JAMB Physics 1998

26 reviewed questions with answers and explanations.

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Question 1

The physical quantity that has the same dimensions as impulse is

  1. Energy
  2. Momentum
  3. Surface tension
  4. Pressure.
Answer and explanation

B: Momentum

Impulse is force multiplied by time, with dimensions MLT⁻² ×T =MLT⁻¹. Momentum is mass multiplied by velocity and has the same dimensions.

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Question 6

Two point masses attract with force 10⁻⁴ N at separation 0.18 m. Find the force when their separation is 0.06 m.

  1. 1.1 ×10⁻⁵ N
  2. 3.3 ×10⁻⁵ N
  3. 3.0 ×10⁻⁴ N
  4. 9.0 ×10⁻⁴ N
Answer and explanation

D: 9.0 ×10⁻⁴ N

Gravitational force varies inversely with separation squared. Reducing distance from 0.18 m to 0.06 m multiplies the force by(0.18/0.06)² =9, giving 9.0 ×10⁻⁴ N.

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Question 7

A fixed net force F is applied to different masses. Which graphs correctly show acceleration a? I. a versus m: straight increasing line through the origin. II. a versus 1/m: straight decreasing line. III. a versus 1/m: straight increasing line through the origin. IV. a versus m: decreasing inverse curve.

  1. III and IV only
  2. II and IV only
  3. I and III only
  4. I and II only
Answer and explanation

A: III and IV only

For fixed force, a =F/m. Thus a plotted against m is a decreasing inverse curve, while a plotted against 1/m is a straight increasing line through the origin.

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Question 12

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?

  1. 20 N
  2. 25 N
  3. 50 N
  4. 80 N
Answer and explanation

B: 25 N

Velocity ratio is wheel diameter divided by axle diameter:40/8 =5. Efficiency equals mechanical advantage divided by velocity ratio, so MA =0.8 ×5 =4. Effort is 100/4 =25 N.

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Question 13

A tendon of original length 0.01 m extends by 2.0 ×10⁻⁵ m under a 5 N force. Find its longitudinal strain.

  1. 5 ×10⁶
  2. 5 ×10²
  3. 2 ×10⁻³
  4. 2 ×10⁻⁷
Answer and explanation

C: 2 ×10⁻³

Strain is extension divided by original length. Thus strain =(2.0 ×10⁻⁵)/0.01 =2.0 ×10⁻³. It has no unit because both lengths use metres.

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Question 14

For uniform liquids at rest, consider pressure measured above each liquid’s free-surface pressure. Which statements are correct? I. Pressure acts equally in all directions at a point. II. Pressure increases with depth. III. Pressure at a given depth depends on container shape. IV. At equal depths, pressure rises are proportional to liquid density.

  1. I, II and III only
  2. I, II and IV only
  3. I, III and IV only
  4. II, III and IV only
Answer and explanation

B: I, II and IV only

For static liquids, pressure at a point is isotropic. The pressure rise below a free surface isρgh: it increases with depth, is proportional to density at a fixed depth, and does not depend on container shape.

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Question 15

Water has surface pressure 1.3 ×10⁶ Pa. Find the total pressure 10 m below the surface, using density 1000 kg/m³ and g =10 m/s².

  1. 1.3 ×10⁷ Pa
  2. 1.4 ×10⁶ Pa
  3. 1.4 ×10⁵ Pa
  4. 1.0 ×10⁵ Pa
Answer and explanation

B: 1.4 ×10⁶ Pa

The water addsρgh =1000 ×10 ×10 =100000 Pa. Adding the specified surface pressure gives 1.3 ×10⁶ +0.1 ×10⁶ =1.4 ×10⁶ Pa.

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Question 18

Brass has linear expansivity 2 ×10⁻⁵ °C⁻¹ and volume 15.00 cm³ at 0 °C. Use the first-order isotropic expansion approximation to find its volume at 100 °C.

  1. 16.03 cm³
  2. 16.00 cm³
  3. 15.09 cm³
  4. 15.03 cm³
Answer and explanation

C: 15.09 cm³

For small isotropic expansion, volume expansivity is 3α. Thus V =15.00[1 +3(2 ×10⁻⁵)(100)] =15.00(1.006) =15.09 cm³.

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Question 21

What minimum mass of water at 17 °C melts 42 g of ice initially at 0 °C? Neglect heat loss and container heat capacity. Use L =3.4 ×10⁵ J/kg and cwater =4200 J/(kg·K).

