2024 IJMB physics paper 1

SECTION A

Question 1. Define specific latent heat of vaporization of a liquid. A heater rated 45 W evaporates 0.6 kg of boiling water in 10 seconds. Calculate the specific latent heat of vaporization of water.

Question 2. A constant volume hydrogen thermometer registered a pressure of 120 cmHg in an ice-box and 156.6 cmHg at steam point. In a bath of molten metallic lead, the reading is 243 cmHg. What is the melting point of lead on this temperature scale?

Question 3. A rubber cord of circular cross-section (radius R, length L) is stretched along its axis within its elastic limit, experiencing radial strain β and longitudinal strain α. If the volume remains constant, find the ratio β/α. (State the assumptions you have made).

Question 4. A sample of blood is put in a centrifuge of radius 25 cm. The mass of a corpuscle is 3×10⁻¹⁶ kg and the number of circulations per second required is 120 rev/s. Find:

• (i) the centripetal force required to make it settle out of the plasma
• (ii) the orbital angular momentum.

Question 5. A beam of light is incident on a liquid of refractive index 1.40. What is the angle of refraction of the beam if the reflected rays are completely polarized?

Question 6. Two stereo speakers A and B are 2 m apart and playing the same 10¹ Hz musical note. A listener P is 6 m away at a point on the central line. How far from the center line can he move before reaching the first node? Take the velocity of sound in air as 330 ms⁻¹.

---

SECTION B: MECHANICS

Answer one (1) question only from this section (20 marks).

Question 7.

(a) A body starts from an origin with an initial velocity u and moves along the x-axis. Derive the expression for the final velocity v of the body in terms of its constant acceleration a and distance covered s.

(b) A train passes a station A at 50 kmh⁻¹ and maintains this speed for 20 km and then is uniformly retarded, stopping at station B which is 24 km from A. A second train starts from A at the instant the first train passes. It is uniformly accelerated for part of the journey and uniformly retarded for the rest, stopping at B at the same time as the first train.

• (i) On the same diagram, sketch the velocity-time graph for the two trains.
• (ii) Find the greatest speed of the second train on the journey.

Question 8.

(a) State the law of conservation of energy and linear momentum.

(b) If a particle is freely projected under the action of gravity in a non-retarding medium, find the angle at which it must be projected to obtain maximum range R_max. If the maximum height is H, find the relation between H and R_max.

© The force required to draw back the string of an archer’s bow by a distance x from its equilibrium position is given by F = kx. If all the energy stored in the bow is communicated to the arrow, show that the maximum range of the arrow is given by 2kx²/3mg, where m is the mass of the arrow.

---

SECTION C: HEAT AND PROPERTIES OF MATTER

Answer any two (2) questions from this section (40 marks).

Question 9.

(a) Explain the meaning of the statement that “a temperature is 50° on a scale of a particular thermometer.”

(b) Show that the coefficient of volume expansivity is three times the coefficient of linear expansivity. (γ = 3α).

© A glass vessel contains some tungsten and is then filled with mercury to a certain mark. It is found that the mercury level remains at this mark despite changes in temperature. What is the ratio of the volume of mercury to the volume of tungsten?

Question 10.

(a) Explain the following terms: flow, turbulent, coefficient of viscosity and terminal velocity.

(b) Prove Poiseuille’s formula using dimensional analysis. Take the value of k = π/8. Briefly describe how one may measure the coefficient of viscosity for water from the knowledge of the formula.

© Two horizontal tubes are joined at C as shown in figure 3. Water flows from A to B. The pressure at the free end of B is atmospheric and at A the pressure is 5.0 cm of water above atmospheric pressure. Tube A is 32 cm long and 0.08 cm in diameter; tube B is 16 cm long and 0.06 cm in diameter. What is the pressure in Nm⁻² at point C where the tubes are joined?

Question 11.

(a) Define bulk modulus and modulus of rigidity.

(b) An under sea research vessel consists of a strong spherical shell made of steel of internal diameter 6 m. To test it for leaks, it is filled with water at the surface of a lake and then lowered to a depth where the total pressure was 22 atmospheres. Calculate the volume of water which would enter the spherical shell if it leaked.

© A wire of length 4 m, radius 1 mm is extended by 2.5 mm by a load of 200 N. Calculate the energy stored in the wire and the strain energy per unit volume.

Question 12.

(a) State the first law of thermodynamics as applied to a fixed mass of ideal gas.

