## SECTION A
**1i).** Explain the terms *"fundamental quantities and derived quantities"*. **(ii).** What is the dimension of *h* in the equation *E = hf*?
**2.** An object of mass 4 kg is suspended from a spring balance which is attached to a ceiling of a lift. If the spring balance is calibrated in SI unit of force. What will be the reading on the balance when (i) the lift is stationary (ii) the lift is moving upwards with an acceleration of 0.15 ms⁻¹. (iii) the lift is moving downwards with an acceleration of 0.20 ms⁻².
**3.** A block of wood of mass 1 kg is held in a vice and a bullet of mass 0.01 kg is fired into it. It penetrates to a depth of 0.10 m. if the block is now suspended so that it can move freely and a second bullet is fired in to it, to what depth will the bullet penetrate?
**4.** An observer stands by the side of a railway truck while a train approaches him at a constant speed of 50 ms⁻¹. If the train is sounding its whistle at a frequency of 500 Hz, what will be the frequency perceived by the observer as the train (i) approaches (ii) recedes. Comment on your results.
**5.** The temperature of three different liquids are maintained at 15°C, 20°C and 25°C respectively. When equal masses of the first two liquids are mixed, the final temperature is 18°C and when equal masses of the last two liquids are mixed, the final temperature is 24°C. What temperature will be achieved by mixing equal masses of the first and the last liquids?
**6.** The solar constant is 0.14 Wcm⁻². Calculate the surface temperature of the sun. the ratio of the earth orbit to the radius of the sun is approximately 215. (Assume the sun as perfectly radiating body).
**7.** Water flows into a water tank of large cross sectional area at a rate of 104 m²s⁻¹ but flows out of from a hole of cross sectional area of 10⁻⁴m² at the base of the tank. How high does the water rise in the tank?
**8.** A sonar device emits waves of frequency 4 × 10⁴ Hz. The velocities of the wave in air and water are 400 ms⁻¹ and 1350 ms⁻¹ respectively. Suppose that the device is fixed to the bottom of a ship. If it emits a signal and the echo from the ocean bed returns to the observer 0.8 s later. Determine the (i) the depth of the ocean at that point. (ii) the wavelength of the signal in air and in water. (HINT: the frequency of the wave emitted is the same in air and in water).
**9.** A steel tyre of diameter 1.50 m at a temperature 10°C is to be fitted on to a train's wheel of diameter 1.51 m. to what temperature must the tyre be heated to just fit the wheel? (Linear expansivity of steel is 11.0 × 10⁻⁶ K⁻¹).
**10. a).** What do you understand by the term evaporation? **b).** evaporation is usually accompanied by cooling. Explain.
---
## SECTION B: MECHANICS
### Answer only one (1) Question from this Section. (20 marks)
**11a). (i)** state the Newton's third law of motion. **(ii)** what do you understand by non conservative forces? **(iii)** state the principle of conservation of linear momentum.
**b).** A small rectangular mirror with a mass of 9.0 × 10⁻⁶ kg is suspended from a thin quartz filament 0.04 m long. A powerful laser beam is emitted in a direction perpendicular to the mirror so that, the system is deflected from the vertical by a certain angle. The energy of the laser beam is 100J. **(i)** Using the principle of conservation of energy and momentum, discuss briefly the interaction between the laser beam and the mirror. **(ii)** Determine the angle of deflection of the mirror. Assuming the mirror is a perfect reflector and undergoes negligible vibration.
**12a). (i)** State the law of universal gravitation. **(ii)** Obtain the expression for mass m of the earth and density p of the earth. hence, deduce their numerical values.
**b).** A fully equipped astronaut in space can jump 2 m vertically on earth using maximum effort. **(i)** if the diameter of the moon is 0.25 that of the earth and its density is 2/3 that of the earth. Show that, the acceleration due to gravity on the moon is 1/6 that of the earth. **(ii)** How far can the astronaut jump on the moon?
---
## SECTION C: HEAT AND PROPERTIES OF MATTER
### Answer any two (2) questions from this section. (40 marks)
**13a).** Explain the terms (i) *laminar flow*, (ii) *turbulent flow*.
**b).** write down the expression of Reynold number, a basic parameter for the characteristic of fluid motion and show that, it is dimensionless.
**c).** For a streamline motion of an incompressible non viscous fluid in a pipe, the Bernoulli's principle states that:
**(P₁ − P₂)Δv = ½ρΔv(v₂² − v₁²) + ρgΔv(h₂ − h₁)**
Where all the symbols retain their usual meanings. Show that, the total work done per unit volume of the fluid is constant.
**d).** Water enters a house through a pipe of internal diameter of 0.02 m at an absolute pressure of 4 × 10⁵ Pa. The pipe leading to the second flow bathroom 5 m above is 0.01 m in diameter. (i). When the flow velocity at the inlet pipe is 4 ms⁻¹. Calculate the flow velocity and the pressure in the bathroom. (ii) What is the pressure in the bathroom if the tap is turned off?
**14a)** Define the following and state their units. (i) specific heat capacity (ii) specific latent heat of fusion.
**b).** describe the method of continuous flow for the determination of the specific heat of liquid.
**c).** In an experiment to determine the specific heat capacity of water using continuous flow method, the following sets of measurements were taken.
| | Exp. 1 | Exp. 2 |
|---|---|---|
| Current in heating coil | 2.0 A | 1.5 A |
| Potential difference across coil | 5.5 V | 4.5 V |
| Mass of water collected | 0.0405 kg | 0.0700 kg |
| Time of flow | 60.0 s | 180.0 s |
| Inlet temperature | 38.0°C | 38.0°C |
| Outlet temperature | 45.0°C | 45.0°C |
Use the given data to obtain the specific heat capacity of water.
**c).** State and explain three factors that affect evaporation of liquids.
**15a).** Define coefficient of viscosity and briefly describe two applications of viscosity of liquids.
**b). (i).** The viscous force opposing the motion of a sphere of radius **r** moving through a liquid of viscosity η is given by **f = krvη**. obtain an expression for the terminal velocity **V_T** of the sphere falling freely in the liquid if the density of the liquid is **ρ** and the density of the sphere is **ρₛ**.
---
## SECTION D: VIBRATION AND WAVES
### ANSWER Only One (1) Question Only From This Section. (20 marks)
**17a).** Explain the meaning of the following: (i) fundamental note. (ii). Overtones, (iii) Harmonic frequencies.
**b). (i).** Draw sketches to show how the first three overtones are produced in a closed pipe.
**(ii).** Show that, a closed pipe produces only odd harmonics.
**c).** Explain the meaning of end correction of resonance tube.
**d).** You are given a closed pipe 0.46 m long and an opened pipe 0.60 m long.. they both have same diameter and each sound its first overtone at the same pitch. What is the end correction?
**18a). (i).** What is simple harmonic motion? **(ii).** State and explain three examples of simple harmonic motion.
**b).** A liquid column of total length L, and density ρ is at rest in a U- tube of uniform cross sectional area A. if the liquid in a U-tube is given a small placement x. (i). Show that, the issuing motion is simple harmonic. (ii) deduce an expression for the period of the water column.
**c).** The displacement x, of of an object oscillating with simple harmonic motion is given by **x = 12cos(5πt)**. Where x is in millimeter and t is in second. Determine the (i) amplitude (ii). The period. (iii) the maximum velocity attained by the object.
---
*— Page 4 continues (Section C Q16) —*
**16a) (i).** State and explain the first law of thermodynamics. **(ii).** Show that, the internal energy of a gas depends only on its temperature.
**b).** With the aid of a sketch diagram, explain the isothermal and adiabatic changes.
**c). (i).** A mass of an ideal gas of volume 6 × 10⁻⁴ m³ at 219 K expands adiabatically and its temperature falls to 273 K. what is the final volume if the ratio of the principal heat capacities of the ideal gas γ = 1.40? **(ii).** In a Wilson cloud chamber at a temperature of 20°C, particles tracks are made visible by causing condensation on ions by an approximately reversible adiabatic expansion of the volume in the ratio 1.375 to 1. Estimate the gas temperature after the expansion.
