48 questions. Try each one, then open the answer to check yourself and read the explanation.
A lead bullet of mass 0.05 kg is fired with a velocity of 200 m s−1 into a lead block of mass 0.95 kg. Given that the lead block can move freely, the final kinetic energy after impact is
Answer: A (50 J)
Momentum is conserved: 0.05 × 200 = (0.05 + 0.95)v, so v = 10 m s−1. Final KE = ½ × 1 × 102 = 50 J.
A ball of mass 0.1 kg is thrown vertically upwards with a speed of 10 m s−1 from the top of a tower 10 m high. Neglecting air resistance, its total energy just before hitting the ground is [g = 10 m s−2]
Answer: C (15 J)
Total energy is conserved, so it equals the starting energy: KE ½ × 0.1 × 102 = 5 J plus PE 0.1 × 10 × 10 = 10 J, giving 15 J.
The figure above represents a frictionless pulley system in which a weight W is in equilibrium with a weight of 40 N. Find the value of W.
Answer: C (40.0 N)
The same rope passes over a frictionless pulley, so the tension is the same on both sides. W balances the 40 N weight, so W = 40 N.
The diagram above shows a solid figure with base PQ and centre of gravity G on an inclined plane. Which of the following statements is correct?
Answer: A (The solid will fall over if the vertical line through G lies outside the base)
A body topples when the vertical line through its centre of gravity falls outside its base, because its weight then gives a turning moment about the edge.
Two bodies have masses in the ratio 3 : 1. They experience forces which impart to them accelerations in the ratio 2 : 9 respectively. Find the ratio of the forces the masses experience.
Answer: C (2:3)
F = ma, so F1/F2 = (3 × 2)/(1 × 9) = 6/9 = 2/3.
The diagram above shows the velocity time graph of a vehicle. Its acceleration and retardation respectively are
Answer: C (4.0 m s−2, 2.0 m s−2)
Acceleration = 80/20 = 4.0 m s−2 (0 to 20 s). Retardation = 80/(90 − 50) = 2.0 m s−2.
A particle of mass 10−2 kg is fixed to the tip of a fan blade which rotates with angular velocity of 100 rad s−1. If the radius of the blade is 0.2 m, the centripetal force is
Answer: B (20 N)
F = mω2r = 10−2 × 1002 × 0.2 = 20 N.
The inner diameter of a small test tube can be measured accurately using a
Answer: D (pair of vernier calipers.)
Vernier calipers have inside jaws for internal diameters and read to 0.01 cm; a micrometer screw gauge cannot reach inside a tube.
A gas at a volume Vo in a container at pressure Po is compressed to one fifth of its volume. What will be its pressure if it maintains its original temperature T?
Answer: D (5 Po)
At constant temperature PV is constant (Boyle's law). If the volume becomes Vo/5, the pressure becomes 5Po.
A piece of substance of specific heat capacity 450 J kg−1 K−1 falls through a vertical distance of 20 m from rest. Calculate the rise in temperature of the substance on hitting the ground when all its energies are converted into heat
Answer: B (4/9 °C)
mgh = mcθ, so θ = gh/c = (10 × 20)/450 = 4/9 °C (taking g = 10 m s−2).
The diagram above shows a lift pump with valves at P and Q. During a downward stroke of the piston
Answer: B (P is open while Q is closed)
On the downstroke the water below the piston is pushed up: it closes the bottom valve Q and opens the piston valve P so the water passes above the piston.
In the J-tube above, Y and X are on the same horizontal level and 30 cm3 of air is trapped above Y when the atmospheric pressure is 75 cm Hg. Calculate the volume of air trapped above Y when 15 cm Hg is now poured into the limb above X.
Answer: B (25 cm3)
The trapped air's pressure rises from 75 to 75 + 15 = 90 cm Hg. By Boyle's law 75 × 30 = 90 × V, so V = 25 cm3.
If the stress on a wire is 107 N m−2 and the wire is stretched from its original length of 10.00 cm to 10.05 cm. The Young's modulus of the wire is
Answer: D (2.0 × 109 N m−2)
Strain = 0.05/10.00 = 0.005. Young's modulus = stress/strain = 107/0.005 = 2.0 × 109 N m−2.
A solid weighs 10.0 N in air, 6.0 N when fully immersed in water and 7.0 N when fully immersed in a certain liquid X. Calculate the relative density of the liquid.
Answer: C (3/4)
Relative density = loss of weight in liquid ÷ loss of weight in water = (10 − 7)/(10 − 6) = 3/4.
I. A liquid boils when its saturated vapour pressure is equal to the external pressure
II. Dissolved substances in pure water lead to increase in the boiling point
III. When the external pressure is increased, the boiling point increases
IV. Dissolved substances in pure water decrease the boiling point.
Which combination of the above are peculiarities of the boiling point of a liquid?
