Step-by-step Derivations:

1. Identify the Governing Principle: The problem describes the diffusion of two different gases (vapors of X and Y) from opposite ends of a tube. The distance traveled by each gas before they meet is determined by their respective rates of diffusion. This scenario is governed by Graham's Law of Diffusion.

2. State Graham's Law of Diffusion: Graham's Law states that the rate of diffusion (r) of a gas is inversely proportional to the square root of its molar mass (M), provided temperature and pressure are constant. $r \propto \frac{1}{\sqrt{M}}$ For two gases, X and Y, the ratio of their diffusion rates is given by: $\frac{r_X}{r_Y} = \sqrt{\frac{M_Y}{M_X}}$

3. Relate Diffusion Rate to Distance and Time: The rate of diffusion can be expressed as the distance traveled (d) per unit time (t). Since the cotton plugs are placed simultaneously, the time (t) taken for the vapors to meet is the same for both X and Y. Let $d_X$ be the distance traveled by vapor X and $d_Y$ be the distance traveled by vapor Y.

   • Rate of diffusion of X: $r_X = d_X / t$
   • Rate of diffusion of Y: $r_Y = d_Y / t$

4. Set up the Equations based on the Problem Statement:

   • Molar mass of X, $M_X = 10$ g mol⁻¹
   • Molar mass of Y, $M_Y = 40$ g mol⁻¹
   • Total length of the tube, L = 24 cm.
   • The distance from the plug soaked in X where the product is first observed is d. So, $d_X = d$.
   • The two vapors meet inside the tube, so the sum of the distances they travel is equal to the length of the tube: $d_X + d_Y = L$ $d + d_Y = 24$ This means the distance traveled by vapor Y is $d_Y = 24 - d$.

5. Apply Graham's Law to the Specifics of the Problem: Substitute the expressions for rates into the Graham's Law equation: $\frac{d_X / t}{d_Y / t} = \sqrt{\frac{M_Y}{M_X}}$ The time t cancels out: $\frac{d_X}{d_Y} = \sqrt{\frac{M_Y}{M_X}}$

6. Substitute the Given Values and Solve for d: Substitute $d_X = d$, $d_Y = 24 - d$, $M_X = 10$, and $M_Y = 40$ into the equation: $\frac{d}{24 - d} = \sqrt{\frac{40}{10}}$ $\frac{d}{24 - d} = \sqrt{4}$ $\frac{d}{24 - d} = 2$ Now, solve for d: $d = 2(24 - d)$ $d = 48 - 2d$ $d + 2d = 48$ $3d = 48$ $d = \frac{48}{3}$ $d = 16 \text{ cm}$

7. Conclusion: The value of d is 16 cm. This means the vapours meet at a distance of 16 cm from the end with substance X. This corresponds to option C.