**(ii).** Spherical particles of pollen are shaken up in water and allowed to stand. The depth of water is 0.02 m. what is the diameter of of the largest particles remaining in suspension 1 hour later? Density of pollen is 1.8 × 10³ kgm⁻³.
# 2025 IJMBE Physics I — Full Solutions
---
## SECTION A
---
### Question 1
**i) Fundamental vs Derived Quantities**
**Fundamental quantities** are independent physical quantities that cannot be expressed in terms of other quantities. Examples: mass (kg), length (m), time (s), temperature (K), current (A).
**Derived quantities** are obtained by combining fundamental quantities through multiplication or division. Examples: velocity (m/s), force (kg·m/s²), energy (J).
---
**ii) Dimension of h in E = hf**
- E = energy → dimensions: [ML²T⁻²]
- f = frequency → dimensions: [T⁻¹]
$$h = \frac{E}{f} = \frac{[ML^2T^{-2}]}{[T^{-1}]} = [ML^2T^{-1}]$$
---
### Question 2
Mass m = 4 kg, g = 10 ms⁻²
**i) Stationary lift:**
$$W = mg = 4 \times 10 = \boxed{40 \text{ N}}$$
**ii) Moving upward, a = 0.15 ms⁻²:**
$$W = m(g + a) = 4(10 + 0.15) = 4 \times 10.15 = \boxed{40.6 \text{ N}}$$
**iii) Moving downward, a = 0.20 ms⁻²:**
$$W = m(g - a) = 4(10 - 0.20) = 4 \times 9.80 = \boxed{39.2 \text{ N}}$$
---
### Question 3
**When block is fixed (bullet embeds):**
- m₁ = 0.01 kg (bullet), m₂ = 1 kg (block)
- Using work-energy theorem, retarding force F:
$$F \cdot d_1 = \frac{1}{2}m_1v^2$$
$$F \times 0.10 = \frac{1}{2}(0.01)v^2 \quad \text{...(1)}$$
**When block is free (bullet embeds, momentum conserved):**
By conservation of momentum:
$$m_1 v = (m_1 + m_2)V$$
$$V = \frac{0.01v}{1.01} \approx \frac{v}{101}$$
KE available to decelerate bullet relative to block:
$$KE_{rel} = \frac{1}{2}\mu v_{rel}^2$$
where reduced mass μ = m₁m₂/(m₁+m₂):
$$\mu = \frac{0.01 \times 1}{1.01} \approx 0.0099 \text{ kg}$$
Relative velocity of bullet w.r.t. block after impact begins:
$$v_{rel} = v - V = v - \frac{0.01v}{1.01} = v\left(\frac{1}{1.01}\right) \approx \frac{v}{1.01}$$
Energy available in free case:
$$KE_2 = \frac{1}{2}\mu v_{rel}^2 = \frac{1}{2} \cdot \frac{m_1 m_2}{m_1+m_2} \cdot \left(\frac{v}{1.01}\right)^2$$
$$= \frac{1}{2} \cdot \frac{0.01 \times 1}{1.01} \cdot \frac{v^2}{1.0201}$$
$$= \frac{1}{2} \cdot \frac{0.01 v^2}{(1.01)^2} = \frac{1}{2} \cdot \frac{0.01 v^2}{1.0201}$$
From (1): ½(0.01)v² = F × 0.10, so ½v² = 10F/0.01 → ½(0.01)v² = 0.1F...
More directly — ratio of penetration depths:
$$\frac{d_2}{d_1} = \frac{KE_2}{KE_1} = \frac{\frac{1}{2}\cdot\frac{m_1 m_2}{m_1+m_2}\cdot v^2}{\frac{1}{2}m_1 v^2} = \frac{m_2}{m_1+m_2}$$
$$\frac{d_2}{0.10} = \frac{1}{1.01} = 0.99$$
$$\boxed{d_2 \approx 0.099 \text{ m} \approx 9.9 \text{ cm}}$$
---
### Question 4
Speed of sound v = 340 ms⁻¹, v_s = 50 ms⁻¹, f₀ = 500 Hz
**Doppler formula:** Observer stationary, source moving.
**i) Train approaching:**
$$f = f_0 \cdot \frac{v}{v - v_s} = 500 \times \frac{340}{340 - 50} = 500 \times \frac{340}{290}$$
$$= 500 \times 1.172 = \boxed{586.2 \text{ Hz}}$$
**ii) Train receding:**
$$f = f_0 \cdot \frac{v}{v + v_s} = 500 \times \frac{340}{340 + 50} = 500 \times \frac{340}{390}$$
$$= 500 \times 0.872 = \boxed{435.9 \text{ Hz}}$$
**Comment:** The perceived frequency is higher when the source approaches and lower when it recedes — this is the **Doppler Effect**. The actual whistle frequency (500 Hz) lies between the two perceived values.
---
### Question 5
Let specific heat capacities be c₁, c₂, c₃ for liquids at 15°C, 20°C, 25°C.
**Mix 1 + 2 → 18°C:**
$$mc_1(18-15) = mc_2(20-18)$$
$$3c_1 = 2c_2 \Rightarrow c_2 = 1.5c_1 \quad \text{...(1)}$$
**Mix 2 + 3 → 24°C:**
$$mc_2(24-20) = mc_3(25-24)$$
$$4c_2 = c_3 \Rightarrow c_3 = 4c_2 = 6c_1 \quad \text{...(2)}$$
**Mix 1 + 3 → T°C:**
$$mc_1(T-15) = mc_3(25-T)$$
$$c_1(T-15) = 6c_1(25-T)$$
$$T - 15 = 150 - 6T$$
$$7T = 165$$
$$\boxed{T = 23.6°C}$$
---
### Question 6
Solar constant S = 0.14 Wcm⁻² = 1400 Wm⁻²
Ratio R_orbit/R_sun = 215
Stefan's Law: Power radiated by sun:
$$P = \sigma T^4 \cdot 4\pi R_s^2$$
Solar constant at Earth:
$$S = \frac{P}{4\pi R_{orbit}^2} = \sigma T^4 \left(\frac{R_s}{R_{orbit}}\right)^2$$
$$T^4 = \frac{S}{\sigma} \times \left(\frac{R_{orbit}}{R_s}\right)^2 = \frac{1400}{5.67\times10^{-8}} \times (215)^2$$
$$T^4 = 2.469 \times 10^{10} \times 46225 = 1.141 \times 10^{15}$$
$$T = (1.141 \times 10^{15})^{0.25} = \boxed{5.81 \times 10^3 \text{ K} \approx 5810 \text{ K}}$$
---
### Question 7
Inflow rate Q_in = 10⁻⁴ m²s⁻¹ *(interpreting as volume flow rate)*
Outflow hole area A = 10⁻⁴ m²
At equilibrium, inflow = outflow:
$$Q_{in} = A\sqrt{2gh}$$
$$10^{-4} = 10^{-4}\sqrt{2 \times 10 \times h}$$
$$1 = \sqrt{20h}$$
$$1 = 20h$$
$$\boxed{h = 0.05 \text{ m}}$$
---
### Question 8
f = 4 × 10⁴ Hz, v_air = 400 ms⁻¹, v_water = 1350 ms⁻¹
Echo time t = 0.8 s (round trip in water)
**i) Depth of ocean:**
$$d = \frac{v_{water} \times t}{2} = \frac{1350 \times 0.8}{2} = \frac{1080}{2} = \boxed{540 \text{ m}}$$
**ii) Wavelength in air:**
$$\lambda_{air} = \frac{v_{air}}{f} = \frac{400}{4\times10^4} = \boxed{0.01 \text{ m} = 1 \text{ cm}}$$
**Wavelength in water:**
$$\lambda_{water} = \frac{v_{water}}{f} = \frac{1350}{4\times10^4} = \boxed{0.03375 \text{ m} \approx 3.375 \text{ cm}}$$
---
### Question 9
d₁ = 1.50 m → r₁ = 0.75 m (tyre)
d₂ = 1.51 m → r₂ = 0.755 m (wheel)
T₁ = 10°C, α = 11.0 × 10⁻⁶ K⁻¹
For the tyre to expand to wheel diameter:
$$d_2 = d_1(1 + \alpha\Delta T)$$
$$1.51 = 1.50(1 + 11\times10^{-6}\times\Delta T)$$
$$\frac{1.51}{1.50} = 1 + 11\times10^{-6}\Delta T$$
$$0.006667 = 11\times10^{-6}\Delta T$$
$$\Delta T = \frac{0.006667}{11\times10^{-6}} = 606.1°C$$
$$T_2 = 10 + 606.1 = \boxed{616.1°C}$$
---
### Question 10
**a) Evaporation:**
Evaporation is the process by which molecules at the surface of a liquid escape into the vapour phase at temperatures below the boiling point. It occurs at any temperature and only at the liquid surface.
