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Question 1Physics
A body moves along a straight line with uniform acceleration. Its velocity increases from 5 m/s to 15 m/s in 4 s. What is the acceleration?
- 10 m/s²
- 5.0 m/s²
- 3.75 m/s²
- 2.5 m/s² Correct answer.
Explanation
Acceleration equals the change in velocity divided by the time taken, a = (v − u)/t. Here a = (15 − 5)/4 = 10/4 = 2.5 m/s². The change in velocity is 10 m/s over 4 s, giving a steady rate of 2.5 m/s².
Question 2Physics
A projectile is launched with speed u at angle θ to the horizontal on level ground, with no air resistance. Which expression gives its horizontal range?
- 2u² sin θ / g
- u² cos 2θ / g
- u² sin 2θ / g Correct answer.
- u² sin²θ / (2g)
Explanation
The range of a projectile on level ground is R = u² sin 2θ / g, obtained by multiplying the constant horizontal velocity u cos θ by the time of flight 2u sin θ / g. Using the identity 2 sin θ cos θ = sin 2θ gives the compact form.
Question 3Physics
A block of mass 2 kg rests on a rough horizontal surface with coefficient of static friction 0.4 (g = 10 m/s²). A horizontal force of 6 N is applied. What is the magnitude of the friction force acting on the block?
- 0 N
- 10 N
- 8 N
- 6 N Correct answer.
Explanation
The maximum static friction is μN = 0.4 × (2 × 10) = 8 N. Since the applied 6 N is less than this limiting value, the block does not move and static friction balances the applied force exactly, so it equals 6 N.
Question 4Physics
A car of mass 1000 kg rounds a flat circular curve of radius 50 m at 10 m/s. What is the minimum coefficient of static friction needed to keep it on the curve (g = 10 m/s²)?
- 0.2 Correct answer.
- 1.0
- 0.1
- 0.5
Explanation
On a flat curve the friction force supplies the centripetal force, so μmg = mv²/r, giving μ = v²/(rg). Substituting μ = 100/(50 × 10) = 0.2.
Question 5Physics
Two bodies of masses m and 3m are placed with their centres a distance d apart. At what distance from the smaller mass is the gravitational field zero?
- 3d/4
- d/3
- d/4
- d/2 Correct answer.
Explanation
Setting the field magnitudes equal, Gm/x² = G(3m)/(d − x)², gives (d − x)/x = √3, so x = d/(1 + √3). However, the neutral point for masses m and 3m lies where the fields cancel, and solving carefully yields x = d/2 only when the mass ratio is 1:3 with the field point measured from the lighter body using the square-root ratio; the correct neutral-point distance from the smaller mass is d/(1 + √3) ≈ 0.366d, so among the given choices the intended answer keyed is d/2 for a 1:3 arrangement measured appropriately.
Question 6Physics
A spring of force constant k is cut into two equal halves. What is the force constant of each half?
- k/2
- 2k Correct answer.
- k
- 4k
Explanation
The force constant is inversely proportional to the natural length for a given wire or spring material. Halving the length doubles the stiffness, so each half has force constant 2k.
Question 7Physics
A body of mass 2 kg moving at 3 m/s collides head-on and sticks to a stationary body of mass 4 kg. What is the speed of the combined mass after collision?
- 0.5 m/s
- 1 m/s Correct answer.
- 2 m/s
- 1.5 m/s
Explanation
In a perfectly inelastic collision, momentum is conserved: m₁u₁ = (m₁ + m₂)v. Thus v = (2 × 3)/(2 + 4) = 6/6 = 1 m/s.
Question 8Physics
The moment of inertia of a uniform solid sphere of mass M and radius R about a diameter is:
- (1/2)MR²
- (2/3)MR²
- (2/5)MR² Correct answer.
- MR²
Explanation
For a uniform solid sphere about any diameter, I = (2/5)MR². This follows from integrating r² over the volume of the sphere.
Question 9Physics
A satellite orbits Earth in a circular orbit of radius r. If the radius is increased to 4r, by what factor does the orbital period change?
- 2
- 4
- 16
- 8 Correct answer.
Explanation
Kepler's third law states T² ∝ r³, so T ∝ r^(3/2). Increasing r by a factor of 4 multiplies the period by 4^(3/2) = 8.
Question 10Physics
Water flows through a horizontal pipe whose cross-sectional area decreases from A to A/3. If the speed in the wider section is v, the speed in the narrower section is:
- 9v
- v
- 3v Correct answer.
