30 reviewed questions with answers and explanations.
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Question 4
In a hydrostatic model, the head-to-feet blood-pressure difference is 1.65 ×10⁴ Pa. Find the vertical height difference using density 1.1 ×10³ kg/m³ and g =10 m/s².
- 1.5 m
- 2.0 m
- 0.6 m
- 0.5 m
Answer and explanation
A: 1.5 m
For the stated hydrostatic model, ΔP =ρgh. Thus h =(1.65 ×10⁴)/(1.1 ×10³ ×10) =1.5 m.
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Question 5
If the net force acting on a particle remains zero, its linear momentum will
- Be constant
- Increase
- Increase then decrease
- Decrease
Answer and explanation
A: Be constant
The net force equals the time rate of change of momentum: F_net =dp/dt. If the net force remains zero, momentum has zero rate of change and stays constant.
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Question 7
On a wire’s elastic loading curve, force rises linearly from 0.10 N at extensionE =0.05 m to 0.20 N at extensionF =0.10 m. Find the additional energy stored fromE toF.
- 1.5 ×10⁻² J
- 7.5 ×10⁻³ J
- 7.5 ×10⁻¹ J
- 2.5 ×10⁻³ J
Answer and explanation
B: 7.5 ×10⁻³ J
Between extensions 0.05 m and 0.10 m, force rises linearly from 0.10 N to 0.20 N. The added elastic energy is the trapezium area:½(0.10 +0.20)(0.10 −0.05) =0.0075 J.
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Question 8
The separation between two 10 kg point masses is tripled. What fraction of their original gravitational attraction remains?
- One ninth
- One quarter
- One third
- One half
Answer and explanation
A: One ninth
Gravitational attraction varies inversely with separation squared. Replacing r by 3 r gives F_new/F_old =r²/(3 r)² =1/9. The masses remain unchanged.
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Question 10
The component of force along motion rises linearly from zero at displacement 0 toF newtons at displacementx metres. Find the work done over that interval.
- F/x J
- Fx² J
- Fx/2 J
- Fx J
Answer and explanation
C: Fx/2 J
Work is the area under the force-distance graph. A force rising linearly from zero toF over distancex forms a triangle, so W =½ ×x ×F =Fx/2.
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Question 12
Two forces act on a body:10 N north and 10 N east. Find the resultant magnitude and its direction measured from north.
- 20 N,45° west of north
- 10√2 N,45° west of north
- 10√2 N,45° east of north
- 20 N,45° east of north
Answer and explanation
C: 10√2 N,45° east of north
North and east components are perpendicular and both 10 N. The magnitude is√(10² +10²) =10√2 N, and equal components give 45° east of north.
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Question 15
Which eye defect is commonly corrected by adding cylindrical power to a spectacle lens?
- Presbyopia
- Chromatic aberration
- Myopia
- Astigmatism
Answer and explanation
D: Astigmatism
Regular astigmatism gives different focusing powers in different directions across the eye. A cylindrical lens supplies direction-dependent power to compensate for that difference.
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Question 16
Which listed quantity is transported from place to place by a travelling wave?
- Amplitude
- Wavelength
- Frequency
- Energy
Answer and explanation
D: Energy
A travelling wave transfers energy from one region to another. Amplitude, wavelength and frequency describe the wave, while energy is what the propagating disturbance transports.
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Question 18
A 0.6 m stretched string fixed at both ends has fundamental frequency 220 Hz. Find the transverse-wave speed.
- 264 ms-1
- 132 ms-1
- 66 ms-1
- 528 ms-1
Answer and explanation
A: 264 ms-1
The fundamental mode of a string fixed at both ends has half a wavelength along its length. Thus λ =2 L =1.2 m and v =fλ =220 ×1.2 =264 m/s.
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Question 19
Heat is carried away from a motor-car radiator’s fins by air flowing over them mainly through
- Radiation and conduction
- Radiation
- Conduction
- Convection
Answer and explanation
D: Convection
Air flowing over radiator fins carries heat away by forced convection. Heat first reaches the fin surfaces through the metal, but its transfer from the surface into moving air is convective.
