A block body at $1227^{\circ}\text{C}$ emits radiations with maximum intensity at a wavelength of $5000\text{ \AA}$. If the temperature of the body is increased by $1000^{\circ}\text{C}$, the maximum intensity will be observed at (2006)
We consider the radiation emitted by the human, body. Which of the following statements is true: (2003)
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.
The Wien’s displacement law express relation between (2002)
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 $T$.
Which of the following is best close to an ideal black body: (2002)
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.
For a black body at temperature $727^{\circ}\text{C}$, its radiating power is $60 \text{ watt}$ and temperature of surrounding is $227^{\circ}\text{C}$. If temperature of black body is changed to $1227^{\circ}\text{C}$ then its radiating power will be: (2002)
From Stefan's law, $E = \sigma T^4$. The unit of emissive power $E$ is $\text{J}/(\text{s m}^2)$ or $\text{W}/\text{m}^2$. Therefore, the unit of $\sigma$ is $\text{W}/(\text{m}^2 \text{K}^4)$.
A black body has wavelength $\lambda_m$ corresponding to maximum energy at $2000 \text{ K}$. Its wavelength corresponding to maximum energy at $3000 \text{ K}$ will be: (2001)
A sphere maintained at temperature $600 \text{ K}$, has cooling rate $R$ in an external environment of $200 \text{ K}$ temperature. If its temperature, falls to $400 \text{ K}$ then its cooling rate will be: (1999)
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