50 questions. Try each one, then open the answer to check yourself and read the explanation.
At what respective values of x, y and z would the unit of force, the newton, be dimensionally equivalent to MxLyTz?
Answer: B (1, 1, −2)
Force = mass × acceleration = kg × m s−2, so its dimensions are M1L1T−2.
The distance x m travelled by a particle in time t seconds is described by the equation x = 10 + 12t2. Find the average speed of the particle between the time interval t = 2 s and t = 5 s.
Answer: C (84 m s−1)
At t = 2 s, x = 10 + 48 = 58 m; at t = 5 s, x = 10 + 300 = 310 m. Average speed = distance ÷ time = 252 ÷ 3 = 84 m s−1.
A 5 kg block is released from rest on a smooth plane inclined at an angle of 30° to the horizontal. What is the acceleration down the plane? [g = 10 m s−2]
Answer: A (5.0 m s−2)
On a smooth slope only the component of weight along the slope acts, so a = g sin 30° = 10 × 0.5 = 5.0 m s−2.
An arrow of mass 0.1 kg moving with a horizontal velocity of 15 m s−1 is shot into a wooden block of mass 0.4 kg lying at rest on a smooth horizontal surface. Their common velocity after impact is
Answer: D (3.0 m s−1)
Momentum is conserved: 0.1 × 15 = (0.1 + 0.4)v, so v = 1.5 ÷ 0.5 = 3.0 m s−1.
Two bodies X and Y are projected on the same horizontal plane, with the same initial speed but at angles 30° and 60° respectively to the horizontal. Neglecting air resistance, the ratio of the range of X to that of Y is
Answer: A (1 : 1)
Range R = u2 sin 2θ/g. sin 60° = sin 120°, so both ranges are equal: 1 : 1.
Which of the following with respect to a body performing simple harmonic motion are in phase?
Answer: D (Force acting on the body and the acceleration)
By F = ma, force and acceleration always point the same way, so they are in phase. Force is opposite to displacement, and velocity is 90° out of phase with both.
A body of mass 2 kg moving vertically upwards has its velocity increased uniformly from 10 m s−1 to 40 m s−1 in 4 s. Neglecting air resistance, calculate the upward vertical force acting on the body. [g = 10 m s−2]
Answer: C (35 N)
a = (40 − 10)/4 = 7.5 m s−2. The force must hold up the weight and also accelerate it: F = m(g + a) = 2 × 17.5 = 35 N.
A planet has mass m1 and is at a distance r1 from the sun. A second planet has mass m2 = 10m1 and is at a distance of r2 = 2r1 from the sun. Determine the ratio of the gravitational forces experienced by the planets.
Answer: B (2 : 5)
F ∝ m/r2. F2/F1 = 10/22 = 5/2, so F1 : F2 = 2 : 5.
For what value of θ are the forces in the diagram above in equilibrium?
Answer: D (60°)
The 12 N string must balance the 6 N downward force vertically: 12 cos θ = 6, so cos θ = 0.5 and θ = 60°. Check: 12 sin 60° = 6√3 N balances the horizontal force.
The figure above shows a block of mass m sliding down a rough inclined plane QP at angle θ. The forces acting on the block along the plane QP are
Answer: B (mg sin θ and the force of friction)
Along the slope there is the component of weight mg sin θ pulling the block down and friction opposing its motion. The normal reaction and mg cos θ act at right angles to the slope.
A uniform light beam XZ is hinged at X and kept in equilibrium by the forces T and F as shown in the diagram above. If XO = 20 cm and OY = 30 cm, express T in terms of F.
Answer: D (T = 5F)
Take moments about X. T acts at O, 20 cm away, at 30° to the beam, so its moment is T sin 30° × 20 = 10T. F acts at Y, 50 cm away: 10T = 50F, so T = 5F.
An object of mass 100 g projected vertically upwards from the ground level has a velocity of 20 m s−1 at a height of 10 m. Calculate its initial kinetic energy at the ground level. [g = 10 m s−2, neglect air resistance]
Answer: C (30 J)
Energy is conserved. At 10 m: KE = ½ × 0.1 × 202 = 20 J and PE = 0.1 × 10 × 10 = 10 J. The initial KE was 20 + 10 = 30 J.
An electric water pump rated 1.5 kW, lifts 200 kg of water through a vertical height of 6 metres in 10 seconds. What is the efficiency of the pump? [g = 10 m s−2, neglect air resistance]
Answer: C (80.0%)
Useful power = mgh/t = 200 × 10 × 6 ÷ 10 = 1200 W. Efficiency = 1200/1500 × 100% = 80%.
For a rough inclined plane on which lies a body of weight W, the angle θ in the diagram above becomes the angle of friction if
Answer: A (tan θ = coefficient of static friction)
When the body is just about to slide, W sin θ = μW cos θ, so μ = tan θ. That angle is called the angle of friction.
A load of 20 N on a wire of cross-sectional area 8 × 10−7 m2, produces an extension of 10−4 m. Calculate Young's modulus for the material of the wire if its length is 3 m.
