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Maharashtra HSC Physics study materials
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Maharashtra HSC Physics practice questions

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Question 1Rotational Dynamics

A rigid body rotates about a fixed axis with constant angular velocity. Which quantity remains the same for every particle of the body?

  1. Tangential velocity
  2. Centripetal acceleration
  3. Angular velocity Correct answer.
  4. Linear speed

Explanation

For a rigid body rotating about a fixed axis, every particle completes the same angle in the same time, so the angular velocity is common to all particles. Linear speed, tangential velocity and centripetal acceleration all depend on the distance from the axis, so they differ from particle to particle.

Question 2Rotational Dynamics

A flywheel rotating at 300 rad/s is brought to rest uniformly in 10 s. What is the magnitude of its angular retardation?

  1. 3 rad/s^2
  2. 0.03 rad/s^2
  3. 30 rad/s^2 Correct answer.
  4. 3000 rad/s^2

Explanation

Using omega = omega0 + alpha*t with final angular velocity zero gives alpha = -omega0/t. The magnitude is therefore 300/10 = 30 rad/s^2. The negative sign only shows direction, opposite to the rotation.

Question 3Rotational Dynamics

The moment of inertia of a rigid body about a given axis depends on which combination of factors?

  1. Only the total mass of the body
  2. Only the angular velocity of rotation
  3. Only the torque acting on the body
  4. The mass and its distribution about the axis Correct answer.

Explanation

Moment of inertia is the sum of m*r^2 over all particles, so it depends on the mass and on how that mass is distributed relative to the chosen axis. The same body can have very different moments of inertia about different axes. Angular velocity and torque are dynamic quantities and do not define the inertia.

Question 4Rotational Dynamics

A torque of 12 N m acts on a body of moment of inertia 4 kg m^2 about a fixed axis. What is the angular acceleration produced?

  1. 3 rad/s^2 Correct answer.
  2. 8 rad/s^2
  3. 48 rad/s^2
  4. 0.33 rad/s^2

Explanation

Newton's second law for rotation states torque equals moment of inertia times angular acceleration. Dividing 12 by 4 gives alpha = 3 rad/s^2 about the axis of the torque.

Question 5Rotational Dynamics

In rotational motion, which quantity plays the same role that force plays in linear motion?

  1. Moment of inertia
  2. Angular momentum
  3. Torque Correct answer.
  4. Angular displacement

Explanation

Torque is the turning effect that changes the state of rotation, just as force changes linear motion. The rotational analogue of Newton's second law is torque equals moment of inertia times angular acceleration. The other quantities correspond to mass, momentum and displacement instead.

Question 6Rotational Dynamics

A solid sphere and a hollow spherical shell of the same mass and radius roll without slipping down an incline from rest. Which reaches the bottom first?

  1. The hollow shell
  2. It depends on the angle of the incline
  3. Both reach together
  4. The solid sphere Correct answer.

Explanation

Rolling acceleration down an incline is g sin(theta)/(1 + I/(MR^2)). The solid sphere has a smaller ratio I/(MR^2) = 2/5 than the hollow shell at 2/3, so it has the larger acceleration and arrives first. The result is independent of the incline angle.

Question 7Rotational Dynamics

A metre rule is pivoted at its centre of mass. A 2 N force acts perpendicular to the rule at 0.30 m from the pivot. What is the magnitude of the torque about the pivot?

  1. 0.60 N m Correct answer.
  2. 6.0 N m
  3. 1.7 N m
  4. 0.15 N m

Explanation

For a force perpendicular to the moment arm, torque equals force times perpendicular distance from the pivot. Multiplying 2 N by 0.30 m gives 0.60 N m. Only the perpendicular component of a force produces torque.

Question 8Rotational Dynamics

A disc of mass 2 kg and radius 0.5 m rotates about an axis through its centre perpendicular to its plane. What is its moment of inertia about this axis?

  1. 1.0 kg m^2
  2. 0.50 kg m^2
  3. 0.125 kg m^2
  4. 0.25 kg m^2 Correct answer.