**b) Evaporation causes cooling:**
The molecules with the highest kinetic energy escape from the liquid surface during evaporation. This lowers the average kinetic energy of the remaining molecules, hence reducing the temperature of the liquid. The liquid thus absorbs latent heat from its surroundings, producing a cooling effect.
---
## SECTION B: MECHANICS
---
### Question 11
**a) i) Newton's Third Law:**
For every action, there is an equal and opposite reaction. When body A exerts a force on body B, body B exerts an equal force in the opposite direction on body A.
**ii) Non-conservative forces:**
These are forces for which the work done depends on the path taken, not just the initial and final positions. The work done in a closed loop is not zero. Examples: friction, air resistance, viscous drag.
**iii) Conservation of Linear Momentum:**
The total linear momentum of a system of bodies remains constant provided no external force acts on the system.
$$\sum p_{before} = \sum p_{after}$$
---
**b) Mirror-laser interaction:**
**i) Discussion:**
When photons strike the mirror, they exert radiation pressure. Each photon carries momentum p = E/c. Since the mirror is a perfect reflector, each photon's momentum changes by 2p (reversal). By Newton's third law, the mirror gains momentum equal to 2E/c. The laser energy is entirely reflected (no absorption), so by conservation of energy, the mirror gains kinetic energy from the recoil and the reflected beam has slightly less energy (negligible for a massive mirror).
**ii) Angle of deflection:**
Momentum of laser beam:
$$p = \frac{E}{c} = \frac{100}{3\times10^8} = 3.33\times10^{-7} \text{ kg·ms}^{-1}$$
Change in momentum of mirror = 2p (reflection):
$$\Delta p_{mirror} = 2 \times 3.33\times10^{-7} = 6.67\times10^{-7} \text{ kg·ms}^{-1}$$
This is the horizontal impulse imparted to the mirror.
The mirror swings as a pendulum. Using energy conservation:
$$\frac{(\Delta p)^2}{2m} = mgl(1-\cos\theta)$$
$$1-\cos\theta = \frac{(\Delta p)^2}{2m^2 gl}$$
$$= \frac{(6.67\times10^{-7})^2}{2\times(9\times10^{-6})^2\times10\times0.04}$$
$$= \frac{4.45\times10^{-13}}{2\times8.1\times10^{-11}\times0.4}$$
$$= \frac{4.45\times10^{-13}}{6.48\times10^{-11}} = 6.87\times10^{-3}$$
$$\theta = \cos^{-1}(1 - 0.00687) \approx \boxed{6.7°}$$
---
### Question 12
**a) i) Law of Universal Gravitation:**
Every particle of matter in the universe attracts every other particle with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between them:
$$F = \frac{Gm_1m_2}{r^2}$$
**ii) Mass and density of Earth:**
At Earth's surface: gravitational force = weight:
$$\frac{GMm}{R^2} = mg \Rightarrow M = \frac{gR^2}{G}$$
$$M = \frac{10 \times (6.4\times10^6)^2}{6.67\times10^{-11}} = \frac{10 \times 4.096\times10^{13}}{6.67\times10^{-11}}$$
$$\boxed{M = 6.14\times10^{24} \text{ kg}}$$
Density:
$$\rho = \frac{M}{\frac{4}{3}\pi R^3} = \frac{6.14\times10^{24}}{\frac{4}{3}\pi(6.4\times10^6)^3}$$
$$= \frac{6.14\times10^{24}}{1.098\times10^{21}} = \boxed{5.59\times10^3 \text{ kgm}^{-3}}$$
---
**b) i) Show g_moon = g_earth/6:**
Given: R_moon = 0.25R_earth, ρ_moon = (2/3)ρ_earth
$$g = \frac{GM}{R^2} = \frac{G \cdot \frac{4}{3}\pi R^3 \rho}{R^2} = \frac{4}{3}\pi G\rho R$$
$$\frac{g_{moon}}{g_{earth}} = \frac{\rho_{moon} \cdot R_{moon}}{\rho_{earth} \cdot R_{earth}} = \frac{2}{3} \times 0.25 = \frac{1}{6} \quad \checkmark$$
$$\boxed{g_{moon} = \frac{g}{6} = \frac{10}{6} = 1.67 \text{ ms}^{-2}}$$
**ii) Height of jump on moon:**
On Earth: h = 2 m, g = 10 ms⁻²
Initial KE = PE gained:
$$v^2 = 2gh = 2\times10\times2 = 40 \text{ m}^2\text{s}^{-2}$$
On moon, same initial velocity v² = 40:
$$h_{moon} = \frac{v^2}{2g_{moon}} = \frac{40}{2\times\frac{10}{6}} = \frac{40}{\frac{20}{6}} = \frac{40\times6}{20} = \boxed{12 \text{ m}}$$
---
## SECTION C: HEAT AND PROPERTIES OF MATTER
---
### Question 13
**a) i) Laminar flow:**
Laminar (streamline) flow is the smooth, orderly flow of fluid in parallel layers with no disruption between layers. Each fluid particle follows a smooth path and layers do not mix.
**ii) Turbulent flow:**
Turbulent flow is an irregular, chaotic flow in which fluid particles move in random directions forming eddies and vortices. It occurs when flow velocity exceeds a critical value.
---
**b) Reynolds Number:**
$$Re = \frac{\rho v d}{\eta}$$
where ρ = density, v = velocity, d = pipe diameter, η = viscosity.
**Dimensionless check:**
$$[Re] = \frac{[kgm^{-3}][ms^{-1}][m]}{[kgm^{-1}s^{-1}]} = \frac{kgm^{-1}s^{-1}}{kgm^{-1}s^{-1}} = \text{dimensionless} \checkmark$$
---
**c) Bernoulli's Equation — Work per unit volume constant:**
The given equation:
$$(P_1-P_2)\Delta v = \frac{1}{2}\rho\Delta v(v_2^2-v_1^2) + \rho g\Delta v(h_2-h_1)$$
Dividing through by Δv:
$$P_1 - P_2 = \frac{1}{2}\rho(v_2^2-v_1^2) + \rho g(h_2-h_1)$$
Rearranging:
$$P_1 + \frac{1}{2}\rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho gh_2$$
Each term has dimensions of pressure = energy per unit volume = work per unit volume. Since this sum is equal at any two points, **total work done per unit volume is constant**. ✓
---
**d) Pipe flow problem:**
Given: d₁ = 0.02 m → A₁ = π(0.01)² = 3.14×10⁻⁴ m²
d₂ = 0.01 m → A₂ = π(0.005)² = 7.85×10⁻⁵ m²
P₁ = 4×10⁵ Pa, v₁ = 4 ms⁻¹, h₂ - h₁ = 5 m
**i) Flow velocity at bathroom (continuity):**
$$A_1v_1 = A_2v_2$$
$$v_2 = \frac{A_1v_1}{A_2} = \frac{3.14\times10^{-4}\times4}{7.85\times10^{-5}} = \frac{1.256\times10^{-3}}{7.85\times10^{-5}} = \boxed{16 \text{ ms}^{-1}}$$
**Pressure at bathroom (Bernoulli):**
$$P_2 = P_1 + \frac{1}{2}\rho(v_1^2-v_2^2) - \rho g(h_2-h_1)$$
$$= 4\times10^5 + \frac{1}{2}(1000)(16-256) - 1000\times10\times5$$
$$= 4\times10^5 + 500(-240) - 50000$$
$$= 400000 - 120000 - 50000$$
$$= \boxed{2.3\times10^5 \text{ Pa}}$$
**ii) If tap is turned off (v₂ = 0 → v₁ = 0 by continuity):**
Using static pressure only:
$$P_2 = P_1 - \rho g h = 4\times10^5 - 1000\times10\times5 = 4\times10^5 - 5\times10^4$$
$$= \boxed{3.5\times10^5 \text{ Pa}}$$
---
### Question 14
**a) i) Specific heat capacity:**
The quantity of heat required to raise the temperature of 1 kg of a substance by 1 K (or 1°C).
