Question

To find the distance d over which a signal can be seen clearly in foggy conditions, a railways engineer uses dimensional analysis and assumes that the distance depends on the mass density $\rho$ of the fog, intensity (power/area) S of the light from the signal and its frequency f. The engineer finds that d is proportional to $S^{1/n}$. The value of n is.

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (C)

3

Answer 2

The distance $d$ is assumed to depend on the mass density $\rho$ of the fog, the intensity $S$ of the light, and its frequency $f$. We can write this relationship in the form:

$d = C S^a \rho^b f^c$

where $C$ is a dimensionless constant and $a$, $b$, and $c$ are exponents.

First, we write down the dimensions of each physical quantity involved:

• Distance, $d$: $[L]$

• Mass density, $\rho$: Mass per unit volume. $[\rho] = [M]/[L]^3 = [ML^{-3}]$

• Intensity, $S$: Power per unit area. Power is Energy per unit time, and Energy is Force times distance. Force is mass times acceleration. $[Force] = [M][LT^{-2}] = [MLT^{-2}]$ $[Energy] = [ML^2T^{-2}]$ $[Power] = [ML^2T^{-2}]/[T] = [ML^2T^{-3}]$ $[Area] = [L^2]$ $[Intensity], [S] = [Power]/[Area] = [ML^2T^{-3}]/[L^2] = [MT^{-3}]$

• Frequency, $f$: 1/Period. $[f] = [T^{-1}]$

Now substitute the dimensions into the assumed relationship $d = C S^a \rho^b f^c$:

$[L] = [MT^{-3}]^a [ML^{-3}]^b [T^{-1}]^c$

$[L^1 M^0 T^0] = [M^a T^{-3a}] [M^b L^{-3b}] [T^{-c}]$

Combine the powers of the fundamental dimensions (M, L, T) on the right side:

$[L^1 M^0 T^0] = [M^{a+b} L^{-3b} T^{-3a-c}]$

Equate the powers of M, L, and T on both sides of the equation:

For M: $0 = a + b$

For L: $1 = -3b$

For T: $0 = -3a - c$

Now solve the system of linear equations for $a$, $b$, and $c$:

From the equation for L: $1 = -3b \implies b = -\frac{1}{3}$.

From the equation for M: $0 = a + b \implies a = -b = -(-\frac{1}{3}) = \frac{1}{3}$.

From the equation for T: $0 = -3a - c \implies c = -3a = -3(\frac{1}{3}) = -1$.

So the relationship is $d = C S^{1/3} \rho^{-1/3} f^{-1}$.

The problem states that the engineer finds that $d$ is proportional to $S^{1/n}$.

From our derived relationship, $d \propto S^{1/3} \rho^{-1/3} f^{-1}$.

The proportionality to $S$ is given by the term $S^{1/3}$.

Comparing this with $S^{1/n}$, we have:

$S^{1/n} = S^{1/3}$

This implies that the exponents must be equal:

$\frac{1}{n} = \frac{1}{3}$

$n = 3$

The value of $n$ is 3.