HINTS & SOLUTIONS

Answer: (b) 4 Explanation: In the number 20340, the trailing zero (0 at the end) without a decimal point is generally not considered significant according to standard rules. The significant figures are 2, 0, 3, 4. Therefore, there are 4 significant figures.

Answer: (b) $1~m/s$ Solution: Position $x = 5t - t^2$. Instantaneous velocity $v = \frac{dx}{dt} = \frac{d}{dt}(5t - t^2) = 5 - 2t$. At $t = 2\text{ s}$, $v = 5 - 2(2) = 5 - 4 = 1\text{ m/s}$. Speed is the magnitude, which is $1\text{ m/s}$.

Answer: (a) $0^{\circ}$ Explanation: Two vectors are defined as "equal" only if they have the exact same magnitude and point in the exact same direction. Since their directions are identical, the angle between them is $0^\circ$.

Answer: (a) Inertia of Motion. Explanation: When a moving bus stops suddenly, the lower part of the passenger's body (in contact with the bus) comes to rest. However, the upper part of the body tends to continue moving forward due to the inertia of motion, causing the passenger to fall forward.

Answer: (b) 150 J Solution: Work done $W = F \cdot d \cos\theta$. Since the object moves in the direction of the force, $\theta = 0^\circ$ and $\cos(0^\circ) = 1$. $W = 10\text{ N} \times 15\text{ m} = 150\text{ Joules}$.

Answer: (c) Radiation. Explanation: Radiation is the transfer of heat in the form of electromagnetic waves. Unlike conduction and convection, which require matter (solids, liquids, or gases) to propagate, radiation can travel through a perfect vacuum.

Answer: (b) Energy Explanation: The First Law of Thermodynamics ($\Delta Q = \Delta U + \Delta W$) is the thermodynamic adaptation of the universal law of conservation of energy. It states that energy can neither be created nor destroyed, only transferred as heat or work.

Answer: (a) $\frac{5}{3}$ Solution: A monoatomic gas has 3 translational degrees of freedom ($f = 3$). The ratio of specific heats $\gamma = \frac{C_p}{C_v} = 1 + \frac{2}{f} = 1 + \frac{2}{3} = \frac{5}{3}$.

Answer: (b) Decrease Explanation: The time period of a swing (acting like a pendulum) is $T = 2\pi\sqrt{\frac{l}{g}}$, where $l$ is the distance from the pivot to the center of gravity (CG) of the child. When the child stands up, their CG shifts upward, effectively decreasing the length $l$. Since $T \propto \sqrt{l}$, the time period decreases.

Answer: (b) Does not depend upon change in pressure. Explanation: The speed of sound in a gas is $v = \sqrt{\frac{\gamma P}{\rho}}$. At a constant temperature, any change in pressure ($P$) causes a proportional change in density ($\rho$), meaning the ratio $P/\rho$ remains strictly constant (Boyle's Law). Thus, velocity is independent of pressure changes.

Answer: (b) Both A and R are true and R is not the correct explanation for A. Explanation: Both statements are factually correct. However, the reason sound cannot travel in a vacuum is because it is a *mechanical wave* (requiring interacting particles to transfer energy), not simply because it is longitudinal. Light travels in a vacuum because it consists of self-propagating electromagnetic fields, irrespective of it being transverse. Therefore, R is not the correct explanation of A.

Answer: (a) Both A and R are true and R is correct explanation of A. Explanation: The hydrostatic pressure of a liquid column is given by $P = h\rho g$. Since the feet are at a lower elevation than the brain, the effective height ($h$) of the blood column supporting them is much greater. This greater fluid height creates a higher blood pressure at the feet.

