48 questions. Try each one, then open the answer to check yourself and read the explanation.
A bead travelling on a straight wire is brought to rest at 0.2m by friction. If the mass of the bead is 0.01kg and the coefficient of friction between the bead and the wire is 0.1, determine the work done by the friction. [g = 10ms−2]
Answer: B (2 × 10−3J)
Friction = μmg = 0.1 × 0.01 × 10 = 0.01 N; work = 0.01 × 0.2 = 2 × 10−3 J.
On top of a spiral spring of force constant 500Nm−1 is placed a mass of 5 × 10−3 kg. If the spring is compressed downwards by a length of 0.02m and then released, calculate the height to which the mass is projected. [g = 10ms−2]
Answer: C (2m)
Energy stored = ½ke2 = ½ × 500 × 0.022 = 0.1 J. This becomes mgh, so h = 0.1 ÷ (5 × 10−3 × 10) = 2 m.
A test tube of radius 1.0cm is loaded to 8.8g. If it is placed upright in water, find the depth to which it would sink. [g = 10ms−2, density of water = 1000kgm−3]
Answer: A (2.8cm)
It floats when it displaces 8.8 g of water, i.e. 8.8 cm3. πr2h = 8.8 with r = 1 cm gives h ≈ 2.8 cm.
A hose of cross-sectional area 0.5m2 is used to discharge water from a water tanker at a velocity of 60ms−1 in 20s into a container. If the container is filled completely, the volume of the container is
Answer: B (600m3)
Volume = area × speed × time = 0.5 × 60 × 20 = 600 m3.
If an object just begins to slide on a surface inclined at 30° to the horizontal, the coefficient of friction is
Answer: D (1/√3)
When sliding just begins, μ = tan θ = tan 30° = 1/√3.
A force of 100N is used to kick a football of mass 0.8kg. Find the velocity with which the ball moves if it takes 0.8s to be kicked.
Answer: D (100ms−1)
Impulse = change in momentum: Ft = mv, so v = 100 × 0.8 ÷ 0.8 = 100 m s−1.
Water does not drop through an open umbrella of silk material unless the inside of the umbrella is touched. This is due to
Answer: D (surface tension.)
Surface tension makes a skin of water across the tiny gaps in the cloth; touching it breaks the film and water leaks through.
The stylus of a phonograph record exerts a force of 7.7 × 10−2N on a groove of radius 10−5m. Compute the pressure exerted by the stylus on the groove.
Answer: A (2.45 × 108 Nm−2)
Area = πr2 = 3.14 × 10−10 m2; P = 7.7 × 10−2 ÷ 3.14 × 10−10 ≈ 2.45 × 108 N m−2.
A piece of stone attached to one end of a string is whirled round in a horizontal circle and the string suddenly cuts. The stone will fly off in a direction
Answer: C (tangential to the circular path)
Once the string breaks there is no centripetal force, so the stone moves straight on along the tangent at that point.
A 90cm uniform lever has a load of 30N suspended at 15cm from one of its ends. If the fulcrum is at the centre of gravity, the force that must be applied at its other end to keep it in horizontal equilibrium is
Answer: B (20N)
Fulcrum at 45 cm. Moments: 30 × (45 − 15) = F × 45, so F = 900/45 = 20 N.
The figure above shows two velocities v1 and v2. Which of the following diagrams correctly represents the vector difference w = v1 − v2?
Answer: C (Diagram C)
w = v1 − v2 means w + v2 = v1: w and v2 placed nose to tail must add up to v1. Diagram C shows exactly this.
A satellite is in a parking orbit if its period is
Answer: C (equal to the period of the earth)
A parking (geostationary) satellite goes round once in the time the earth takes to rotate once (24 h), so it stays over the same place.
T°C: 0, 5, 10, 15, 20, 40, 60
S.V.P (mmHg): 4.58, 6.51, 8.94, 12.67, 17.50, 55.10, 149.00
The table above shows the saturation vapour pressure against temperature in a certain town. If the vapour pressure in this town at 20°C is 10mmHg, what is the relative humidity?
Answer: B (57.0%)
RH = actual vapour pressure ÷ SVP at that temperature × 100 = 10 ÷ 17.5 × 100 ≈ 57%.
A tuning fork of frequency 340Hz is vibrated just above a cylindrical tube of height 1.2m. If water is slowly poured into the tube, at what maximum height will resonance occur? [speed of sound in air = 340ms−1]
Answer: A (0.95 m)
λ = 340/340 = 1 m. The shortest resonating air column is λ/4 = 0.25 m, so the water can rise to 1.2 − 0.25 = 0.95 m.
By what factor will the size of an object placed 10cm from a convex lens be increased if the image is seen on a screen placed 25cm from the lens?
Answer: B (2.5)
Magnification = v/u = 25/10 = 2.5.
