From Wien's displacement law, $T \propto 1/\lambda_{max}$. Thus $T'/T = \lambda_0 / (3\lambda_0/4) = 4/3$. According to Stefan's law, Power $P \propto T^4$. So, $P'/P = (T'/T)^4 = (4/3)^4 = 256/81$.

Power radiated $P = \sigma A T^4 = \sigma (4\pi r^2) T^4 \Rightarrow P \propto r^2 T^4$. Given $r' = r/2$ and $T' = 2T$. Thus, $P' = P(1/2)^2 (2)^4 = P(1/4)(16) = 4P = 4 \times 450 = 1800\text{ W}$.

From Wien's displacement law, $\lambda_{max} = \frac{b}{T} = \frac{2.88 \times 10^6}{5760} = 500\text{ nm}$. This means maximum energy is radiated at $500\text{ nm}$. Therefore, the energy $U_2$ is maximum, so $U_2 > U_1$ and $U_2 > U_3$.

The wavelengths corresponding to maximum intensity are $\lambda_P < \lambda_R < \lambda_Q$ because violet has the shortest wavelength and red has the longest. From Wien's law, $T \propto 1/\lambda_{max}$. Thus, the temperatures are in the reverse order: $T_P > T_R > T_Q$.

As the temperature of the iron increases, the wavelength at which it emits maximum energy decreases, according to Wien's displacement law ($lambda_{max} T = text{constant}$). This causes the color to shift from longer wavelengths (red) to shorter wavelengths (yellow, then all visible mixing to white).

Convection involves the macroscopic movement of fluid which relies on density differences. These differences in density lead to buoyant forces, which require gravity to operate.

Using average form of Newton's law of cooling: $\frac{90-80}{t} = K(\frac{90+80}{2}-20) \Rightarrow \frac{10}{t} = K(65)$. For second case: $\frac{80-60}{t'} = K(\frac{80+60}{2}-20) \Rightarrow \frac{20}{t'} = K(50)$. Dividing the two equations yields $t' = \frac{13}{5}t$.

By Newton's law of cooling: $\frac{3T-2T}{10} = K(\frac{3T+2T}{2}-T) \Rightarrow \frac{T}{10} = K(1.5T)$. For next 10 mins: $\frac{2T-T'}{10} = K(\frac{2T+T'}{2}-T)$. Substituting $K = \frac{1}{15}$, we get $2T - T' = \frac{1}{15}(0.5T' + T) \times 10$. Solving gives $T' = \frac{3}{2}T$.

According to Wien's displacement law, $\lambda_{m1} T_1 = \lambda_{m2} T_2$. $\lambda_m (2000) = \lambda_{m2} (3000) \Rightarrow \lambda_{m2} = \frac{2000}{3000} \lambda_m = \frac{2}{3} \lambda_m$.

Cooling rate $\propto (T^4 - T_0^4)$. $\frac{R'}{R} = \frac{400^4 - 200^4}{600^4 - 200^4} = \frac{2^4 - 1^4}{3^4 - 1^4} = \frac{15}{80} = \frac{3}{16}$. Since $\frac{3}{16} = \frac{9}{48}$, the correct cooling rate would be $\frac{3}{16}R$. The closest option is incorrectly printed as 9/27, the answer is 3/16 R