Answer: A (I, II and III)
Statements I, II and III are true. IV is false: dissolved substances raise the boiling point, they do not lower it.
When the temperature of a liquid is increased, its surface tension
Answer: A (decreases)
Heating weakens the attraction between liquid molecules, so surface tension falls as temperature rises.
When the brakes in a car are applied, the frictional force on the tyres is
Answer: D (an advantage because it is in the opposite direction of motion of the car.)
Friction acts opposite to the car's motion, and that is what slows the car down, so here it is useful.
A man stands 4 m in front of a plane mirror. If the mirror is moved 1 m towards the man, the distance between him and his new image is
Answer: C (6 m)
The man is now 3 m from the mirror, and his image is 3 m behind it, so they are 3 + 3 = 6 m apart.
If a sound wave goes from a cold-air region to a hot-air region, its wavelength will
Answer: A (increase)
Sound travels faster in hot air while the frequency stays the same, so λ = v/f increases.
The inside portion of part of a hollow metal sphere of diameter 20 cm is polished. The portion will therefore form a
Answer: A (concave mirror of focal length 5 cm)
A polished inside surface curves inwards, so it is a concave mirror. Its radius of curvature is 10 cm, so f = r/2 = 5 cm.
The equation of a wave travelling along the positive x-direction is given by y = 0.25 × 10−3 sin (500 t − 0.025x). Determine the angular frequency of the wave motion.
Answer: C (5.00 × 102 rad s−1)
Compare with y = A sin(ωt − kx): the coefficient of t is ω = 500 = 5.00 × 102 rad s−1.
Calculate the mass of ice that would melt when 2 kg of copper is quickly transferred from boiling water to a block of ice without heat loss [Specific heat capacity of copper = 400 J kg−1 K−1; Latent heat of fusion of ice = 3.3 × 105 J kg−1]
Answer: A (8/33 kg)
Heat given by copper cooling from 100 °C to 0 °C = 2 × 400 × 100 = 80 000 J. Mass melted = 80 000/(3.3 × 105) = 8/33 kg.
The diagram above represents a transverse electromagnetic wave travelling with speed 3.0 × 108 m s−1. What is the frequency of the wave?
Answer: D (3.0 × 109 Hz)
Three complete waves fit in 0.3 m, so λ = 0.1 m. f = v/λ = 3.0 × 108/0.1 = 3.0 × 109 Hz.
The temperature gradient across a copper rod of thickness 0.02 m, maintained at two temperature junctions of 20 °C and 80 °C respectively is
Answer: B (3.0 × 103 K m−1)
Temperature gradient = temperature difference ÷ thickness = (80 − 20)/0.02 = 3000 = 3.0 × 103 K m−1.
Which of the following electromagnetic waves is least energetic?
Answer: A (Infra-red rays)
Photon energy rises with frequency. Of these, infra-red has the lowest frequency, so it is the least energetic; gamma rays are the most energetic.
The diagram above shows a magnet X with its South pole moved along a soft-iron bar PQ in the direction shown. After some time, the poles at P and Q respectively are
Answer: B (North-South)
In single-touch stroking the end where each stroke finishes gets the opposite pole to the stroking pole. The S pole is stroked from Q to P, so P becomes North and Q South.
Four cells each of e.m.f. 1.5 V and internal resistance of 4 Ω are connected in parallel. What is the effective e.m.f. and internal resistance of the combination?
Answer: D (1.5 V, 1 Ω.)
Identical cells in parallel keep the e.m.f. of one cell (1.5 V), and their internal resistances combine as 4/4 = 1 Ω.
In the circuit above, the ammeter reads a current of 5.0 A when R = 8 Ω and reads 7.0 A when R = 5 Ω. The value of the unknown resistance X is
Answer: D (2.5 Ω)
E = I(R + X) in both cases: 5(8 + X) = 7(5 + X), so 40 + 5X = 35 + 7X and X = 2.5 Ω.
An astronomical telescope is said to be in normal adjustment when the
Answer: D (final image is at infinity.)
In normal adjustment the final image is formed at infinity, so the eye views it relaxed.
Steel is more suitable for permanent magnet than iron because the former
Answer: D (retains magnetism more than iron.)
Steel is harder to magnetise but keeps its magnetism much longer than soft iron, which is what a permanent magnet needs.
The diagram above shows the prism arrangement in a
Answer: C (periscope)
Two right-angled prisms each turn the light through 90° by total internal reflection, so light entering at the top leaves lower down in the same direction: a periscope.
