## 2021 A-level Physics (9749) Paper 1 Suggested Solutions

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# Answer Key to 2021 A Level Physics Paper 1:

Ans: C

Mass of a smartphone \sim200g

W=mg=(\frac{200}{1000})(10)=2 \text{ N}

Ans: A

F_{B}=BIL\therefore[B]=[\frac{F_{B}}{IL}]=[\frac{ma}{IL}]=\frac{\text{kg (m s)}^{-2}}{\text{A m}}=\text{kg A}^{-1}\text{ s}^{-2}

Ans: B

Ans: B

v_{2}-v_{1}=-(u_{2}-u_{1})0.67v-v_{X}=-(0-v)

\left | v_{X} \right |=\left |0.67v-v \right |=\left | -0.33v \right |=0.33v

Ans: A

Ans: B

Initial Reading = mg

Final Reading = mg-U=mg-\rho_{\text{f}}V_{\text{disp}}g=0.75mg

V_{\text{disp}}=\frac{V_{\text{ball}}}{2}=\frac{0.25m}{\rho_{\text{f}}}=\frac{0.25(100)}{1}V_{\text{ball}}=50\text{ cm}^{3}Ans: D

Ans: A

Work Done=F_{\text{eng}}\cdot d=(400)(1000)=0.4 \text{ MJ}

\eta =\frac{W_{\text{out}}}{W_{\text{m}}}

W_{\text{in}}=\frac{W_{\text{out}}}{\eta}=\frac{0.4}{0.16}=2.5\text{ MJ}

48 MJ ⇒ 1 kg

2.5MJ = \frac{1}{48}\times 2.5=0.052kg=52\text{ g}

Ans: A

w=\frac{v}{r}=\frac{v}{\frac{mv}{Be}}=\frac{Be}{m}\therefore w\propto BWhen B is increased, w increases.

Ans: B

Work Done =mgh=m(\frac{\Delta \phi}{\Delta d})h=(2)(\frac{6}{10})(2.5)=3\text{ J}

Ans: D

v=rw=r(\frac{2\pi}{T})=(3.6\times 10^{7}+6.4\times 10^{6})(\frac{2\pi}{24\times 3600})=3083.4\text{ m s}^{-1}Ans: C

pV=\frac{1}{3}Nm<c^{2}>=\frac{1}{3}\cdot N\cdot\frac{M}{N}\cdot<c^{2}>c_{\text{rms}}=\sqrt{\frac{3pV}{M_{\text{gas}}}}=\sqrt{\frac{3(1\times10^{5})(10)(3)(4)}{(\frac{29}{1000})(5000)}}=498.2\text{ m s}^{-1}

Ans: D

Q=mc\theta+ml_{\text{v}}=5(4190)(100-30)+(5)(2260000)=1.276\times 10^{7}\text{ J}Ans: B

\Delta U=Q_{\text{to}}+W_{\text{on}}=0=\Delta \text{KE}Ans: C

Ans: D

\frac{\Delta \phi}{2\pi}=\frac{\Delta x}{\lambda}=\frac{1.5\times 10^{-6}}{\frac{3\times10^{8}}{5\times 10^{14}}}=\frac{1.5\times 10^{-6}}{1}\cdot \frac{5\times 10^{14}}{3\times 10^{8}}=5\pi – 4\pi=\piAns: B

sin2\theta=2sin\theta cos\theta(sin\theta cos\theta)^{2}=\frac{1}{4}=sin^{2}2\theta

I_{\text{y}}=I\frac{1}{4}sin^{2}2\theta

if \theta=45^{\circ}, \sin2(45^{\circ})=0,I_{\text{y}}=0

Ans: D

\theta_{\text{min}}=\frac{\lambda}{b},\tan(\frac{\theta_{\text{min}}}{2})=\frac{\frac{D}{2}}{5.7\times 10^{16}}

\tan(\frac{\lambda}{2b})=\frac{\lambda}{26}=\frac{620\times 10^{-9}}{50\times10^{12}}=\frac{D}{2(5.7\times10^{16})}

D=7.068\times 10^{10}\text{ m}

Ans: D

Ans: C

F_{\text{old}}=\frac{Q_{1}Q_{2}}{4\pi\varepsilon_{0}d^{2}}=2 \text{ N}

F_{\text{new}}=\frac{(2Q_{1})(Q_{2})}{4\pi\varepsilon_{0}(2d)^{2}}=1 \text{ N}

Ans: B

\frac{R}{l}=G\therefore R=lGIf l=L, then R=LG

The wires are arranged in parallel.

R_{\text{eff}}=[\frac{1}{R}+\frac{1}{R}+\frac{1}{R}+\frac{1}{R}]^{-1}=\frac{R}{4}=\frac{GL}{4}Ans: A

The sum of the voltage across resistor and thermistor should be 3.0 V.

Ans: C

For filament lamp to glow more brightly, power across lamp increases.

The resistance of LDR and thermistor is inversely proportional to the light intensity and temperature respectively.

By the potential divider rule, when light intensity increases and thermistor temperature decreases, voltage across lamp will increase. Hence, power P=\frac{V^{2}}{R}, increases and lamp glow more brightly.

Ans: A

Using RHGR, the direction of magnetic field due to circular coil is into the page.

Using FLHR, the direction of magnetic force is upwards.

Ans: D

\phi=BAThe unit for B is Tesla (T) and unit for A is m^{2}.

Hence, unit for magnetic flux \phi is \text{tesla metre}^{2}

Ans: A

cos\theta_{\text{i}}=0^{\circ}, cos\theta_{\text{f}}=90^{\circ}

\left | E \right |=\left | \frac{\Delta BAN\cos\theta}{\Delta t} \right |=\left | BAN\left ( \frac{\cos\theta_{\text{f}}-\cos\theta{\text{i}}]}{\Delta t} \right ) \right |=\left | (1.8)(\pi)(0.01)^{2}(3000)\left [ \frac{\cos90^{\circ}-\cos0^{\circ}}{0.06} \right ] \right |=28.27 \text{ V}Ans: B

\frac{\text{V}_{\text{s}}}{\text{V}_{\text{p}}}=\frac{\text{I}_{\text{p}}}{\text{I}_{\text{s}}}

\text{I}_{\text{s}}=\text{I}_{\text{p}}\cdot \frac{\text{V}_{\text{p}}}{\text{V}_{\text{s}}}=\frac{(0.32)(1.4)}{12}=0.0373\simeq 38\text{ mA}

Ans: B

E=\frac{hc}{\lambda}

\frac{1}{\lambda_{1}}+\frac{1}{\lambda_{2}}=\frac{1}{\lambda_{3}}

\frac{1}{440}+\frac{1}{590}=\frac{1}{\lambda_{3}}

\lambda_{3}=252.03\text{ nm}

Ans: D

I=\frac{P}{A}=\frac{E}{tA}=\frac{nhf}{tA}Frequency of light is constant. Hence, greater intensity is explained by \frac{n}{tA} (no. of photons passing per unit cross-sectional area per unit time) increasing.

Ans: B

_{}^{137}\textrm{Cs}\rightarrow _{}^{137}\textrm{Ba}+_{-1}^{0}\textrm{e}

\text{M}_{\text{LHS}}=136.90709u

\text{M}_{\text{RHS}}=136.90709+5.49\times 10^{-4}=136.906379u

Mass Defect, \Delta m=7.11\times 10^{-4}u, 1u=933.75MeV

7.11\times 10^{-4}u=0.66389625 MeV=1.0622\times 10^{-13}\text{ J}

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