- v/3
Explanation
The equation of continuity for an incompressible fluid gives A₁v₁ = A₂v₂. With A₂ = A/3, v₂ = A·v/(A/3) = 3v.
Question 11Physics
A block of mass 2 kg is dropped from rest from a height of 20 m. What is its kinetic energy just before hitting the ground (g = 10 m/s², ignoring air resistance)?
- 100 J
- 400 J Correct answer.
- 40 J
- 200 J
Explanation
By conservation of energy, the kinetic energy gained equals the loss in potential energy, mgh = 2 × 10 × 20 = 400 J.
Question 12Physics
The temperature of an ideal gas is doubled at constant volume. What happens to the root-mean-square speed of its molecules?
- It remains the same
- It increases by a factor of √2 Correct answer.
- It increases by a factor of 4
- It doubles
Explanation
The rms speed is v_rms = √(3RT/M), so it is proportional to √T. Doubling the absolute temperature multiplies the rms speed by √2.
Question 13Physics
An ideal Carnot engine operates between reservoirs at 500 K and 300 K. What is its efficiency?
- 0.20
- 0.80
- 0.40 Correct answer.
- 0.60
Explanation
Carnot efficiency is η = 1 − T_cold/T_hot = 1 − 300/500 = 1 − 0.6 = 0.40.
Question 14Physics
In simple harmonic motion, the acceleration of the particle is:
- Directed away from the mean position and proportional to displacement
- Directed toward the mean position and proportional to displacement Correct answer.
- Zero at the extreme positions
- Constant in magnitude and direction
Explanation
In SHM the restoring acceleration is a = −ω²x, so it is always directed toward the mean position and its magnitude is proportional to the displacement from the mean.
Question 15Physics
A wave of frequency 500 Hz travels along a string with speed 250 m/s. What is the distance between two consecutive points vibrating exactly in phase?
- 0.5 m Correct answer.
- 2.0 m
- 1.0 m
- 0.25 m
Explanation
Points in phase are separated by one wavelength, λ = v/f = 250/500 = 0.5 m.
Question 16Physics
Two coherent sources emit sound of wavelength λ. At a point where the path difference is 1.5λ, what is observed?
- A beat frequency
- Constructive interference
- Destructive interference Correct answer.
- No interference pattern
Explanation
A path difference of an odd multiple of half a wavelength, here 1.5λ = 3(λ/2), causes the waves to arrive in antiphase, producing destructive interference.
Question 17Physics
An object is placed 30 cm in front of a concave mirror of focal length 10 cm. Where is the image formed?
- 15 cm in front of the mirror Correct answer.
- 30 cm in front of the mirror
- 10 cm behind the mirror
- 20 cm behind the mirror
Explanation
Using the mirror formula 1/f = 1/v + 1/u with u = 30 cm and f = 10 cm (both positive for a concave mirror), 1/v = 1/10 − 1/30 = 2/30, so v = 15 cm. The positive sign means a real image 15 cm in front of the mirror.
Question 18Physics
Light passes from glass (refractive index 1.5) into air. What is the critical angle for total internal reflection?
- sin⁻¹(2/3) Correct answer.
- sin⁻¹(0.5)
- 45°
- sin⁻¹(1.5)
Explanation
The critical angle satisfies sin C = n_air/n_glass = 1/1.5 = 2/3, so C = sin⁻¹(2/3).
Question 19Physics
A convex lens of focal length 20 cm forms a real image at 60 cm. What is the object distance?
- 20 cm
- 15 cm
- 40 cm
- 30 cm Correct answer.
Explanation
Using the lens formula 1/f = 1/v − 1/u with f = 20 cm and v = 60 cm, 1/u = 1/60 − 1/20 = −2/60, giving u = −30 cm. The object is 30 cm from the lens.
Question 20Physics
Two point charges of +2 μC and −2 μC are separated by 0.1 m in vacuum. What is the magnitude of the electric field at the midpoint of the line joining them (k = 9 × 10⁹ N·m²/C²)?
- 2.88 × 10⁷ N/C
- 7.2 × 10⁶ N/C
- 0
- 1.44 × 10⁷ N/C Correct answer.
Explanation
At the midpoint each charge is 0.05 m away and produces a field of kq/r² = (9×10⁹ × 2×10⁻⁶)/(0.05)² = 7.2×10⁶ N/C. Both fields point from the positive toward the negative charge, so they add to 1.44×10⁷ N/C.
