MITIHANI POPOTE PHYSICS 2 SERIES 1 (With detailed Solutions)

MITIHANI POPOTE
MITIHANI POPOTE EXAMIANTION SERIES

PHYSICS 2 - SERIES 1

131/02
TIME: 2:30 HRS
2025

INSTRUCTIONS

1. This paper consists of six (06) questions.
2. Answer five (5) questions.
3. Each question carries twenty (20) marks.
4. Mathematical tables and non-programmable calculators may be used.
5. Cellular phones and any unauthorized materials are not allowed in the examination room.
6. Write your examination number on every page of your answer sheet/booklet(s).

The following information may be useful:

• Acceleration due to gravity \( g = 9.8 \text{m/s}^2 \)
• Density of water, \( pw = 10^3 \text{kg} \text{m}^{-3} \)
• Density of air, \( \rho_{\text{air}} = 1.29 \text{kg/m}^3 \)
• Density of mercury \( = 13.6 \times 10^3 \text{kg/m}^3 \)
• Coefficient of viscosity of water \( = 1.0 \times 10^{-3} \text{Ns} \text{m}^{-2} \)
• Velocity of sound in air, \( V = 330 \text{m/s} \)
• Coefficient of linear expansion of copper \( = 8 \times 10^{-6} \text{°C} \)
• Bulk modulus of elasticity of cooper \( = 3.6 \times 10^{11} \text{N/m}^2 \)
• Surface tension of water \( = 7.2 \times 10^{-2} \text{N/m} \)
• Permeability of free space, \( \mu_0 = 4 \pi \times 10^{-7} \text{Hm}^{-1} \)
• Permittivity of free space, \( \epsilon_0 = 8.854 \times 10^{-12} \text{Fm}^{-1} \)
• Mass of \( \frac{3}{2} \text{Ca} = 39.962589u \)
• Mass of neutron, \( \text{Mn} = 1.008665u \)
• Mass of proton, \( \text{Mp} = 1.007825u \)
• Universal gas constant, \( R = 8.31 \text{J/molK} \)
• Young modulus of aluminium, \( \text{Yal} = 7 \times 10^{10} \text{Nm}^{-2} \)
• Boltzman constant \( (K_p) = 1.38 \times 10^{-23} \text{J/K} \)
• Electronic charge, \( e = 1.6 \times 10^{-19} \text{C} \)
• Avogadro’s number, \( NA = 6.0 \times 10^{23} \text{mole} \)
• Mass of electron, \( Me = 9.1 \times 10^{-31} \text{kg} \)
• Planck’s constant, \( h = 6.62 \times 10^{-34} \text{s} \)
• Pie, \( \pi = 3.14 \)

Page 1 of 4

Answer five questions.

1.

(a)

(i) Why rain drop falling under gravity do not acquire very high velocity? (02 marks)

(ii) A sphere is dropped under gravity through a fluid of viscosity, \(\eta\). Taking average acceleration as half of the initial acceleration show that the time taken to attain terminal velocity is independent of fluid density (05 marks)

(b)

(i) Deep water runs slowly. Why? (02 marks)

(ii) If the velocity head of stream of water is equal to 10cm, what is the speed of flow? (02 marks)

(c)

(i) Water flows steadily along the horizontal pipe at a rate of \(8 \times 10^3 m^3/s\). If the area of cross-section of the pipe is \(40 \times 10^{-4} m^2\). Calculate the flow velocity of water. (01 marks)

(ii) Find the total pressure in the pipe if the static pressure in the horizontal pipe is \(3 \times 10^4 Pa\). What is the new velocity if the total pressure is \(3.6 \times 10^4 Pa\)? (03 marks)

(d) Water flows steadily through a tube which consists of two parts joined together end to end, one part is 21cm long and the other is 7cm long and has a diameter of 0.07cm. If the pressure difference between the ends of the tube is 14cm of water. Find the pressure difference between end of each part if the other part has a diameter of 0.225cm (05 marks)

2.

(a)

(i) State Doppler principle concerning with sound waves. (01 marks)

(ii) We cannot hear echo in a room. Explain. (02 marks)

(iii) A man riding a merry - go - round emits a sound of a certain frequency. If another man is at the centre of merry - go - round, will he observe Doppler effect? Explain. (02 marks)

(b)

(i) Identify two distinct examples between transverse waves and longitudinal waves. (02 marks)

(ii) The third overtone of a closed pipe is found to be unison with the first overtone of an open pipe. Find the ratio of the length of the pipe. (03 marks)

(c)

(i) State four conditions for sustainable interference. (02 marks)

(ii) Monochromatic light of wavelength 600nm is incident normally on an optical transmission of spacing 200\(\mu m\). Calculate the number of diffracted beams which can be observed and the maximum order possible. (03 marks)

(d)

(i) Briefly explain, what happen to fringe separation if the slit separation is doubled? (02 marks).

(ii) What is the Brewster angle for air to glass transmission? (03 marks)

Page 2 of 4

3.

(a)

(i) Why does not water comes out of a Doppler unless the rubber bulb is pressed?

(ii) There is a soap bubble of radius \(2.4 \times 10^{-4} m\) in an air cylinder which is originally at a pressure of \(10^5 N/m^2\). The air in the cylinder is now compressed isothermally until the radius of the bubble is halved. Calculate now the pressure of air in the cylinder. The surface tension of soap film is 0.08Nm\(^{-1}\) (04 marks)

(b)

(i) How will the rise of liquid be affected if the top capillary tube is closed? (01 marks)

(ii) Water rises in a capillary tube to a certain height such that the upward force due to surface tension is balanced by \(75 \times 10^{-4} N\) force due to the weight of the liquid. What is the inner radius of the capillary tube? (03 marks)

(c)

(i) What happen to work done during stretching of wire?