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Question 21
At fixed liquid temperature, blowing relatively dry air across its surface aids evaporation mainly by
- Reducing vapour partial pressure in the air just above the liquid
- Reducing the liquid’s density
- Increasing the liquid’s exposed surface area
- Increasing the liquid’s temperature
Answer and explanation
A: Reducing vapour partial pressure in the air just above the liquid
Moving relatively dry air sweeps vapour away from just above the liquid. This lowers the local vapour partial pressure and reduces condensation back onto the liquid, increasing net evaporation.
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Question 22
Find the pressure of 3 mol of ideal gas at 27°C in volume 10⁻³ m³. Use R =8.3 J/(mol·K) and 27°C ≈300 K.
- 7.47 ×10⁵ Pa
- 2.49 ×10⁶ Pa
- 7.47 ×10⁶ Pa
- 2.49 ×10⁵ Pa
Answer and explanation
C: 7.47 ×10⁶ Pa
Using 27°C ≈300 K, the ideal-gas law gives P =nRT/V =(3 ×8.3 ×300)/10⁻³ =7.47 ×10⁶ Pa. Temperature must be in kelvin.
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Question 23
To produce an enlarged and erect image with a concave mirror, the object must be positioned
- Between the principal focus and the pole
- Between the principal focus and centre of curvature
- Beyond the centre of curvature
- At the principal focus
Answer and explanation
A: Between the principal focus and the pole
With an object inside the focal length of a concave mirror, reflected rays diverge and their backward extensions form an upright virtual image. Its magnification is greater than one.
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Question 24
The colours seen in soap bubbles are due to
- Refraction
- Diffraction
- Interference
- Dispersion
Answer and explanation
C: Interference
Light reflected from the front and back surfaces of the thin soap film takes different optical paths. Their interference reinforces some wavelengths and suppresses others, producing colours that vary with film thickness.
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Question 25
The fading persistence of sound in a room through many overlapping reflections after the source stops is called
- Reverberation
- Acoustic vibration
- Rarefaction
- Echo
Answer and explanation
A: Reverberation
Reverberation is the fading persistence of sound caused by many overlapping reflections after the source stops. A distinct delayed repetition is called an echo.
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Question 29
Under Newton’s law of cooling with a constant heat-transfer coefficient, the heat-loss rate of a warmer body is proportional to
- Temperature of its surroundings
- Difference in temperature between the body and its surrounding
- Temperature of the body
- Ratio of the temperature of the body to that of its surrounding.
Answer and explanation
B: Difference in temperature between the body and its surrounding
Newton’s law of cooling models heat-loss rate as H =k(T_body −T_surroundings) when the heat-transfer coefficient is effectively constant. It is the temperature difference that drives the heat flow.
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Question 30
An electric iron’s resistive element is rated 1000 W at 230 V. Find its resistance at the rated operating condition.
- 57.6 Ohms
- 55.9 Ohms
- 51.9 Ohms
- 52.9 Ohms
Answer and explanation
D: 52.9 Ohms
For a resistive element, P =V²/R. Therefore R =230²/1000 =52.9 Ω at its rated operating condition.
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Question 31
An electric-field sketch shows arrows pointing away from each of two point chargesX andY. What are their signs?
- Both X and Y are positive
- X is positive and Y is negative
- X is negative and Y is positive
- Both X and Y are negative.
Answer and explanation
A: Both X and Y are positive
Electric field lines point away from positive charges and toward negative charges. Since the arrows emerge from bothX andY, both charges are positive.
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Question 32
Which part of the eye changes the pupil’s size to control the amount of light reaching the retina?
- Iris
- Optic nerve
- Cornea
- Retina
Answer and explanation
A: Iris
Muscles in the iris change the pupil’s diameter, controlling how much light enters the eye and reaches the retina. The iris therefore acts as the adjustable aperture.
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Question 36
A 10 μF capacitor carries charge 100 μC. Find its stored energy.
- 5 ×10⁻⁴ J
- 4 ×10⁻³ J
- 4 ×10² J
- 5 ×10⁴ J
Answer and explanation
A: 5 ×10⁻⁴ J
Capacitor energy is U =Q²/(2 C). With Q =100 ×10⁻⁶ C and C =10 ×10⁻⁶ F, U =(10⁻⁴)²/(2 ×10⁻⁵) =5 ×10⁻⁴ J.