Explanation

For a uniform disc about its central perpendicular axis, I = (1/2)MR^2. Substituting gives 0.5 x 2 x 0.25 = 0.25 kg m^2. Using MR^2 instead would give the value for a ring of the same mass and radius.

Question 9Rotational Dynamics

A dancer spinning with arms outstretched pulls the arms close to the body. Assuming no external torque, what happens?

  1. Angular velocity increases and kinetic energy increases Correct answer.
  2. Angular velocity increases while kinetic energy stays constant
  3. Angular velocity decreases and kinetic energy decreases
  4. Angular velocity stays constant because angular momentum is conserved

Explanation

With no external torque, angular momentum L = I*omega is conserved. Pulling the arms in reduces the moment of inertia, so omega must rise. Rotational kinetic energy L^2/(2I) also rises because I falls, and the extra energy comes from the work done by the dancer's muscles.

Question 10Rotational Dynamics

The radius of gyration of a body of mass 3 kg about an axis is 0.4 m. What is its moment of inertia about that axis?

  1. 7.5 kg m^2
  2. 1.2 kg m^2
  3. 0.48 kg m^2 Correct answer.
  4. 0.16 kg m^2

Explanation

Moment of inertia equals mass times the square of the radius of gyration, I = MK^2. Here 3 x (0.4)^2 = 0.48 kg m^2. The radius of gyration is the distance at which the whole mass could be concentrated to give the same inertia.

Question 11Rotational Dynamics

A particle of mass 0.2 kg moves in a circle of radius 0.5 m at a constant speed of 4 m/s. What is its angular momentum about the centre?

  1. 0.4 kg m^2/s Correct answer.
  2. 0.8 kg m^2/s
  3. 4.0 kg m^2/s
  4. 1.6 kg m^2/s

Explanation

Angular momentum for circular motion is L = m*v*r when velocity is perpendicular to the radius. Substituting gives 0.2 x 4 x 0.5 = 0.4 kg m^2/s. Equivalently, I = mr^2 = 0.05 and omega = v/r = 8 rad/s give the same product.

Question 12Rotational Dynamics

A solid cylinder rolls without slipping on a horizontal surface. What fraction of its total kinetic energy is rotational?

  1. One quarter
  2. One half
  3. One third Correct answer.
  4. Two thirds

Explanation

For a solid cylinder I = (1/2)MR^2 and rolling gives v = omega*R. The rotational energy is (1/4)Mv^2 while translational is (1/2)Mv^2, so rotational is one third of the total. The ratio depends only on the shape factor of the body.

Question 13Rotational Dynamics

Two forces act on a rigid body in translational equilibrium but form a couple. What is the net effect on the body?

  1. Rotation only, with no translation Correct answer.
  2. Neither equilibrium condition can be satisfied
  3. No translation and no rotation
  4. Translation only, with no rotation

Explanation

A couple consists of two equal, opposite, non-collinear forces, so the vector sum is zero and the centre of mass does not accelerate. However, the two forces produce torques that add rather than cancel, so the body acquires angular acceleration and rotates.

Question 14Rotational Dynamics

A wheel starting from rest attains 20 rad/s in 5 s with uniform angular acceleration. What angular displacement does it cover in this interval?

  1. 100 rad
  2. 50 rad Correct answer.
  3. 4 rad
  4. 40 rad

Explanation

For uniform angular acceleration, displacement is the average angular velocity times time. The average of 0 and 20 rad/s is 10 rad/s, so theta = 10 x 5 = 50 rad. This also follows from theta = omega0*t + (1/2)*alpha*t^2 with alpha = 4 rad/s^2.

Question 15Rotational Dynamics

The moment of inertia of a body about an axis through its centre of mass is 6 kg m^2. What is its moment of inertia about a parallel axis 0.5 m away if the mass is 8 kg?