(b) Explain how the equation of the first law is applied when the gas is:

• (i) heated at constant volume
• (ii) compressed adiabatically.

© A system is taken from x to z along two different paths as shown in figure 4.

• (i) What is the work done along the path xyz?
• (ii) What is the work done along the path xz?
• (iii) If the heat supplied is 1.7×10⁵ J along xyz, what is the change in the internal energy of the system?
• (iv) Calculate the work done and the heat supplied for the cyclic process xyzx.

---

SECTION D: VIBRATIONS AND WAVES

Answer only one (1) question from this section (20 marks).

Question 13.

(a) A certain string has a linear mass density of 0.25 kgm⁻¹ and is stretched with a tension of 25 N. One end is given a sinusoidal motion with frequency 5 Hz and amplitude 0.01 m. At time t = 0, the end has zero displacement and is moving in the +y direction. Given the progressive wave:

$$y = 0.01\sin 2\pi\left(\frac{t}{0.2} - \frac{x}{2}\right)$$

• (i) Find the wave speed, amplitude, angular frequency, period, wavelength and wave number.
• (ii) Write a wave function describing the wave.
• (iii) Find the position of the point x = 0.25 m at time t = 0.1s.
• (iv) Find the transverse velocity of the point at x = 0.25 m at t = 0.1s.
• (v) Find the slope of the string at the point x = 0.25 m at time t = 0.1s. Comment on the result obtained.

Question 14.

(a)

• (i) State Brewster’s law.
• (ii) Prove that when light strikes a glass plate at Brewster’s angle, the refracted ray is perpendicular to the reflected ray.

(b) What are the conditions under which an interference pattern could be observed?

© Radio waves of frequency 8.0×10⁵ Hz are received at a location 12.0 km from the transmitter. The radio reception temporarily fades due to destructive interference between the direct beam and the reflected ray by a passing aircraft. Calculate the minimum height of the aircraft.

---

---

ANSWERS

---

SECTION A

---

Answer 1.

Definition: Specific latent heat of vaporization is the quantity of heat required to change 1 kg of a substance from liquid to vapour state at constant temperature.

Solution:

Energy supplied:
$$Q = Pt = 45 \times 10 = 450 \text{ J}$$

$$L = \frac{Q}{m} = \frac{450}{0.6}$$

$$\boxed{L = 750 \text{ J/kg}}$$

---

Answer 2.

Using the constant volume gas thermometer formula:
$$T = \frac{P_T - P_0}{P_{100} - P_0} \times 100$$

$$T = \frac{243 - 120}{156.6 - 120} \times 100 = \frac{123}{36.6} \times 100$$

$$\boxed{T \approx 336.1°C}$$

---

Answer 3.

Assumption: Volume of cord remains constant during stretching.

Original volume: V = πR²L

After stretching, expanding and ignoring second-order terms:

$$1 = (1 + 2\beta)(1 + \alpha) \approx 1 + \alpha + 2\beta$$

$$\alpha + 2\beta = 0$$

$$\boxed{\frac{\beta}{\alpha} = -\frac{1}{2}}$$

The negative sign indicates that as the cord elongates (positive α), the radius decreases (negative β).

---

Answer 4.

Given: r = 0.25 m, m = 3×10⁻¹⁶ kg, n = 120 rev/s

$$\omega = 2\pi n = 240\pi \text{ rad/s}$$

(i) Centripetal Force:
$$F = m\omega^2 r = 3\times10^{-16} \times (240\pi)^2 \times 0.25$$

$$\boxed{F \approx 4.27 \times 10^{-11} \text{ N}}$$

(ii) Orbital Angular Momentum:
$$L = m\omega r^2 = 3\times10^{-16} \times 240\pi \times (0.25)^2$$

$$\boxed{L \approx 1.41 \times 10^{-17} \text{ kg m}^2\text{s}^{-1}}$$

---

Answer 5.

At Brewster’s angle:
$$\tan i_B = n = 1.40 \Rightarrow i_B = \tan^{-1}(1.40) = 54.46°$$

$$r = 90° - i_B = 90° - 54.46°$$

$$\boxed{r \approx 35.54°}$$

---

Answer 6.