Thermal equilibrium between two objects exists when
Answer: A (the temperatures of both objects are equal)
Heat flows only from hot to cold; when the two temperatures are equal there is no net flow, which is thermal equilibrium.
An open pipe closed at one end produces its first fundamental note. If the velocity of sound in air is v and l the length of the pipe, the frequency of the note is
Answer: C (v/4l)
For a pipe closed at one end the fundamental has λ = 4l, so f = v/(4l).
Given that Young's modulus for aluminium is 7.0 × 1010Nm−2 and density is 2.7 × 103kgm−3, find the speed of the sound produced if a solid bar is struck at one end with a hammer.
Answer: A (5.1 × 103ms−1)
v = √(E/ρ) = √(7.0 × 1010 ÷ 2.7 × 103) ≈ 5.1 × 103 m s−1.
On a fairly cool rainy day when the temperature is 20°C, the length of a steel railroad track is 20m. What will be its length on a hot dry day when the temperature is 40°C? [coefficient of linear expansion of steel = 11 × 10−6 K−1]
Answer: C (20.004m)
Increase = αLΔθ = 11 × 10−6 × 20 × 20 = 0.0044 m, so the length is about 20.004 m.
If an object is placed between two parallel plane mirrors with their reflecting surfaces facing each other, how many images of the object will be formed?
Answer: A (Infinite)
For parallel mirrors the angle between them is 0°, so n = 360/0 − 1 is infinite: the reflections go on forever.
The phenomenon whereby the water droplets in the atmosphere combine with dust particles in the air to reduce visibility is
Answer: D (fog.)
Fog forms when water vapour condenses on dust particles near the ground, forming droplets that cut down visibility.
At what position will an object be placed in front of a concave mirror in order to obtain an image at infinity?
Answer: B (At the principal focus)
Rays from an object at the principal focus leave the concave mirror parallel, so the image is at infinity.
If the distance from a point source of sound is doubled, by what factor does the intensity decrease?
Answer: D (0.25)
Intensity follows the inverse-square law, I ∝ 1/r2: doubling r makes the intensity (½)2 = 0.25 of its first value.
A 2000W electric heater is used to heat a metal object of mass 5kg initially at 10°C. If a temperature rise of 30°C is obtained after 10min, the heat capacity of the material is
Answer: B (4.0 × 104 J °C−1)
Heat supplied = 2000 × 600 = 1.2 × 106 J. Heat capacity = 1.2 × 106 ÷ 30 = 4.0 × 104 J °C−1.
If 1.2 × 106J of heat energy is given off in 1 sec. from a vessel maintained at a temperature gradient of 30Km−1, the surface area of the vessel is [thermal conductivity of the vessel = 400Wm−1K−1]
Answer: A (1.0 × 102m2)
Q/t = kA × (temperature gradient), so A = 1.2 × 106 ÷ (400 × 30) = 100 m2.
In the diagram above, if each of the resistors can dissipate a maximum of 18W without becoming excessively heated, what is the maximum power that circuit can dissipate?
Answer: A (27W)
The series 2 Ω resistor carries the full current: 18 = I2 × 2 gives I = 3 A. Total resistance = 2 + 1 = 3 Ω, so P = I2R = 9 × 3 = 27 W.
The operation of an optical fibre is based on the principle of
Answer: C (refraction of light)
Optical fibres work by total internal reflection, which happens when light inside the glass meets the boundary at more than the critical angle; of the choices only refraction of light is related.
In the diagram above, what would happen to the current, I, if another resistor, R2, is connected in parallel to R1?
Answer: C (It will increase because the effective resistance will decrease.)
Adding any resistor in parallel lowers the total resistance, so more current flows from the same supply.
A ray of light is incident on an equilateral triangular glass prism of refractive index 3/2. Calculate the angle through which the ray is minimally deviated in the prism.
Answer: B (37.2°)
n = sin((A + D)/2) ÷ sin(A/2). With A = 60° and n = 1.5: sin((60 + D)/2) = 0.75, so (60 + D)/2 = 48.6° and D ≈ 37.2°.
A wire of 5Ω resistance is drawn out so that its new length is two times the original length. If the resistivity of the wire remains the same and the cross-sectional area is halved, the new resistance is
Answer: B (20Ω)
R = ρL/A. Doubling L and halving A each double R, so R = 5 × 2 × 2 = 20 Ω.
An electron of charge 1.6 × 10−19C is accelerated between two metal plates. If the kinetic energy of the electron is 4.8 × 10−17J, the potential difference between the plates is
Answer: C (300V)
eV = KE, so V = 4.8 × 10−17 ÷ 1.6 × 10−19 = 300 V.
