Question 61:
moderateA solid cube of side \(4\text{ m}\) having coefficient of areal expansion \(2 \times 10^{-5}/^\circ\text{C}\). If temperature is changed by \(40^\circ\text{C}\) then the change in side length of the cube will be
Coefficient of linear expansion is \(\alpha = \frac{\beta}{2} = 1 \times 10^{-5}/^\circ\text{C}\). Change in length \(\Delta L = L \alpha \Delta T = 4 \times (1 \times 10^{-5}) \times 40 = 1.6 \times 10^{-3}\text{ m} = 1.6\text{ mm}\).
Question 62:
moderateThermal capacity of \(40\text{ g}\) of aluminium of specific heat \(0.2\text{ cal/(g }^\{circ}\text{C)}\) is
Thermal capacity is given by the formula \(C = m \cdot s\). Substituting the values, \(C = 40\text{ g} \times 0.2\text{ cal/(g }^\circ\text{C)} = 8\text{ cal/}^\circ\text{C}\.
Question 63:
moderateThe equation of state for 14 g nitrogen gas at a pressure P and temperature T, when occupying a volume V will be
Number of moles \(n = \frac{m}{M} = \frac{14\text{ g}}{28\text{ g/mol}} = 0.5\text{ mol}\). Substituting \(n = \frac{1}{2}\) in \(PV = nRT\) gives \(PV = \frac{1}{2}RT\).
Question 64:
moderate| Column I | Column II |
| (A) Total translational kinetic energy | (P) $\frac{5}{2} K_B T$ |
| (B) Total rotational kinetic energy | (Q) $nRT$ |
| (C) Total kinetic energy per mole | (R) $\frac{3}{2} nRT$ |
| (D) Total kinetic energy per molecule | (S) $\frac{5}{2} RT$ |
Translational KE is \(\frac{3}{2}nRT\). Rotational KE of a rigid diatomic gas is \(nRT\). Total KE per mole is \(\frac{5}{2}RT\) and total KE per molecule is \(\frac{5}{2}K_B T\).
Question 65:
moderateConsider a sample of \(n\) moles of rigid diatomic gas. Match the columns and tick the correct option (symbols have their usual meanings):
| Column I | Column II |
| (A) Total translational kinetic energy | (P) $\frac{5}{2}K_B T$ |
| (B) Total rotational kinetic energy | (Q) $nRT$ |
| (C) Total kinetic energy per mole | (R) $\frac{3}{2}nRT$ |
| (D) Total kinetic energy per molecule | (S) $\frac{5}{2}RT$ |
Translational KE of \(n\) moles is \(\frac{3}{2}nRT\) (R). Rotational KE is \(nRT\) (Q). KE per mole of diatomic gas is \(\frac{5}{2}RT\) (S). KE per molecule is \(\frac{5}{2}k_B T\) (P). Thus, (A)-(R), (B)-(Q), (C)-(S), (D)-(P).
Question 66:
moderateMatch the columns and tick the correct option. (Symbols have their usual meanings)
\begin{array}{|l|l|}
\hline
\textbf{Column-I} & \textbf{Column-II} \\ \hline
\text{a. } \gamma = \frac{5}{3} & \text{(i) Diatomic gas} \\ \hline
\text{b. } C_v = \frac{5}{2}R & \text{(ii) Triatomic non-linear gas} \\ \hline
\text{c. } C_v = 3R & \text{(iii) Monoatomic gas} \\ \hline
\end{array}
Monoatomic gas has \(\gamma = 5/3\) (a-iii). Diatomic gas has \(C_v = 5/2 R\) (b-i). Triatomic non-linear gas has \(C_v = 3R\) (c-ii). Thus, the correct matching is a(iii), b(i), c(ii).
Question 67:
moderateFour moles of helium are mixed with two moles of oxygen. The molar specific heat capacity of the mixture at constant volume is
For helium (monoatomic), \(C_{v1} = \frac{3}{2}R\) and \(n_1 = 4\). For oxygen (diatomic), \(C_{v2} = \frac{5}{2}R\) and \(n_2 = 2\). The mixture molar specific heat is \(C_{v,\text{mix}} = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2} = \frac{4 \times 1.5R + 2 \times 2.5R}{4 + 2} = \frac{11R}{6}\).
Question 68:
moderateIn thermodynamic processes, correct match of column-I with column-II is:
| Column-I (Type of process) | Column-II (Feature) |
| a. Isothermal | (iv) Temperature constant |
| b. Isobaric | (ii) Pressure constant |
| c. Isochoric | (i) Volume constant |
| d. Adiabatic | (iii) No heat flow between system and surroundings |
Isothermal process has constant temperature (a-iv). Isobaric has constant pressure (b-ii). Isochoric has constant volume (c-i). Adiabatic has no heat flow (d-iii). Matching these gives option D.
Question 69:
moderate\(15\text{ gm}\) of ice at \(0^\circ\text{C}\) is mixed with \(300\text{ gm}\) of water at \(50^\circ\text{C}\) in a container. There is no heat loss due to radiation and water equivalent of container is ignored. What will be final temperature of water?
Heat absorbed to melt ice: \(Q_1 = 15 \times 80 = 1200\text{ cal}\). Let final temperature be \(T\). Heat gained by melted ice: \(15 T\). Heat lost by hot water: \(300(50 - T)\). Equilibrium: \(1200 + 15T = 300(50-T) \implies 315T = 13800 \implies T \approx 43.8^\circ\text{C}\).
Question 70:
moderateMercury thermometer can be used to measure temperature upto: (1992)
The boiling point of mercury is approximately $356.7^\circ\text{C}$. Therefore, a standard mercury thermometer can measure temperatures up to around $360^\circ\text{C}$.