A wave is set up in a stretched string by plucking it. The velocity "v" of the wave depend up on tension "f" in the string, its length "L" and its mass "m". Making use of knowledge of dimensions, prove that V proportional to sqrt(Tension / Mass per Unit Length).

The velocity of the wave is $V$. Its dimensions are:

$$[V] = [L T^{-1}]$$

The tension in the string is $F$. Tension is a force, so its dimensions are:

$$[F] = [M L T^{-2}]$$

The mass per unit length is $\mu$. Its dimensions are:

$$\left[\frac{\text{Mass}}{\text{Length}}\right] = \left[\frac{M}{L}\right] = [M L^{-1}]$$

Let's assume that the velocity $V$ depends on tension $F$ and mass per unit length $\mu$ in the following form, where $k$ is a dimensionless constant and $a, b$ are exponents to be determined:

$$V = k \cdot F^a \mu^b$$

Equating the dimensions on both sides of the equation:

$$[L T^{-1}] = [M L T^{-2}]^a [M L^{-1}]^b$$ $$[L T^{-1}] = [M^a L^a T^{-2a}] [M^b L^{-b}]$$ $$[M^0 L^1 T^{-1}] = [M^{a+b} L^{a-b} T^{-2a}]$$

Comparing the powers of M:

$$0 = a+b \quad (1)$$

Comparing the powers of L:

$$1 = a-b \quad (2)$$

Comparing the powers of T:

$$-1 = -2a \quad (3)$$

From equation (3):

$$-1 = -2a \implies a = \frac{1}{2}$$

Substitute $a = \frac{1}{2}$ into equation (1):

$$0 = \frac{1}{2} + b \implies b = -\frac{1}{2}$$

Verify with equation (2):

$$1 = a-b = \frac{1}{2} - \left(-\frac{1}{2}\right) = \frac{1}{2} + \frac{1}{2} = 1$$

The values are consistent.

$$V = k \cdot F^{1/2} \mu^{-1/2}$$ $$V = k \cdot \frac{F^{1/2}}{\mu^{1/2}}$$ $$V = k \cdot \sqrt{\frac{F}{\mu}}$$

Since $k$ is a dimensionless constant, we can write the proportionality:

$$\boxed{V \propto \sqrt{\frac{\text{Tension}}{\text{Mass per Unit Length}}}}$$