  # 2019 O Level Physics 6091 Paper 2

## 2019 O-level Physics (6091) Paper 2 Suggested Solutions

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

Concept First is Singapore’s Most Comprehensive Physics Tuition Centre that specialises in Secondary Science Tuition, O Level Pure Physics Tuition, O Level Combined Physics Tuition, IP and A Level H2 Physics Tuition since 2016. Find out about our classes here!

# Answer Key to 2019 O Level Physics Paper 2:

[Maximum Marks: 6]

(a)

Accept any time between 1.1s and 1.8s

(b)

Since athlete is always travelling in the positive direction, his velocity is always positive. For t<4.8s, gradient of the velocity time graph and hence acceleration is positive (i.e. athlete speeds up).

For t>7s, gradient and hence acceleration is negative (athlete slows down).

(c)

Student can show that the area under the graph is equal to 100 m.

(d)

Average speed = \displaystyle \frac{{\text{total distance}}}{{\text{time}}}=\frac{{\text{100 m}}}{{\text{9}\text{.8 s}}} = 10.2 m/s

[Maximum Marks: 5]

(a)

The centre of gravity is the point through which the entire weight of the object appears to act on.

(b) Mass of object = 250 g
Mass of object = 0.25 kg

Weight of object = 0.25 kg × 10 N/kg
Weight of object = 2.5 N

Moment of object about pin at A
= Force × ⊥ distance AX
= 2.5 N × 0.30 m
= 0.75 Nm

(c)

When G is below A, the line of action of weight directly passes through pivot A. Perpendicular distance is zero and hence, there is no net moment of the metal sheet’s weight about pivot A, hence the object stops rotating.

[Maximum Marks: 7]

(a)

Kinetic energy is the energy of an object due to its motion.

Gravitational potential energy is the stored energy in an object due to its vertical height from the ground.

(bi)

Kinetic energy \displaystyle =\frac{1}{2}m{{v}^{2}}=\frac{1}{2}\times 1.5\,\text{kg}\times {{(20\,\text{m/s})}^{2}}

KE = 300 J

(bii)

1. Gain in GPE = Loss in KE

Gain in GPE = 300 J − 180 J

Gain in GPE = 120 J

2. Gain in GPE = mgh

120 J = 1.5 kg × 10 N/kg × h

h = 8.0 m

[Maximum Marks: 7]

(a)

When the volume decreases, the number of particles per unit volume increases. This increases the frequency of collision of the air molecules with the inner walls of the pump, causing the pressure to increase.

(b)

In the liquid state, the water molecules are closely packed, so the distance between water molecules is very small. When a force is exerted, the distance between particles cannot be reduced further, and hence volume of water does not change.

(ci) (cii)

Difference in pressure = 1.8\times10^{5}-1.0\times10^{5}\textup{ Pa}=0.8\times10^{5}\textup{ Pa}

Difference in pressure =\rho gh

\displaystyle 0.8\times {{10}^{5}}\,\text{Pa}=14000\,\text{kg/}{{\text{m}}^{\text{3}}}\times 10\,\text{N/kg}\times h

h = 0.57 m

[Maximum Marks: 5]

(a)

Difference in time taken by sound = 3.6 s − 2.3 s = 1.3 s

Difference in distance travelled by sound = 200 m × 2 = 400 m

Speed of sound = \frac{400\textup{ m}}{1.3\textup{ s}} = 310 m/s

(bi)

The sound becomes softer when amplitude decreases.

(bii)

The pitch of the sound becomes higher when wavelength decreases.

Explanation: The frequency of the sound increases when it wavelength decreases.

[Maximum Marks: 7]

(a)

current; potential difference

(bi) (bii)

1. p.d. across 1000 Ω resistor = 6.0 V − 1.4 V = 4.6 V

Current through 1000 Ω resistor = \frac{4.6\textup{ V}}{1000\Omega }=4.6\times10^{-3}\textup{ A}

Since current through 1000 Ω resistor = current through thermistor, resistance of thermistor {{R}_{{XY}}}=\frac{{1.4\,\text{V}}}{{4.8\times {{{10}}^{{-3}}}\text{A}}}=304\,\Omega (3 s.f.)

2. When the p.d. across the 1000 Ω resistor increases, the p.d. across the thermistor decreases as the sum of the p.d. = e.m.f. of the battery. The p.d. of the thermistor decreases because its resistance decreases with higher temperature when placed in hot water.

[Maximum Marks: 6]

(a)

(Any one) Wind / Hydroelectricity / Geothermal

(bi)

1. Amount received in one year = 1500 kWh × $0.26 =$390

Number of years to equal initial cost = $6000 ÷$390 = 15.4 years (3 s.f.)

2. Energy output every year is consistent. / No maintenance cost was incurred in that time period.

(bii)

1. 150 kWh is the 15% that was converted by the solar cell.