**Unit:** Jkg⁻¹K⁻¹
**ii) Specific latent heat of fusion:**
The quantity of heat required to change 1 kg of a solid to liquid at constant temperature.
**Unit:** Jkg⁻¹
---
**b) Continuous flow method:**
- Liquid flows at a steady rate through a calorimeter heated by an electrical coil
- At steady state, inlet and outlet temperatures are constant
- Power supplied = heat gained by liquid per second
- Two experiments with different flow rates but same temperature rise eliminate heat losses:
$$IV_1 - IV_2 = (m_1 - m_2)c\Delta\theta/t$$
(Heat losses cancel since same temperature difference maintained)
---
**c) Calculating specific heat capacity:**
Using the two-experiment method to eliminate heat losses (Q_loss):
$$I_1V_1t_1 - Q_{loss} = m_1c\Delta\theta$$
$$I_2V_2t_2 - Q_{loss} = m_2c\Delta\theta$$
Subtracting:
$$I_1V_1t_1 - I_2V_2t_2 = (m_1 - m_2)c\Delta\theta$$
Calculate:
- Exp 1: P₁t₁ = 2.0 × 5.5 × 60 = 660 J
- Exp 2: P₂t₂ = 1.5 × 4.5 × 180 = 1215 J
- m₁ = 0.0405 kg, m₂ = 0.0700 kg
- Δθ = 45 - 38 = 7°C
$$c = \frac{I_1V_1t_1 - I_2V_2t_2}{(m_1-m_2)\Delta\theta}$$
Note: since m₂ > m₁ and P₂t₂ > P₁t₁, use correct sign:
$$c = \frac{1215 - 660}{(0.0700 - 0.0405)\times7} = \frac{555}{0.0295\times7} = \frac{555}{0.2065}$$
$$\boxed{c \approx 2688 \text{ Jkg}^{-1}\text{K}^{-1} \approx 2690 \text{ Jkg}^{-¹K}^{-1}}$$
*(Small deviation from 4200 suggests experimental losses — the method eliminates systematic heat loss but measurement values give this result.)*
---
**c) Factors affecting evaporation:**
**1. Temperature:** Higher temperature → more molecules have sufficient KE to escape → faster evaporation.
**2. Surface area:** Larger exposed surface area → more molecules at the surface able to escape → faster evaporation.
**3. Humidity (concentration of vapour above liquid):** Lower humidity above the liquid → steeper concentration gradient → faster evaporation. High humidity retards evaporation.
*(Other valid factors: wind/air movement, nature of liquid)*
---
### Question 15
**a) Coefficient of viscosity:**
The coefficient of viscosity η of a fluid is defined as the ratio of the shear stress to the velocity gradient:
$$\eta = \frac{F/A}{dv/dx}$$
**Unit:** Pa·s (or Nsm⁻²)
**Applications:**
1. **Lubrication in engines** — viscous oil reduces friction between moving engine parts
2. **Medical use (blood flow)** — viscosity of blood is monitored to diagnose circulatory disorders; IV fluid viscosity must match for safe delivery
---
**b) i) Terminal velocity expression:**
At terminal velocity, weight = upthrust + viscous drag:
$$\rho_s \cdot \frac{4}{3}\pi r^3 g = \rho \cdot \frac{4}{3}\pi r^3 g + kv_T r\eta$$
$$(\rho_s - \rho)\frac{4}{3}\pi r^3 g = kv_T r\eta$$
$$\boxed{v_T = \frac{4\pi r^2(\rho_s-\rho)g}{3k\eta}}$$
*(With Stokes' law k = 6π: V_T = 2r²(ρₛ-ρ)g / 9η)*
---
## SECTION C (continued) — Question 16
**a) i) First Law of Thermodynamics:**
The total internal energy of a system increases by the amount of heat added to the system minus the work done by the system on its surroundings:
$$\Delta U = Q - W$$
Heat added to a system either increases its internal energy or does work on surroundings.
**ii) Internal energy depends only on temperature:**
For an ideal gas, internal energy U = nCᵥT.
From the first law, for an isothermal process (ΔT = 0):
If we show that (∂U/∂V)_T = 0 for an ideal gas using PV = nRT, then U depends only on T, not on V or P. This is Joule's experiment — free expansion of ideal gas causes no temperature change, confirming internal energy depends only on temperature. ✓
---
**b) Isothermal vs Adiabatic changes:**
**Isothermal:** Temperature is constant (T = const). On a P-V diagram, it follows the curve PV = constant. Heat is exchanged with surroundings.
**Adiabatic:** No heat exchange with surroundings (Q = 0). Temperature changes. Follows PVᵞ = constant. The curve is steeper than isothermal on P-V diagram.
*(Sketch: Two curves from the same initial point — adiabatic falls more steeply than isothermal)*
---
**c) i) Adiabatic expansion:**
Given: V₁ = 6×10⁻⁴ m³, T₁ = 219 K, T₂ = 273 K, γ = 1.40
For adiabatic: TV^(γ-1) = constant
$$T_1V_1^{\gamma-1} = T_2V_2^{\gamma-1}$$
$$V_2^{\gamma-1} = V_1^{\gamma-1}\cdot\frac{T_1}{T_2}$$
$$V_2^{0.4} = (6\times10^{-4})^{0.4}\times\frac{219}{273}$$
$$(6\times10^{-4})^{0.4}:$$
$$= (6)^{0.4}\times(10^{-4})^{0.4} = 2.048\times10^{-1.6} = 2.048\times0.02512 = 0.05145$$
$$V_2^{0.4} = 0.05145\times0.8022 = 0.04127$$
$$V_2 = (0.04127)^{1/0.4} = (0.04127)^{2.5}$$
$$= (0.04127)^2 \times (0.04127)^{0.5} = 1.703\times10^{-3}\times0.2031 = 3.46\times10^{-4}$$
$$\boxed{V_2 \approx 3.46\times10^{-4} \text{ m}^3}$$
---
**c) ii) Wilson cloud chamber — temperature after expansion:**
Initial T₁ = 20°C = 293 K
Expansion ratio V₂/V₁ = 1.375, γ = 1.40
$$T_2 = T_1\left(\frac{V_1}{V_2}\right)^{\gamma-1} = 293\times\left(\frac{1}{1.375}\right)^{0.4}$$
$$= 293\times(0.7273)^{0.4}$$
$$(0.7273)^{0.4} = e^{0.4\ln(0.7273)} = e^{0.4\times(-0.3185)} = e^{-0.1274} = 0.8804$$
$$T_2 = 293\times0.8804 = \boxed{257.9 \text{ K} \approx 258 \text{ K} (-15°C)}$$
---
**Stokes' settling (pollen particles):**
Density of pollen ρₛ = 1.8×10³ kgm⁻³, ρ_water = 1000 kgm⁻³
Depth h = 0.02 m, time t = 3600 s
η_water = 1×10⁻³ Pa·s
Terminal velocity needed:
$$v_T = \frac{h}{t} = \frac{0.02}{3600} = 5.56\times10^{-6} \text{ ms}^{-1}$$
Stokes' law:
$$v_T = \frac{2r^2(\rho_s-\rho)g}{9\eta}$$
$$r^2 = \frac{9\eta v_T}{2(\rho_s-\rho)g} = \frac{9\times10^{-3}\times5.56\times10^{-6}}{2\times800\times10}$$
$$= \frac{5.0\times10^{-8}}{16000} = 3.125\times10^{-12} \text{ m}^2$$
$$r = 1.77\times10^{-6} \text{ m}$$
$$\boxed{d = 2r \approx 3.5\times10^{-6} \text{ m} = 3.5 \text{ μm}}$$
---
## SECTION D: VIBRATION AND WAVES
---
### Question 17
**a) i) Fundamental note:**
The lowest frequency (first harmonic) produced by a vibrating body. It is the note with the simplest standing wave pattern — one loop in an open pipe, half a loop in closed pipe.
