A Level Physics Diagnostic Test

A Level Physics — Diagnostic Test

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Instructions This diagnostic test covers the entire A Level Physics syllabus. Attempt all 45 questions. Each question has an answer and a revision redirect. Time allowed: 90 minutes.

Section A: Mechanics (Questions 1–5)

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Q1. A projectile is launched at 30 m s$^{-1}$ at 60° to the horizontal. What is its maximum height? ($g = 9.81$ m s$^{-2}$)

Answer. $v_y = 30\sin 60° = 25.98$ m s$^{-1}$. $h_{\max} = v_y^2/(2g) = 675/19.62 = 34.4$ m.

If you get this wrong, revise: Kinematics

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Q2. A 2.0 kg object moving at 5.0 m s$^{-1}$ collides elastically with a stationary 3.0 kg object. Calculate the velocity of the 3.0 kg object after the collision.

Answer. $v_2 = \frac{2m_1 u_1}{m_1 + m_2} = \frac◆LB◆2 \times 2.0 \times 5.0◆RB◆◆LB◆5.0◆RB◆ = 4.0$ m s$^{-1}$.

If you get this wrong, revise: Momentum

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Q3. A spring of constant 200 N m$^{-1}$ is compressed by 0.05 m. How much elastic potential energy is stored?

Answer. $E = \frac{1}{2}kx^2 = \frac{1}{2} \times 200 \times 0.0025 = 0.25$ J.

If you get this wrong, revise: Work, Energy, Power

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Q4. A car of mass 1200 kg travels around a bend of radius 50 m at 15 m s$^{-1}$. Calculate the centripetal force required.

Answer. $F = mv^2/r = 1200 \times 225/50 = 5400$ N.

If you get this wrong, revise: Circular Motion

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Q5. A mass on a spring oscillates with period 0.80 s and amplitude 4.0 cm. Calculate its maximum velocity.

Answer. $v_{\max} = \omega A = \frac◆LB◆2\pi◆RB◆◆LB◆T◆RB◆ \times A = \frac◆LB◆2\pi◆RB◆◆LB◆0.80◆RB◆ \times 0.040 = 0.314$ m s$^{-1}$.

If you get this wrong, revise: Oscillations

Section B: Waves (Questions 6–10)

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Q6. Light of wavelength 580 nm passes through double slits 0.40 mm apart onto a screen 1.5 m away. What is the fringe spacing?

Answer. $\Delta w = \frac◆LB◆\lambda D◆RB◆◆LB◆s◆RB◆ = \frac◆LB◆580 \times 10^{-9} \times 1.5◆RB◆◆LB◆0.40 \times 10^{-3}◆RB◆ = 2.175 \times 10^{-3}$ m $= 2.18$ mm.

If you get this wrong, revise: Superposition and Interference

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Q7. Unpolarised light of intensity 200 W m$^{-2}$ passes through two polarisers at 45° to each other. What is the transmitted intensity?

Answer. After first polariser: $I = 100$ W m$^{-2}$. After second: $I = 100\cos^2 45° = 100 \times 0.5 = 50$ W m$^{-2}$.

If you get this wrong, revise: Wave Properties

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Q8. The critical angle for a glass-air boundary is 42°. What is the refractive index of the glass?

Answer. $n = 1/\sin 42° = 1/0.669 = 1.49$.

If you get this wrong, revise: Refraction and TIR

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Q9. A stationary wave on a string has nodes 0.40 m apart. If the wave speed is 320 m s$^{-1}$, what is the frequency?

Answer. Node separation $= \lambda/2 = 0.40$ m, so $\lambda = 0.80$ m. $f = v/\lambda = 320/0.80 = 400$ Hz.

If you get this wrong, revise: Superposition and Interference

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Q10. A diffraction grating has 600 lines per mm. What is the maximum order visible with light of wavelength 500 nm?

Answer. $d = 1/600$ mm $= 1.667 \times 10^{-6}$ m. $n_{\max} = d/\lambda = 1.667 \times 10^{-6}/500 \times 10^{-9} = 3.33$. Maximum order $= 3$.

If you get this wrong, revise: Superposition and Interference

Section C: Electricity (Questions 11–15)

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Q11. A 6 $\Omega$ and a 3 $\Omega$ resistor are in parallel. What is the equivalent resistance?

Answer. $R = \frac◆LB◆6 \times 3◆RB◆◆LB◆6 + 3◆RB◆ = 2.0\,\Omega$.

If you get this wrong, revise: DC Circuits

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Q12. A capacitor of 47 $\mu$F is charged to 12 V. How much energy does it store?

