## 2021 O-level Physics (6091) Paper 2 Suggested Solutions

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# Answer Key to 2021 O Level Physics Paper 2:

[Maximum Marks: 7]

(a)

kelvin / mole / candela / ampere

(bi)

10^{-9}

(bii)

M, k, d, c

(ci)

\textup{average speed}=\frac{\textup{total distance}}{\textup{total time}}=\frac{1}{2}(2\pi\times20)\div 7.5=8.3775\approx 8.38 \textup{ m/s}

(cii)

40.0 m

(ciii)

Average velocity = total displacement / total time. Since total displacement is zero as athlete reaches back same point, average velocity is zero.

[Maximum Marks: 7]

(a)

W=mg

m=\frac{W}{g}=\frac{61}{10}=6.10\textup{ kg}

(b)

F_{\textup{net}}=ma

95-61=(6.1)(a)

\therefore a=5.5737\approx 5.57 \textup{ m/s}^{2}

(ci)

Any two:

1. Surface area of the balloon

2. Speed of balloon

3. Density of surrounding air

(cii)

When air resistance = 34 N, total downward force = total upward force.

As forces are balanced, net force = 0. Since F_{net}=ma, a = 0. Hence, balloon continues moving upwards with uniform velocity.

[Maximum Marks: 7]

(a)

When a body is in equilibrium, the sum of clockwise moments about a pivot is equal to the sum of anti-clockwise moments about the same pivot.

(b)

Perpendicular distance from pivot P to the line of action of his weight

(ci)

Apply Principle of Moments,

Sum of clockwise moments = Sum of anti-clockwise moments

300 × 0.9 + 600 × 3.4 = F × 1.6

F = 1443.75 N = 1440 N (to 3 s.f.)

(cii)

Total upwards force = 600 + 300 + 1443.75 = 2343.75 ≈ 2340 N (to 3 s.f.)

[Maximum Marks: 4]

(a)

Pressure is the force acting per unit area.

(b)

1. Measure the vertical height of oil column using a long measuring tape.

2. P_{\textup{atm}} can then be found using the formula, P=h\rho g

where h = vertical height in m

\rho = density of oil in \textup{kg/m}^{3}

g = gravitational field strength in N/kg

and P = P_{\text{atm}} in Pa

[Maximum Marks: 7]

(a)

parallel, perpendicular, matter

(bi)

f=\frac{1}{T}=\frac{1}{4\times10^{-3}}=250\textup{ Hz}

(bii)

v=f\lambda

\lambda=\frac{v}{f}=\frac{340}{250}=1.36 \textup{ m}

(biii)

[Maximum Marks: 6]

(a)

focal length is the distance between the focal point / principal focus and the optical centre of the lens.

(bi)

The image cannot be captured on a screen.

(bii)

(biii)

3.80 cm

[Maximum Marks: 7]

(a)

(bi)

R=\frac{V}{I}=\frac{18.0}{1.50}=12.0 Ω

(bii)

Since resistance of variable resistor is maxed, total R cannot increase further. Hence, current cannot be smaller than than the minimum of 0.40 A.

(biii)

R_{2}=\frac{2}{2^{2}}\times R_{1}=6.00 Ω

[Maximum Marks: 5]

(a)

1. Magnetic field strength at P > Magnetic field strength at Q since P is nearer.

2. Magnetic field at P is out of paper while magnetic field at Q is into paper.

(bi)

Into the paper

(bii)

Using FLHR, where the index finger which represents direction of magnetic field, points downwards from N to S pole; middle finger which represents current points from left to right, causing thumb which represents the force to point into the paper.

[Maximum Marks: 12]

(a)

\textup{K.E.}=\frac{1}{2}mv^{2}=\frac{1}{2}(1400)(\frac{90000\textup{m}}{(60\times60)\textup{ s}})^{2}=437500≈438000 \textup{ J}

(bi)

Average energy used = \frac{140+160}{2} = 150 W h

(bii)

distance = \frac{\frac{80}{100}\times30000}{160} = 150 km

(biii)

mass = \frac{30000}{130} = 230.76 = 231 kg

(biv)

Claim: mass of petrol needed = \frac{1}{25}\times 230.76 = 9.2304 kg

Useful energy needed to travel max distance for electric car = \frac{80}{100}\times 30000 = 24000 W h

Useful energy provided by 9.2304 kg of petrol = (9.2304 \times 13000)\times \frac{20}{100} = 23999.04 = 24000 W h which means it can travel the max distance.