**ii) Overtones:**
Frequencies higher than the fundamental that are produced simultaneously with it. First overtone = second harmonic (or third harmonic for closed pipe), etc.
**iii) Harmonic frequencies:**
Whole-number multiples of the fundamental frequency. First harmonic = f₀, second harmonic = 2f₀, etc.
---
**b) i) Overtones in closed pipe:**
*(Sketches — standing wave diagrams):*
- **1st overtone (3rd harmonic):** 3 quarter-wavelengths — node at closed end, antinode at open end, one extra node-antinode pair.
- **2nd overtone (5th harmonic):** 5 quarter-wavelengths
- **3rd overtone (7th harmonic):** 7 quarter-wavelengths
**ii) Closed pipe produces only odd harmonics:**
For a closed pipe of length L:
$$L = \frac{n\lambda}{4}, \quad n = 1, 3, 5...$$
$$f_n = \frac{nv}{4L}, \quad n = 1, 3, 5...$$
Only odd values of n satisfy the boundary conditions (node at closed end, antinode at open end), therefore **only odd harmonics are produced**. ✓
---
**c) End correction:**
The antinode in an open pipe or the open end of a resonance tube does not form exactly at the physical end of the tube but slightly beyond it. The end correction **e** is the small extra length that must be added to the physical length to get the effective acoustic length. It is approximately 0.3d where d = diameter.
---
**d) Finding end correction:**
Closed pipe: L_c = 0.46 m, Open pipe: L_o = 0.60 m (same diameter d, same end correction e)
Let e = end correction.
First overtone of **closed pipe** = 3rd harmonic:
$$f = \frac{3v}{4(L_c + e)}$$
First overtone of **open pipe** = 2nd harmonic:
$$f = \frac{2v}{2(L_o + 2e)} = \frac{v}{L_o + 2e}$$
Setting equal:
$$\frac{3v}{4(L_c+e)} = \frac{v}{L_o+2e}$$
$$3(L_o+2e) = 4(L_c+e)$$
$$3L_o + 6e = 4L_c + 4e$$
$$2e = 4L_c - 3L_o$$
$$2e = 4(0.46) - 3(0.60) = 1.84 - 1.80 = 0.04$$
$$\boxed{e = 0.02 \text{ m} = 2 \text{ cm}}$$
---
### Question 18
**a) i) Simple Harmonic Motion (SHM):**
SHM is a periodic motion in which the restoring force is directly proportional to displacement from the equilibrium position and always directed towards that position:
$$F = -kx \quad \Rightarrow \quad a = -\omega^2 x$$
**ii) Three examples:**
1. **Simple pendulum** (small oscillations) — gravity provides restoring force proportional to displacement
2. **Mass on a spring** — elastic restoring force obeys Hooke's law
3. **Liquid in a U-tube** — weight of displaced liquid provides linear restoring force
---
**b) Liquid in U-tube:**
**i) Show motion is SHM:**
When displaced by x, the height difference = 2x.
Net restoring force = weight of excess liquid column:
$$F = -\rho A(2x)g$$
Mass of liquid: m = ρAL
$$ma = -2\rho Agx$$
$$a = -\frac{2g}{L}x$$
Since a = −ω²x where ω² = 2g/L → motion is **SHM** ✓
**ii) Period:**
$$T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{L}{2g}} = \boxed{\pi\sqrt{\frac{2L}{g}}}$$
---
**c) x = 12cos(5πt)**
**i) Amplitude:**
$$\boxed{A = 12 \text{ mm}}$$
**ii) Period:**
$$\omega = 5\pi \text{ rad/s}$$
$$T = \frac{2\pi}{\omega} = \frac{2\pi}{5\pi} = \boxed{0.4 \text{ s}}$$
**iii) Maximum velocity:**
$$v_{max} = A\omega = 12\times10^{-3}\times5\pi = 0.06\pi$$
$$= \boxed{0.1885 \text{ ms}^{-1} \approx 0.19 \text{ ms}^{-1}}$$
*(Converting A from mm: 12 mm = 0.012 m)*
# 2025 IJMBE Physics I — Full Solutions
---
## SECTION A
---
### Question 1
**i) Fundamental vs Derived Quantities**
**Fundamental quantities** are independent physical quantities that cannot be expressed in terms of other quantities. Examples: mass (kg), length (m), time (s), temperature (K), current (A).
**Derived quantities** are obtained by combining fundamental quantities through multiplication or division. Examples: velocity (m/s), force (kg·m/s²), energy (J).
---
**ii) Dimension of h in E = hf**
- E = energy → dimensions: [ML²T⁻²]
- f = frequency → dimensions: [T⁻¹]
$$h = \frac{E}{f} = \frac{[ML^2T^{-2}]}{[T^{-1}]} = [ML^2T^{-1}]$$
---
### Question 2
Mass m = 4 kg, g = 10 ms⁻²
**i) Stationary lift:**
$$W = mg = 4 \times 10 = \boxed{40 \text{ N}}$$
**ii) Moving upward, a = 0.15 ms⁻²:**
$$W = m(g + a) = 4(10 + 0.15) = 4 \times 10.15 = \boxed{40.6 \text{ N}}$$
**iii) Moving downward, a = 0.20 ms⁻²:**
$$W = m(g - a) = 4(10 - 0.20) = 4 \times 9.80 = \boxed{39.2 \text{ N}}$$
---
### Question 3
**When block is fixed (bullet embeds):**
- m₁ = 0.01 kg (bullet), m₂ = 1 kg (block)
- Using work-energy theorem, retarding force F:
$$F \cdot d_1 = \frac{1}{2}m_1v^2$$
$$F \times 0.10 = \frac{1}{2}(0.01)v^2 \quad \text{...(1)}$$
**When block is free (bullet embeds, momentum conserved):**
By conservation of momentum:
$$m_1 v = (m_1 + m_2)V$$
$$V = \frac{0.01v}{1.01} \approx \frac{v}{101}$$
KE available to decelerate bullet relative to block:
$$KE_{rel} = \frac{1}{2}\mu v_{rel}^2$$
where reduced mass μ = m₁m₂/(m₁+m₂):
$$\mu = \frac{0.01 \times 1}{1.01} \approx 0.0099 \text{ kg}$$
Relative velocity of bullet w.r.t. block after impact begins:
$$v_{rel} = v - V = v - \frac{0.01v}{1.01} = v\left(\frac{1}{1.01}\right) \approx \frac{v}{1.01}$$
Energy available in free case:
$$KE_2 = \frac{1}{2}\mu v_{rel}^2 = \frac{1}{2} \cdot \frac{m_1 m_2}{m_1+m_2} \cdot \left(\frac{v}{1.01}\right)^2$$
$$= \frac{1}{2} \cdot \frac{0.01 \times 1}{1.01} \cdot \frac{v^2}{1.0201}$$
$$= \frac{1}{2} \cdot \frac{0.01 v^2}{(1.01)^2} = \frac{1}{2} \cdot \frac{0.01 v^2}{1.0201}$$
From (1): ½(0.01)v² = F × 0.10, so ½v² = 10F/0.01 → ½(0.01)v² = 0.1F...
More directly — ratio of penetration depths:
$$\frac{d_2}{d_1} = \frac{KE_2}{KE_1} = \frac{\frac{1}{2}\cdot\frac{m_1 m_2}{m_1+m_2}\cdot v^2}{\frac{1}{2}m_1 v^2} = \frac{m_2}{m_1+m_2}$$
$$\frac{d_2}{0.10} = \frac{1}{1.01} = 0.99$$
$$\boxed{d_2 \approx 0.099 \text{ m} \approx 9.9 \text{ cm}}$$
---
### Question 4
Speed of sound v = 340 ms⁻¹, v_s = 50 ms⁻¹, f₀ = 500 Hz
**Doppler formula:** Observer stationary, source moving.
**i) Train approaching:**
$$f = f_0 \cdot \frac{v}{v - v_s} = 500 \times \frac{340}{340 - 50} = 500 \times \frac{340}{290}$$
$$= 500 \times 1.172 = \boxed{586.2 \text{ Hz}}$$
**ii) Train receding:**
$$f = f_0 \cdot \frac{v}{v + v_s} = 500 \times \frac{340}{340 + 50} = 500 \times \frac{340}{390}$$
$$= 500 \times 0.872 = \boxed{435.9 \text{ Hz}}$$
**Comment:** The perceived frequency is higher when the source approaches and lower when it recedes — this is the **Doppler Effect**. The actual whistle frequency (500 Hz) lies between the two perceived values.
