Thermal Physics - NEET Physics Questions
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Thermal Physics

Practice Questions:
Question 51: moderate

Match list-I with list-II about effect of various factors on speed of sound in a gas.

\[\begin{array}{|l|l|} \hline
\text{List-I (Factor)} & \text{List-II (Speed)} \\ \hline
\text{a. With increase in temperature} & \text{(ii) Increases} \\ \hline
\text{b. With increase in molecular weight of gas (temp const)} & \text{(iii) Decreases} \\ \hline
\text{c. With increase in pressure at constant temperature} & \text{(i) Same} \\ \hline
\end{array}\]

 

1. a(ii), b(i), c(iii)
2. a(ii), b(iii), c(i)
3. a(iii), b(ii), c(i)
4. a(i), b(ii), c(ii)
View Answer

Speed of sound is \(v = \sqrt{\frac{\gamma RT}{M}}\). Temperature increase increases speed, molecular weight increase decreases speed, and pressure has no effect at constant temperature.

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Question 52: moderate

In a thermodynamic process, pressure (in Pa) varies as, \(P = a + bV\) [where \(V\) represents volume in \(\text{m}^3\), \(a\) and \(b\) are constants]. The work done by gas during expansion from \(2 \text{m}^3\) to \(3 \text{m}^3\) will be

1. \(\left(\frac{3a}{2} + b \right) \text{J}\)
2. \(\left(a + \frac{5b}{2}\right) \text{J}\)
3. \(\left(b + \frac{5a}{2}\right) \text{J}\)
4. Zero
View Answer

Work done \(W = \int_{V_1}^{V_2} P dV = \int_2^3 (a+bV) dV = \left[aV + \frac{bV^2}{2}\right]_2^3 = a(3-2) + \frac{b}{2}(9-4) = a + \frac{5b}{2}\).

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Question 53: moderate

Four particles have velocities 1, 0, 2 and 3 m/s. The root mean square velocity of the particles is

1. \(\sqrt{3.5} \text{m/s}\)
2. \(\sqrt{5.3} \text{m/s}\)
3. \(\sqrt{2.8} \text{m/s}\)
4. \(\sqrt{4.7} \text{m/s}\)
View Answer

The RMS velocity is given by \[v_{\text{rms}} = \sqrt{\frac{v_1^2 + v_2^2 + v_3^2 + v_4^2}{4}}\]. Substituting the values, \[v_{text{rms}} = \sqrt{\frac{1^2 + 0^2 + 2^2 + 3^2}{4}} = \sqrt{\frac{14}{4}} = \sqrt{3.5} \text{m/s}\].

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Question 54: moderate

Consider the following statements:

A. The temperature at which the liquid and solid state of substance is in thermal equilibrium with each other is called its melting point.

B. The temperature at which the liquid and the vapour states of the substance coexist is called its boiling point.

C. The change from solid state to vapour state without passing through liquid state is called sublimation.

Based on above information pick correct option.

1. Only A and B correct
2. Only C and B are correct
3. Only C is incorrect
4. All statements A, B and C are correct
View Answer

All statements correctly define physical principles: statement A defines melting point, statement B defines boiling point, and statement C defines sublimation.

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Question 55: moderate

The root mean square speed of \(\text{H}_2\) molecules contained in a vessel is \(300\text{ m/s}\). If half of the gas leaks out at constant temperature, then the rms speed of the remaining molecules in the vessel will be

1. \(150\text{ m/s}\)
2. \(600\text{ m/s}\)
3. \(300\text{ m/s}\)
4. \(900\text{ m/s}\)
View Answer

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 \(300\text{ m/s}\).

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Question 56: moderate

A faulty thermometer shows \(40^\circ{C}\) at ice point and \(80^{\circ}{C}\) 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. \(75^{\circ} C\)
View Answer

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 $T = \frac{200}{3}\,^\circ\text{C}$.

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Question 57: moderate

A blackbody and a real body of identical dimensions are heated to same temperature. If ratio of rates of radiation of the blackbody and the real body is \(4 : 3\), then emissivity of the real body is equal to

1. \(0.25\)
2. \(0.50\)
3. \(0.75\)
4. \(0.67\)
View Answer

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 \(e = \frac{3}{4} = 0.75\).

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Question 58: moderate

Water equivalent of a metallic bar is \(100\text{ g}\), the energy required to increase its temperature by \(1^\circ\text{C}\) is

1. \(100\text{ cal}\)
2. \(1000\text{ cal}\)
3. \(500\text{ cal}\)
4. \(50\text{ cal}\)
View Answer

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.

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Question 59: moderate

A gas mixture consists of \(2\) moles of \(\text{O}_2\) and \(4\) moles of \(\text{He}\) at temperature \(T\). Neglecting all vibrational modes, total internal energy of the system is

1. \(11\text{ }RT\)
2. \(13 RT\)
3. \(15 RT\)
4. \(9 RT\)
View Answer

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, \(U = 2 \left(\frac{5}{2}\right) RT + 4 \left(\frac{3}{2}\right) RT = 5RT + 6RT = 11RT\).

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Question 60: moderate

A body cools from \(80^\circ\text{ C}\) to \(70^\circ\text{ C}\) in \(12\text{minutes}\) and from \(70^\circ\text{ C}\) to \(60^\circ\text{ C}\) in \(t \text{ minutes}\). The value of \(t\) is (temperature of surrounding is \(40^\circ\text{ C}\).

1. \(13.8\)
2. \(16.8\)
3. \(18.8\)
4. \(20.8\)
View Answer

According to Newton's Law of Cooling, \(\frac{T_i - T_f}{t} = K\left(\frac{T_i + T_f}{2} - T_s\right)\). For the first interval, \(\frac{80-70}{12} = K(75-40) ⇒ K = \frac{1}{42}\). For the second interval, \(\frac{70-60}{t} = K(65-40) ⇒ \frac{10}{t} = \frac{25}{42}\), which gives \(t = 16.8\text{ minutes}\).

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