48 questions. Try each one, then open the answer to check yourself and read the explanation.
The physical quantity that has the same dimensions as impulse is
Answer: B (momentum)
Impulse = force × time = change in momentum, so both have dimensions MLT−1.
A ball is moving at 18 m s−1 in a direction inclined at 60° to the horizontal. The horizontal component of its velocity is
Answer: D (9 m s−1)
Horizontal component = 18 cos 60° = 18 × 0.5 = 9 m s−1.
A body on the ground is acted on by a force of 10 N at a point P as shown in the diagram above. What force is needed to stop the body from moving eastward?
Answer: B (5 N in the direction of West)
The 10 N force makes 60° with East, so its eastward component is 10 cos 60° = 5 N. A 5 N force towards the West cancels it.
In free fall, a body of mass 1 kg drops from a height of 125 m from rest in 5 s. How long will it take another body of mass 2 kg to fall from rest from the same height? [g = 10 m s−2]
Answer: A (5 s)
In free fall every body has the same acceleration g whatever its mass, so it also takes 5 s.
A ball of mass 0.15 kg is kicked against a rigid vertical wall with a horizontal velocity of 50 m s−1. If it rebounced with a horizontal velocity of 30 m s−1, calculate the impulse of the ball on the wall.
Answer: D (12.0 N s)
The velocity reverses, so the change is 50 − (−30) = 80 m s−1. Impulse = 0.15 × 80 = 12.0 N s.
The force of attraction between two point masses is 10−4 N when the distance between them is 0.18 m. If the distance is reduced to 0.06 m, calculate the force.
Answer: D (9.0 × 10−4 N)
F ∝ 1/r2. The distance is divided by 3, so the force is multiplied by 9: 9.0 × 10−4 N.
In an experiment, a constant force is applied to several masses (m) and the corresponding accelerations (a) are measured. Which of the following graphs correctly represents the experiment?
Answer: A (III and IV only)
With F constant, a = F/m. So a against 1/m is a straight line through the origin (III), and a against m is a curve falling as m grows (IV).
A uniform metre rule weighing 0.5 N is to be pivoted on a knife-edge at the 30 cm-mark. Where will a force of 2 N be placed from the pivot to balance the metre rule?
Answer: D (5 cm)
The weight acts at the 50 cm mark, 20 cm from the pivot: moment = 0.5 × 20 = 10 N cm. The 2 N force must be 10 ÷ 2 = 5 cm from the pivot on the other side.
A bullet fired at a wooden block of thickness 0.15 m manages to penetrate the block. If the mass of the bullet is 0.025 kg and the average resisting force of the wood is 7.5 × 103 N, calculate the speed of the bullet just before it hits the wooden block.
Answer: C (300 m s−1)
The bullet's kinetic energy is used up against the wood: ½ × 0.025 × v2 = 7.5 × 103 × 0.15 = 1125 J, so v2 = 90 000 and v = 300 m s−1.
A man whose mass is 80 kg climbs a staircase in 20 s and expends a power of 120 W. Find the height of the staircase. [g = 10 m s−2]
Answer: D (3.0 m)
Work done = power × time = 120 × 20 = 2400 J = mgh, so h = 2400 ÷ (80 × 10) = 3.0 m.
A parachute attains a terminal velocity when
Answer: B (the viscous force of the air and the upthrust completely counteract its weight)
At terminal velocity there is no resultant force: the upward forces (air resistance and upthrust) exactly balance the weight, so the speed stays constant.
In a wheel and axle mechanism, the diameters of the wheel and axle are 40 cm and 8 cm respectively. Given that the machine is 80% efficient, what effort is required to lift a load of 100 N?
Answer: B (25 N)
Velocity ratio = 40/8 = 5. Mechanical advantage = 0.8 × 5 = 4. Effort = load ÷ MA = 100 ÷ 4 = 25 N.
The tendon in a man's leg is 0.01 m long. If a force of 5 N stretches the tendon by 2.0 × 10−5 m, calculate the strain on the muscle.
Answer: C (2 × 10−3)
Strain = extension ÷ original length = 2.0 × 10−5 ÷ 0.01 = 2 × 10−3.
Which of these statements are correct for the pressures in liquids?
I. Pressure in a liquid at a point acts equally in all directions
II. Pressure increases with depth
III. Pressure at a depth depends on the shape of the container
IV. Pressure at the same depth in different liquids are proportional to the densities of the liquids.
Answer: B (I, II and IV only)
Liquid pressure p = hρg acts equally in all directions, grows with depth h and is proportional to density ρ. It does not depend on the container's shape, so III is false.
The atmospheric pressure due to water is 1.3 × 106 N m−2. What is the total pressure at the bottom of an ocean 10 m deep? [Density of water = 1000 kg m−3, g = 10 m s−2]
Answer: B (1.4 × 106 N m−2)
Water pressure at 10 m = hρg = 10 × 1000 × 10 = 1.0 × 105 N m−2. Adding the given surface pressure: 1.3 × 106 + 0.1 × 106 = 1.4 × 106 N m−2.