  1. 14 kg m^2
  2. 8.0 kg m^2 Correct answer.
  3. 10 kg m^2
  4. 2.0 kg m^2

Explanation

The parallel axis theorem gives I = I_cm + M*d^2. Here 6 + 8 x (0.5)^2 = 6 + 2 = 8 kg m^2. The added term is always positive, so a parallel axis away from the centre of mass always has a larger moment of inertia.

Question 16Rotational Dynamics

A torque of 5 N m acts on a flywheel of moment of inertia 2 kg m^2 for 3 s, starting from rest. What is the angular momentum gained?

  1. 15 kg m^2/s Correct answer.
  2. 1.67 kg m^2/s
  3. 30 kg m^2/s
  4. 0.83 kg m^2/s

Explanation

The angular impulse equals the change in angular momentum, so L = torque x time = 5 x 3 = 15 kg m^2/s. This is the rotational analogue of impulse equals change in linear momentum. The moment of inertia is not needed when torque and time are both known.

Question 17Rotational Dynamics

A uniform rod of length L and mass M is pivoted at one end and released from rest in the horizontal position. What is its initial angular acceleration?

  1. 2g/(3L)
  2. g/(2L)
  3. g/L
  4. 3g/(2L) Correct answer.

Explanation

About an end, the rod has I = ML^2/3 and the weight acts at the centre of mass, giving torque MgL/2. Dividing torque by moment of inertia gives alpha = (MgL/2)/(ML^2/3) = 3g/(2L). The result is independent of the mass.

Question 18Rotational Dynamics

A solid sphere of mass M and radius R rolls without slipping down an incline of angle theta. What is the magnitude of the friction force needed to maintain rolling?

  1. (2/5)Mg sin(theta)
  2. (1/3)Mg sin(theta)
  3. (2/7)Mg sin(theta) Correct answer.
  4. (5/7)Mg sin(theta)

Explanation

Newton's laws for rolling give Mg sin(theta) - f = Ma and fR = I*alpha with a = alpha*R and I = (2/5)MR^2. Solving yields f = (2/7)Mg sin(theta). Static friction acts up the incline and is essential for rolling without slipping.

Question 19Rotational Dynamics

A horizontal disc of moment of inertia I rotates freely with angular velocity omega. A second disc of moment of inertia 2I, initially at rest, is dropped gently onto it and the two rotate together. What fraction of the initial kinetic energy is lost?

  1. One ninth
  2. Two thirds Correct answer.
  3. One half
  4. One third

Explanation

Angular momentum is conserved, so the common angular velocity is I*omega/(3I) = omega/3. Initial energy is (1/2)I*omega^2 and final energy is (1/2)(3I)(omega/3)^2 = (1/6)I*omega^2. The lost fraction is therefore 1 - 1/3 = 2/3.

Question 20Rotational Dynamics

A thin uniform ring and a solid disc have the same mass and radius and are spun up by identical torques from rest. After the same time, which has the greater angular velocity, and why?

  1. The disc, because its moment of inertia is smaller Correct answer.
  2. The ring, because its mass is concentrated farther out
  3. They are equal, because the torques are equal
  4. The ring, because it has the smaller moment of inertia

Explanation

The ring has I = MR^2 and the disc has I = (1/2)MR^2, so the disc has the smaller moment of inertia. With equal torques for equal times, the disc gains the larger angular acceleration and hence the larger angular velocity. Concentrating mass farther from the axis increases inertia, so the ring is harder to spin up.

Question 21Mechanical Properties of Fluids

A hydraulic lift uses a small piston of area 5 cm² and a large piston of area 500 cm². If a force of 200 N is applied to the small piston, what load can be lifted at the large piston, assuming the pistons are at the same height?

  1. 20000 N Correct answer.
  2. 200 N
  3. 200000 N
  4. 2000 N

Explanation

By Pascal's law the pressure transmits undiminished through the enclosed fluid, so F₂/A₂ = F₁/A₁. The area ratio is 500/5 = 100, hence F₂ = 200 N × 100 = 20000 N. The height condition removes any hydrostatic pressure difference between the two pistons.

Question 22Mechanical Properties of Fluids

A body floats with 40% of its volume above the surface of a liquid. What is the density of the liquid if the density of the body is 600 kg/m³?