Answer. $E = \frac{1}{2}CV^2 = \frac{1}{2} \times 47 \times 10^{-6} \times 144 = 3.38 \times 10^{-3}$ J $= 3.38$ mJ.

If you get this wrong, revise: Capacitance

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Q13. A cell of e.m.f. 9.0 V and internal resistance 1.5 $\Omega$ is connected to a 4.5 $\Omega$ load. What is the terminal p.d.?

Answer. $I = 9.0/(4.5 + 1.5) = 1.5$ A. $V = \mathcal{E} - Ir = 9.0 - 1.5 \times 1.5 = 6.75$ V.

If you get this wrong, revise: DC Circuits

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Q14. A 100 $\mu$F capacitor discharges through a 50 k$\Omega$ resistor. What is the time constant?

Answer. $\tau = RC = 50 \times 10^3 \times 100 \times 10^{-6} = 5.0$ s.

If you get this wrong, revise: Capacitance

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Q15. A copper wire has resistivity $1.7 \times 10^{-8}\,\Omega$ m, length 10 m, and diameter 1.0 mm. What is its resistance?

Answer. $A = \pi(0.5 \times 10^{-3})^2 = 7.85 \times 10^{-7}$ m$^2$. $R = \rho L/A = 1.7 \times 10^{-8} \times 10 / 7.85 \times 10^{-7} = 0.217\,\Omega$.

If you get this wrong, revise: Current and Resistance

Section D: Fields (Questions 16–22)

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Q16. Two charges of $+3\,\mu$C and $-5\,\mu$C are 0.20 m apart. What is the force between them?

Answer. $F = kq_1q_2/r^2 = 8.99 \times 10^9 \times 3 \times 10^{-6} \times 5 \times 10^{-6}/0.04 = 3.37$ N (attractive).

If you get this wrong, revise: Electric Fields

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Q17. An electron enters a 0.50 T magnetic field perpendicular to its velocity of $4.0 \times 10^6$ m s$^{-1}$. What is the radius of its path?

Answer. $r = mv/(Be) = 9.11 \times 10^{-31} \times 4.0 \times 10^6/(0.50 \times 1.60 \times 10^{-19}) = 4.56 \times 10^{-5}$ m.

If you get this wrong, revise: Magnetic Fields

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Q18. A velocity selector has $E = 5.0 \times 10^5$ V m$^{-1}$ and $B = 0.20$ T. What velocity is selected?

Answer. $v = E/B = 5.0 \times 10^5/0.20 = 2.5 \times 10^6$ m s$^{-1}$.

If you get this wrong, revise: Magnetic Fields

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Q19. Derive the speed of light from $\mu_0 = 4\pi \times 10^{-7}$ T m A$^{-1}$ and $\varepsilon_0 = 8.85 \times 10^{-12}$ F m$^{-1}$.

Answer. $c = 1/\sqrt◆LB◆\mu_0\varepsilon_0◆RB◆ = 1/\sqrt◆LB◆4\pi \times 10^{-7} \times 8.85 \times 10^{-12}◆RB◆ = 1/\sqrt◆LB◆1.113 \times 10^{-17}◆RB◆ = 3.00 \times 10^8$ m s$^{-1}$.

If you get this wrong, revise: Electromagnetism Unification

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Q20. What is the electric potential at 0.05 m from a $+2\,\mu$C charge?

Answer. $V = kQ/r = 8.99 \times 10^9 \times 2 \times 10^{-6}/0.05 = 3.60 \times 10^5$ V.

If you get this wrong, revise: Electric Fields

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Q21. A coil of 150 turns experiences a flux change of 0.04 Wb in 0.02 s. What is the average induced e.m.f.?

Answer. $\mathcal{E} = N\Delta\Phi/\Delta t = 150 \times 0.04/0.02 = 300$ V.

If you get this wrong, revise: Magnetic Fields

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Q22. State Faraday's law and Lenz's law.

Answer. Faraday's law: the induced e.m.f. equals the negative rate of change of flux linkage: $\mathcal{E} = -d(N\Phi)/dt$. Lenz's law: the direction of the induced current opposes the change in flux that produced it.

If you get this wrong, revise: Magnetic Fields

Section E: Thermal Physics (Questions 23–27)

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Q23. Calculate the rms speed of helium atoms ($M_r = 0.004$ kg mol$^{-1}$) at 300 K.

Answer. $v_{\mathrm{rms}} = \sqrt{3RT/M_r} = \sqrt◆LB◆3 \times 8.31 \times 300/0.004◆RB◆ = \sqrt{1869750} = 1367$ m s$^{-1}$.