---
### Question 5
Let specific heat capacities be c₁, c₂, c₃ for liquids at 15°C, 20°C, 25°C.
**Mix 1 + 2 → 18°C:**
$$mc_1(18-15) = mc_2(20-18)$$
$$3c_1 = 2c_2 \Rightarrow c_2 = 1.5c_1 \quad \text{...(1)}$$
**Mix 2 + 3 → 24°C:**
$$mc_2(24-20) = mc_3(25-24)$$
$$4c_2 = c_3 \Rightarrow c_3 = 4c_2 = 6c_1 \quad \text{...(2)}$$
**Mix 1 + 3 → T°C:**
$$mc_1(T-15) = mc_3(25-T)$$
$$c_1(T-15) = 6c_1(25-T)$$
$$T - 15 = 150 - 6T$$
$$7T = 165$$
$$\boxed{T = 23.6°C}$$
---
### Question 6
Solar constant S = 0.14 Wcm⁻² = 1400 Wm⁻²
Ratio R_orbit/R_sun = 215
Stefan's Law: Power radiated by sun:
$$P = \sigma T^4 \cdot 4\pi R_s^2$$
Solar constant at Earth:
$$S = \frac{P}{4\pi R_{orbit}^2} = \sigma T^4 \left(\frac{R_s}{R_{orbit}}\right)^2$$
$$T^4 = \frac{S}{\sigma} \times \left(\frac{R_{orbit}}{R_s}\right)^2 = \frac{1400}{5.67\times10^{-8}} \times (215)^2$$
$$T^4 = 2.469 \times 10^{10} \times 46225 = 1.141 \times 10^{15}$$
$$T = (1.141 \times 10^{15})^{0.25} = \boxed{5.81 \times 10^3 \text{ K} \approx 5810 \text{ K}}$$
---
### Question 7
Inflow rate Q_in = 10⁻⁴ m²s⁻¹ *(interpreting as volume flow rate)*
Outflow hole area A = 10⁻⁴ m²
At equilibrium, inflow = outflow:
$$Q_{in} = A\sqrt{2gh}$$
$$10^{-4} = 10^{-4}\sqrt{2 \times 10 \times h}$$
$$1 = \sqrt{20h}$$
$$1 = 20h$$
$$\boxed{h = 0.05 \text{ m}}$$
---
### Question 8
f = 4 × 10⁴ Hz, v_air = 400 ms⁻¹, v_water = 1350 ms⁻¹
Echo time t = 0.8 s (round trip in water)
**i) Depth of ocean:**
$$d = \frac{v_{water} \times t}{2} = \frac{1350 \times 0.8}{2} = \frac{1080}{2} = \boxed{540 \text{ m}}$$
**ii) Wavelength in air:**
$$\lambda_{air} = \frac{v_{air}}{f} = \frac{400}{4\times10^4} = \boxed{0.01 \text{ m} = 1 \text{ cm}}$$
**Wavelength in water:**
$$\lambda_{water} = \frac{v_{water}}{f} = \frac{1350}{4\times10^4} = \boxed{0.03375 \text{ m} \approx 3.375 \text{ cm}}$$
---
### Question 9
d₁ = 1.50 m → r₁ = 0.75 m (tyre)
d₂ = 1.51 m → r₂ = 0.755 m (wheel)
T₁ = 10°C, α = 11.0 × 10⁻⁶ K⁻¹
For the tyre to expand to wheel diameter:
$$d_2 = d_1(1 + \alpha\Delta T)$$
$$1.51 = 1.50(1 + 11\times10^{-6}\times\Delta T)$$
$$\frac{1.51}{1.50} = 1 + 11\times10^{-6}\Delta T$$
$$0.006667 = 11\times10^{-6}\Delta T$$
$$\Delta T = \frac{0.006667}{11\times10^{-6}} = 606.1°C$$
$$T_2 = 10 + 606.1 = \boxed{616.1°C}$$
---
### Question 10
**a) Evaporation:**
Evaporation is the process by which molecules at the surface of a liquid escape into the vapour phase at temperatures below the boiling point. It occurs at any temperature and only at the liquid surface.
**b) Evaporation causes cooling:**
The molecules with the highest kinetic energy escape from the liquid surface during evaporation. This lowers the average kinetic energy of the remaining molecules, hence reducing the temperature of the liquid. The liquid thus absorbs latent heat from its surroundings, producing a cooling effect.
---
## SECTION B: MECHANICS
---
### Question 11
**a) i) Newton's Third Law:**
For every action, there is an equal and opposite reaction. When body A exerts a force on body B, body B exerts an equal force in the opposite direction on body A.
**ii) Non-conservative forces:**
These are forces for which the work done depends on the path taken, not just the initial and final positions. The work done in a closed loop is not zero. Examples: friction, air resistance, viscous drag.
**iii) Conservation of Linear Momentum:**
The total linear momentum of a system of bodies remains constant provided no external force acts on the system.
$$\sum p_{before} = \sum p_{after}$$
---
**b) Mirror-laser interaction:**
**i) Discussion:**
When photons strike the mirror, they exert radiation pressure. Each photon carries momentum p = E/c. Since the mirror is a perfect reflector, each photon's momentum changes by 2p (reversal). By Newton's third law, the mirror gains momentum equal to 2E/c. The laser energy is entirely reflected (no absorption), so by conservation of energy, the mirror gains kinetic energy from the recoil and the reflected beam has slightly less energy (negligible for a massive mirror).
**ii) Angle of deflection:**
Momentum of laser beam:
$$p = \frac{E}{c} = \frac{100}{3\times10^8} = 3.33\times10^{-7} \text{ kg·ms}^{-1}$$
Change in momentum of mirror = 2p (reflection):
$$\Delta p_{mirror} = 2 \times 3.33\times10^{-7} = 6.67\times10^{-7} \text{ kg·ms}^{-1}$$
This is the horizontal impulse imparted to the mirror.
The mirror swings as a pendulum. Using energy conservation:
$$\frac{(\Delta p)^2}{2m} = mgl(1-\cos\theta)$$
$$1-\cos\theta = \frac{(\Delta p)^2}{2m^2 gl}$$
$$= \frac{(6.67\times10^{-7})^2}{2\times(9\times10^{-6})^2\times10\times0.04}$$
$$= \frac{4.45\times10^{-13}}{2\times8.1\times10^{-11}\times0.4}$$
$$= \frac{4.45\times10^{-13}}{6.48\times10^{-11}} = 6.87\times10^{-3}$$
$$\theta = \cos^{-1}(1 - 0.00687) \approx \boxed{6.7°}$$
---
### Question 12
**a) i) Law of Universal Gravitation:**
Every particle of matter in the universe attracts every other particle with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between them:
$$F = \frac{Gm_1m_2}{r^2}$$
**ii) Mass and density of Earth:**
At Earth's surface: gravitational force = weight:
$$\frac{GMm}{R^2} = mg \Rightarrow M = \frac{gR^2}{G}$$
$$M = \frac{10 \times (6.4\times10^6)^2}{6.67\times10^{-11}} = \frac{10 \times 4.096\times10^{13}}{6.67\times10^{-11}}$$
$$\boxed{M = 6.14\times10^{24} \text{ kg}}$$
Density:
$$\rho = \frac{M}{\frac{4}{3}\pi R^3} = \frac{6.14\times10^{24}}{\frac{4}{3}\pi(6.4\times10^6)^3}$$
$$= \frac{6.14\times10^{24}}{1.098\times10^{21}} = \boxed{5.59\times10^3 \text{ kgm}^{-3}}$$
---
**b) i) Show g_moon = g_earth/6:**
Given: R_moon = 0.25R_earth, ρ_moon = (2/3)ρ_earth
$$g = \frac{GM}{R^2} = \frac{G \cdot \frac{4}{3}\pi R^3 \rho}{R^2} = \frac{4}{3}\pi G\rho R$$
$$\frac{g_{moon}}{g_{earth}} = \frac{\rho_{moon} \cdot R_{moon}}{\rho_{earth} \cdot R_{earth}} = \frac{2}{3} \times 0.25 = \frac{1}{6} \quad \checkmark$$
$$\boxed{g_{moon} = \frac{g}{6} = \frac{10}{6} = 1.67 \text{ ms}^{-2}}$$
**ii) Height of jump on moon:**
On Earth: h = 2 m, g = 10 ms⁻²
Initial KE = PE gained:
$$v^2 = 2gh = 2\times10\times2 = 40 \text{ m}^2\text{s}^{-2}$$
On moon, same initial velocity v² = 40:
$$h_{moon} = \frac{v^2}{2g_{moon}} = \frac{40}{2\times\frac{10}{6}} = \frac{40}{\frac{20}{6}} = \frac{40\times6}{20} = \boxed{12 \text{ m}}$$
---
## SECTION C: HEAT AND PROPERTIES OF MATTER
---
### Question 13
**a) i) Laminar flow:**
Laminar (streamline) flow is the smooth, orderly flow of fluid in parallel layers with no disruption between layers. Each fluid particle follows a smooth path and layers do not mix.