Amount of solar energy incident =\frac{1500\textup{ kWh}}{15}\times1000=10000\textup{ kWh}

2. (Any one) Some of the solar energy could be reflected off the surface of the solar cells./ Some of the solar energy is used to heat up the cells.

[Maximum Marks: 7]

(a) (b)

There will be an increase in the number of field lines and the field lines will be more closely spaced together (indicating a stronger magnetic field).

(ci)

A: magnet

B: iron

C: maget

(cii)

For A and C to repel each other, both of them must be magnets so that when their like poles are put together, the repulsion occurs. When the same pole A (a magnet) is placed next to B on both ends, attraction occurs. This implies that B is a piece of unmagnetized iron that becomes an induced magnet when placed near a magnet and hence can only be attracted to the magnet.

[Maximum Marks: 12]

(ai)

Resistance is inversely proportional to the cross-sectional area of the wire.

(aii)

Since R=\frac{\textup{k}}{A}, hence R\times A will be a constant. Checking every row, RA=1.7\times10^{-7}

(bi)

E=11\times 10^{6}\times R

E=I^{2}Rt=(30000A)^{2}\times R\times (12\times 10^{-3}s)=11\times 10^{6}\times R (shown)

(bii)

Mass of copper rod = Volume × Density = \displaystyle (2.5\times {{10}^{{-5}}}\,{{\text{m}}^{\text{2}}}\times 10\,\text{m})\times 9000\,\text{kg/}{{\text{m}}^{\text{3}}} = 2.25 kg

From Table 9.1, when A=2.5\times {{10}^{{-5}}}{{\text{m}}^{\text{2}}}, R=6.8\times 10^{-3}\Omega

Since E=mc\theta, 11\times 10^{6}\times R = 2.25 \textup{kg}\times 390 J/(kg^{\circ}C)\times \theta

11\times 10^{6}\times 6.8\times 10^{-3} = 2.25 \textup {kg}\times 390 J/(kg^{\circ}C)\times \theta\theta =85{}^\circ \text{C}

(biii)

Iron has a higher resistance, and by E = (11\times10^{6})R, there will be heat generated in iron than in copper. Since both have similar specific heat capacity, this will mean that for iron, there will be a higher temperature rise since \theta=\frac{11\times 10^{6}\times R}{mc}.

Although iron has a higher melting point, the increase in temperature during a lighting strike outweighs its rise in melting point, hence, iron rod is more likely to melt.

(c)

When a high current flows through the fuse exceeding the rating, it will melt the fuse breaking the circuit. However, lightning rod provides a path of least resistance for large current in a lighting strike to flow safely to the ground.

(d)

When lighting strikes, current through each rod causes a circular clockwise magnetic field when viewed from top. The magnetic field interact and result in a stronger magnetic field surrounding the wires and a weaker magnetic field between the rods. This causes an attractive magnetic force between the 2 rods. [Maximum Marks: 8]

(ai)

Infra-red

(aii)

Ultraviolet

(b)

Effective output power of the microwave = 850 W × \frac{45}{100} = 382.5 W

Total energy supplied in 120 s = P × t = 382.5 W × 120 s = 45 900 J

Since E = ml,

45 900 J = m × 2.3 × 10^{6} J/kg

m = 0.020 kg

(c)

1. Killing of cells

2. Heating up of cellular tissue

3. Ionisation of cells which can lead to abnormal cell division / Mutations

[Maximum Marks: 10]

(a)

Unbalanced forces can cause an object to accelerate. This can cause either a change in its speed or direction of motion or both.

(b) (ci)

1. They have the same magnitude.

2. They have the same nature.

Note: The weight of the book is a non-contact force while the reaction force on the book is the contact force. These two forces are not of the same nature and hence, they are not an action-reaction pair.

(cii)

1. They act on different bodies

2. They act in opposite directions

(ciii) The attractive pull on the book on Earth with a force of 10 N. [Maximum Marks: 10]

(a)

Direct the ray to the center of the semicircle as indicated by O (shown below). A refracted ray will emerge from the side AB. Rotate the ray of light about O towards A and observe the refracted ray rotating towards B. When the refracted ray of light is along the side OB, measure the angle θ with a protractor. This is the critical angle for the glass. (bi) (bii)

1. Since n =\frac{\sin i}{\sin r} , and the angle of incidence i is the same, if the refractive index n is larger, the angle of refraction r will be smaller. Blue light will bend towards the normal more than red light in the drop of water.

2. Refer to (bi)

(biii)

\frac{c_{a}}{c_{w}}=1.3 (since refractive index = \frac{\textup{speed of light in vacuum}}{\textup{speed of light in medium}})

c_{w}=\frac{3\times 10^{8}\textup{ m/s}}{1.3}

4.3\times 10^{14}\textup {Hz}\times \lambda =\frac{3\times 10^{8}m/s}{1.3} (since c_{w}=f \lambda)

\lambda=5.4\times 10^{-7} m

Concept First is the First and Only Physics Tuition Centre with
Fully-Equipped Physics Lab Sessions in Singapore.

## Keen to find out more?   