Properties Of Matter – NEET UG PhysicsPractice Questions & PYQs

The Stefan-Boltzmann law states that the heat radiated per unit surface area of a black body is directly proportional to the fourth power of its absolute temperature, expressed as $E \propto T^4$. To apply this relationship, we first convert the given temperatures to Kelvin by adding 273, resulting in initial and final absolute temperatures of 500 K and 1000 K, respectively. Since the temperature doubles from 500 K to 1000 K, the rate of heat radiation increases by a factor of $2^4$, which equals 16. Multiplying the initial radiation rate of 7 cal/cm²s by this factor of 16 yields a final rate of 112 cal/cm²s.

Because the W scale is linear, the change in temperature maintains a constant proportion relative to the Celsius scale. Between the freezing and boiling points of water, the temperature rises by 100 degrees Celsius, while on the W scale, it increases by $239 - 39 = 200$ degrees W. This defines a relationship where every Celsius degree corresponds to a change of $\frac{200}{100} = 2$ units on the W scale. Finding the equivalent temperature for 39 degrees Celsius requires adding this scaled value to the freezing point, calculated as $39 + (39 \times 2) = 117$ degrees W.

According to Stefan's law, the rate at which a black body radiates energy is directly proportional to the fourth power of its absolute temperature, which is expressed as $E \propto T^4$. Because this law requires temperature in Kelvin, we convert the given 727 degrees Celsius by adding 273, resulting in 1000 Kelvin. Substituting this absolute temperature into the proportionality relationship demonstrates that the rate of energy emission is proportional to $1000^4$.

Wien's displacement law establishes that the wavelength of maximum intensity emitted by a black body is inversely proportional to its absolute temperature, which leads to the relationship $\lambda_1 T_1 = \lambda_2 T_2$. First, convert the temperatures from Celsius to Kelvin by adding 273, resulting in an initial absolute temperature of $1500\text{ K}$ and a final absolute temperature of $2500\text{ K}$. Using the rearranged formula $\lambda_2 = \frac{\lambda_1 T_1}{T_2}$, substitute the known values to get $\frac{5000 \times 1500}{2500}$. This calculation yields a final wavelength of $3000\,\text{Å}$.

The amount of heat conducted through a rod is defined by the relationship $Q = \frac{K A (T_1 - T_2) t}{l}$, indicating that for a fixed material and temperature difference, efficiency depends entirely on the geometric ratio $\frac{A}{l}$. By substituting the cross-sectional area formula $A = \pi r^2$, we see that the heat flow is proportional to $\frac{r^2}{l}$. Among the given choices, the rod with $r=2r_0$ and $l=l_0$ provides the largest value of this ratio, specifically $\frac{4r_0^2}{l_0}$, ensuring it transfers heat more rapidly than any other configuration.

The behavior of black body radiation is described by Wien's displacement law, which reveals an inverse relationship between the temperature of a body and the wavelength at which its emission reaches a maximum. Mathematically, this relationship is denoted by $\lambda_m T = b$, where $b$ is the Wien's displacement constant, which has a value of approximately $2.898 \times 10^{-3}$ m·K. Because the product of the peak wavelength and temperature remains constant, we find that $\lambda_m \propto T^{-1}$. Consequently, increasing the temperature of a black body shifts the peak radiation toward shorter, higher-energy wavelengths, while a decrease in temperature shifts the peak toward longer, lower-energy wavelengths.

When materials are arranged in series, the total thermal resistance of the composite slab is the sum of the individual thermal resistances. Since each component has a thickness of $L$, the composite slab has a total thickness of $2L$ and an equivalent thermal conductivity $K_{eq}$, which corresponds to a total thermal resistance of $\frac{2L}{K_{eq}A}$. Setting this total equal to the sum of the individual resistances for the two slabs gives the equation $\frac{2L}{K_{eq}A} = \frac{L}{KA} + \frac{L}{2KA}$. By cancelling the common factors of $L$ and $A$ from each term, the expression simplifies to $\frac{2}{K_{eq}} = \frac{1}{K} + \frac{1}{2K}$, which becomes $\frac{2}{K_{eq}} = \frac{3}{2K}$ when the right side is added. Solving for $K_{eq}$ yields the final equivalent thermal conductivity of $\frac{4}{3} K$.

Stefan's constant is defined by the Stefan-Boltzmann law, which states that the total power radiated by a black body is directly proportional to its surface area and the fourth power of its absolute temperature. This relationship is expressed as $P = \sigma A T^4$, where $P$ is power, $A$ is surface area, and $T$ is absolute temperature. By rearranging this equation to isolate the constant as $\sigma = \frac{P}{A T^4}$, it becomes clear that the unit is watt/m^2 K^4.

