Tungsten Filament Non Uniform Evaporation and Electrical Behavior

To determine the correct statements, let us analyze the physical and electrical behavior of the tungsten filament as it evaporates over time.

1. Temperature Distribution (Option A)

Consider a small section of the filament of length $dx$ and radius $r(x)$. The cross-sectional area is $A(x) = \pi r(x)^2$. Since the filament is connected in series, the same electric current $I$ flows through every section.

The electrical power dissipated as heat in this section of length $dx$ is: $dP_{\text{in}} = I^2 dR = I^2 \frac{\rho \, dx}{\pi r(x)^2}$

In steady state, assuming thermal radiation dominates, the power radiated away by the surface according to Stefan-Boltzmann law is: $dP_{\text{out}} = \sigma \cdot (2\pi r(x) \, dx) \cdot T(x)^4$

Equating $dP_{\text{in}}$ and $dP_{\text{out}}$: $I^2 \frac{\rho \, dx}{\pi r(x)^2} = 2\pi \sigma r(x) \, dx \, T(x)^4$

Solving for temperature $T(x)$: $T(x)^4 = \frac{I^2 \rho}{2\pi^2 \sigma r(x)^3} \implies T(x) \propto \frac{1}{r(x)^{3/4}}$

Since tungsten evaporates non-uniformly, $r(x)$ varies along the length of the filament. Consequently, the local temperature $T(x)$ is higher at thinner sections, making the temperature distribution across the filament non-uniform.
Therefore, Option A is incorrect.

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2. Resistance of Small Sections (Option B)

The resistance of a small section of length $dx$ is given by: $dR = \frac{\rho \, dx}{A(x)}$

As tungsten evaporates over time, the cross-sectional area $A(x)$ of the section decreases. Since $A(x)$ is in the denominator, the resistance $dR$ of any small section where evaporation occurs increases with time.
Therefore, Option B is incorrect.

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3. Frequency of Emitted Light before Snapping (Option C)

As time progresses, non-uniform evaporation creates a region where the filament becomes extremely thin (a "hot spot"). At this section, $r(x) \to 0$, leading to a sharp local increase in temperature $T$.

According to Wien's Displacement Law: $\lambda_{\text{max}} T = b \implies \lambda_{\text{max}} \propto \frac{1}{T}$

As the temperature $T$ at the thinnest section increases dramatically right before snapping, the peak wavelength $\lambda_{\text{max}}$ shifts toward smaller values, which corresponds to higher frequencies ($\nu = \frac{c}{\lambda}$). Thus, the filament emits more radiation in the higher frequency band right before it snaps.
Therefore, Option C is correct.

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4. Total Electrical Power Consumed (Option D)

The total resistance of the filament is given by integrating over its entire length: $R_{\text{total}} = \int_0^L \frac{\rho \, dx}{\pi r(x)^2}$

As evaporation reduces the cross-sectional area $A(x) = \pi r(x)^2$ along the filament over time, the total resistance $R_{\text{total}}$ of the filament increases.

Since the bulb operates under a constant voltage supply $V$, the total electrical power consumed is: $P = \frac{V^2}{R_{\text{total}}}$

Because $R_{\text{total}}$ increases toward the end of the bulb's lifespan, the overall power consumed $P$ decreases.
Therefore, Option D is correct.

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Conclusion

The correct options are C and D.