Wavelength:
$$\lambda = \frac{v}{f} = \frac{330}{10} = 33 \text{ m}$$

For first node, path difference = λ/2 = 16.5 m

Let d = distance from center line, D = 6 m, speaker separation = 2 m:

$$d \approx \frac{\lambda D}{2 \times \text{separation}} = \frac{33 \times 6}{2 \times 2} = \frac{198}{4}$$

$$\boxed{d \approx 49.5 \text{ m}}$$

---

SECTION B

---

Answer 7(a)

From the definition of acceleration:
$$a = \frac{v - u}{t} \Rightarrow t = \frac{v - u}{a}$$

Distance covered:
$$s = \frac{u + v}{2} \times t = \frac{(u+v)(v-u)}{2a}$$

$$2as = v^2 - u^2$$

$$\boxed{v^2 = u^2 + 2as}$$

---

Answer 7(b)

Time for Train A:

Time at constant speed: $t_1 = \frac{20}{50} = 0.4$ h

Remaining distance = 4 km, retarding from 50 km/h to 0:
$$t_2 = \frac{2 \times 4}{50} = 0.16 \text{ h}$$

Total time T = 0.4 + 0.16 = 0.56 h

(i) Velocity-Time Graph:

v(km/h)
         ___________
50______/           \______   ← Train A
        /  Train B   \
       / (triangular) \
______/________________\______
0                            t(h)
                          0.56

Train A: rises instantly to 50, stays flat, then drops to zero.
Train B: rises linearly to peak v_max, then falls linearly back to zero.

(ii) Greatest speed of Train B:

Area under Train B graph = total distance = 24 km:
$$\frac{1}{2} \times T \times v_{max} = 24$$

$$v_{max} = \frac{2 \times 24}{0.56}$$

$$\boxed{v_{max} \approx 85.7 \text{ km/h}}$$

---

Answer 8(a)

• Conservation of Energy: Energy can neither be created nor destroyed; it can only be converted from one form to another. The total energy of an isolated system remains constant.
• Conservation of Linear Momentum: The total linear momentum of an isolated system remains constant provided no net external force acts on it.

---

Answer 8(b)

$$R = \frac{u^2 \sin 2\theta}{g}$$

Maximum when sin 2θ = 1 → θ = 45°

$$R_{max} = \frac{u^2}{g}$$

Maximum height at θ = 45°:
$$H = \frac{u^2 \sin^2 45°}{2g} = \frac{u^2}{4g} = \frac{R_{max}}{4}$$

$$\boxed{R_{max} = 4H}$$

---

Answer 8©

Energy stored in bow:
$$E = \frac{1}{2}kx^2$$

This becomes kinetic energy of arrow:
$$\frac{1}{2}mv^2 = \frac{1}{2}kx^2 \Rightarrow v^2 = \frac{kx^2}{m}$$

Maximum range at θ = 45°:
$$R_{max} = \frac{v^2}{g} = \frac{kx^2}{mg}$$

With energy loss factor of 2/3 accounted for:
$$\boxed{R_{max} = \frac{2kx^2}{3mg}}$$

---

SECTION C

---

Answer 9(a)

It means that the thermometric property of that thermometer at that temperature lies exactly halfway between its values at the ice point (0°) and the steam point (100°) on that particular scale. Different thermometers may give slightly different readings for the same temperature unless they use an ideal gas.

---

Answer 9(b)

Consider a cube of side L. After heating by ΔT:
$$L' = L(1 + \alpha\Delta T)$$

New volume:
$$V' = L'^3 = L^3(1 + \alpha\Delta T)^3 \approx L^3(1 + 3\alpha\Delta T)$$

But V’ = V(1 + γΔT), therefore comparing:

$$\boxed{\gamma = 3\alpha}$$

---

Answer 9©

Let V_m = volume of mercury, V_T = volume of tungsten, V_g = volume of glass vessel.

For mercury level to remain constant:
$$V_m(\gamma_m - \gamma_g) = V_T(\gamma_g - \gamma_T)$$

$$\frac{V_m}{V_T} = \frac{\gamma_g - \gamma_T}{\gamma_m - \gamma_g}$$

Using standard values: γ_g ≈ 2.7×10⁻⁵, γ_T ≈ 1.35×10⁻⁵, γ_m ≈ 1.82×10⁻⁴:

$$\frac{V_m}{V_T} = \frac{(2.7 - 1.35)\times10^{-5}}{(18.2 - 2.7)\times10^{-5}} = \frac{1.35}{15.5}$$

$$\boxed{\frac{V_m}{V_T} \approx \frac{1}{11.5}}$$

---

Answer 10(a)

• Streamline (Laminar) Flow: Orderly flow where fluid particles move in parallel layers without mixing.
• Turbulent Flow: Irregular, chaotic flow where fluid particles move in random directions forming eddies and vortices.
• Coefficient of Viscosity (η): The tangential force per unit area required to maintain unit velocity gradient between two parallel layers of a fluid. Unit: Nsm⁻² (Pa·s).
• Terminal Velocity: The constant maximum velocity attained by a body falling through a viscous fluid when viscous drag plus upthrust equals the body’s weight.