A magnetic field is said to exist at a point if a force is
Answer: B (exerted on a moving charge at the point)
A magnetic field acts only on moving charges (or currents); a stationary charge feels no magnetic force.
I. It retains its magnetism much longer than steel.
II. It is more easily magnetized than steel.
III. It is more easily demagnetized than steel.
IV. It produces a stronger magnet than steel.
Which combination of the above makes iron preferable to steel in the making of electromagnets?
Answer: B (II and III only)
An electromagnet must switch on and off, so its core must magnetise and demagnetise easily: soft iron does both, unlike steel.
Which of the following pairs of light rays shows the widest separation in the spectrum of white light?
Answer: A (Blue and red)
The order is red, orange, yellow, green, blue, indigo, violet; blue and red are furthest apart of the pairs given.
The most suitable cell used for short interval switches in electric bells is a
Answer: C (Leclanche cell)
The Leclanché cell is cheap and polarises when used continuously but recovers between short uses, so it suits electric bells.
A ray incident on a glass prism undergoes minimum deviation when the
Answer: D (incident angle is equal to the angle of emergence.)
At minimum deviation the ray passes through the prism symmetrically, so the angle of incidence equals the angle of emergence.
In the calibration of an ammeter using Faraday's laws of electrolysis, the ammeter reading is kept constant at 1.20A. If 0.990g of copper is deposited in 40 minutes, the correction to be applied to the ammeter is [e.c.e. of copper = 3.3 × 10−4gC−1]
Answer: C (0.05A)
m = zIt: 0.990 = 3.3 × 10−4 × I × 2400, so the true current I = 1.25 A. Correction = 1.25 − 1.20 = +0.05 A.
The count rate of a radioactive material is 800 count/min. If the half-life of the material is 4 days, what would the count rate be 16 days later?
Answer: C (50 count/min.)
16 days is 4 half-lives: 800 → 400 → 200 → 100 → 50 count/min.
A circuit has an area of 0.4m2 and consists of 50 loops of wire. If the loops are twisted and allowed to rotate at a constant angular velocity of 10 rad s−1 in a uniform magnetic field of 0.4T, the amplitude of the induced voltage is
Answer: D (80V)
Peak e.m.f. = NBAω = 50 × 0.4 × 0.4 × 10 = 80 V.
If the uncertainty in the measurement of the position of a particle is 5 × 10−10m, the uncertainty in the momentum of the particle is [h = 6.6 × 10−34Js]
Answer: C (1.32 × 10−24Ns)
Using Δx × Δp ≈ h: Δp = 6.6 × 10−34 ÷ 5 × 10−10 = 1.32 × 10−24 N s.
The force on a charge moving with velocity v in a magnetic field B is half of the maximum force when the angle between v and B is
Answer: C (30°)
F = qvB sin θ, which is greatest at 90°. Half of that needs sin θ = ½, so θ = 30°.
The maximum kinetic energy of the photoelectrons emitted from a metal surface is 0.34eV. If the work function of the metal surface is 1.83eV, find the stopping potential.
Answer: D (0.34V)
The stopping potential must cancel the maximum kinetic energy: eVs = 0.34 eV, so Vs = 0.34 V. The work function is not needed.
When an alternating current given by I = 10sin(120π)t passes through a 12Ω resistor, the power dissipated in the resistor is
Answer: B (600W)
Irms = 10/√2 A, so P = Irms2R = 50 × 12 = 600 W.
In a fission process, the decrease in mass is 0.01%. How much energy could be obtained from the fission of 1.0g of the material? [c = 3.0 × 108ms−1]
Answer: A (9.0 × 109J)
Mass lost = 0.01% of 1 g = 10−7 kg. E = mc2 = 10−7 × 9 × 1016 = 9.0 × 109 J.
In a semiconductor junction diode, as the depletion or barrier layer is forward-biased, the layer
Answer: B (narrows)
Forward bias pushes electrons and holes towards the junction, so the depletion layer becomes narrower and current flows.
An electron makes a transition from a certain energy level EK to the ground state E0. If the frequency of emission is 8.0 × 1014 Hz, the energy emitted is [h = 6.6 × 10−34Js]
Answer: B (5.28 × 10−19J)
E = hf = 6.6 × 10−34 × 8.0 × 1014 = 5.28 × 10−19 J.
The major difference between a pure semiconductor and a pure metal is that
Answer: C (the resistance of metals increases with temperature, while for semiconductors, it is the reverse)
Heating a metal makes its atoms vibrate more, raising resistance; heating a semiconductor frees more charge carriers, lowering resistance.
When a nucleus is formed by bringing protons and neutrons together, the actual mass of the formed nucleus is less than the sum of the masses of the constituent protons and neutrons. The energy equivalent of this mass difference is the
Answer: C (building energy)
The missing mass (mass defect) corresponds to the binding energy that holds the nucleus together.