2025 IJMB physics paper 3

QUESTION 1A

The following data was obtained in an experiment to verify some certain relationship by the IJMB students in table 1 according to the empirical formula:

p = ne^λx …(1)

where n and λ are constants to be determined.

x	0.21	0.46	0.61	0.8	0.99	1.25	1.3	1.6	1.81	1.9
P	1.61	1.75	1.73	1.96	2.08	2.27	2.26	2.51	2.90	3.00

(i) Transform equation (1) into a suitable straight line graph to determine the constants n and λ.

(ii) Prepare a composite table to plot a suitable graph to determine the constants n and λ.

(iii) Use your graph to determine the values of n and λ.

(iv) Use your graph to estimate the value of P when x = 0.92.

(v) Substitute your values of n and λ obtained from your graph in (iii) above into equation to determine the value of p when x = 0.92.

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1b. Given that z = √(r − y³). Find the percentage error in z when r = 2.01 ± 0.071 and y = 0.77 ± 0.0031.

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SOLUTIONS

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1A(i) — Linearising the equation

Starting with:

p = ne^λx

Take natural log of both sides:

ln p = ln n + λx

This is in the form Y = C + mX, where:

• Y = ln p
• X = x
• Gradient m = λ
• Y-intercept C = ln n, so n = e^C

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1A(ii) — Composite Table

Compute ln P for each value:

x	P	ln P
0.21	1.61	0.476
0.46	1.75	0.559
0.61	1.73	0.548
0.80	1.96	0.673
0.99	2.08	0.732
1.25	2.27	0.820
1.30	2.26	0.815
1.60	2.51	0.920
1.81	2.90	1.065
1.90	3.00	1.099

Plot ln P (y-axis) against x (x-axis). The graph should be a straight line.

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1A(iii) — Finding n and λ from the graph

Using two well-separated points on the best-fit line, e.g.:

• Point 1: (0.21, 0.476)
• Point 2: (1.90, 1.099)

Gradient = λ:

$$\lambda = \frac{1.099 - 0.476}{1.90 - 0.21} = \frac{0.623}{1.69} \approx \boxed{0.369 \text{ per unit x}}$$

Y-intercept (ln n):

Using Y = mX + C:

0.476 = 0.369(0.21) + C
C = 0.476 − 0.077 = 0.399

$$n = e^{0.399} \approx \boxed{1.49}$$

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1A(iv) — Estimate P when x = 0.92 (from graph)

Read off the graph at x = 0.92:

ln P ≈ 0.399 + 0.369(0.92) ≈ 0.399 + 0.340 = 0.739

$$P = e^{0.739} \approx \boxed{2.09}$$

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1A(v) — Substitute into equation

$$p = ne^{\lambda x} = 1.49 \times e^{0.369 \times 0.92}$$

$$= 1.49 \times e^{0.340} = 1.49 \times 1.404 \approx \boxed{2.09}$$

This confirms the graphical estimate.

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1B — Percentage Error in z

Given: z = √(r − y³), r = 2.01 ± 0.071, y = 0.77 ± 0.0031

Step 1: Find z

$$y^3 = 0.77^3 = 0.4565$$

$$r - y^3 = 2.01 - 0.4565 = 1.5535$$

$$z = \sqrt{1.5535} \approx 1.2464$$

Step 2: Find the error propagation

Let u = r − y³, so z = u^(1/2)

$$\delta u = \delta r + 3y^2 \cdot \delta y$$

$$= 0.071 + 3(0.77)^2(0.0031)$$

$$= 0.071 + 3(0.5929)(0.0031)$$

$$= 0.071 + 0.00551 = 0.07651$$

Then for z = √u:

$$\frac{\delta z}{z} = \frac{1}{2} \cdot \frac{\delta u}{u} = \frac{1}{2} \times \frac{0.07651}{1.5535}$$

$$= \frac{1}{2} \times 0.04924 = 0.02462$$

Percentage error in z:

$$= 0.02462 \times 100 \approx \boxed{2.46\%}$$