  1. 200 g
  2. 300 g
  3. 320 g
  4. 400 g
Answer and explanation

A: 200 g

Melting the ice requires 0.042 ×3.4 ×10⁵ =14280 J. Water cooling from 17 °C to 0 °C supplies 71400 J per kilogram, so the minimum mass is 14280/71400 =0.200 kg =200 g.

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Question 23

Which is an assumption of the elementary kinetic theory for a pure ideal gas?

  1. Molecules move constantly and their collision rate can never change
  2. Molecule number must increase whenever pressure increases
  3. Collision rate remains fixed as temperature increases
  4. Molecules are identical and very small compared with their separation
Answer and explanation

D: Molecules are identical and very small compared with their separation

In the elementary kinetic model of a pure ideal gas, particles are identical and their individual size is negligible compared with their separation. Their collision rate can change with temperature and density.

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Question 24

Which pair describes behaviours of ordinary longitudinal sound waves in air?

  1. Reflection and diffraction
  2. Polarization and reflection
  3. Polarization and diffraction
  4. Polarization and refraction
Answer and explanation

A: Reflection and diffraction

Ordinary sound in air is longitudinal. It reflects from boundaries and diffracts around obstacles, but it cannot be polarized like a transverse wave.

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Question 25

Sound travels at 360 m/s in a gas at 27 °C. With unchanged gas composition and heat-capacity ratio, find its speed at 127 °C. Use K = °C +273.

  1. 120√3 m/s
  2. 240 m/s
  3. 240√3 m/s
  4. 720√3 m/s
Answer and explanation

C: 240√3 m/s

For the same ideal gas, sound speed is proportional to√T. The kelvin temperatures are 300 and 400, so v₂ =360√(400/300) =240√3 m/s.

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Question 26

In an ideal organ pipe closed at one end, neglect end correction. An air-displacement antinode occurs at

  1. The closed end
  2. The open end
  3. The middle
  4. All the parts of the pipe.
Answer and explanation

B: The open end

At the open end, air can move most freely, giving a displacement antinode. The closed end prevents air movement and is a displacement node; pressure nodes and antinodes are reversed.

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Question 29

A real object is between a concave mirror and its principal focus. Which description applies to its image?

  1. Virtual and diminished
  2. Magnified and erect
  3. Real and inverted
  4. Diminished and erect.
Answer and explanation

B: Magnified and erect

An object between a concave mirror and its focus forms an upright virtual image behind the mirror. The image is larger than the object.

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Question 31

Which specific mirror shape is most suitable for collimating light from a small source into a narrow searchlight beam?

  1. Concave mirror
  2. Convex mirror
  3. Spherical mirror
  4. Parabolic mirror.
Answer and explanation

D: Parabolic mirror.

A paraboloid sends rays from a small source at its focus into parallel rays, avoiding the spherical aberration of a spherical reflector. This makes the parabolic shape suitable for a narrow searchlight beam.

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Question 32

An object touches the bottom of a glass block of thickness t and refractive index n>1. Viewed nearly normally from air through the top face, its apparent upward shift d is

  1. t − n
  2. t(1 + 1/n)
  3. t(1 − 1/n)
  4. t(1/n − 1)
Answer and explanation

C: t(1 − 1/n)

Near normal viewing through glass gives apparent depth t/n. The apparent upward shift is actual depth minus apparent depth: d =t −t/n =t(1 −1/n).

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Question 35

Which statements describe infrared radiation? I. It is invisible to the human eye. II. It is commonly called heat radiation. III. Its frequency is higher than blue light. IV. In free space it is transverse.

  1. I, II, III and IV
  2. I, II and IV only
  3. I, III and IV only
  4. II, III and IV only
Answer and explanation

B: I, II and IV only

Infrared is outside human visible wavelengths and is commonly associated with thermal radiation. It is a transverse electromagnetic wave, with lower frequency than blue light.

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Question 36

In vacuum, point charges +5 μC and +8 μC are 0.02 m apart. Find their repulsive force using k =9 ×10⁹ N·m²/C².

  1. 1.8 ×10⁻¹⁰ N
  2. 9.0 ×10⁻⁸ N
  3. 9.0 ×10² N
  4. 4.5 ×10³ N
Answer and explanation

C: 9.0 ×10² N

Coulomb’s law gives F =kq₁q₂/r². Thus F =(9 ×10⁹)(5 ×10⁻⁶)(8 ×10⁻⁶)/(0.02)² =900 N. Both charges are positive, so the force is repulsive.