Answer: B (7.5 × 1011 N m−2)
E = (F/A) ÷ (e/L) = FL/(Ae) = (20 × 3)/(8 × 10−7 × 10−4) = 7.5 × 1011 N m−2.
A cube of sides 0.1 m hangs freely from a string. What is the upthrust on the cube when totally immersed in water? [Density of water is 1000 kg m−3, g = 10 m s−2]
Answer: D (10 N)
Upthrust = weight of water displaced = 0.001 m3 × 1000 kg m−3 × 10 = 10 N.
A piece of wire gauze can be made to lie on water because
Answer: C (the water surface has the effect of an elastic thin skin)
Surface tension makes the water surface behave like a stretched elastic skin, which can support a light gauze even though the metal is denser than water.
A liquid of mass 1.0 × 103 kg fills a rectangular tank of length 2.5 m and width 2.0 m. If the tank is 4 m high, what is the pressure at the middle of the tank? [g = 10 m s−2]
Answer: D (1.0 × 103 N m−2)
Volume = 2.5 × 2.0 × 4 = 20 m3, so density = 1000/20 = 50 kg m−3. At the middle (2 m deep), p = hρg = 2 × 50 × 10 = 1.0 × 103 N m−2.
An empty 60 litre petrol tank has a mass of 10 kg. Its mass when full of fuel of relative density 0.72 is
Answer: D (53.2 kg)
Fuel density = 0.72 × 1000 = 720 kg m−3, so 60 litres (0.060 m3) has mass 43.2 kg. Adding the 10 kg tank gives 53.2 kg.
The length of mercury thread when it is at 0 °C, 100 °C and at an unknown temperature θ is 25 mm, 225 mm and 175 mm respectively. The value of θ is
Answer: C (75.0 °C)
θ = (175 − 25)/(225 − 25) × 100 = 150/200 × 100 = 75.0 °C.
Equal masses of copper and rubber are raised to the same temperature. After sometime, the copper was observed to be at a lower temperature because
Answer: A (the specific heat capacity of copper is lower than that of rubber)
Copper has a lower specific heat capacity, so losing the same amount of heat makes its temperature fall more than rubber's.
Before starting a journey, the tyre pressure of a car was 3 × 105 N m−2 at 27 °C. At the end of the journey, the pressure rose to 4 × 105 N m−2. Calculate the temperature of the tyre after the journey assuming the volume is constant.
Answer: D (127 °C)
At constant volume P/T is constant (T in kelvin): T2 = 300 × 4/3 = 400 K = 127 °C.
A mass of a liquid at 30 °C is mixed with a mass of the same liquid at 70 °C and the temperature of the mixture is 45 °C. Find the ratio of the mass of the cold liquid to the mass of the other liquid.
Answer: B (5 : 3)
Heat gained = heat lost: mc × 15 = mh × 25, so mc : mh = 25 : 15 = 5 : 3.
A heating coil rated at 1000 W is used to boil off 0.5 kg of boiling water. The time taken to boil off the water is [Specific latent heat of vaporization of water = 2.3 × 106 J kg−1]
Answer: D (1.15 × 103 s)
Heat needed = mL = 0.5 × 2.3 × 106 = 1.15 × 106 J. Time = 1.15 × 106 ÷ 1000 = 1.15 × 103 s.
What happens when a certain quantity of pure ice is completely changed to water at 0 °C?
Answer: A (Latent heat is absorbed, the mass remains constant and the volume decreases)
Melting needs latent heat to be absorbed. The mass does not change, and because ice is less dense than water, the volume gets smaller.
The correct cooling curve for a molten substance such as naphthalene is represented by
Answer: D (Graph D)
The liquid cools, then the temperature stays constant while it solidifies (latent heat is given out), then the solid cools. Graph D shows the flat part.
What happens when a gas expands at constant temperature?
Answer: A (Its pressure decreases)
By Boyle's law, PV is constant at fixed temperature, so a larger volume means a lower pressure. The temperature, and so the kinetic energy of the molecules, stays the same.
Ripples on water and light waves are similar because both
Answer: B (can be refracted and diffracted)
Both are waves, so both can be refracted and diffracted. Both are transverse, and their speeds and frequencies are very different.
Under constant tension and constant mass per unit length, the note produced by a plucked string is 500 Hz when the length of the string is 0.90 m. At what length is the frequency 150 Hz?
Answer: A (3 m)
Frequency is inversely proportional to length: f1L1 = f2L2. 500 × 0.90 = 150 × L, so L = 3 m.
The colours seen in thin films of oil on the road and in soap bubbles are due to
Answer: B (interference)
Light reflected from the top and bottom of the thin film overlaps. Some colours add up and others cancel, depending on the film's thickness: this is interference.
The diagram above shows two wave forms P and Q at a particular instant of time. The two waves will interfere
Answer: C (constructively to produce a wave of a larger amplitude)
P and Q have crests and troughs at the same places (in phase), so they add up constructively, giving a larger amplitude.
If the load at the end of a sonometer wire is immersed in a bucket of water, the original fundamental frequency of the wire could be restored by
Answer: A (decreasing the length of the wire)
In water the load feels an upthrust, so the tension and the frequency fall. Since f = (1/2L)√(T/m), making L smaller raises the frequency back to its original value.