Solution: Let the time period $T$ depend on the length of the pendulum $l$ and acceleration due to gravity $g$. We can write: $T \propto l^a g^b \implies T = k \cdot l^a \cdot g^b$ (where $k$ is a dimensionless constant). Writing the dimensional formulas for both sides: $[M^0 L^0 T^1] = [L]^a \cdot [L T^{-2}]^b$ $[M^0 L^0 T^1] = [L^{a+b} \cdot T^{-2b}]$ Applying the principle of homogeneity of dimensions, we equate the powers of $L$ and $T$: For T: $1 = -2b \implies b = -1/2$. For L: $0 = a + b \implies a - 1/2 = 0 \implies a = 1/2$. Substituting the values of $a$ and $b$ back into the first equation: $T = k \cdot l^{1/2} \cdot g^{-1/2} = k \sqrt{\frac{l}{g}}$. (Experimentally, the constant $k = 2\pi$, so the formula is $T = 2\pi\sqrt{\frac{l}{g}}$).

Solution: Instantaneous Velocity: It is defined as the velocity of an object at a specific, exact instant of time or at a specific point in its path.
How it is calculated: It is mathematically calculated by taking the limit of the average velocity $\left(\frac{\Delta x}{\Delta t}\right)$ as the time interval ($\Delta t$) approaches zero. This is equivalent to the first derivative of the position ($x$) with respect to time ($t$). Formula: $\vec{v} = \lim_{\Delta t \to 0} \frac{\Delta \vec{x}}{\Delta t} = \frac{d\vec{x}}{dt}$.

Solution: Acceleration due to gravity ($g$): It is the uniform acceleration produced in a freely falling body due to the gravitational pull of the Earth.
Derivation: Consider a body of mass $m$ lying on the surface of the Earth. Let the mass of the Earth be $M$ and its radius be $R$. According to Newton's Law of Universal Gravitation, the force of attraction between the Earth and the body is: $F = \frac{G M m}{R^2}$ --- (Equation 1) According to Newton's Second Law of Motion, the weight of the body (gravitational force) is: $F = mg$ --- (Equation 2) Equating both expressions for force: $mg = \frac{G M m}{R^2}$ $g = \frac{GM}{R^2}$. This is the required expression.

Solution: Principle of Superposition of Waves: It states that when two or more waves travel simultaneously through a medium, the resultant displacement of any particle of the medium at any instant is equal to the vector sum of the individual displacements produced by the constituent waves at that instant. ($\vec{y} = \vec{y}_1 + \vec{y}_2 + \dots + \vec{y}_n$). Solution (Alternative): Differences: 1. Direction of Vibration: In transverse waves, the particles of the medium vibrate perpendicular to the direction of wave propagation. In longitudinal waves, particles vibrate parallel to the direction of propagation. 2. Formation: Transverse waves travel in the form of alternate crests and troughs. Longitudinal waves travel in the form of alternate compressions and rarefactions. 3. Polarisation: Transverse waves can be polarised, whereas longitudinal waves cannot be polarised.

Solution: Let a projectile be fired with initial velocity $u$ at an angle $\theta$ with the horizontal ground. We can resolve the initial velocity into two rectangular components: Horizontal component: $u_x = u \cos\theta$ Vertical component: $u_y = u \sin\theta$
Maximum Height ($H$): At the highest point of the trajectory, the final vertical velocity $v_y = 0$. The vertical acceleration is $a_y = -g$. Using the equation of motion: $v_y^2 - u_y^2 = 2a_y H$. $0 - (u\sin\theta)^2 = 2(-g)H \implies -u^2\sin^2\theta = -2gH$. $H = \frac{u^2\sin^2\theta}{2g}$.

Horizontal Range ($R$): It is the horizontal distance covered during the entire Time of Flight ($T$). Time of flight $T = \frac{2u\sin\theta}{g}$. Since there is no horizontal acceleration ($a_x = 0$), horizontal velocity $u_x$ remains constant. $R = u_x \times T = (u\cos\theta) \times \left(\frac{2u\sin\theta}{g}\right)$. $R = \frac{u^2 (2\sin\theta\cos\theta)}{g}$. Using the trigonometric identity $\sin(2\theta) = 2\sin\theta\cos\theta$, we get: $R = \frac{u^2\sin(2\theta)}{g}$.

