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2020 O-level Physics (6091) Paper 2 Suggested Solutions

All solutions here are suggested. Concept First will hold no liability for any errors.

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Answer Key to 2020 O Level Physics Paper 2:

[Maximum Marks: 5]

(a)

Vector: Displacement/ Velocity/ Acceleration/ Force/ Weight

Scalar: Distance/ Mass/ Volume/ Speed/ Time

(bi)

Speed of this point remains constant but its direction changes continuously throughout the rotation, therefore, its velocity changes continuously. Its velocity is +0.24 m/ s at one point and -0.24 m/ s after one complete revolution.

(bii)

Total distance travelled by the point in one rotation = 0.24 m/ s x 30 min x 60 s = 430 m

Circumference 430 m = πD

D = 137 m

Radius of circular path = \frac{137}{2} = 69 m

[Maximum Marks: 4]

(a)

Acceleration of the ball

(bi)

(bii)

See the diagram for X – when the velocity is zero at t = 0.2s

[Maximum Marks: 7]

(a)

The jetpack exerts an upward force on the man while the man’s weight exerts an equal but opposite downward force on the man. The 2 forces balance and there is no resultant force. Thus, man remains stationary in mid-air.

(bi)

Mass is the measure of the amount of matter in a body while weight refers to the amount of gravitational force acting on the body.

(bii)

Mass of the jetpack = \frac{{160\text{N}}}{{10\text{N/kg}}}=16 kg
Total mass of man and jetpack = 75 kg + 16 kg = 91 kg

(biii)

Resultant force on the man and jetpack = 91\text{kg}\times 0.20\text{m/}{{\text{s}}^{\text{2}}} = 18.2 N
Resultant force = Upward force – Total weight
18.2 N = Upward force – (91 kg) (10 N/ kg)
Upward force = 928.2 N

[Maximum Marks: 7]

(a)

Work done by a constant force is the product of the constant force and the distance moved by object in the direction of this force.

(b)

At the stringed area, the shuttlecock has the highest amount of kinetic energy.

As the shuttlecock moves to a lower position at Y, it loses gravitational potential (GPE -> KE).

Also, as the shuttlecock leaves the stringed area, work is done against air resistance as it moves from the stringed area to point Y (some KE -> heat)

(c)

kinetic energy = \frac{1}{2}mv^{2}

0.36 J = \frac{1}{2}(5.0\times 10^{-3})v^{2}

v  = 12m/s

[Maximum Marks: 6]

(a)

Filament has small cross-sectional area compared to connecting wire, causing it to have much larger resistance. As current flows through wires, power loss in filament is higher than in connecting wires due to power dissipated being proportional to resistance P=I^{2}R.

(b)

Filament heats up due to current passing through it. This causes the gas surrounding the filament to heat up by conduction, expand (particles move further apart) & become less dense.

When gas becomes less dense, gas rises. The cooler gas around will sink, & take place of displaced gas & becomes heated in turn, setting up convection current above filament.

(c)

Glass is solid and heat transfer by conduction occurs most effectively through solids as the particles are closely packed together in fixed positions.

[Maximum Marks: 7]

(ai)

ultraviolet; infrared

(aii)

Ultraviolet/ X-ray/ Gamma rays

(aiii)

To disinfect equipment/ To kill cancer cells in cancer treatment

(bi)

All are transverse waves./ Obey law of reflection and refraction.

(bii)

Using v=f\lambda ,

3.0\times {{10}^{8}}\text{m/s}=f\times 2.0\times {{10}^{{-2}}}\text{m}

f=1.5\times 10^{10}\textup{ Hz}

[Maximum Marks: 6]

(ai)

The screen becomes positively charged by friction as the glass screen loses electrons to soft cloth.

(aii)

The dust particles can become charged by induction, with the side closer to the screen having more electrons since unlike charges attract, becoming negative. This causes the side further from the screen to have equal number of excess positive charges becoming positive. The attractive force between the electrons and the TV screen is greater than the repulsive force between the positive charges and the TV screen because of the shorter distance between the electrons and the TV. Hence, dust particles are attracted to the screen.