When the sun, the moon and the earth are as shown in the diagram above, an observer standing at X is in
Answer: A (penumbra and sees a partial eclipse)
X is in the partly shaded region (penumbra), where the moon hides only part of the sun, so the observer sees a partial eclipse.
A boy on looking into a mirror observed that his face appeared to have grown bigger. The boy must have been looking at a
Answer: B (concave mirror with his face between the focus and the mirror)
Only a concave mirror magnifies, and it gives an upright, enlarged image when the object is between the focus and the mirror (as in a shaving mirror).
Light of velocity 3.0 × 108 m s−1 is incident on a material of refractive index n. If the velocity of light is reduced to 2.4 × 108 m s−1 in the material, what is n?
Answer: D (1.25)
n = speed in air ÷ speed in material = 3.0 × 108 ÷ 2.4 × 108 = 1.25.
A patient with a sight defect has a least distance of distinct vision of 150 cm. For him to be able to read a material placed at a distance of 25 cm, what is the focal length of the glasses he should wear?
Answer: D (30.0 cm)
The converging lens must make a virtual image of the page (u = 25 cm) at his near point, 150 cm away on the same side. 1/f = 1/25 − 1/150 = 5/150, so f = 30.0 cm.
The electromagnetic waves that are sensitive to temperature changes are
Answer: D (infra-red rays)
Infra-red rays are heat radiation: they are given out by warm bodies and are detected by their heating effect, e.g. with a thermopile or blackened thermometer.
In the diagram above, which of the angles θ1, θ2, θ3 and θ4 is the angle of deviation of a ray of light passing through the glass prism, XYZ?
Answer: A (θ3)
The angle of deviation is between the incident ray extended forward (dashed line) and the emergent ray, which is θ3.
Which of the figures below shows the correct distribution of charges on an isolated positively charged hollow metal?
Answer: C (Figure C)
Charges on a hollow conductor repel each other and settle only on the outside surface, with none inside. Figure C shows this.
Three electric cells each of e.m.f. 1.5 V and internal resistance 1.0 Ω are connected in parallel across an external resistance of 2/3 Ω. Calculate the value of the current in the resistor.
Answer: C (1.5 A)
Identical cells in parallel give e.m.f. 1.5 V and internal resistance 1/3 Ω. I = 1.5 ÷ (2/3 + 1/3) = 1.5 A.
A galvanometer of internal resistance 50 Ω has a full scale deflection for a current of 5 mA. What is the resistance required to convert it to a voltmeter with full scale deflection of 10 V?
Answer: B (1950 Ω)
Total resistance = 10 ÷ 0.005 = 2000 Ω. Subtract the galvanometer's 50 Ω: a 1950 Ω resistor is needed in series.
If 8 × 10−2 J of work is required to move 100 μC of charge from a point X to a point Y in an electrical circuit, the potential difference between X and Y is
Answer: C (8.0 × 102 V)
V = W/Q = 8 × 10−2 ÷ (100 × 10−6) = 800 V = 8.0 × 102 V.
In a metre bridge experiment, two resistors 2 Ω and 3 Ω occupy the left and right gaps respectively. Find the balance point from the left side of the bridge.
Answer: B (40 cm)
At balance 2/3 = l/(100 − l), so 3l = 200 − 2l and l = 40 cm.
Two 50 μF parallel plate capacitors are connected in series. The combined capacitor is then connected across a 100-V battery. The charge on each plate of the capacitor is
Answer: B (2.50 × 10−3 C)
Two equal capacitors in series give 50/2 = 25 μF. Q = CV = 25 × 10−6 × 100 = 2.50 × 10−3 C.
What is the total electrical energy consumed by using an electric cooker rated 1000 W for 5 hrs?
Answer: C (1.8 × 107 J)
Energy = power × time = 1000 W × (5 × 3600 s) = 1.8 × 107 J.
The power dissipated in an a.c. circuit with an r.m.s. current of 5 A, r.m.s. voltage of 10 V and a phase angle of 60° is
Answer: A (25 W)
Average a.c. power P = IV cos φ = 5 × 10 × cos 60° = 25 W.
The voltage of the domestic electric supply is represented by the equation V = 311 sin 314.2t. Determine the frequency of the a.c. supply. [π = 3.142]
Answer: A (50.0 Hz)
Compare with V = V0 sin 2πft: 2πf = 314.2, so f = 314.2 ÷ 6.284 = 50.0 Hz.
In a purely inductive circuit, the current
Answer: A (lags behind the voltage in phase by 90°)
An inductor opposes changes in current, so in a pure inductor the current lags the voltage by 90°.
One of the features of the fission process is that
Answer: B (it leads to chain reaction)
Each fission releases extra neutrons that can split more nuclei, so fission can lead to a chain reaction. Its products are radioactive and some mass is turned into energy.
The graphite rods surrounding the uranium fuel rods in a nuclear reactor, are used to
Answer: C (slow down the neutrons and hence slow the nuclear process)
Graphite is a moderator: it slows down fast neutrons. Of the choices, only C says the graphite slows the neutrons; absorbing neutrons is the job of the control rods.