Solution: Kinetic Energy: It is the energy possessed by a body by virtue of its state of motion. It is measured by the amount of work the body can do before coming to rest.
Derivation: Consider a body of mass $m$ initially at rest ($u = 0$). Let a constant force $F$ be applied to it, producing an acceleration $a$ and causing a displacement $s$ until its velocity becomes $v$. The work done by the force is $W = F \times s$. According to Newton's second law, $F = ma$, so $W = ma \times s$. Using the third equation of motion: $v^2 - u^2 = 2as$. Since $u = 0$, we have $v^2 = 2as \implies as = \frac{v^2}{2}$. Substitute the value of '$as$' into the work equation: $W = m \times \left(\frac{v^2}{2}\right) = \frac{1}{2}mv^2$. This work done is stored in the body as its Kinetic Energy ($K$). Therefore, $K = \frac{1}{2}mv^2$.

Solution: Escape Velocity: The minimum initial velocity with which a body must be projected from the surface of a planet so that it just overcomes the gravitational pull of the planet and never returns.
Derivation: Let a planet have mass $M$ and radius $R$. An object of mass $m$ is on its surface. The gravitational potential energy of the object on the surface is $U = -\frac{GMm}{R}$. Let the required escape velocity be $v_e$. The kinetic energy imparted to it is $K = \frac{1}{2}mv_e^2$. At infinity, the object is free from the gravitational field, so its total energy is zero. By conservation of mechanical energy: Total Energy at surface = Total Energy at infinity. $K + U = 0 \implies \frac{1}{2}mv_e^2 + \left(-\frac{GMm}{R}\right) = 0$. $\frac{1}{2}mv_e^2 = \frac{GMm}{R} \implies v_e^2 = \frac{2GM}{R} \implies v_e = \sqrt{\frac{2GM}{R}}$. We know the relation between acceleration due to gravity $g$ and $G$ is $g = \frac{GM}{R^2} \implies GM = gR^2$. Substitute $GM$ in the escape velocity equation: $v_e = \sqrt{\frac{2(gR^2)}{R}} = \sqrt{2gR}$. Hence proved.

Solution: (a) Elastic limit: In the stress-strain curve, it is the maximum stress up to which a material completely regains its original dimensions (length, shape, volume) upon the removal of the deforming force. If stretched beyond this limit, the material acquires a permanent set.

(b) Ultimate tensile strength (UTS): It corresponds to the highest point on the stress-strain curve. It is the maximum possible stress a material can withstand before it begins to neck (narrow locally) and eventually fracture. Breaking stress is determined just after this point.

(c) Elastomers: Materials that can be elastically stretched to extremely large values of strain (sometimes over 1000%) without breaking. They do not strictly obey Hooke's Law for most of their extension (the stress-strain curve is largely non-linear). Example: Rubber, tissue of the aorta.

Solution: Thermal Expansion: It is the phenomenon of the increase in dimensions (length, area, or volume) of a body due to an increase in its temperature.
Derivation: Consider a solid cube of original side length $L$. Its original volume is $V = L^3$. If temperature increases by $\Delta T$, the new length becomes $L' = L(1 + \alpha\Delta T)$, where $\alpha$ is the coefficient of linear expansion. The new volume $V' = (L')^3 = [L(1 + \alpha\Delta T)]^3 = L^3(1 + \alpha\Delta T)^3$. Expanding using the binomial theorem: $V' = V[1 + 3\alpha\Delta T + 3(\alpha\Delta T)^2 + (\alpha\Delta T)^3]$. Since $\alpha$ is very small, higher powers of $\alpha$ can be neglected. $V' \approx V(1 + 3\alpha\Delta T) = V + V(3\alpha)\Delta T$. Change in volume $\Delta V = V' - V = V(3\alpha)\Delta T$. By definition, volume expansion $\Delta V = V\gamma\Delta T$ (where $\gamma$ is the coefficient of cubical expansion). Comparing both, we get $\gamma = 3\alpha \implies \alpha = \frac{\gamma}{3}$. Similarly, for area expansion of a face $A = L^2$, $\Delta A = A(2\alpha)\Delta T = A\beta\Delta T \implies \beta = 2\alpha \implies \alpha = \frac{\beta}{2}$. Therefore, $\alpha = \frac{\beta}{2} = \frac{\gamma}{3}$.

