According to Wien's displacement law, $\lambda_{max} \propto \frac{1}{T}$. The wavelength of red is greater than yellow, which is greater than blue ($\lambda_B > \lambda_C > \lambda_A$). Therefore, the temperatures will be in the reverse order: $T_A > T_C > T_B$.

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).

From Wien's displacement law, $\lambda_m T = \text{constant}$. $\lambda_1 T_1 = \lambda_2 T_2 \Rightarrow 5000 \times (1227+273) = \lambda_2 \times (1227+1000+273) \Rightarrow \lambda_2 = \frac{5000 \times 1500}{2500} = 3000\text{ \AA}$.

The temperature of the human body is about $310 \text{ K}$. According to Wien's law, the maximum emission wavelength is around $9.3 \mu\text{m}$, which falls in the infrared region.

Wien's displacement law states that the wavelength $\lambda_m$ corresponding to maximum spectral emissive power of a black body is inversely proportional to its absolute temperature $T$. So, it relates $\lambda_m$ and $T$.

Ferry's black body consists of a hollow double-walled sphere with a small opening. Radiation entering it suffers multiple reflections and gets absorbed. So a cavity maintained at constant temperature is the closest to an ideal black body.