(bi)

The plastic rod is also positively charged. When the plastic rod is brought near the glass rod, the two rods repel one another, and caused the glass rod to exert a larger force on balance (Weight + Repulsive electrostatic force), causing the reading on the balance to increase.

(bii)

When there is an electric field around the plastic rod, a charged object (glass rod) will experience a force when it is in the electric field, and the force can be detected on the balance.

[Maximum Marks: 7]

(a)

(bi)

Effective resistance =\frac{12\textup{V}}{6.0\textup{A}}=2.0\Omega

Resistor R is calculated by using, 2.0=(\frac{1}{1.0}+\frac{1}{1.0})^{-1}+R

R=1.5\,\Omega

(bii)

Circuit breaker does not need to be replaced when the current exceeds the rating. It can simply be reset to work again once the problem is rectified.

[Maximum Marks: 11]

(a)

Point P: closed; closed

Point R: open; open/closed

(bi)

Maximum output voltage = 0.59 V x 4 = 2.36 V

(bii)

Maximum current = \frac{2.36\textup{ V}}{2.0\Omega} = 1.18 A

(ci)

Efficiency refers to the percentage of solar energy converted to electrical energy within the solar cell.

(cii)

At point P, while the current is the maximum, the potential difference is zero. And at point Q, while the voltage is maximum, the current is zero.

Since Power = Current x Voltage, the power dissipated is zero at both P and Q, no energy is produced and hence zero efficiency.

(ciii)

1. Since maximum output power of the cell = 2.0 W or 2.0 J/ s (at brightness = 1000 \text{W/}{{\text{m}}^{\text{2}}})

Efficiency of solar cell =\frac{18}{100}

Joules per second falling on the cell = 2.0\div\frac{18}{100} = 11.1 J/ s or 11.1 W

2. Surface area of cell can be calculated by proportion as,

\frac{{1000\,\text{W}}}{{11.1\,\text{W}}}=\frac{{1{{\text{m}}^{\text{2}}}}}{{\text{Area}}}

Area = 0.011 {{\text{m}}^{\text{2}}}

(civ)

For a single solar cell, ratio of maximum power output and brightness of the light is always a consistent value of 2.0\times10^{-3} regardless of the brightness of the brightness of the incident light.

Since total power input is directly proportional to brightness, its ratio to maximum power output remains constant regardless of brightness.

[Maximum Marks: 9]

(ai)

An alternating current refers to a current value that varies constantly from a maximum value in 1 direction to 0 and then to the same maximum value in the opposite direction and back to 0 many times in 1 second.

(aii)

The alternating current in the primary coil produces a constantly changing magnetic field within the soft iron core. The secondary coil around the soft iron core experiences a constantly changing magnetic flux linkage.

By Faraday’s Law of Electromagnetic Induction, an induced e.m.f. is produced in the secondary coil proportional to the rate of change of electromagnetic flux linkage experienced by the coil.

(bi)

Period of alternating current = 4 division x 10 ms/division = 4\times10^{-2}s

Frequency of alternating current =\frac{1}{\textup{period}}=\frac{1}{4\times10^{-2}}=25\textup{Hz}

(bii)

\frac{V_{s}}{V_{p}}=\frac{N_{s}}{N_{p}}

(biii)

Amplitude = 2 division x 5 V/division = 10 V

[Maximum Marks: 10]

(a)

Refractive Index =\frac{\textup{speed of light in vacuum}}{\textup{speed of light in the medium}}

(bi)

(bii)

Since 1.5=\frac{\sin r}{\sin 30^{\circ}}

\sin r=\sin 30^{\circ}\times1.5=0.75

r=48.6^{\circ}\textup{ or }49^{\circ}

Deviation from the original path = 49^{\circ}- 30^{\circ}=19^{\circ}

(ci)

(cii)

The critical angle of light in the prism is c=\sin^{-1}(\frac{1}{1.5})=41.8^{\circ}. The ray of light incident on the diagonal surface has an incident angle of 45^{\circ} which exceeds the critical angle of the prism.

Since light is travelling from optically denser medium to less dense medium, and the angle of incidence exceeds the critical angle, light undergoes total internal reflection.

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