  1. 1500 kg/m³
  2. 600 kg/m³
  3. 400 kg/m³
  4. 1000 kg/m³ Correct answer.

Explanation

For a floating body, the weight of displaced liquid equals the weight of the body, so ρ_body·V = ρ_liquid·V_submerged. Since 60% of the volume is submerged, ρ_liquid = 600/0.6 = 1000 kg/m³.

Question 23Mechanical Properties of Fluids

Water flows steadily through a horizontal pipe whose cross-sectional radius at one section is twice that at another. The speed of flow at the wider section is v. What is the speed at the narrower section?

  1. v/4
  2. 2v
  3. 4v Correct answer.
  4. v/2

Explanation

The equation of continuity gives A₁v₁ = A₂v₂, and area scales as the square of the radius. If the radius halves, the area becomes one quarter, so the speed must become four times larger to keep the volume flow rate constant.

Question 24Mechanical Properties of Fluids

According to Bernoulli's principle for an ideal fluid in steady, irrotational flow along a horizontal streamline, which quantity remains constant?

  1. Static pressure alone
  2. Kinetic energy per unit volume alone
  3. The sum of pressure, kinetic energy per unit volume and potential energy per unit volume Correct answer.
  4. The product of pressure and velocity

Explanation

Bernoulli's equation states that P + ½ρv² + ρgh is constant along a streamline for steady, non-viscous, incompressible flow. The three terms represent pressure energy, kinetic energy and potential energy per unit volume respectively.

Question 25Mechanical Properties of Fluids

The terminal velocity of a small spherical drop falling through a viscous liquid is found to be v. If the radius of the drop is doubled while all other factors remain the same, the new terminal velocity becomes

  1. 2v
  2. v/2
  3. v
  4. 4v Correct answer.

Explanation

Stokes' law gives terminal velocity proportional to the square of the radius, v_t = 2r²(ρ−σ)g/(9η). Doubling the radius therefore multiplies the terminal velocity by four.

Question 26Mechanical Properties of Fluids

Which property of a liquid is responsible for the rise of liquid in a narrow capillary tube?

  1. The compressibility of the liquid
  2. The buoyant force of displaced air
  3. Surface tension acting along the meniscus Correct answer.
  4. Viscosity of the liquid

Explanation

Capillary rise occurs because surface tension acts along the liquid meniscus and its vertical component supports the weight of the raised liquid column. The balance yields h = 2T cosθ/(rρg), showing rise increases as the tube radius decreases.

Question 27Mechanical Properties of Fluids

A U-tube contains mercury. Water is poured into one limb so that a column of water 13.6 cm high stands above the mercury level in that limb. How far does the mercury surface in the other limb rise above the water-mercury interface? (Density of mercury = 13.6 g/cm³, water = 1 g/cm³)

  1. 13.6 cm
  2. 0.5 cm
  3. 1 cm Correct answer.
  4. 10 cm

Explanation

At the same horizontal level in the connected mercury the pressures are equal, so ρ_w·g·h_w = ρ_Hg·g·h_Hg. Thus h_Hg = (1 × 13.6)/13.6 = 1 cm. The mercury column that balances the water column is only 1 cm high.

Question 28Mechanical Properties of Fluids

Two soap bubbles of radii r₁ and r₂ are joined by a narrow tube. If r₁ > r₂, what happens to the bubbles?

  1. The larger bubble shrinks and the smaller bubble grows
  2. Both bubbles remain unchanged in size
  3. The larger bubble grows and the smaller bubble shrinks Correct answer.
  4. Both bubbles merge into a single bubble of average radius

Explanation

The excess pressure inside a soap bubble is 4T/r, so the smaller bubble has the higher internal pressure. Air therefore flows from the smaller bubble into the larger one, making the larger bubble grow and the smaller one shrink until the smaller disappears.