Wien's displacement law establishes a fundamental inverse relationship between the absolute temperature of a black body and the specific wavelength at which it radiates the greatest amount of energy. This principle is mathematically expressed as $\lambda_{max} T = \text{constant}$, signifying that as an object becomes hotter, its peak radiation emission shifts toward shorter wavelengths. By defining this inverse proportionality, the law specifically correlates the wavelength corresponding to maximum energy with the temperature of the source.

The net heat radiated by a black body is determined by the Stefan-Boltzmann law, which indicates that the power output is proportional to the difference between the fourth powers of the body's absolute temperature and the surrounding environment's temperature. The relationship is expressed as $P = \sigma A(T^4 - T_0^4)$, where $T$ is the temperature of the black body and $T_0$ is the temperature of the surroundings.

Converting the temperatures to Kelvin, the initial state corresponds to $T = 1000\text{ K}$ and $T_0 = 500\text{ K}$, with a power of $60\text{ W}$. When the black body is heated to $1227^{\circ}\text{C}$, the new temperature becomes $1500\text{ K}$. By comparing the second state to the first using the ratio $\frac{P'}{60} = \frac{1500^4 - 500^4}{1000^4 - 500^4}$, we can eliminate the constants $\sigma$ and $A$. Factoring out $500^4$ simplifies the expression to $\frac{3^4 - 1^4}{2^4 - 1^4}$, which is $\frac{81 - 1}{16 - 1}$, resulting in a fraction of $\frac{80}{15}$. Multiplying this ratio by the initial $60\text{ W}$ yields $P' = 60 \times \frac{16}{3}$, which equals $320\text{ W}$.

An ideal black body is strictly defined as a surface that absorbs all incoming radiation, allowing for no reflection or transmission. While materials such as a black lamp absorb roughly 96% of light, they fail to achieve complete absorption, whereas a cavity featuring a small aperture forces radiation to undergo numerous internal reflections. Because each subsequent reflection within the enclosure further absorbs energy, these multiple bounces effectively trap all incoming radiation, making the cavity a significantly better approximation of an ideal black body than any individual substance.

The steady-state rate of heat transfer through a material by conduction is governed by the relation $Q = \frac{K A (T_1 - T_2)}{l}$, where $K$ represents thermal conductivity, $A$ is the cross-sectional area, and $(T_1 - T_2)$ is the temperature difference across length $l$. Since both rods possess identical lengths and operate between the same temperature endpoints, the variables $l$ and $(T_1 - T_2)$ remain consistent for both physical systems. Because the problem specifies that the rate of heat loss is equal for both rods, setting the heat flow equations equal to each other allows the common length and temperature terms to cancel out. This calculation reveals that the product of the thermal conductivity and cross-sectional area must be balanced between the two rods, resulting in the relationship $K_1 A_1 = K_2 A_2$. Notably, the specific heat capacity does not influence the rate of heat flow under steady-state conditions, as that parameter relates to energy storage during transient heating rather than the instantaneous rate of transfer through the rod.

The steady-state rate of heat flow through a cylindrical rod is determined by the formula
$Q = \frac{KA(T_{1}-T_{2})}{L}$
which indicates that the heat transfer rate is directly proportional to the cross-sectional area and inversely proportional to the length. When all linear dimensions are doubled, the radius doubles, increasing the cross-sectional area $A$ to $4A$, and the length $L$ becomes $2L$. Substituting these changes into the expression gives $Q_{2} = \frac{K(4A)(T_{1}-T_{2})}{2L}$. Simplifying this calculation confirms that the new rate is twice the original, resulting in $Q_{2} = 2Q_{1}$.

Wien's displacement law dictates that the wavelength at which a black body emits the maximum intensity of radiation is inversely proportional to its absolute temperature. This relationship is defined by the equation $\lambda_{m} T = \text{constant}$, which implies that for any change in temperature, the product of the peak wavelength and temperature remains unchanged. Given an initial temperature of $2000 \,\text{K}$ and a final temperature of $3000 \,\text{K}$, we relate the two states using the expression:

$\lambda_{m1} T_1 = \lambda_{m2} T_2$

Substituting the known values, where the initial wavelength is $\lambda_{m}$ at $2000 \,\text{K}$, we solve for the new wavelength by calculating $\lambda_{m2} = \lambda_{m} \times \frac{2000}{3000}$. Simplifying this expression results in a final wavelength of $\frac{2}{3} \lambda_{m}$.