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Question 37

In the standard dry electrostatic experiment, an ebonite rod rubbed with fur attracts a glass rod rubbed with silk because

  1. Ebonite has a negative charge while glass has a positive charge
  2. Ebonite has a positive charge while glass has a negative charge
  3. Both have negative charges
  4. Both have positive charges.
Answer and explanation

A: Ebonite has a negative charge while glass has a positive charge

In the usual dry charging experiment, ebonite gains electrons from fur and becomes negative; glass loses electrons to silk and becomes positive. Opposite charges attract.

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Question 40

Terminal P connects through 1 Ω to node A, and terminal Q through 1 Ω to node B. Between A and B are two parallel paths: one 3 Ω resistor, and three 2 Ω resistors in series. Find the resistance between P and Q.

  1. 18.0 Ω
  2. 11.0 Ω
  3. 4.0 Ω
  4. 2.0 Ω
Answer and explanation

C: 4.0 Ω

The three 2 Ω resistors make a 6 Ω series path. This is in parallel with 3 Ω, giving 2 Ω. The two 1 Ω lead resistors are in series with it, so total resistance is 1 +2 +1 =4 Ω.

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Question 41

Find the external work required to move a +2 C charge slowly from X to Y when Y is 100 V above X.

  1. 50 J
  2. 100 J
  3. 200 J
  4. 400 J
Answer and explanation

C: 200 J

Work needed per charge is the potential rise. Moving 2 C through 100 V requires W =qΔV =2 ×100 =200 J.

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Question 43

An induction-coil iron core is subdivided into bundles of electrically insulated wires mainly to

  1. Minimize eddy-currents
  2. Generate eddy-currents
  3. Prevent sparking at the contact breaker
  4. Get the greatest possible secondary voltage.
Answer and explanation

A: Minimize eddy-currents

Separate insulated iron wires interrupt large circulating current paths in the core. This reduces eddy currents and the heating losses they would cause as magnetic flux changes.

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Question 44

A supply cable has total resistance 0.5 Ω. Its maximum RMS current is 100 A. Find the maximum heat dissipated in one hour if that current is maintained.

  1. 3.6 ×10³ J
  2. 5.0 ×10³ J
  3. 3.0 ×10⁵ J
  4. 1.8 ×10⁷ J
Answer and explanation

D: 1.8 ×10⁷ J

At 100 A, cable heating power is I²R =100² ×0.5 =5000 W. Sustaining this for 3600 s gives energy 5000 ×3600 =1.8 ×10⁷ J.

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Question 45

A sinusoidal source supplies a series circuit containing a 15 Ω resistor, 5 μF capacitor and 8 mH ideal inductor. Find the current-resonance frequency.

  1. 5000π Hz
  2. 2500π Hz
  3. 5000/π Hz
  4. 2500/π Hz
Answer and explanation

D: 2500/π Hz

In a series RLC circuit, current resonance occurs when ωL =1/(ωC). Here LC =(8 ×10⁻³)(5 ×10⁻⁶) =4 ×10⁻⁸, so f =1/(2π√LC) =2500/π Hz.

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Question 47

A metal has electrochemical equivalent 1.3 ×10⁻⁷ kg/C. What mass is deposited by 2.0 ×10⁴ C through a suitable electrolyte?

  1. 6.5 ×10⁻² kg
  2. 2.6 ×10⁻² kg
  3. 6.5 ×10⁻³ kg
  4. 2.6 ×10⁻³ kg
Answer and explanation

D: 2.6 ×10⁻³ kg

Deposited mass equals electrochemical equivalent multiplied by charge: m =zQ =(1.3 ×10⁻⁷)(2.0 ×10⁴) =2.6 ×10⁻³ kg.

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Question 49

A fusion reaction has a decrease in total rest mass of 1.0 ×10⁻⁶ kg. Find the released energy using c =3.0 ×10⁸ m/s.

  1. 3.0 ×10⁻⁴ J
  2. 3.0 ×10⁻¹ J
  3. 9.0 ×10⁴ J
  4. 9.0 ×10¹⁰ J
Answer and explanation

D: 9.0 ×10¹⁰ J

Released energy is the decrease in rest mass multiplied by c². Thus E =10⁻⁶(3 ×10⁸)² =9 ×10¹⁰ J.

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