Question 29Mechanical Properties of Fluids

The excess pressure inside a liquid drop of radius r and surface tension T is given by

  1. T/r
  2. 4T/r
  3. 2T/r Correct answer.
  4. T/(2r)

Explanation

For a liquid drop with a single surface, the excess pressure is ΔP = 2T/r. A soap bubble has two surfaces, which is why its excess pressure is 4T/r instead.

Question 30Mechanical Properties of Fluids

A stream of water from a horizontal pipe is directed vertically upward and rises to a maximum height of 5 m. Taking g = 10 m/s², what is the speed of efflux at the pipe opening?

  1. 5 m/s
  2. 10 m/s Correct answer.
  3. 50 m/s
  4. 25 m/s

Explanation

Applying Bernoulli's equation between the pipe opening and the top of the jet gives v = √(2gh). Substituting h = 5 m and g = 10 m/s² yields v = √(2 × 10 × 5) = 10 m/s.

Question 31Mechanical Properties of Fluids

A rectangular tank contains water to a depth of 4 m. What is the gauge pressure at the bottom due to the water column? (ρ = 1000 kg/m³, g = 10 m/s²)

  1. 40000 Pa Correct answer.
  2. 4000 Pa
  3. 400000 Pa
  4. 400 Pa

Explanation

The gauge pressure due to a liquid column is P = ρgh. Substituting gives 1000 × 10 × 4 = 40000 Pa. This is the pressure due to the water alone, above atmospheric pressure.

Question 32Mechanical Properties of Fluids

A steel needle can float on the surface of water even though steel is denser than water. This is best explained by

  1. The viscosity of water holding the needle in place
  2. The buoyant force of the water exceeding the weight of the needle
  3. Surface tension supporting the needle on a slightly depressed water surface Correct answer.
  4. The needle being hollow and containing air

Explanation

A small dense object can rest on water if it does not break the surface, because surface tension acts upward along the depressed surface and balances its weight. The effect works only for objects light enough that the required surface deformation stays small.

Question 33Mechanical Properties of Fluids

In a venturimeter fitted to a horizontal pipe, the pressure is lowest at the

  1. throat, where the cross-section is smallest Correct answer.
  2. point where the pipe is vertical
  3. wide inlet section
  4. outlet section

Explanation

By the equation of continuity the fluid speed is greatest at the narrow throat, and by Bernoulli's principle the pressure is correspondingly lowest there. The measured pressure difference between the inlet and throat gives the flow rate.

Question 34Mechanical Properties of Fluids

The coefficient of viscosity η of a liquid has the SI unit

  1. N/m²
  2. N·s/m² Correct answer.
  3. N·m²/s
  4. kg/(m·s²)

Explanation

From Newton's law of viscosity, F = ηA(dv/dx), the coefficient η has units of N·s/m², equivalently Pa·s. It represents the tangential force per unit area per unit velocity gradient.

Question 35Mechanical Properties of Fluids

Water flows through a horizontal pipe of varying cross-section. At a point where the speed is 2 m/s, the pressure is 100 kPa. What is the pressure at another point where the speed is 6 m/s? (Density of water = 1000 kg/m³)

  1. 116 kPa
  2. 84 kPa Correct answer.
  3. 68 kPa
  4. 132 kPa

Explanation

Bernoulli's equation for a horizontal pipe gives P₁ + ½ρv₁² = P₂ + ½ρv₂². The change in dynamic pressure is ½ × 1000 × (36 − 4) = 16000 Pa, so P₂ = 100000 − 16000 = 84000 Pa = 84 kPa.

Question 36Mechanical Properties of Fluids

A hydraulic press has input and output piston diameters in the ratio 1:10. To lift a load of 5000 N at the output piston, the force required at the input piston is

  1. 50 N Correct answer.
  2. 500 N
  3. 50000 N
  4. 5000 N

Explanation

The force ratio equals the ratio of piston areas, which is the square of the diameter ratio, so 1:100. The input force is therefore 5000/100 = 50 N.

Question 37Kinetic Theory of Gases and Radiation

According to the kinetic theory of gases, the pressure exerted by an ideal gas on the walls of its container is given by P = (1/3)(N/V)m v̄², where v̄² is the mean square speed. Which physical quantity does the factor (1/3) arise from?