Solution: Isothermal process: A thermodynamic process in which the pressure and volume of a system change, but the temperature of the system remains perfectly constant ($\Delta T = 0$).
Essential conditions: (i) The walls of the container must be perfectly conducting so heat can easily exchange with the surroundings to maintain a constant temperature. (ii) The process of expansion or compression must be done very slowly to allow sufficient time for heat exchange.
Work done derivation: Consider 1 mole of an ideal gas expanding isothermally from volume $V_1$ to $V_2$ at constant temperature $T$. Small work done $dW = P \cdot dV$. Total work done $W = \int_{V_1}^{V_2} P \cdot dV$. From the ideal gas equation, $PV = RT \implies P = \frac{RT}{V}$. Substituting $P$: $W = \int_{V_1}^{V_2} \left(\frac{RT}{V}\right) dV = RT \int_{V_1}^{V_2} \frac{1}{V} dV$. $W = RT \left[ \log_e V \right]_{V_1}^{V_2} = RT (\log_e V_2 - \log_e V_1)$. $W = RT \log_e \left(\frac{V_2}{V_1}\right)$. Hence proved.

Solution: Simple Harmonic Motion (SHM): It is a special type of oscillatory motion in which the restoring force acting on a particle is directly proportional to its displacement from the mean position and is always directed towards that mean position ($F \propto -x$).
Motion of a Simple Pendulum: Consider a pendulum of length $l$ and bob mass $m$, displaced by a small angle $\theta$. The weight $mg$ acts vertically down. It has two components: $mg\cos\theta$ balances the tension, and $mg\sin\theta$ provides the restoring force towards the mean position. Restoring force $F = -mg\sin\theta$. For small angular displacements, $\sin\theta \approx \theta$ (in radians). From geometry, arc length $x = l\theta \implies \theta = x/l$. $F = -mg\left(\frac{x}{l}\right) = -\left(\frac{mg}{l}\right)x$. Since $m, g$, and $l$ are constant, $F \propto -x$. This strict proportionality proves that the pendulum's motion is SHM.
Time Period: In standard SHM, $F = -kx$. Comparing this with our equation, the force constant $k = \frac{mg}{l}$. The time period of SHM is $T = 2\pi\sqrt{\frac{m}{k}}$. Substituting $k$: $T = 2\pi\sqrt{\frac{m}{mg/l}} = 2\pi\sqrt{\frac{l}{g}}$.

Solution: (A) (b) $Kg~m^{2}$ Reasoning: Moment of Inertia $I = \sum mr^2$. The unit of mass ($m$) is kg, and distance squared ($r^2$) is m$^2$. Hence, the SI unit is $\text{kg m}^2$.

(B) (b) Mass Reasoning: The paragraph explicitly states, "M.I. plays the same role in rotational motion as the force plays in the translation motion" - wait, let's re-read the paragraph closely. "M.I. plays the same role in rotational motion as the mass plays in the translation motion". (The PDF likely meant mass, as M.I. is rotational inertia. Inertia in translation is Mass). Therefore, the rotational analogue of mass is Moment of Inertia.

(C) (b) $2~Kg~m^{2}$ Reasoning: The Moment of Inertia of a solid sphere about its diameter is $I = \frac{2}{5} M R^2$. Given $M = 5\text{ kg}$, $R = 1\text{ m}$. $I = \frac{2}{5} \times 5 \times (1)^2 = 2 \text{ kg m}^2$.