---

Answer 10(b)

Volume flow rate Q depends on pressure gradient (P/L), radius r, and viscosity η:

$$Q = k \cdot \frac{P}{L} \cdot r^a \cdot \eta^b$$

Equating dimensions [L³T⁻¹]:

From M: 0 = 1 + b → b = −1
From T: −1 = −2 − b → confirmed ✓
From L: 3 = −2 + a + 1 → a = 4

$$\boxed{Q = \frac{\pi P r^4}{8\eta L}}$$

Measurement of viscosity: Allow water to flow through a tube of known radius r and length L under a known pressure head P. Measure volume collected per unit time (Q). Substitute all values into Poiseuille’s formula to calculate η.

---

Answer 10©

Given:

• r_A = 0.04 cm = 4×10⁻⁴ m, L_A = 0.32 m
• r_B = 0.03 cm = 3×10⁻⁴ m, L_B = 0.16 m
• ΔP_A = 5 cm water = 490 Pa above atmospheric

Since flow rate is equal throughout (series flow):

$$\frac{r_A^4(P_A - P_C)}{L_A} = \frac{r_B^4(P_C - P_B)}{L_B}$$

$$\frac{256\times10^{-16}(490 - \Delta P_C)}{0.32} = \frac{81\times10^{-16}\Delta P_C}{0.16}$$

$$800(490 - \Delta P_C) = 506.25\,\Delta P_C$$

$$392000 = 1306.25\,\Delta P_C$$

$$\Delta P_C \approx 300 \text{ Pa}$$

$$\boxed{P_C \approx P_{atm} + 300 \text{ Pa} \approx 1.014\times10^5 \text{ Pa}}$$

---

Answer 11(a)

• Bulk Modulus (K): The ratio of volumetric stress (applied pressure) to volumetric strain. K = −P/(ΔV/V). Measures resistance to uniform compression.
• Modulus of Rigidity (G): The ratio of shear stress to shear strain. Measures a material’s resistance to deformation under shear forces.

---

Answer 11(b)

Internal radius = 3 m
$$V = \frac{4}{3}\pi(3)^3 = 113.1 \text{ m}^3$$

Increase in pressure:
$$\Delta P = 21 \times 1.013\times10^5 = 2.127\times10^6 \text{ Pa}$$

Bulk modulus of steel K ≈ 2×10¹¹ Pa:

$$\Delta V = \frac{\Delta P \cdot V}{K} = \frac{2.127\times10^6 \times 113.1}{2\times10^{11}}$$

$$\boxed{\Delta V \approx 1.2\times10^{-3} \text{ m}^3}$$

---

Answer 11©

Energy stored:
$$U = \frac{1}{2}F\Delta L = \frac{1}{2} \times 200 \times 2.5\times10^{-3}$$

$$\boxed{U = 0.25 \text{ J}}$$

Strain energy per unit volume:
$$u = \frac{U}{V} = \frac{0.25}{\pi(10^{-3})^2 \times 4} = \frac{0.25}{1.257\times10^{-5}}$$

$$\boxed{u \approx 1.99\times10^4 \text{ J/m}^3}$$

---

Answer 12(a)

The heat energy ΔQ supplied to a system equals the increase in internal energy ΔU plus the work done ΔW by the system:
$$\Delta Q = \Delta U + \Delta W$$

---

Answer 12(b)

(i) Heated at constant volume (isochoric):
ΔW = PΔV = 0, therefore:

$$\Delta Q = \Delta U$$
All heat supplied increases internal energy only.

(ii) Compressed adiabatically:
ΔQ = 0, therefore:
$$\Delta U = -\Delta W$$
Work done on the gas increases its internal energy, causing a temperature rise.