Hence, claim is true.

(bv)

E=Pt

\therefore t=\frac{E}{P}=\frac{30\textup{ kWh}}{7.4\textup{ kW}}=4.054=4.05\textup{ h}

(bvi)

P=VI \therefore I=\frac{P}{V}=\frac{50000\textup{ W}}{230\textup{ V}}=217 \textup{ A} which will exceed the maximum current in an ordinary home. Hence, this causes the circuit breaker to trip and results in a blackout.

[Maximum Marks: 8]

(ai)

Specific latent heat is the amount of thermal energy needed to change a unit mass of substance from either a solid to liquid or a liquid to gas, vice-versa, without a change in temperature.

(aii)

Latent heat of fusion is the amount of thermal energy required to change a substance from solid to liquid but latent heat of vaporisation is the amount of thermal energy required to change a liquid to gas without a change in temperature.

(bi)

melting occurs: B

boiling occurs: D

(bii)

1. kinetic energy increases; potential energy remains the same

2. kinetic energy remians the same; potential energy increases

(biii)

More time and more energy is needed to change the substance from liquid to gas than from solid to liquid state. Hence, latent heat of vaporisation is greater than the latent heat of fusion. Since mass of substance is constant, l_{v}=\frac{L_{v}}{\textup{m}}>\frac{L_{f}}{\textup{m}}=l_{f}

[Maximum Marks: 10]

(a)

e.m.f. is the work done by an electrical source to drive a unit charge through the entire circuit.

(bi)

\textup{R}_{\textup{T}}=\left (\frac{1}{60}+\frac{1}{60}+\frac{1}{30} \right )^{-1}=15.0\textup{ }\Omega

(bii)

potential difference

(ci)

resistance of R: constant

resistance of L: increases

(cii)

1. \varepsilon = \textup{V}_{\textup{R}} + \textup{V}_{\textup{L}} = 6 \textup{ V} + 6 \textup{ V}= 12.0 \textup{ V}

2. When e.m.f. of power supply = 2.0 V, then \textup{I}_{\textup{R}} = 0.20 A , \textup{I}_{\textup{L}} = 0.40 A based on Fig. 11.4. Since they are in parallel, current through battery = \textup{I}_{\textup{R}}+\textup{I}_{\textup{L}} = 2.0 A + 0.40 A = 0.60 A.

[Maximum Marks: 10]

(a)

1. Allow the changing magnetic flux generated in primary coil to be linked to secondary coil.

2. Helps strengthen the magnetic flux generated / reduce magnetic flux leakage between primary and secondary coil for greater efficiency

(b)

\frac{\textup{N}_{\textup{s}}}{\textup{V}_{\textup{s}}}=\frac{\textup{N}_{\textup{p}}}{\textup{V}_{\textup{p}}}

\textup{N}_{\textup{s}}=\frac{\textup{N}_{\textup{p}}}{\textup{V}_{\textup{p}}} \times \textup{V}_{\textup{s}}=\frac{920}{230} \times 12 =48

(ci)

(cii)

A fuse melts and opens the circuit when current above its rating passes through it. To prevent excessive current in primary coil, the fuse needs to be in series with primary coil since current in a series circuit is same throughout.

(ciii)

By Principle of Conservation of Energy,

\frac{90}{100}\left ( \textup{V}_{\textup{p}}\textup{I}_{\textup{p}} \right )=\textup{V}_{\textup{s}}\textup{I}_{\textup{s}}, \textup{I}_{\textup{p}} max = 200 mA due to fuse

(0.9)\left ( 230\times\frac{200}{1000} \right )=(12)\textup{I}_{\textup{s}}\textup{I}_{\textup{s}} = 3.45 A (max)

n\left ( \frac{\textup{P}}{\textup{V}} \right )=3.45, where n = no. of lamps

n\left ( \frac{8}{12} \right )=3.45n=5.175 \approx 5 (rounded down)

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