(D) (a) Distribution of Mass from axis of Rotation. Reasoning: The paragraph states "Its value depends upon the axis of rotation and the distribution of mass of the body". It is independent of kinematic variables like angular velocity or acceleration.

Solution: Need for Banking of Road: When a vehicle turns on a flat road, the necessary centripetal force is provided entirely by static friction between tires and the road. Friction is unreliable (e.g., wet or icy roads) and heavy reliance on it causes rapid tire wear. By raising the outer edge of the road (banking), a component of the vehicle's normal reaction contributes to the centripetal force, allowing safe, high-speed turns without solely depending on friction.
Expression for Maximum Velocity: Consider a vehicle of mass $m$ on a road banked at angle $\theta$ with radius $R$. The forces acting are: weight $mg$ (downward), normal reaction $N$ (perpendicular to road), and friction $f$ (down the incline). Resolve $N$: Vertical $= N\cos\theta$, Horizontal $= N\sin\theta$ (towards center). Resolve $f$: Vertical $= f\sin\theta$ (downward), Horizontal $= f\cos\theta$ (towards center). Balancing vertical forces: $N\cos\theta = mg + f\sin\theta \implies mg = N\cos\theta - f\sin\theta$. Centripetal force is provided by the net horizontal force: $\frac{mv^2}{R} = N\sin\theta + f\cos\theta$. Divide the centripetal equation by the vertical equation: $\frac{v^2}{Rg} = \frac{N\sin\theta + f\cos\theta}{N\cos\theta - f\sin\theta}$. For maximum safe speed $v_{max}$, friction reaches its limiting value $f = \mu_s N$. $\frac{v_{max}^2}{Rg} = \frac{N\sin\theta + \mu_s N\cos\theta}{N\cos\theta - \mu_s N\sin\theta} = \frac{\sin\theta + \mu_s\cos\theta}{\cos\theta - \mu_s\sin\theta}$. Dividing numerator and denominator by $\cos\theta$: $\frac{v_{max}^2}{Rg} = \frac{\tan\theta + \mu_s}{1 - \mu_s\tan\theta}$. $v_{max} = \sqrt{Rg \left(\frac{\mu_s + \tan\theta}{1 - \mu_s\tan\theta}\right)}$. Solution (Alternative): Newton's Second Law of Motion: The rate of change of linear momentum of a body is directly proportional to the applied unbalanced external force, and takes place in the direction of the force. ($\vec{F} = \frac{d\vec{p}}{dt} = m\vec{a}$).
Real Law of Motion (Proof): To show the 2nd law is the "real" law, we must prove it contains both the 1st and 3rd laws within it.
1. Second Law contains First Law: According to the 2nd law, $F = ma$. If no external force acts on the body, $F = 0$. Then $ma = 0$. Since mass $m \neq 0$, acceleration $a$ must be $0$. $a = 0$ means the body is either at rest or moving with uniform velocity. This is exactly the statement of the First Law (Law of Inertia). Thus, the 1st law is a special case of the 2nd law.
2. Second Law contains Third Law: Consider an isolated system of two bodies A and B colliding. Let force on A by B be $F_{AB}$ and force on B by A be $F_{BA}$. By the 2nd law, change in momentum of A is $\Delta p_A = F_{AB} \times \Delta t$, and for B is $\Delta p_B = F_{BA} \times \Delta t$. Since no external force acts on the isolated system, total momentum is conserved ($\Delta p_A + \Delta p_B = 0$). $F_{AB} \Delta t + F_{BA} \Delta t = 0 \implies (F_{AB} + F_{BA})\Delta t = 0$. Since $\Delta t \neq 0$, we get $F_{AB} = -F_{BA}$. This means action and reaction are equal and opposite, which is Newton's Third Law. Because the 2nd law encompasses both the 1st and 3rd laws, it is known as the real (fundamental) law of motion.