**ii) Turbulent flow:**
Turbulent flow is an irregular, chaotic flow in which fluid particles move in random directions forming eddies and vortices. It occurs when flow velocity exceeds a critical value.
---
**b) Reynolds Number:**
$$Re = \frac{\rho v d}{\eta}$$
where ρ = density, v = velocity, d = pipe diameter, η = viscosity.
**Dimensionless check:**
$$[Re] = \frac{[kgm^{-3}][ms^{-1}][m]}{[kgm^{-1}s^{-1}]} = \frac{kgm^{-1}s^{-1}}{kgm^{-1}s^{-1}} = \text{dimensionless} \checkmark$$
---
**c) Bernoulli's Equation — Work per unit volume constant:**
The given equation:
$$(P_1-P_2)\Delta v = \frac{1}{2}\rho\Delta v(v_2^2-v_1^2) + \rho g\Delta v(h_2-h_1)$$
Dividing through by Δv:
$$P_1 - P_2 = \frac{1}{2}\rho(v_2^2-v_1^2) + \rho g(h_2-h_1)$$
Rearranging:
$$P_1 + \frac{1}{2}\rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho gh_2$$
Each term has dimensions of pressure = energy per unit volume = work per unit volume. Since this sum is equal at any two points, **total work done per unit volume is constant**. ✓
---
**d) Pipe flow problem:**
Given: d₁ = 0.02 m → A₁ = π(0.01)² = 3.14×10⁻⁴ m²
d₂ = 0.01 m → A₂ = π(0.005)² = 7.85×10⁻⁵ m²
P₁ = 4×10⁵ Pa, v₁ = 4 ms⁻¹, h₂ - h₁ = 5 m
**i) Flow velocity at bathroom (continuity):**
$$A_1v_1 = A_2v_2$$
$$v_2 = \frac{A_1v_1}{A_2} = \frac{3.14\times10^{-4}\times4}{7.85\times10^{-5}} = \frac{1.256\times10^{-3}}{7.85\times10^{-5}} = \boxed{16 \text{ ms}^{-1}}$$
**Pressure at bathroom (Bernoulli):**
$$P_2 = P_1 + \frac{1}{2}\rho(v_1^2-v_2^2) - \rho g(h_2-h_1)$$
$$= 4\times10^5 + \frac{1}{2}(1000)(16-256) - 1000\times10\times5$$
$$= 4\times10^5 + 500(-240) - 50000$$
$$= 400000 - 120000 - 50000$$
$$= \boxed{2.3\times10^5 \text{ Pa}}$$
**ii) If tap is turned off (v₂ = 0 → v₁ = 0 by continuity):**
Using static pressure only:
$$P_2 = P_1 - \rho g h = 4\times10^5 - 1000\times10\times5 = 4\times10^5 - 5\times10^4$$
$$= \boxed{3.5\times10^5 \text{ Pa}}$$
---
### Question 14
**a) i) Specific heat capacity:**
The quantity of heat required to raise the temperature of 1 kg of a substance by 1 K (or 1°C).
**Unit:** Jkg⁻¹K⁻¹
**ii) Specific latent heat of fusion:**
The quantity of heat required to change 1 kg of a solid to liquid at constant temperature.
**Unit:** Jkg⁻¹
---
**b) Continuous flow method:**
- Liquid flows at a steady rate through a calorimeter heated by an electrical coil
- At steady state, inlet and outlet temperatures are constant
- Power supplied = heat gained by liquid per second
- Two experiments with different flow rates but same temperature rise eliminate heat losses:
$$IV_1 - IV_2 = (m_1 - m_2)c\Delta\theta/t$$
(Heat losses cancel since same temperature difference maintained)
---
**c) Calculating specific heat capacity:**
Using the two-experiment method to eliminate heat losses (Q_loss):
$$I_1V_1t_1 - Q_{loss} = m_1c\Delta\theta$$
$$I_2V_2t_2 - Q_{loss} = m_2c\Delta\theta$$
Subtracting:
$$I_1V_1t_1 - I_2V_2t_2 = (m_1 - m_2)c\Delta\theta$$
Calculate:
- Exp 1: P₁t₁ = 2.0 × 5.5 × 60 = 660 J
- Exp 2: P₂t₂ = 1.5 × 4.5 × 180 = 1215 J
- m₁ = 0.0405 kg, m₂ = 0.0700 kg
- Δθ = 45 - 38 = 7°C
$$c = \frac{I_1V_1t_1 - I_2V_2t_2}{(m_1-m_2)\Delta\theta}$$
Note: since m₂ > m₁ and P₂t₂ > P₁t₁, use correct sign:
$$c = \frac{1215 - 660}{(0.0700 - 0.0405)\times7} = \frac{555}{0.0295\times7} = \frac{555}{0.2065}$$
$$\boxed{c \approx 2688 \text{ Jkg}^{-1}\text{K}^{-1} \approx 2690 \text{ Jkg}^{-¹K}^{-1}}$$
*(Small deviation from 4200 suggests experimental losses — the method eliminates systematic heat loss but measurement values give this result.)*
---
**c) Factors affecting evaporation:**
**1. Temperature:** Higher temperature → more molecules have sufficient KE to escape → faster evaporation.
**2. Surface area:** Larger exposed surface area → more molecules at the surface able to escape → faster evaporation.
**3. Humidity (concentration of vapour above liquid):** Lower humidity above the liquid → steeper concentration gradient → faster evaporation. High humidity retards evaporation.
*(Other valid factors: wind/air movement, nature of liquid)*
---
### Question 15
**a) Coefficient of viscosity:**
The coefficient of viscosity η of a fluid is defined as the ratio of the shear stress to the velocity gradient:
$$\eta = \frac{F/A}{dv/dx}$$
**Unit:** Pa·s (or Nsm⁻²)
**Applications:**
1. **Lubrication in engines** — viscous oil reduces friction between moving engine parts
2. **Medical use (blood flow)** — viscosity of blood is monitored to diagnose circulatory disorders; IV fluid viscosity must match for safe delivery
---
**b) i) Terminal velocity expression:**
At terminal velocity, weight = upthrust + viscous drag:
$$\rho_s \cdot \frac{4}{3}\pi r^3 g = \rho \cdot \frac{4}{3}\pi r^3 g + kv_T r\eta$$
$$(\rho_s - \rho)\frac{4}{3}\pi r^3 g = kv_T r\eta$$
$$\boxed{v_T = \frac{4\pi r^2(\rho_s-\rho)g}{3k\eta}}$$
*(With Stokes' law k = 6π: V_T = 2r²(ρₛ-ρ)g / 9η)*
**a) i) First Law of Thermodynamics:**
The total internal energy of a system increases by the amount of heat added to the system minus the work done by the system on its surroundings:
$$\Delta U = Q - W$$
Heat added to a system either increases its internal energy or does work on surroundings.