  1. The fraction of molecular collisions that are perfectly elastic rather than inelastic
  2. The reduction of molecular speed caused by intermolecular attractive forces
  3. The equal distribution of molecular velocities among the three mutually perpendicular directions in space Correct answer.
  4. The ratio of the number of molecules striking a wall to the total number of molecules in the container

Explanation

In the derivation, molecular velocities are resolved along three perpendicular axes and isotropy of the gas makes each direction contribute equally, so only one-third of the mean square velocity is associated with motion normal to any chosen wall. The factor therefore reflects the three-dimensional randomness of molecular motion, not collision statistics or intermolecular forces.

Question 38Kinetic Theory of Gases and Radiation

The average translational kinetic energy of one molecule of an ideal gas at absolute temperature T is

  1. 3k_BT
  2. (3/2)k_BT Correct answer.
  3. (2/3)k_BT
  4. (1/2)k_BT

Explanation

By the equipartition theorem a molecule free to translate in three dimensions has three translational degrees of freedom, each contributing (1/2)k_BT, giving a total of (3/2)k_BT per molecule. This result is independent of the molecular mass and depends only on absolute temperature.

Question 39Kinetic Theory of Gases and Radiation

A gas is enclosed in a rigid vessel at constant volume. Its temperature is raised from 300 K to 600 K. If the mean free path at 300 K is λ, what is the mean free path at 600 K?

  1. λ Correct answer.
  2. 4λ
  3. λ/2
  4. 2λ

Explanation

For an ideal gas the mean free path is λ = 1/(√2 π n d²), which depends only on the number density n and the molecular diameter d, not on temperature. Since the vessel volume is fixed and the number of molecules is unchanged, n is constant and so the mean free path stays λ.

Question 40Kinetic Theory of Gases and Radiation

The root-mean-square speed of the molecules of a gas is proportional to which expression, where T is absolute temperature and M is molar mass?

  1. M/T
  2. √(M/T)
  3. √(T/M) Correct answer.
  4. T/M

Explanation

From (1/2)M v_rms² = (3/2)RT one obtains v_rms = √(3RT/M), so the rms speed varies as the square root of T/M. Lighter gases therefore have higher rms speeds at the same temperature.

Question 41Kinetic Theory of Gases and Radiation

At a given temperature, the rms speed of hydrogen molecules is about four times that of oxygen molecules. What is the ratio of the average translational kinetic energy of a hydrogen molecule to that of an oxygen molecule at the same temperature?

  1. 1:1 Correct answer.
  2. 4:1
  3. 16:1
  4. 1:4

Explanation

Average translational kinetic energy per molecule is (3/2)k_BT and depends only on absolute temperature, not on molecular mass. At the same temperature the two gases therefore have identical average kinetic energy per molecule, even though their speeds differ.

Question 42Kinetic Theory of Gases and Radiation

Which statement best expresses the physical meaning of the absolute zero of temperature in the kinetic theory of gases?

  1. Translational kinetic energy of the molecules would become zero, so molecular motion ceases in the classical picture Correct answer.
  2. The gas becomes an ideal gas with the maximum possible mean free path
  3. The molecules stop colliding with each other but continue to move with unchanged speed
  4. The pressure of every real gas becomes exactly zero while molecular motion continues unchanged

Explanation

Since average translational kinetic energy equals (3/2)k_BT, extrapolating to T = 0 K makes this energy vanish in the classical model, meaning translational motion stops. Real gases liquefy and quantum zero-point motion persists, but the classical kinetic-theory statement is as given.

Question 43Kinetic Theory of Gases and Radiation

A vessel contains a mixture of 2 mol of helium and 3 mol of argon at temperature T. Treating both as ideal monatomic gases, the average translational kinetic energy per molecule of the mixture is

  1. (3/5)k_BT
  2. (3/2)k_BT Correct answer.
  3. (2/3)k_BT
  4. (5/2)k_BT

Explanation

In thermal equilibrium every molecule, regardless of mass or species, has the same average translational kinetic energy (3/2)k_BT because energy is shared equally among translational degrees of freedom. The molar composition affects only the total energy of the mixture, not the per-molecule average.