Solution: Bernoulli's Theorem: It states that for an ideal (incompressible, non-viscous) fluid undergoing steady, streamline flow, the sum of pressure energy, kinetic energy, and potential energy per unit volume remains constant at all cross-sections of the flow. $P + \frac{1}{2}\rho v^2 + \rho gh = \text{Constant}$.
Proof: Consider a tube of varying cross-section. Let fluid enter at end 1 (area $A_1$, height $h_1$, velocity $v_1$, pressure $P_1$) and leave at end 2 ($A_2, h_2, v_2, P_2$). Let density be $\rho$. In time $\Delta t$, mass of fluid passing is $m = \rho A_1 v_1 \Delta t = \rho A_2 v_2 \Delta t$. Volume $\Delta V = m/\rho$. Work done on the fluid at end 1 is $W_1 = P_1 A_1 (v_1 \Delta t) = P_1 \Delta V$. Work done by the fluid at end 2 is $W_2 = P_2 A_2 (v_2 \Delta t) = P_2 \Delta V$. Net work done $\Delta W = W_1 - W_2 = (P_1 - P_2) \Delta V$. This work changes the fluid's kinetic and potential energy. Change in K.E. ($\Delta K$) = $\frac{1}{2}m v_2^2 - \frac{1}{2}m v_1^2$. Change in P.E. ($\Delta U$) = $mgh_2 - mgh_1$. By Work-Energy Theorem: $\Delta W = \Delta K + \Delta U$. $(P_1 - P_2) \Delta V = (\frac{1}{2}m v_2^2 - \frac{1}{2}m v_1^2) + (mgh_2 - mgh_1)$. Divide entire equation by $\Delta V$ (since $m/\Delta V = \rho$): $P_1 - P_2 = \frac{1}{2}\rho v_2^2 - \frac{1}{2}\rho v_1^2 + \rho gh_2 - \rho gh_1$. Rearranging terms: $P_1 + \frac{1}{2}\rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho gh_2$. Hence proved.
Rapids in a river: No, Bernoulli's equation cannot be used to describe flow through rapids. Rapids exhibit highly turbulent and rotational flow with immense energy dissipation due to internal viscous friction, violating the core assumption of steady, non-viscous streamline flow. Solution (Alternative): Surface Tension: The property of a liquid at rest by virtue of which its free surface behaves like a stretched elastic membrane trying to contract to have minimum surface area. It is measured as force per unit length drawn on the surface ($T = F/l$).
Surface Energy: Molecules on a liquid surface have extra potential energy compared to bulk molecules because work must be done against inward cohesive forces to bring a molecule to the surface. The total extra energy possessed by surface molecules per unit area is called surface energy.
Relationship: Consider a rectangular wire frame with a sliding wire of length $L$. Dip it in soap solution to form a film. The film has two free surfaces, exerting an inward pull $F = T \times 2L$. To move the sliding wire outward by a distance $x$, external work must be done: $W = F \times x = (T \times 2L) \times x$. The increase in total surface area is $\Delta A = 2L \times x$. Surface Energy ($E$) is work done per unit increase in area: $E = \frac{W}{\Delta A} = \frac{T \times 2Lx}{2Lx} = T$. Thus, surface energy is numerically equal to surface tension.
Variation with Temperature: The surface tension of a liquid generally decreases with an increase in temperature. This is because higher temperature increases the kinetic energy of molecules, weakening the cohesive intermolecular forces responsible for surface tension.