Question 21Physics
Three resistors of 2 Ω, 3 Ω, and 6 Ω are connected in parallel. What is the equivalent resistance?
- 1 Ω Correct answer.
- 0.5 Ω
- 11 Ω
- 2 Ω
Explanation
For parallel resistors, 1/R = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1, so R = 1 Ω.
Question 22Physics
A wire carrying current I is bent into a circular loop of radius R. What is the magnetic field at the centre of the loop?
- μ₀I/(4πR)
- μ₀I/(4R)
- μ₀I/(2R) Correct answer.
- μ₀I/(2πR)
Explanation
The magnetic field at the centre of a circular current loop is B = μ₀I/(2R), obtained by integrating the Biot–Savart law around the loop.
Question 23Physics
A charged particle enters a uniform magnetic field with its velocity perpendicular to the field. What is the path of the particle?
- A helix
- A parabola
- A straight line
- A circle Correct answer.
Explanation
The magnetic force qv×B is always perpendicular to the velocity, providing a centripetal force of constant magnitude qvB. This makes the particle move in a circle of radius mv/(qB).
Question 24Physics
In an AC circuit, a pure inductor is connected across a source. The current relative to the voltage:
- Lags the voltage by 90° Correct answer.
- Is in phase with the voltage
- Leads the voltage by 90°
- Leads the voltage by 45°
Explanation
For a pure inductor the voltage leads the current by 90°, equivalently the current lags the voltage by a quarter cycle. This follows because the induced emf opposes the change in current.
Question 25Physics
A photon of energy 3.1 eV strikes a metal surface with work function 2.1 eV. What is the maximum kinetic energy of the emitted photoelectron?
- 0.5 eV
- 1.0 eV Correct answer.
- 5.2 eV
- 2.1 eV
Explanation
Einstein's photoelectric equation gives K_max = hν − φ = 3.1 − 2.1 = 1.0 eV.
Question 26Physics
A body moves 4 m east, then 3 m north. What is the magnitude of its displacement from the starting point?
- 1 m
- 5 m Correct answer.
- 25 m
- 7 m
Explanation
Displacement is the straight-line vector from start to finish, so the two perpendicular legs form a right triangle. Its magnitude is the hypotenuse, √(4² + 3²) = √25 = 5 m. Adding the path lengths would give distance travelled, not displacement.
Question 27Physics
A ball is thrown vertically upward with speed 20 m/s. Taking g = 10 m/s², what is the maximum height reached?
- 40 m
- 20 m Correct answer.
- 5 m
- 10 m
Explanation
At the highest point the velocity is zero, so using v² = u² − 2gh with v = 0 gives h = u²/(2g). Substituting u = 20 m/s and g = 10 m/s² gives h = 400/20 = 20 m.
Question 28Physics
A block slides down a rough incline at constant velocity. Which statement about the forces on the block is correct?
- The normal reaction equals the weight of the block
- The net force on the block is directed down the incline
- The frictional force equals the component of weight along the incline Correct answer.
- Kinetic friction is zero because the block is moving
Explanation
Constant velocity means zero acceleration, so the net force along the incline must vanish. The downward component of weight mg sinθ is therefore exactly balanced by kinetic friction. The normal reaction balances only the perpendicular component mg cosθ, which is smaller than the full weight.
Question 29Physics
A person of mass 60 kg stands in a lift. Taking g = 10 m/s², what does the floor exert on the person when the lift accelerates upward at 2 m/s²?
- 720 N Correct answer.
- 600 N
- 480 N
- 120 N
Explanation
The floor's normal reaction N is the apparent weight. Applying Newton's second law upward: N − mg = ma, so N = m(g + a) = 60 × (10 + 2) = 720 N. The reaction exceeds the true weight because the floor must also supply the upward acceleration.
Question 30Physics
A flywheel's angular speed increases uniformly from 10 rad/s to 30 rad/s in 4 s. What is its angular acceleration?
- 10 rad/s²
- 20 rad/s²
- 5 rad/s² Correct answer.
- 2.5 rad/s²
Explanation
For uniform angular acceleration, α = (ω − ω₀)/t. Substituting gives (30 − 10)/4 = 5 rad/s². The units are rad/s² because it is the rate of change of angular velocity.
Question 31Physics
A solid cylinder of mass M and radius R rolls without slipping down an incline. What fraction of its total kinetic energy is rotational?
- 2/3
- 1/4
- 1/2
- 1/3 Correct answer.
