Question 41:
Temperature of a body rises by $2^circtext{C}$, the corresponding temperature rise in Kelvin will be
Since the size of a degree on Celsius and Kelvin scales is the same, any change in temperature has equal numerical value in both scales: $Delta T_C = Delta T_K$.
Question 42:
The physical quantity that determines whether or not the given system $A$ is in thermal equilibrium with another system $B$ is called
By definition, temperature is the measure of hotness or coldness of a system and determines thermal equilibrium.
Question 43:
A gas mixture consists of $2text{ moles of }text{O}_2text{ and }4text{ moles of He}$ at temperature $T$. Neglecting all vibrational modes, total internal energy of the system is
Internal energy $U = sum n_i frac{f_i}{2} RT$. For diatomic $text{O}_2$, $f=5 implies U_1 = 5RT$. For monatomic $text{He}$, $f=3 implies U_2 = 6RT$. Total $U = 11RT$.
Question 44:
The root mean square speed of $text{H}_2$ molecules contained in a vessel is $300text{ m/s}$. If half of the gas leaks out at constant temperature, then the rms speed of the remaining molecules in the vessel will be
The rms speed is $v_{text{rms}} = sqrt{frac{3RT}{M}}$, which depends only on temperature and molecular weight. Since the temperature is constant, $v_{text{rms}}$ remains $300text{ m/s}$.
Question 45:
On increasing the number density for a gas in a vessel, mean free path of the gas will
Mean free path is given by the formula \(\lambda = \frac{1}{\sqrt{2}\pi d^2 n}\), where \(n\) is the number density. Thus, as \(n\) increases, \(\lambda\) decreases.
Question 46:
Equal masses of an ideal gas are sealed in two vessels one of pressure \(P_0\) and other of pressure \(2P_0\). If first vessel is at temperature of \(400\text{ K}\) and the other is at \(600\text{ K}\). Find the ratio of volume of two container.
From \(PV = nRT = \frac{m}{M}RT\), we get \(V \propto \frac{T}{P}\. Therefore, \(\frac{V_1}{V_2} = \frac{T_1}{T_2} \times \frac{P_2}{P_1} = \frac{400}{600} \times \frac{2P_0}{P_0} = \frac{4}{3}\).
Question 47:
Degrees of freedom of a rigid diatomic molecule is
A rigid diatomic molecule has 3 translational and 2 rotational degrees of freedom, giving a total of 5 degrees of freedom.
Question 48:
The amplitude of oscillations is reduced to half of its initial value of 40 cm due to damping by resistive force in time \(t_0\), then after time \(3t_0\), its amplitude will be
Damped amplitude is given by \(A = A_0 e^{-\gamma t}\). Here \(e^{-\gamma t_0} = 1/2\). For \(t = 3t_0\), \(A = A_0 (e^{-\gamma t_0})^3 = 40 \times (1/2)^3 = 5\text{ cm}\).
Question 49:
Match List-I with List-II.
**List-I (Complex)**
a. \([\text{Fe}(\text{CN})_6]^{3-}\)
b. \([\text{Co}(\text{H}_2\text{O})_6]^{3+}\)
c. \([\text{FeF}_6]^{3-}\)
d. \([\text{NiCl}_4]^{2-}\)
**List-II (Number of unpaired electrons)**
(i) Zero
(ii) 5
(iii) 2
(iv) 1
\([\text{Fe}(\text{CN})_6]^{3-}\) has strong field ligands and 1 unpaired electron. \([\text{Co}(\text{H}_2\text{O})_6]^{3+}\) is low spin with zero unpaired electrons. \([\text{FeF}_6]^{3-}\) has weak field ligands and 5 unpaired electrons, while \([\text{NiCl}_4]^{2-}\) has 2 unpaired electrons.
Question 50:
If a body A of mass M is thrown with velocity v at an angle of \(30°\) to the horizontal and another body B of the same mass is thrown with the same speed at an angle of \(60°\) to the horizontal, the ratio of horizontal range of A to B will be: (1992, 90)
Concept: Horizontal range for complementary angles.
Formula: \(R = (u^2 sin 2theta) / g\).
For A, \(theta_A = 30°\). For B, \(theta_B = 60°\).
Since \(theta_A + theta_B = 90°\), they are complementary angles. Range is same if initial speed is same. Thus, \(R_A : R_B = 1:1\).