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Question 37
Across 12 V, a circuit draws 2 A through a 2 Ω resistor, a parallel pair of 3 Ω andX, and a 1.5 Ω resistor, in that series order. FindX.
- 15 Ω
- 12 Ω
- 9 Ω
- 6 Ω
Answer and explanation
A: 15 Ω
Total resistance is 12/2 =6 Ω. The parallel pair therefore has 6 −2 −1.5 =2.5 Ω. Solving 1/2.5 =1/3 +1/X gives 1/X =1/15, so X =15 Ω.
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Question 39
Which particle normally triggers the fission chain reaction in a nuclear fission reactor?
- Electron
- Neutron
- Photon
- Proton
Answer and explanation
B: Neutron
A neutron absorbed by a suitable fissile nucleus can trigger fission. The resulting fission releases more neutrons, which can sustain the chain reaction in a reactor.
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Question 40
At what frequency does a 2.5 μF capacitor have reactance 250 Ω?
- 200π Hz
- π/800 Hz
- 2000π Hz
- 800/π Hz
Answer and explanation
D: 800/π Hz
Capacitive reactance is X_C =1/(2πfC). Thus f =1/(2π ×250 ×2.5 ×10⁻⁶) =800/π Hz.
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Question 41
What percentage of the original undecayed nuclei remains after five half-lives, rounded to the nearest whole percent?
- 1 %
- 3 %
- 5 %
- 8 %
Answer and explanation
B: 3 %
After five half-lives, the fraction of original undecayed nuclei is(1/2)⁵ =1/32. As a percentage this is 3.125%, which rounds to 3% to the nearest whole percent.
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Question 43
In the nuclear reaction ²³₁₁Na +X → ²⁰₉F +⁴₂He, identify particleX.
- Neutron
- Alpha
- Gamma
- Beta
Answer and explanation
A: Neutron
Conservation of nucleon number gives 23 +A_X =20 +4, so A_X =1. Conservation of charge gives 11 +Z_X =9 +2, so Z_X =0. A particle with mass number 1 and charge number 0 is a neutron.
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Question 45
Which feature helps make a moving-coil galvanometer sensitive to small currents?
- Few turns in its coil
- A strong permanent magnet producing a strong field
- A small coil area
- Stiff control springs producing a large restoring couple
Answer and explanation
B: A strong permanent magnet producing a strong field
A stronger magnetic field gives a larger turning effect on the current-carrying coil. This improves current sensitivity when coil turns, area and spring stiffness are unchanged.
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Question 46
Pure silicon can be converted to a p-type material by adding a controlled amount of
- Pentavalent atoms
- Trivalent atoms
- Hexavalent atoms
- Tetravalent atoms
Answer and explanation
B: Trivalent atoms
Trivalent dopants have one fewer valence electron than silicon. They introduce acceptor states and holes, making holes the majority carriers in p-type silicon.
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Question 47
For a fixed power transmitted through a line, using a higher voltage mainly reduces
- Resistive heating in the transmission conductors
- Magnetic flux leakage
- Hysteresis loss
- Eddy-current loss
Answer and explanation
A: Resistive heating in the transmission conductors
For the same transmitted power, increasing voltage reduces current because I =P/V. Resistive heating in the transmission conductors is I²R, so reducing current sharply reduces this loss.
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Question 49
Find the energy of a photon of frequency 3.0 ×10⁵ Hz to two significant figures. Use h =6.63 ×10⁻³⁴ J·s.
- 2.0 ×10⁻²⁹ J
- 2.0 ×10⁻²⁸ J
- 1.3 ×10⁻²⁹ J
- 1.3 ×10⁻²⁸ J
Answer and explanation
B: 2.0 ×10⁻²⁸ J
Photon energy is E =hf =(6.63 ×10⁻³⁴)(3.0 ×10⁵) =1.989 ×10⁻²⁸ J. To two significant figures this is 2.0 ×10⁻²⁸ J.
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Question 50
The carbon-granule microphone works on the principle of change in
- Capacitance
- Voltage
- Inductance
- Resistance
Answer and explanation
D: Resistance
Sound pressure changes the contact between carbon granules, varying their electrical resistance. With an applied bias, this resistance change produces a corresponding electrical signal.
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