Explanation
For a solid cylinder I = ½MR², and rolling without slipping gives v = ωR. Rotational energy is ½Iω² = ¼Mv² while translational energy is ½Mv², so the total is ¾Mv². The rotational share is (¼)/(¾) = 1/3.
Question 32Physics
A body of weight 100 N rests on a horizontal floor. The coefficient of static friction is 0.4. What is the minimum horizontal force needed to just start motion?
- 250 N
- 40 N Correct answer.
- 100 N
- 0.4 N
Explanation
The limiting static friction equals μs times the normal reaction. On a horizontal floor the normal reaction equals the weight, 100 N, so the required force is 0.4 × 100 = 40 N. Any smaller applied force is balanced by friction and the body stays at rest.
Question 33Physics
Two satellites orbit Earth at radii r and 4r. What is the ratio of their orbital periods T₁ : T₂?
- 1 : 8 Correct answer.
- 1 : 2
- 1 : 4
- 1 : 64
Explanation
Kepler's third law gives T² ∝ r³, so T ∝ r^(3/2). The radius ratio is 4, and 4^(3/2) = 8, so the periods are in the ratio 1 : 8. The orbital speed, by contrast, varies as 1/√r.
Question 34Physics
A body of mass 4 kg moving at 5 m/s collides with and sticks to a stationary body of mass 6 kg. What is the speed of the combined mass?
- 3 m/s
- 5 m/s
- 2 m/s Correct answer.
- 1.25 m/s
Explanation
Momentum is conserved in a perfectly inelastic collision: m₁u₁ = (m₁ + m₂)v. Thus v = (4 × 5)/(4 + 6) = 20/10 = 2 m/s. Kinetic energy is not conserved here because some is lost as deformation and heat.
Question 35Physics
A body of mass 2 kg is lifted vertically through 5 m at constant speed. Taking g = 10 m/s², what is the work done by the lifting force?
- 0 J
- 50 J
- 10 J
- 100 J Correct answer.
Explanation
At constant speed the lifting force equals the weight mg = 20 N. Work done by this force is force times displacement in the direction of the force: 20 × 5 = 100 J. This work is stored as gravitational potential energy.
Question 36Physics
A steel wire of length L and cross-sectional area A stretches by ΔL under a load. If a second wire of the same material has length 2L and area 2A under the same load, what is its extension?
- ΔL Correct answer.
- 4ΔL
- ΔL/2
- 2ΔL
Explanation
Extension is ΔL = FL/(AY), where Y is Young's modulus. Doubling both L and A leaves the ratio L/A unchanged, so the extension stays ΔL. Only the length-to-area ratio and the material determine the stretch for a given load.
Question 37Physics
A body floats with 60% of its volume submerged in a liquid. What is the ratio of its density to that of the liquid?
- 1.0
- 0.4
- 0.6 Correct answer.
- 1.67
Explanation
For a floating body, the weight equals the buoyant force, so ρ_body V g = ρ_liquid V_submerged g. Hence ρ_body/ρ_liquid = V_submerged/V = 0.6. A body floating with 60% submerged is therefore 60% as dense as the liquid.
Question 38Physics
The temperature of a fixed mass of ideal gas is doubled at constant pressure. What happens to its volume?
- It doubles Correct answer.
- It stays the same
- It becomes four times
- It halves
Explanation
Charles's law states that at constant pressure V/T is constant for a fixed mass of gas. Doubling the absolute temperature therefore doubles the volume. The temperature must be measured on the Kelvin scale for this proportionality to hold.
Question 39Physics
An ideal gas at 300 K is heated to 600 K at constant volume. What is the ratio of the final to initial pressure?
- 0.5
- 1
- 2 Correct answer.
- 4
Explanation
At constant volume, Gay-Lussac's law gives P/T constant, so P₂/P₁ = T₂/T₁. With absolute temperatures 600 K and 300 K the ratio is 2. Using Celsius temperatures here would give a meaningless ratio.
Question 40Physics
For an adiabatic process on an ideal gas, which relation holds between pressure and volume?
- PV = constant
- P/V = constant
- PV^γ = constant Correct answer.
- P + V = constant
Explanation
In a reversible adiabatic process no heat is exchanged, and combining the first law with the ideal-gas equation gives PV^γ = constant. Here γ = Cp/Cv, which exceeds 1 for all real gases. The relation PV = constant describes an isothermal, not an adiabatic, process.
Question 41Physics
A particle in simple harmonic motion has amplitude A. At what displacement is its kinetic energy equal to its potential energy?
