Questions
PHYSICS PAPER III (ALTERNATIVE A)
1A. The current I flowing through a discharging capacitor decreases with time according to:
I = I₀e^(−t/RC) ...(1)
where I₀ and RC are constants (RC is the time constant of the circuit).
| t/s | 0.20 | 0.25 | 0.30 | 0.35 | 0.40 | 0.45 | 0.50 | 0.60 | 0.70 | 0.80 |
|---|---|---|---|---|---|---|---|---|---|---|
| I/mA | 15.20 | 12.60 | 10.80 | 9.10 | 7.85 | 6.60 | 5.70 | 4.15 | 3.05 | 2.25 |
i. Transform equation (1) into a straight-line form to determine I₀ and the time constant RC.
ii. Form a composite table containing all necessary parameters to determine I₀ and RC graphically.
iii. Use your graph to determine the approximate values of I₀ and RC.
iv. From the graph, find the approximate value of I when t = 0.55 s.
v. From your values of I₀ and RC, use equation (1) to find the value of t when I = 8.40 mA.
b. If a = 4.6 ± 0.006 and b = 5.29 ± 0.0015, find the error in determining the value of the term a² − √b.
2A. Attach a small pendulum bob of mass m to a light inextensible string of length ℓ suspended from a rigid clamp as shown in Figure 1. Displace the bob through a small angle and release it gently so that it swings with simple harmonic motion. Using a stopwatch, time 20 complete oscillations. Find the period T and evaluate T². Repeat the experiment for five other different values of ℓ.
i. Tabulate your readings.
ii. Plot a graph of T² against ℓ. Find the slope S of the graph.
iii. If T = 2π√(ℓ/g), use your slope to find the value of g.
iv. State two precautions taken to ensure accurate results.
3A. Use the diagram in Figure 2.1 to carry out the following instructions. For the ray KLMN, the angle of incidence is i and the angle of refraction is r.
i. Trace the outline of a rectangular glass block UVWX. Set the angle of incidence i = 20°. With the help of optical pins, find the angle of refraction r.
ii. Measure and record the lateral displacement marked 'd'. Evaluate sin i and sin r.
iii. Repeat the experiment with i = 30°, 40°, 50°, 60° and 70° respectively.
iv. Prepare a composite table for your readings. Plot a graph of sin i against sin r. Find the slope S of the graph and evaluate S⁻¹.
v. State two precautions taken to ensure accurate results. Attach the traces to your answer booklet.
b. A rectangular glass block has a refractive index of 1.50. If light strikes the block at an angle of incidence of 45°, calculate the angle of refraction.
Solutions
1A i. Taking natural log of both sides of I = I₀e^(−t/RC):
ln I = ln I₀ − t/RC
Plotting ln I (y-axis) against t (x-axis) gives a straight line with slope = −1/RC and y-intercept = ln I₀.
1A ii. Composite table:
| t/s | 0.20 | 0.25 | 0.30 | 0.35 | 0.40 | 0.45 | 0.50 | 0.60 | 0.70 | 0.80 |
|---|---|---|---|---|---|---|---|---|---|---|
| I/mA | 15.20 | 12.60 | 10.80 | 9.10 | 7.85 | 6.60 | 5.70 | 4.15 | 3.05 | 2.25 |
| ln I | 2.72 | 2.53 | 2.38 | 2.21 | 2.06 | 1.89 | 1.74 | 1.42 | 1.11 | 0.81 |
1A iii. Slope using end points (0.20, 2.72) and (0.80, 0.81):
slope = (0.81 − 2.72)/(0.80 − 0.20) = −1.91/0.60 ≈ −3.18 s⁻¹
So 1/RC ≈ 3.18 s⁻¹ → RC ≈ 0.31 s
Intercept: ln I₀ = 2.72 + 3.18(0.20) = 3.36 → I₀ = e^3.36 ≈ 28.8 mA
1A iv. At t = 0.55: ln I = 3.36 − 3.18(0.55) = 1.61 → I = e^1.61 ≈ 5.0 mA
1A v. I = 8.40: ln(8.40) = 2.13
2.13 = 3.36 − 3.18t
t = 1.23/3.18 ≈ 0.39 s
b. a² = 21.16, d(a²) = 2a·da = 2(4.6)(0.006) = 0.0552
√b = 2.30, d(√b) = db/(2√b) = 0.0015/4.60 = 0.00033
z = a² − √b = 18.86
dz = d(a²) + d(√b) = 0.0552 + 0.0003 = 0.0555
z = 18.86 ± 0.056
2A i–ii. (Requires actual lab readings: time 20 oscillations at 6 values of ℓ, compute T = time/20, then T², plot T² vs ℓ.)
2A iii. T² = 4π²ℓ/g → slope S = 4π²/g
g = 4π²/S
2A iv.
- Avoid parallax error by viewing the string/scale at eye level when measuring the length ℓ.
- Keep the angle of swing small (below about 15°) so the motion remains simple harmonic.
3A i–iv. (Requires actual ray-tracing readings for each i; tabulate i, r, d, sin i, sin r; plot sin i vs sin r; slope S and S⁻¹ read from graph.)
3A v.
- Use sharp pencils and well-separated optical pins to minimize parallax error in sighting the ray.
- Hold the glass block firmly in place so its outline does not shift while tracing the rays.
3A b. n = sin i / sin r
1.50 = sin 45° / sin r
sin r = sin 45° / 1.50 = 0.7071/1.50 = 0.4714
r = sin⁻¹(0.4714)
r ≈ 28.1°