A solid of weight 0.600 N is totally immersed in oil and water respectively. If the upthrust in oil is 0.210 N and the relative density of oil is 0.875 find the upthrust in water.
Answer: C (0.240 N)
Relative density = upthrust in oil ÷ upthrust in water, so upthrust in water = 0.210 ÷ 0.875 = 0.240 N.
A platinum resistance thermometer records 3.0 Ω at 0 °C and 8.0 Ω at 100 °C. If it records 6.0 Ω in a certain environment, the temperature of the medium is
Answer: B (60 °C)
θ = (6.0 − 3.0)/(8.0 − 3.0) × 100 = 3/5 × 100 = 60 °C.
The linear expansivity of brass is 2 × 10−5 °C−1. If the volume of a piece of brass is 15.00 cm3 at 0 °C, what is the volume at 100 °C?
Answer: C (15.09 cm3)
Cubic expansivity = 3 × 2 × 10−5 = 6 × 10−5 °C−1. Increase = 15.00 × 6 × 10−5 × 100 = 0.09 cm3, so the volume is 15.09 cm3.
A gas has a volume of 100 cm3 at 27 °C. If it is heated to temperature T until a final volume of 120 cm3 is attained, calculate T.
Answer: C (87 °C)
Charles' law: V/T is constant (T in kelvin). T = 300 × 120/100 = 360 K = 87 °C.
A 500 W heater is used to heat 0.6 kg of water from 25 °C to 100 °C in t1 seconds. If another 1000 W heater is used to heat 0.2 kg of water from 10 °C to 100 °C in t2 seconds, find t1/t2.
Answer: B (5)
t = mcΔθ/P. t1 = 0.6 × c × 75/500 = 0.09c and t2 = 0.2 × c × 90/1000 = 0.018c, so t1/t2 = 5.
How many grams of water at 17 °C must be added to 42 g of ice at 0 °C to melt the ice completely? [Specific latent heat of fusion of ice = 3.4 × 105 J kg−1, Specific heat capacity of water = 4200 J kg−1 K−1]
Answer: A (200 g)
Heat to melt the ice = 0.042 × 3.4 × 105 = 14 280 J. The water cools from 17 °C to 0 °C: m × 4200 × 17 = 14 280, so m = 0.2 kg = 200 g.
One valid assumption of the kinetic theory of gases is that
Answer: D (the molecules of gas are all identical and are very small in size)
Kinetic theory assumes a gas is made of identical molecules that are tiny compared with the distances between them.
The physical properties of sound waves can best be described by
Answer: A (reflection and diffraction)
Sound is a longitudinal wave, so it cannot be polarized. It does show reflection (echoes) and diffraction (bending round corners).
The velocity of a sound wave at 27 °C is 360 m s−1. Its velocity at 127 °C is
Answer: C (240√3 m s−1)
v ∝ √T (in kelvin). v = 360 × √(400/300) = 360 × 2/√3 = 240√3 m s−1.
In a closed organ pipe producing a musical note, an antinode will always be produced at
Answer: B (the open end)
Air is free to move at the open end, so there is always an antinode there; the closed end is always a node.
A steel wire of length 0.50 m is stretched between two fixed points and its fundamental frequency is 200 Hz. The speed of the wave in the wire is
Answer: C (200 m s−1)
For the fundamental, λ = 2L = 1.0 m. v = fλ = 200 × 1.0 = 200 m s−1.
In a resonance tube experiment, if the fundamental frequency of the vibrating air column is 280 Hz, the frequency of the third overtone is
Answer: D (1960 Hz)
A resonance tube is closed at one end, so only odd harmonics occur: 280, 840, 1400, 1960 Hz. The third overtone is the 7th harmonic, 7 × 280 = 1960 Hz.
An object O lies at a distance m in front of concave mirror of focal length f. If m < f, then the final image obtained will be
Answer: B (magnified and erect)
An object inside the focal length of a concave mirror gives a virtual, erect and magnified image.
An object is placed in front of two plane mirrors inclined at an angle of θ°. If the total number of images formed is 7, find the value of θ.
Answer: B (45°)
Number of images n = 360/θ − 1. 7 = 360/θ − 1 gives 360/θ = 8, so θ = 45°.
The most suitable type of mirror used for the construction of a searchlight is the
Answer: D (parabolic mirror)
A parabolic mirror with the lamp at its focus sends out a strong parallel beam, which is what a searchlight needs.
The displacement d produced in a glass block of thickness t and refractive index n when an object is viewed through it is
Answer: C (t(1 − 1/n))
The apparent depth is t/n, so the object seems raised by d = t − t/n = t(1 − 1/n).
The correct shape of the graph of uv against (u + v) for an object distance u and image distance v in an experiment to find the focal length of a convex lens is given as
Answer: A (Graph A)
From 1/f = 1/u + 1/v, uv = f(u + v). So uv against (u + v) is a straight line through the origin with slope f, which is graph A.
