Thermal Physics - NEET Physics Questions
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Thermal Physics

Question 211: easy

If \(M\) is the molar mass of a gas, then the average speed of its molecules at temperature \(T\) is

1. \(\sqrt{\frac{3RT}{\pi M}}\)
2. \(\sqrt{\frac{8RT}{\pi M}}\)
3. \(\sqrt{\frac{2RT}{M}}\)
4. \(\sqrt{\frac{3RT}{M}}\)
View Answer

According to Maxwell-Boltzmann distribution, the average speed of molecules of an ideal gas is given by the expression \(v_{\text{avg}} = \sqrt{\frac{8RT}{\pi M}}\).

Question 212: easy

If transmittance of a surface is \(\frac{1}{7}\), reflectance is \(\frac{1}{8}\), then the absorptance of the surface will be

1. \(\frac{1}{9}\)
2. \(\frac{15}{56}\)
3. \(\frac{9}{56}\)
4. \(\frac{41}{56}\)
View Answer

By conservation of energy, the sum of absorptance \(a\), reflectance \(r\), and transmittance \(t\) is equal to \(1\): \(a + r + t = 1\). Therefore, \(a = 1 - \frac{1}{8} - \frac{1}{7} = 1 - \frac{15}{56} = \frac{41}{56}\).

Question 213: easy

Statement I: Internal energy of an ideal gas remains constant in an adiabatic process.


Statement II: In an adiabatic process, change in internal energy of a gas is equal to work done on or by the gas in the process.

1. Statement I is correct and statement II is incorrect
2. Statement I is incorrect and statement II is correct
3. Both statements are correct
4. Both statements are incorrect
View Answer

In an adiabatic process, \(Q = 0\), so \(\Delta U = -W\), which means internal energy changes, so Statement I is incorrect. Statement II is correct since change in internal energy corresponds directly to the work done on or by the gas.

Question 214: easy

Temperature of a body rises by \(2^{\circ}C\), the corresponding temperature rise in Kelvin will be

1. \(273.15\text{ K}\)
2. \(4\text{ K}\)
3. \(2\text{ K}\)
4. \(260\text{ K}\)
View Answer

The change in temperature on the Celsius scale is equal to the change in temperature on the Kelvin scale because the size of one degree Celsius is equal to one Kelvin: \(\Delta T_{text{C}} = \Delta T_{\text{K}} = 2\text{ K}\).

Question 215: easy

The physical quantity that determines whether or not the given system A is in thermal equilibrium with another system B is called

1. Work
2. Heat
3. Pressure
4. Temperature
View Answer

According to the Zeroth Law of Thermodynamics, temperature is the physical quantity that determines if systems are in thermal equilibrium with each other.

Question 216: easy

The physical quantity that determines whether or not the given system A is in thermal equilibrium with another system B is called

1. Work
2. Heat
3. Pressure
4. Temperature
View Answer

Temperature is the physical quantity that determines thermal equilibrium. Two systems are in thermal equilibrium if and only if they are at the same temperature.

Question 217: easy

On increasing the number density for a gas in a vessel, mean free path of the gas will

1. Decrease
2. Increase
3. Remain same
4. Become double
View Answer

The mean free path is given by \(\lambda = \frac{1}{\sqrt{2} n \pi d^2}\). Since \(\lambda\) is inversely proportional to the number density \(n\), increasing the number density decreases the mean free path.

Question 218: easy

Degrees of freedom of a rigid diatomic molecule is

1. 3
2. 5
3. 6
4. 7
View Answer

A rigid diatomic molecule has 3 translational and 2 rotational degrees of freedom, giving a total of 5 degrees of freedom.

Question 219: easy

The unit of emissive power is

1. \(\text{J m}^{-2}\)
2. \(\text{W s}^{-1}\)
3. \(\text{J m}^{-2} \text{s}^{-1}\)
4. \(\text{W m}^{2} \text{s}^{-1}\)
View Answer

Emissive power is defined as the thermal energy radiated per unit area per unit time. Its SI unit is \(\text{J m}^{-2}\text{s}^{-1}\) (or \(\text{W m}^{-2}\)).

Question 220: easy

A faulty thermometer shows $40^\circ\text{C}$ at ice point and $80^\circ\text{C}$ at steam point. The temperature at which its reading would be correct is

1. $\frac{100}{3} ^\circ\text{C}$
2. $\frac{200}{3} ^\circ\text{C}$
3. $60^\circ\text{C}$
4. $75^\circ\text{C}$
View Answer

Using the relation $\frac{T - \text{LFP}}{\text{UFP} - \text{LFP}} = \frac{C - 0}{100 - 0}$, we substitute $T = C$ for correct reading. This gives $\frac{C - 40}{80 - 40} = \frac{C}{100}$, which simplifies to $C = \frac{200}{3} ^\circ\text{C}$.