NEET Thermal Physics MCQs & PYQs

Using the relation $\frac{T - \text{LFP}}{\text{UFP} - \text{LFP}} = \text{constant}$, we write $\frac{T - 0}{100 - 0} = \frac{T - 40}{80 - 40}$. This simplifies to $\frac{T}{100} = \frac{T - 40}{40}$, which gives $40T = 100T - 4000$ or $60T = 4000$, hence $T = \frac{200}{3}\,^\circ\text{C}$.

The rate of radiation of a blackbody is \(E_b = \sigma A T^4\) and for a real body is \(E = e \sigma A T^4\). Given \(\frac{E_b}{E} = \frac{4}{3}\), we have \(\frac{1}{e} = \frac{4}{3}\), which gives emissivity \(e = \frac{3}{4} = 0.75\).

Heat required is given by \(Q = W \cdot c \cdot \Delta T\), where \(W\) is the water equivalent. Here, \(Q = 100\text{ g} \times 1\text{ cal/(g }^\circ\text{C)} \times 1^\circ\text{C} = 100\text{ cal}\) of energy.

According to the first law of thermodynamics, \(Q = \Delta U + W\). For an isobaric process, the work done is \(W = P\Delta V = nR\Delta T\). Therefore, both equations (1) and (2) are correct representation.

Heat and work depend on the path taken by the system during a thermodynamic process, making them path functions. Internal energy is a state function as it depends only on the initial and final states of the system.

The maximum efficiency is given by the Carnot efficiency formula: \(\eta = 1 - \frac{T_2}{T_1}\). Here, \(T_1 = 600\text{ K}\) and \(T_2 = 200\text{ K}\), so \(\eta = 1 -\frac{200}{600} = 1 - \frac{1}{3} = \frac{2}{3} \approx 66.67%\).

One of the postulates of the kinetic theory of gases is that gas molecules collide elastically with each other as well as with the walls of the container. Hence, the assumption that there are no collisions among molecules is incorrect.

Internal energy \(U = n_1 \frac{f_1}{2} RT + n_2 \frac{f_2}{2} RT\). For diatomic \(\text{O}_2\), \(f_1 = 5\), and for monoatomic \(\text{He}\), \(f_2 = 3\). Thus, \(U = 2 \left(\frac{5}{2}\right) RT + 4 \left(\frac{3}{2}\right) RT = 5RT + 6RT = 11RT\).

The root mean square speed of molecules is given by \(v_{\text{rms}} = \sqrt{\frac{3RT}{M}}\). Since the temperature \(T\) and molecular mass \(M\) of the remaining gas remain constant, the rms speed does not change and remains \(300\text{ m/s}\).

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