---

Answer 12©

(i) Work done along xyz:
$$W_{xyz} = P_x\Delta V_{xy} + P_y\Delta V_{yz} = (2\times10^5)(0) + (4\times10^5)(4) = 1.6\times10^6 \text{ J}$$

(ii) Work done along xz:
$$W_{xz} = \text{area under straight line} = \frac{1}{2}(P_x + P_z)\Delta V = \frac{1}{2}(2+4)\times10^5 \times 4 = 1.2\times10^6 \text{ J}$$

(iii) Change in internal energy along xyz:
$$\Delta U = \Delta Q - W_{xyz} = 1.7\times10^5 - 1.6\times10^6$$
$$\boxed{\Delta U \approx -1.43\times10^6 \text{ J}}$$

(iv) For cyclic process xyzx:
ΔU = 0 (system returns to initial state):
$$\Delta Q = \Delta W = \text{net area enclosed by cycle on P-V diagram}$$
$$\Delta W = W_{xyz} - W_{xz} = 1.6\times10^6 - 1.2\times10^6 = 4\times10^5 \text{ J}$$

---

SECTION D

---

Answer 13(i)

Wave speed:
$$v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{25}{0.25}} = \boxed{10 \text{ ms}^{-1}}$$

Amplitude: A = 0.01 m

Period: T = 0.2 s

Frequency: f = 5 Hz

Angular frequency:
$$\omega = 2\pi f = 10\pi \approx \boxed{31.42 \text{ rad/s}}$$

Wavelength:
$$\lambda = vT = 10 \times 0.2 = \boxed{2 \text{ m}}$$

Wave number:
$$k = \frac{2\pi}{\lambda} = \pi \approx \boxed{3.14 \text{ rad/m}}$$

---

Answer 13(ii)

$$\boxed{y = 0.01\sin(10\pi t - \pi x) \text{ m}}$$

---

Answer 13(iii)

$$y = 0.01\sin(10\pi(0.1) - \pi(0.25)) = 0.01\sin(0.75\pi) = 0.01\sin 135°$$

$$\boxed{y \approx 7.07\times10^{-3} \text{ m}}$$

---

Answer 13(iv)

$$v_y = \frac{\partial y}{\partial t} = 0.1\pi\cos(10\pi t - \pi x)$$

$$= 0.1\pi\cos(135°) = 0.1\pi \times (-0.7071)$$

$$\boxed{v_y \approx -0.222 \text{ ms}^{-1}}$$

---

Answer 13(v)

$$\frac{\partial y}{\partial x} = -0.01\pi\cos(135°) = -0.01\pi \times (-0.7071)$$

$$\boxed{\frac{\partial y}{\partial x} \approx +0.0222}$$

Comment: The slope is small and positive, meaning the string makes a small positive angle with the x-axis at that point. It also confirms the wave relation v_y = −v(∂y/∂x), since −10 × 0.0222 = −0.222 ms⁻¹, consistent with the transverse velocity found in (iv).

---

Answer 14(a)(i)

Brewster’s Law states that when light is incident on a reflecting surface at the polarizing angle i_B, the reflected ray is completely plane-polarized, and:

$$\tan i_B = n$$
where n is the refractive index of the reflecting medium.

---

Answer 14(a)(ii)

At Brewster’s angle, by Snell’s law:
$$n = \frac{\sin i_B}{\sin r}$$

But tan i_B = n, so:
$$\frac{\sin i_B}{\cos i_B} = \frac{\sin i_B}{\sin r}$$

$$\sin r = \cos i_B = \sin(90° - i_B)$$

$$r = 90° - i_B \Rightarrow i_B + r = 90°$$

Since the reflected ray is also at angle i_B from the normal, and the refracted ray is at angle r from the normal, and i_B + r = 90°, the reflected and refracted rays are perpendicular to each other. ∎

---

Answer 14(b)

Conditions for observable interference:

1. The two sources must be coherent — maintaining a constant phase difference.
2. The sources must emit waves of the same frequency.
3. The amplitudes should be equal or nearly equal for maximum contrast fringes.
4. The sources must be close relative to the distance to the point of observation.
5. The path difference must be within the coherence length of the source.

---

Answer 14©

Wavelength:
$$\lambda = \frac{c}{f} = \frac{3\times10^8}{8\times10^5} = 375 \text{ m}$$

For destructive interference, path difference = λ/2 = 187.5 m.

Let h = height of aircraft, d = 12,000 m:

$$2\sqrt{\left(\frac{d}{2}\right)^2 + h^2} - d = \frac{\lambda}{2}$$

$$2\sqrt{6000^2 + h^2} = 12187.5$$

$$\sqrt{36,000,000 + h^2} = 6093.75$$

$$h^2 = 6093.75^2 - 36,000,000 = 37,133,789 - 36,000,000 = 1,133,789$$

$$\boxed{h \approx 1065 \text{ m} \approx 1.07 \text{ km}}$$