with increase in area. Produce the equation in terms of T.
So I was able to form one equation from the information given above.
$$ t = KV^aA^bC^c $$
where V = Volume , A = Surface Area, C = velocity of sound, K = a constant
$$ [t] = [K][V^a][A^b][C^c] $$
$$ T = L^{3a}L^{2b}\frac{L^{c}}{T^{c}}$$
$$ T = L^{3a+2b+c}T^{-c}$$
so now
$$ -c = 1$$
$$ c = -1$$
Also
$$ 3a+2b+c = 0$$
$$ 3a+2b-1 = 0$$
$$ 3a+2b = 1$$
I don't know what to do next because there are two unknowns left. Is there any other equation that can be formed by the information given above?
+Gaurav, I have tried to relate the Surface area to volume. Since the question states that time period increases with volume and decrease with increase in area, one could argue that:-
$$ T = \frac{k.V^a}{A^b}$$ where k is a constant
so $$ T = L^{3a-2b} $$
$ 3a-2b = 0$
We already know that
$$ 3a+2b = 1$$
Adding both the equations
$$ (3a-2b) + (3a+2b) = 1$$
$$ 6a = 1$$
$$ a = \frac{1}{6}$$
But the value of a is supposed to be 1, so the third equation I formulated must be incorrect.
 A: I'm not sure entirely what you mean by a "dimensional relation" here, but in general we cannot take an arbitrary function of anything other than a dimensionless parameter because functions have Taylor expansions which sum together arbitrary powers of their arguments. Dimensional analysis therefore restrains you when you're about to say, "This variable is some unknown arbitrary function of these parameters": it says, "no, that function must have a very special form, so that the units work out to something other than nonsense." And the first thing it says is that the only truly-arbitrary functions involve dimensionless parameters. So, the first thing to look at is whether there are any dimensionless parameters in your case, and there is, because $$[[V^2]] = l^6 = [[A^3]].$$
This means that something with units of $t$ will have the form$$T = \frac{A^{1/2}}{C} ~ f(V^2 / A^3)$$for some arbitrary function $f$.
What you've run into with the $3a + 2b = 1$ equation is the arbitrariness of f. If you'll pardon the pun, $f$ is ineffable, we can't know it. We don't even have a form for it: $\sin(V/A^{3/2}) + \cos(V^2/A^3)$ is perfectly acceptable as far as dimensional analysis is concerned!
However, maybe "a dimensional relation" means that $f(x)$ must take the form of $k x^n$ for one $k$ and $n$. Even then, the criterion you gave ("increases with V, decreases with A") only states that $n > 1/6$, which is not enough to specify $n$.
But, supposing that we add some criterion like $T \propto V$ or so, then we'd have $n = 1/2$ and $T = k V / (C A)$, which looks pretty plausible and compelling. For example, $V/(c A)$ is the time it takes for a tank of volume $V$ to empty out if it all drains out of a pipe of cross-sectional area $A$ at constant speed $c$. But if you say instead that, say $T \propto V^2$, we can accomodate that too, with $n = 1$. There is a ton of freedom here.