A man wears convex lens glasses of focal length 30 cm in order to correct his eye defect. Instead of the optimum 25 cm, his least distance of distinct vision is
Answer: D (150 cm)
The lens must form a virtual image of an object at 25 cm at his near point: 1/30 = 1/25 − 1/v gives 1/v = 1/150, so his near point is 150 cm.
Which of the following are true about infra-red radiation?
I. It is invisible
II. It is called heat ray
III. Its frequency is higher than that of blue light.
IV. It travels as a transverse wave.
Answer: B (I, II and IV only)
Infra-red is invisible, is known as heat radiation and, like all electromagnetic waves, is transverse. Its frequency is lower than that of visible light, so III is false.
The force of repulsion between two point positive charges 5 μC and 8 μC separated at a distance of 0.02 m apart is [1/(4πε0) = 9 × 109 N m2 C−2]
Answer: C (9.0 × 102 N)
F = kq1q2/r2 = 9 × 109 × 5 × 10−6 × 8 × 10−6 ÷ (0.02)2 = 0.36 ÷ 0.0004 = 900 N.
An ebonite rod rubbed with fur attracts a glass rod rubbed with silk because
Answer: A (ebonite has a negative charge while glass has a positive charge)
Ebonite rubbed with fur gains electrons (negative); glass rubbed with silk loses electrons (positive). Unlike charges attract.
Which of the following can be used to reduce local action in a Leclanche cell?
Answer: B (Pure zinc as the negative pole)
Local action is caused by impurities in the zinc forming tiny cells. Using pure (or amalgamated) zinc prevents it.
The total resistance measured at PQ in the diagram above is
Answer: C (4.0 Ω)
The three 2 Ω resistors in series (6 Ω) are in parallel with the 3 Ω: 6 × 3/(6 + 3) = 2 Ω. Adding the two 1 Ω resistors gives 1 + 2 + 1 = 4.0 Ω.
Find the work done in moving a 2 C charge between two points X and Y in an electric field if the potential difference is 100 volts.
Answer: C (200 J)
W = QV = 2 × 100 = 200 J.
A parallel plate capacitor has a common plate area of 5 × 10−8 m2 and plate separation of 2 × 10−3 m. Assuming free space, what is the capacitance? [ε0 = 9.0 × 10−12 C2 N−1 m−2]
Answer: C (2.25 × 10−16 F)
C = ε0A/d = 9.0 × 10−12 × 5 × 10−8 ÷ (2 × 10−3) = 2.25 × 10−16 F.
The iron core of an induction coil is made from bundles of wires so as to
Answer: A (minimize eddy-currents)
Splitting the core into thin insulated wires breaks up the paths of eddy currents, so less energy is wasted as heat in the core.
Electricity is supplied to a school along a cable of total resistance 0.5 Ω with the maximum current drawn from the mains as 100 A. The maximum energy dissipated as heat for 1 hr is
Answer: D (1.8 × 107 J)
H = I2Rt = 1002 × 0.5 × 3600 = 1.8 × 107 J.
In the a.c. circuit diagram above, the resonance frequency is
Answer: D (2500/π Hz)
f0 = 1/(2π√(LC)) = 1/(2π√(8 × 10−3 × 5 × 10−6)) = 1/(2π × 2 × 10−4) = 2500/π Hz.
A transformer which can produce 8 V from a 240 V a.c. supply, has an efficiency of 80%. If the current in the secondary coil is 15 A, calculate the current in the primary coil.
Answer: A (0.625 A)
Output power = 8 × 15 = 120 W. Input power = 120 ÷ 0.8 = 150 W, so the primary current = 150 ÷ 240 = 0.625 A.
The electrochemical equivalent of a metal is 1.3 × 10−7 kg C−1. The mass of the metal which 2.0 × 104 C of electricity will deposit from a suitable electrolyte is
Answer: D (2.6 × 10−3 kg)
Faraday's first law: m = zQ = 1.3 × 10−7 × 2.0 × 104 = 2.6 × 10−3 kg.
A radioactive substance has a half-life of 80 days. If the initial number of atoms in the sample is 6.00 × 1010, how many atoms would remain at the end of 320 days?
Answer: A (3.75 × 109)
320 days is 4 half-lives, so (½)4 = 1/16 remains: 6.00 × 1010 ÷ 16 = 3.75 × 109.
In a nuclear fusion experiment, the loss of mass amounts to 1.0 × 10−6 kg. The amount of energy obtained from the fusion is [Speed of light = 3.0 × 108 m s−1]
Answer: D (9.0 × 1010 J)
E = mc2 = 1.0 × 10−6 × (3.0 × 108)2 = 9.0 × 1010 J.
In photoelectric effect, electrons will leave the metal surface when illuminated by light of appropriate frequency if the photon energy is
Answer: A (greater than the work function)
An electron escapes only if the photon gives it more energy than the work function; the extra energy becomes its kinetic energy.