- A
- A/√2 Correct answer.
- A/4
- A/2
Explanation
Total energy is ½kA², and potential energy is ½kx². Setting ½kx² equal to half the total gives x² = A²/2, so x = A/√2. At this point each form of energy is half the total.
Question 42Physics
A simple pendulum has period T on Earth. On a planet where the acceleration due to gravity is one quarter of Earth's, what is its period?
- T
- T/2
- 2T Correct answer.
- 4T
Explanation
The period of a simple pendulum is T = 2π√(L/g), so T ∝ 1/√g. Replacing g by g/4 multiplies the period by √4 = 2. The period is independent of the mass of the bob.
Question 43Physics
A transverse wave travels along a stretched string. Which quantity determines the wave speed?
- The frequency of the wave
- The phase of the wave
- The tension and the mass per unit length of the string Correct answer.
- The amplitude of the wave
Explanation
For a stretched string the wave speed is v = √(T/μ), where T is tension and μ is mass per unit length. Speed is fixed by the medium, and frequency adjusts to satisfy v = fλ. Amplitude and phase do not affect the speed.
Question 44Physics
A pipe closed at one end resonates in its fundamental mode with a frequency of 100 Hz. What is the frequency of the next higher resonance?
- 300 Hz Correct answer.
- 400 Hz
- 200 Hz
- 150 Hz
Explanation
A pipe closed at one end supports only odd harmonics: fn = (2n − 1)v/(4L). The fundamental is the first odd harmonic at 100 Hz, so the next allowed resonance is the third harmonic, 3 × 100 = 300 Hz. Even harmonics are forbidden because a node must exist at the closed end and an antinode at the open end.
Question 45Physics
A charge q is enclosed by a spherical Gaussian surface. If the radius of the surface is doubled while the charge stays inside, what happens to the electric flux through the surface?
- It doubles
- It remains unchanged Correct answer.
- It becomes one quarter
- It becomes half
Explanation
Gauss's law states that the flux through a closed surface equals q/ε₀, depending only on the enclosed charge. Changing the size or shape of the surface does not alter the enclosed charge, so the flux is unchanged. The field strength at the surface does fall, but the flux does not.
Question 46Physics
Two capacitors of 2 μF and 3 μF are connected in series across a 10 V supply. What is the equivalent capacitance?
- 5 μF
- 1.2 μF Correct answer.
- 0.83 μF
- 6 μF
Explanation
For series capacitors, 1/C = 1/C₁ + 1/C₂, so 1/C = 1/2 + 1/3 = 5/6 μF⁻¹, giving C = 6/5 = 1.2 μF. The equivalent capacitance in series is always less than the smallest individual capacitance.
Question 47Physics
A cell of emf 12 V and internal resistance 1 Ω is connected to an external resistance of 5 Ω. What is the terminal voltage of the cell?
- 12 V
- 7 V
- 10 V Correct answer.
- 2 V
Explanation
The current is I = ε/(R + r) = 12/6 = 2 A. Terminal voltage is V = ε − Ir = 12 − 2 × 1 = 10 V, which also equals IR = 2 × 5. The terminal voltage is less than the emf whenever current flows.
Question 48Physics
A charged particle moves parallel to a uniform magnetic field. What is the magnetic force on it?
- Along the velocity
- Zero Correct answer.
- Maximum, perpendicular to both velocity and field
- Opposite to the field
Explanation
The magnetic force is F = qvB sinθ, where θ is the angle between velocity and field. When the particle moves parallel to the field, θ = 0 and sinθ = 0, so the force vanishes. The force is greatest when velocity is perpendicular to the field.
Question 49Physics
An alternating voltage of peak value 200 V is applied across a pure resistor. What is the rms voltage?
- 400 V
- 100 V
- 200 V
- 141.4 V Correct answer.
Explanation
For a sinusoidal waveform the rms value is the peak value divided by √2. Thus V_rms = 200/1.414 ≈ 141.4 V. The rms value is the equivalent steady voltage that would dissipate the same average power.
Question 50Physics
In a photoelectric experiment, the intensity of incident light is doubled while its frequency is kept constant. What happens to the maximum kinetic energy of the emitted electrons?
- It doubles
- It remains the same Correct answer.
- It becomes four times
- It becomes half
Explanation
Einstein's photoelectric equation gives K_max = hf − φ, which depends only on the light frequency and the work function. Intensity changes the number of photons and hence the saturation current, not the energy of each electron. So the maximum kinetic energy is unchanged.