**ii) Internal energy depends only on temperature:**
For an ideal gas, internal energy U = nCᵥT.
From the first law, for an isothermal process (ΔT = 0):
If we show that (∂U/∂V)_T = 0 for an ideal gas using PV = nRT, then U depends only on T, not on V or P. This is Joule's experiment — free expansion of ideal gas causes no temperature change, confirming internal energy depends only on temperature. ✓
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**b) Isothermal vs Adiabatic changes:**
**Isothermal:** Temperature is constant (T = const). On a P-V diagram, it follows the curve PV = constant. Heat is exchanged with surroundings.
**Adiabatic:** No heat exchange with surroundings (Q = 0). Temperature changes. Follows PVᵞ = constant. The curve is steeper than isothermal on P-V diagram.
*(Sketch: Two curves from the same initial point — adiabatic falls more steeply than isothermal)*
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**c) i) Adiabatic expansion:**
Given: V₁ = 6×10⁻⁴ m³, T₁ = 219 K, T₂ = 273 K, γ = 1.40
For adiabatic: TV^(γ-1) = constant
$$T_1V_1^{\gamma-1} = T_2V_2^{\gamma-1}$$
$$V_2^{\gamma-1} = V_1^{\gamma-1}\cdot\frac{T_1}{T_2}$$
$$V_2^{0.4} = (6\times10^{-4})^{0.4}\times\frac{219}{273}$$
$$(6\times10^{-4})^{0.4}:$$
$$= (6)^{0.4}\times(10^{-4})^{0.4} = 2.048\times10^{-1.6} = 2.048\times0.02512 = 0.05145$$
$$V_2^{0.4} = 0.05145\times0.8022 = 0.04127$$
$$V_2 = (0.04127)^{1/0.4} = (0.04127)^{2.5}$$
$$= (0.04127)^2 \times (0.04127)^{0.5} = 1.703\times10^{-3}\times0.2031 = 3.46\times10^{-4}$$
$$\boxed{V_2 \approx 3.46\times10^{-4} \text{ m}^3}$$
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**c) ii) Wilson cloud chamber — temperature after expansion:**
Initial T₁ = 20°C = 293 K
Expansion ratio V₂/V₁ = 1.375, γ = 1.40
$$T_2 = T_1\left(\frac{V_1}{V_2}\right)^{\gamma-1} = 293\times\left(\frac{1}{1.375}\right)^{0.4}$$
$$= 293\times(0.7273)^{0.4}$$
$$(0.7273)^{0.4} = e^{0.4\ln(0.7273)} = e^{0.4\times(-0.3185)} = e^{-0.1274} = 0.8804$$
$$T_2 = 293\times0.8804 = \boxed{257.9 \text{ K} \approx 258 \text{ K} (-15°C)}$$
---
**Stokes' settling (pollen particles):**
Density of pollen ρₛ = 1.8×10³ kgm⁻³, ρ_water = 1000 kgm⁻³
Depth h = 0.02 m, time t = 3600 s
η_water = 1×10⁻³ Pa·s
Terminal velocity needed:
$$v_T = \frac{h}{t} = \frac{0.02}{3600} = 5.56\times10^{-6} \text{ ms}^{-1}$$
Stokes' law:
$$v_T = \frac{2r^2(\rho_s-\rho)g}{9\eta}$$
$$r^2 = \frac{9\eta v_T}{2(\rho_s-\rho)g} = \frac{9\times10^{-3}\times5.56\times10^{-6}}{2\times800\times10}$$
$$= \frac{5.0\times10^{-8}}{16000} = 3.125\times10^{-12} \text{ m}^2$$
$$r = 1.77\times10^{-6} \text{ m}$$
$$\boxed{d = 2r \approx 3.5\times10^{-6} \text{ m} = 3.5 \text{ μm}}$$
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## SECTION D: VIBRATION AND WAVES
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### Question 17
**a) i) Fundamental note:**
The lowest frequency (first harmonic) produced by a vibrating body. It is the note with the simplest standing wave pattern — one loop in an open pipe, half a loop in closed pipe.
**ii) Overtones:**
Frequencies higher than the fundamental that are produced simultaneously with it. First overtone = second harmonic (or third harmonic for closed pipe), etc.
**iii) Harmonic frequencies:**
Whole-number multiples of the fundamental frequency. First harmonic = f₀, second harmonic = 2f₀, etc.
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**b) i) Overtones in closed pipe:**
*(Sketches — standing wave diagrams):*
- **1st overtone (3rd harmonic):** 3 quarter-wavelengths — node at closed end, antinode at open end, one extra node-antinode pair.
- **2nd overtone (5th harmonic):** 5 quarter-wavelengths
- **3rd overtone (7th harmonic):** 7 quarter-wavelengths
**ii) Closed pipe produces only odd harmonics:**
For a closed pipe of length L:
$$L = \frac{n\lambda}{4}, \quad n = 1, 3, 5...$$
$$f_n = \frac{nv}{4L}, \quad n = 1, 3, 5...$$
Only odd values of n satisfy the boundary conditions (node at closed end, antinode at open end), therefore **only odd harmonics are produced**. ✓
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**c) End correction:**
The antinode in an open pipe or the open end of a resonance tube does not form exactly at the physical end of the tube but slightly beyond it. The end correction **e** is the small extra length that must be added to the physical length to get the effective acoustic length. It is approximately 0.3d where d = diameter.
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**d) Finding end correction:**
Closed pipe: L_c = 0.46 m, Open pipe: L_o = 0.60 m (same diameter d, same end correction e)
Let e = end correction.
First overtone of **closed pipe** = 3rd harmonic:
$$f = \frac{3v}{4(L_c + e)}$$
First overtone of **open pipe** = 2nd harmonic:
$$f = \frac{2v}{2(L_o + 2e)} = \frac{v}{L_o + 2e}$$
Setting equal:
$$\frac{3v}{4(L_c+e)} = \frac{v}{L_o+2e}$$
$$3(L_o+2e) = 4(L_c+e)$$
$$3L_o + 6e = 4L_c + 4e$$
$$2e = 4L_c - 3L_o$$
$$2e = 4(0.46) - 3(0.60) = 1.84 - 1.80 = 0.04$$
$$\boxed{e = 0.02 \text{ m} = 2 \text{ cm}}$$
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### Question 18
**a) i) Simple Harmonic Motion (SHM):**
SHM is a periodic motion in which the restoring force is directly proportional to displacement from the equilibrium position and always directed towards that position:
$$F = -kx \quad \Rightarrow \quad a = -\omega^2 x$$
**ii) Three examples:**
1. **Simple pendulum** (small oscillations) — gravity provides restoring force proportional to displacement
2. **Mass on a spring** — elastic restoring force obeys Hooke's law
3. **Liquid in a U-tube** — weight of displaced liquid provides linear restoring force
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**b) Liquid in U-tube:**
**i) Show motion is SHM:**
When displaced by x, the height difference = 2x.
Net restoring force = weight of excess liquid column:
$$F = -\rho A(2x)g$$
Mass of liquid: m = ρAL
$$ma = -2\rho Agx$$
$$a = -\frac{2g}{L}x$$
Since a = −ω²x where ω² = 2g/L → motion is **SHM** ✓
**ii) Period:**
$$T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{L}{2g}} = \boxed{\pi\sqrt{\frac{2L}{g}}}$$
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**c) x = 12cos(5πt)**
**i) Amplitude:**
$$\boxed{A = 12 \text{ mm}}$$
**ii) Period:**
$$\omega = 5\pi \text{ rad/s}$$
$$T = \frac{2\pi}{\omega} = \frac{2\pi}{5\pi} = \boxed{0.4 \text{ s}}$$
**iii) Maximum velocity:**
$$v_{max} = A\omega = 12\times10^{-3}\times5\pi = 0.06\pi$$
$$= \boxed{0.1885 \text{ ms}^{-1} \approx 0.19 \text{ ms}^{-1}}$$
*(Converting A from mm: 12 mm = 0.012 m)*