Solution: Basic Assumptions of Kinetic Theory of Gases: 1. A gas consists of a very large number of identical, tiny spherical particles called molecules. 2. Molecules are in constant, rapid, random motion in all possible directions. 3. The actual volume occupied by the molecules is negligible compared to the volume of the container. 4. There are no intermolecular forces of attraction or repulsion between molecules. 5. All collisions between molecules and with the walls are perfectly elastic.
Expression for Pressure: Consider an ideal gas in a cubical box of side $L$. Volume $V = L^3$. Let a molecule of mass $m$ move with velocity component $v_x$ towards a face. Initial momentum $= mv_x$. It collides elastically and rebounds with velocity $-v_x$. Final momentum $= -mv_x$. Change in momentum of the molecule $= -mv_x - mv_x = -2mv_x$. Momentum imparted to the wall $= +2mv_x$. The time taken by the molecule to travel to the opposite face and back is $t = \frac{2L}{v_x}$. Rate of change of momentum (Force) by one molecule $f = \frac{2mv_x}{2L/v_x} = \frac{m v_x^2}{L}$. Total force on the wall by all $N$ molecules is $F = \frac{m}{L} (v_{x1}^2 + v_{x2}^2 + \dots + v_{xN}^2)$. Mean square velocity $v_{rms}^2 = v_x^2 + v_y^2 + v_z^2$. Since motion is completely random and isotropic, $v_x^2 = v_y^2 = v_z^2 = \frac{1}{3}v_{rms}^2$. Total force $F = \frac{m}{L} N (\frac{1}{3}v_{rms}^2) = \frac{1}{3} \frac{Nm}{L} v_{rms}^2$. Pressure $P = \frac{F}{\text{Area}} = \frac{F}{L^2} = \frac{1}{3} \frac{Nm}{L^3} v_{rms}^2$. Since $L^3 = V$ (volume) and $Nm = M$ (total mass), and density $\rho = M/V$: $P = \frac{1}{3} \rho v_{rms}^2$. (a) Solution: From kinetic theory, pressure $P = \frac{1}{3}\frac{M}{V} v_{rms}^2$. Multiplying by $V$: $PV = \frac{1}{3} M v_{rms}^2$. Multiply and divide by 2: $PV = \frac{2}{3} \left(\frac{1}{2} M v_{rms}^2\right) = \frac{2}{3} E$ (where $E$ is the total translational kinetic energy of the gas). For 1 mole of an ideal gas, $PV = RT$. Therefore, $\frac{2}{3} E = RT \implies E = \frac{3}{2} RT$. Dividing by Avogadro's number ($N_A$) gives the average kinetic energy per molecule: $K_{avg} = \frac{E}{N_A} = \frac{3}{2} \left(\frac{R}{N_A}\right) T = \frac{3}{2} k_B T$ (where $k_B$ is the Boltzmann constant). Since $\frac{3}{2} k_B$ is a constant, $K_{avg} \propto T$. Hence, the average kinetic energy of a gas molecule is directly proportional to its absolute temperature.

(b) Solution: Given: Temperature $T = 27^\circ\text{C} = 300\text{ K}$ (Same for both gases since they are in a mixture). Mass of Argon ($M_{Ar}$) $= 39.9\text{ u}$. Mass of Chlorine ($M_{Cl_2}$) $= 70.9\text{ u}$. (i) Ratio of average kinetic energy per molecule: As proved above, $K_{avg} = \frac{3}{2} k_B T$. It depends *only* on the absolute temperature, independent of the mass or nature of the gas. Since both gases are at the same temperature in the flask, their average kinetic energy per molecule is identical. Ratio = $1 : 1$. (ii) Ratio of Root mean square speed ($V_{rms}$): We know $V_{rms} = \sqrt{\frac{3RT}{M}}$. Since $R$ and $T$ are constant, $V_{rms} \propto \frac{1}{\sqrt{M}}$. $\frac{V_{rms}(Ar)}{V_{rms}(Cl_2)} = \sqrt{\frac{M_{Cl_2}}{M_{Ar}}} = \sqrt{\frac{70.9}{39.9}} = \sqrt{1.7769} \approx 1.33$. Ratio = $1.33 : 1$.