Question 44Kinetic Theory of Gases and Radiation

The pressure of an ideal gas is doubled while its absolute temperature is also doubled. What happens to its volume?

  1. It doubles
  2. It becomes one-half
  3. It remains unchanged Correct answer.
  4. It becomes one-quarter

Explanation

For an ideal gas PV = nRT, so V = nRT/P. Doubling both P and T multiplies the numerator and denominator by the same factor, leaving V unchanged.

Question 45Kinetic Theory of Gases and Radiation

According to Maxwell's law of distribution of molecular speeds, which feature of the distribution curve changes when the temperature of a fixed mass of gas is increased?

  1. The curve shifts to higher speeds while its peak height and area both increase
  2. The peak becomes higher and narrower while the most probable speed decreases
  3. The whole curve shifts to lower speeds with the area under it decreasing
  4. The most probable speed increases and the curve becomes broader and lower, keeping the area under it constant Correct answer.

Explanation

Raising T increases the most probable speed √(2RT/M), so the peak moves right and flattens, and the distribution broadens. Because the total number of molecules is fixed, the area under the curve, representing that number, must stay the same.

Question 46Kinetic Theory of Gases and Radiation

The ratio of the specific heat at constant pressure to that at constant volume, γ = C_P/C_V, for an ideal monatomic gas is

  1. 7/5
  2. 3/2
  3. 4/3
  4. 5/3 Correct answer.

Explanation

A monatomic gas has three translational degrees of freedom, giving C_V = (3/2)R and, by Mayer's relation, C_P = (5/2)R. The ratio is therefore 5/3, approximately 1.67.

Question 47Kinetic Theory of Gases and Radiation

A gas at 27 °C is heated at constant pressure until its volume doubles. Assuming ideal behaviour, what is its final temperature?

  1. 150 °C
  2. 54 °C
  3. 600 °C
  4. 327 °C Correct answer.

Explanation

At constant pressure, V/T is constant, so the absolute temperature must double: T₂ = 2 × 300 K = 600 K. Converting back gives 600 − 273 = 327 °C, which shows why temperatures must be used in kelvin.

Question 48Kinetic Theory of Gases and Radiation

Two containers of equal volume hold the same ideal gas at the same temperature. Container A holds N molecules and container B holds 2N molecules. What is the ratio of the mean free path in A to that in B?

  1. 4:1
  2. 2:1 Correct answer.
  3. 1:2
  4. 1:1

Explanation

Mean free path λ = 1/(√2 π n d²) is inversely proportional to number density n. Container B has twice the density of A, so its mean free path is half, making the ratio λ_A : λ_B equal to 2:1.

Question 49Kinetic Theory of Gases and Radiation

The pressure of a gas is doubled at constant temperature. If the mean free path before the change was λ, and the gas behaves ideally, the mean free path after the change is

  1. λ/2 Correct answer.
  2. 2λ
  3. λ/4
  4. λ

Explanation

At constant temperature, doubling the pressure halves the volume and hence doubles the number density n. Since λ varies as 1/n, doubling n halves the mean free path, giving λ/2.

Question 50Kinetic Theory of Gases and Radiation

Which of the following is a direct consequence of the equipartition of energy for a diatomic gas at ordinary temperatures, where vibrational modes are not excited?

  1. C_V = (7/2)R because all vibrational degrees of freedom are active
  2. C_V = (3/2)R because only translational degrees of freedom are active
  3. C_V = 3R because each molecule has six active degrees of freedom
  4. C_V = (5/2)R because three translational and two rotational degrees of freedom are active Correct answer.

Explanation

A rigid diatomic molecule has three translational and two rotational degrees of freedom, and each active degree contributes (1/2)R per mole to C_V. This gives C_V = (5/2)R at ordinary temperatures, with vibrational contributions appearing only at high temperature.