If you get this wrong, revise: Thermal Properties

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Q24. A gas at 300 K and 100 kPa occupies 0.025 m$^3$. If the temperature increases to 450 K at constant pressure, what is the new volume?

Answer. $V_1/T_1 = V_2/T_2$. $V_2 = 0.025 \times 450/300 = 0.0375$ m$^3$.

If you get this wrong, revise: Thermal Properties

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Q25. What is the average kinetic energy of a gas molecule at 400 K?

Answer. $\langle E_k \rangle = \frac{3}{2}k_BT = 1.5 \times 1.381 \times 10^{-23} \times 400 = 8.29 \times 10^{-21}$ J.

If you get this wrong, revise: Thermal Properties

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Q26. A heat engine operates between 600 K and 300 K. What is the maximum possible efficiency?

Answer. $\eta_{\mathrm{Carnot}} = 1 - T_C/T_H = 1 - 300/600 = 50\%$.

If you get this wrong, revise: Thermodynamics

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Q27. 300 J of heat is added to a gas and it does 100 J of work on its surroundings. What is the change in internal energy?

Answer. $\Delta U = Q + W = 300 + (-100) = 200$ J. (Work done on the gas is $-100$ J.)

If you get this wrong, revise: Thermodynamics

Section F: Nuclear Physics (Questions 28–35)

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Q28. Write the nuclear equation for $\beta^-$ decay of carbon-14.

Answer. $\prescript{14}{}{6}\mathrm{C} \to \prescript{14}{}{7}\mathrm{N} + \prescript{0}{}{-1}\beta^- + \bar{\nu}_e$.

If you get this wrong, revise: Radioactivity

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Q29. A sample has a half-life of 6.0 hours. What fraction remains after 18 hours?

Answer. $n = 18/6 = 3$ half-lives. Fraction remaining $= 1/2^3 = 1/8 = 0.125 = 12.5\%$.

If you get this wrong, revise: Radioactivity

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Q30. The work function of caesium is 2.14 eV. What is the threshold wavelength?

Answer. $\lambda_0 = hc/\phi = 1240\,\mathrm{eV nm}/2.14\,\mathrm{eV} = 579$ nm.

If you get this wrong, revise: Quantum Physics

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Q31. Calculate the de Broglie wavelength of an electron accelerated through 200 V.

Answer. $\lambda = h/\sqrt{2m_e eV} = 6.63 \times 10^{-34}/\sqrt◆LB◆2 \times 9.11 \times 10^{-31} \times 1.60 \times 10^{-19} \times 200◆RB◆ = 6.63 \times 10^{-34}/7.64 \times 10^{-24} = 8.68 \times 10^{-11}$ m.

If you get this wrong, revise: Quantum Physics

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Q32. A hydrogen electron transitions from $n=3$ to $n=1$. Calculate the photon wavelength.

Answer. $E_3 = -13.6/9 = -1.51$ eV. $E_1 = -13.6$ eV. $\Delta E = 12.09$ eV $= 1.934 \times 10^{-18}$ J. $\lambda = hc/\Delta E = 6.63 \times 10^{-34} \times 3.0 \times 10^8/1.934 \times 10^{-18} = 1.03 \times 10^{-7}$ m $= 103$ nm.

If you get this wrong, revise: Quantum Physics

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Q33. Explain what is meant by the binding energy per nucleon and why iron-56 is the most stable nucleus.

Answer. The binding energy per nucleon is the total binding energy divided by the mass number — the average energy needed to remove one nucleon. Iron-56 has the highest binding energy per nucleon ($\sim 8.8$ MeV), meaning it requires the most energy per nucleon to disassemble. This corresponds to the maximum nuclear stability. Nuclei lighter than iron release energy by fusion; nuclei heavier release energy by fission — both processes move towards the iron peak.

If you get this wrong, revise: Nuclear Energy

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Q34. The activity of a sample decreases from 400 Bq to 50 Bq in 30 minutes. What is the decay constant?

Answer. $50/400 = 1/8 = 1/2^3$, so 3 half-lives. $t_{1/2} = 30/3 = 10$ min $= 600$ s. $\lambda = \ln 2/t_{1/2} = 0.693/600 = 1.16 \times 10^{-3}$ s$^{-1}$.

If you get this wrong, revise: Radioactivity

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Q35. Light of wavelength 450 nm is incident on a metal with work function 2.0 eV. Calculate the maximum kinetic energy of photoelectrons.

Answer. $E = hc/\lambda = 1240/450 = 2.76$ eV. $E_{k,\max} = 2.76 - 2.0 = 0.76$ eV.