The velocities of light in air and glass are 3.0 × 108 m s−1 and 2.0 × 108 m s−1 respectively. If the angle of refraction is 30°, the sine of the angle of incidence is
Answer: D (0.75)
sin i/sin r = speed in air ÷ speed in glass = 1.5, so sin i = 1.5 × sin 30° = 1.5 × 0.5 = 0.75.
In the series a.c circuit shown above, the p.d across the inductor is 8 V r.m.s. and that across the resistor is 6 V r.m.s. The effective voltage is
Answer: B (10 V)
The inductor and resistor voltages are 90° out of phase, so V = √(82 + 62) = 10 V.
The core of an efficient transformer should consist of laminated pieces of metal in order to
Answer: D (reduce the heat produced by reducing the eddy current.)
Insulated laminations break up the paths of eddy currents, so less current flows in the core and less energy is wasted as heat.
In Fleming's right-hand rule, the thumb, the forefinger and the middle finger if held mutually at right angles represent respectively, the
Answer: A (motion, the field and the induced current)
In the right-hand (dynamo) rule: thuMb = Motion, First finger = Field, seCond finger = induced Current.
In the circuit above, the potential across each capacitor is 100 V. The total energy stored in the two capacitors is
Answer: C (2.5 × 10−2 J)
The 2 μF and 3 μF capacitors are in parallel: C = 5 μF. Energy = ½CV2 = ½ × 5 × 10−6 × 1002 = 2.5 × 10−2 J.
At what frequency would a 10 H inductor have a reactance of 2000 Ω?
Answer: C (100/π Hz)
XL = 2πfL, so f = 2000/(2π × 10) = 100/π Hz.
A 3000-W electric cooker is to be used on a 200 V mains circuit. Which of the fuses below can be used safely with the cooker?
Answer: D (20 A)
The cooker draws I = P/V = 3000/200 = 15 A. The fuse must be rated just above this, so only the 20 A fuse will not blow in normal use.
A galvanometer has a resistance of 5 Ω. By using a shunt wire of resistance 0.05 Ω, the galvanometer could be converted to an ammeter capable of reading 2 A. What is the current through the galvanometer?
Answer: C (20 mA)
The galvanometer and shunt share the same p.d.: I × 5 = 0.05 × (2 − I), so I = 0.1/5.05 ≈ 0.020 A = 20 mA.
In the diagram above, if the internal resistance of the cell is zero, the ratio of the powers P1 and P2 dissipated by R1 and R2 respectively, is
Answer: A (R2/R1)
R1 and R2 are in parallel across the cell, so both have the same voltage V. P = V2/R, so P1/P2 = R2/R1.
If Δx is the uncertainty in the measurement of the position of a particle along the x-axis and ΔPx is the uncertainty in the measurement of the linear momentum along the x-axis, then the uncertainty principle relation is given as
Answer: A (Δx ΔPx ≥ h)
Heisenberg's uncertainty principle says the product of the uncertainties in position and momentum cannot be smaller than about Planck's constant.
2311Na + Proton → pqX + alpha particle. What are the values of p and q respectively in the equation above?
Answer: C (20 and 10)
Mass numbers balance: 23 + 1 = p + 4, so p = 20. Atomic numbers balance: 11 + 1 = q + 2, so q = 10.
A semi-conductor diode is used in rectifying alternating current into direct current mainly because it
Answer: D (allows current to flow only in one direction.)
A diode conducts in one direction only, so it blocks the reverse half of each a.c. cycle.
In semi-conductors, the carriers of current at room temperature are
Answer: B (electrons and holes)
In a semiconductor, current is carried by free electrons and by holes (missing electrons) moving the other way.
What is the speed of a particle of mass 10−27 kg whose wavelength is 10−8 m? [h = 6.63 × 10−34 J s]
Answer: B (66.30 m s−1)
de Broglie: λ = h/(mv), so v = h/(mλ) = 6.63 × 10−34/(10−27 × 10−8) = 66.3 m s−1.
In a series R-L-C circuit at resonance, the voltages across the resistor and the inductor are 30 V and 40 V respectively. What is the voltage across the capacitor?
Answer: B (40 V)
At resonance XL = XC, and the same current flows through both, so VC = VL = 40 V.
Gamma rays are produced when
Answer: B (energy changes occur within the nucleus of atoms)
Gamma rays come from the nucleus when it drops from an excited energy state; stopping fast electrons produces X-rays instead.
In the Rutherford scattering experiment, a beam of alpha particles was fired at a thin gold film with some of the particles being considerably deflected. This shows that
Answer: B (the gold nucleus is positively charged and is concentrated in a tiny volume)
Only a few alpha particles were strongly repelled, which shows the positive charge and most of the mass are packed into a very small nucleus.