If you get this wrong, revise: Quantum Physics

Section G: Practical Skills (Questions 36–45)

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Q36. A student measures a voltage five times: 4.52, 4.54, 4.53, 4.55, 4.52 V. Calculate the mean and absolute uncertainty.

Answer. Mean $= 22.66/5 = 4.532$ V. Range $= 4.55 - 4.52 = 0.03$ V. $\Delta V = 0.015$ V. Result: $4.53 \pm 0.02$ V.

If you get this wrong, revise: Measurements and Error Analysis

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Q37. $z = x^2 y$ where $x = 3.0 \pm 0.2$ and $y = 4.0 \pm 0.3$. Calculate the percentage uncertainty in $z$.

Answer. $z = 9.0 \times 4.0 = 36.0$. Fractional uncertainty: $2(0.2/3.0) + 0.3/4.0 = 0.133 + 0.075 = 0.208 = 20.8\%$.

If you get this wrong, revise: Measurements and Error Analysis

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Q38. What is the difference between a systematic error and a random error?

Answer. A systematic error is a consistent deviation from the true value, affecting all readings in the same direction (e.g., a zero error on a balance). A random error causes unpredictable scatter in repeated readings (e.g., reaction time with a stopwatch). Systematic errors affect accuracy; random errors affect precision.

If you get this wrong, revise: Measurements and Error Analysis

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Q39. To linearise $y = ax^n$, what should you plot?

Answer. Taking logarithms: $\ln y = \ln a + n\ln x$. Plot $\ln y$ vs $\ln x$. The gradient is $n$ and the $y$-intercept is $\ln a$.

If you get this wrong, revise: Measurements and Error Analysis

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Q40. A line of best fit has gradient $2.5 \pm 0.3$. What is the percentage uncertainty in the gradient?

Answer. Percentage uncertainty $= (0.3/2.5) \times 100 = 12\%$.

If you get this wrong, revise: Measurements and Error Analysis

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Q41. When measuring the diameter of a wire with a micrometer, why should you take readings at several positions along the wire?

Answer. The wire may not have a perfectly uniform diameter. Taking readings at several positions and averaging reduces the effect of any local variations, giving a more representative value for the cross-sectional area.

If you get this wrong, revise: Experimental Design

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Q42. Explain why timing 20 oscillations of a pendulum gives a more precise result than timing one oscillation.

Answer. Human reaction time ($\sim 0.2$ s) is a constant uncertainty regardless of the total time measured. For one oscillation ($\sim 2$ s), the percentage uncertainty is $\sim 20\%$. For 20 oscillations ($\sim 40$ s), the uncertainty in the total time is $\sim 0.4$ s (start and stop), giving $\sim 1\%$ uncertainty in the total time, which translates to $\sim 1\%$ uncertainty in the period.

If you get this wrong, revise: Experimental Design

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Q43. A result is quoted as $9.6 \pm 0.4$ m s$^{-2}$. The accepted value is $9.81$ m s$^{-2}$. Does the accepted value lie within the uncertainty range?

Answer. Yes. The range is $9.2$ to $10.0$ m s$^{-2}$. Since $9.81$ lies within this range, the result is consistent with the accepted value at this level of uncertainty.

If you get this wrong, revise: Measurements and Error Analysis

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Q44. In a circuit experiment, a student uses an ammeter with resistance $0.5\,\Omega$ to measure current in a branch with resistance $5\,\Omega$. What percentage error does the ammeter introduce?

Answer. The actual branch resistance is $5.0\,\Omega$. With the ammeter: total $= 5.5\,\Omega$. The current is reduced by a factor of $5.0/5.5 = 0.909$, an error of $\sim 9.1\%$. (The ammeter should ideally have zero resistance.)

If you get this wrong, revise: DC Circuits

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Q45. A graph of $\ln I$ vs $t$ for a discharging capacitor gives a straight line with gradient $-0.125$ s$^{-1}$. What is the time constant?

Answer. For discharging: $I = I_0 e^{-t/RC}$, so $\ln I = \ln I_0 - t/(RC)$. Gradient $= -1/(RC) = -0.125$. $\tau = RC = 1/0.125 = 8.0$ s.

If you get this wrong, revise: Capacitance

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Scoring

Score	Grade	Recommendation
40–45	A*	Excellent. Focus on extension problems.
35–39	A	Strong. Review topics where you lost marks.
28–34	B	Good foundation. Systematic revision needed.
20–27	C	Gaps exist. Prioritise weak sections.
< 20	D/U	Significant revision required. Start with fundamentals.

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