(ii) A stone of 0.5kg mass is attached to one end of a 0.8m long aluminium wire of 0.7mm diameter and suspended vertically. The stone is now rotated in a horizontal plane at a rate such that the wire makes an angle of 85° with the vertical. Find the increase in length of the wire. (03 marks)

(d)

(i) In terms of kinetic theory of gases, explain why the pressure of a gas in a container increases when a gas is heated? (03 marks)

(ii) Find the mean free path and collision frequency for nitrogen molecules at a temperature of 20°C and a pressure of 1 a.t.m. Assume a molecular diameter of \(2 \times 10^{-10} m\). Given that the average speed of nitrogen molecule at 20°C is 511m/s.

4.

(a)

(i) What is equipotential surface? (02 marks)

(ii) Give four properties of an equipotential surface. (04 marks)

(b)

(i) State Coulomb’s law of electrostatics. (02 marks)

(ii) Two electric charges q and 4q are placed at a distance of 6cm apart on a horizontal plane. Find the locus of the point on the line joining the two charges where the electric field is zero. (03 marks)

(c) Give reason(s) for the following statements:

i. When we introduce a dielectric slab between plates of a charged air capacitor, its capacitance increases (03 marks)

ii. Any conducting object connected to earth is said to be grounded (03 marks)

(d) A 500μF capacitor with a charge of 300μC is discharged through 200kΩ resistor. What is the current after 20s? (03 marks)

Page 3 of 4

5.

(a)

(i) State the Biot – savart law and use it to derive the expression of magnetic field at the centre of current carrying circular coil (04 marks)

(ii) The electron of hydrogen atom moves along a circular pass of radius 0.5nm with a uniform speed of \(4.0 \times 10^6 m/s\). Calculate the magnetic field produced by the electron at the centre. (02 marks)

(b)

(i) How does “one ampere” is defined in the concept of magnetic force? (02 marks)

(ii) State the rule which gives the direction of force experienced by a straight current carrying conductor placed in a magnetic field (02 marks)

(c) Calculate the force per unit length on a long straight wire carrying a current of 6A in opposite direction. Distance between the wires is 3cm. State the direction of the force. (04 marks)

(d)

(i) State Faraday’s law of electromagnetic induction. (02 marks)

(ii) Which factors determine the magnitude of e.m.f induced in a coil? \(\left(1 \frac{1}{2}\right)\)

6.

(a)

(i) Define activity of radioactive decay (01 marks)

(ii) Derive an expression for the activity constant in terms of half-life. (03 marks)

(iii) A radioactive isotope Cobalt 60 is widely used in medicine and has a half-life of about 5.3 years. What mass of the isotope is required for the activity of 1 millicurie? (04 marks)

(b)

(i) What is the meaning of negative energy of orbiting electron? (01 marks)

(ii) Differentiate between Rutherford’s model and Bohr’s model of an atom. Give two differences. (03 marks)

(iii) Calculate the time taken by an electron to travel the first Bohr orbit in the hydrogen atom. (03 marks)

(c)

(i) How do you account for the stability of a nucleus? (02 mark)

(ii) It is proposed that to use the nuclear fusion reaction; \(1^{H^2} + 1^{H^2} \rightarrow 2^{He^4}\) in a nuclear reactor of 200MW rating. If the energy from the above reaction is used with 25% efficiency in the reactor, how many grams of deuterium fuel will be needed per day? The masses of \(1^{H^2}\) and \(2^{He^4}\) are 2.0141a.m.u and 4.0026a.m.u respectively. (04 marks)

Page 4 of 4

End of Examination Paper

MITIHANI POPOTE

EXAMIANTION SERIES

SERIES PHYSICS 2 - SERIES 1

Code: 131/02

TIME: 2:30 HRS

JANUARY-MAY, 2023

INSTRUCTIONS

1. This paper consists of six (06) questions.
2. Answer five (5) questions.
3. Each question carries twenty (20) marks.
4. Mathematical tables and non-programmable calculators may be used.
5. Cellular phones and any unauthorized materials are not allowed in the examination room.
6. Write your examination number on every page of your answer sheet/booklet(s).

The following information may be useful:

Acceleration due to gravity \( g = 9.8 \text{m/s}^2 \)

Density of water, \( \rho_w = 10^3 \text{kg} \text{m}^{-3} \)

Density of air, \( \rho_{\text{air}} = 1.29 \text{kg/m}^3 \)

Density of mercury \( = 13.6 \times 10^3 \text{kg/m}^3 \)

Coefficient of viscosity of water \( = 1.0 \times 10^{-3} \text{Ns} \text{m}^{-2} \)

Velocity of sound in air, \( V = 330 \text{m/s} \)

Electronic charge, \( e = 1.6 \times 10^{-19} \text{C} \)

Mass of electron, \( m_e = 9.1 \times 10^{-31} \text{kg} \)

Planck's constant, \( h = 6.62 \times 10^{-34} \text{Js} \)

Answer five questions.

Question 1

20 marks

(a) (i) Why rain drop falling under gravity do not acquire very high velocity? (02 marks)

Answer:

Raindrops do not acquire very high velocity due to air resistance. As a raindrop falls, the force of gravity accelerates it downward. However, as its velocity increases, the air resistance (drag force) opposing its motion also increases. Eventually, the drag force equals the gravitational force, and the raindrop reaches a constant velocity known as terminal velocity.

(a) (ii) A sphere is dropped under gravity through a fluid of viscosity, \(\eta\). Taking average acceleration as half of the initial acceleration show that the time taken to attain terminal velocity is independent of fluid density (05 marks)

Answer:

Let's denote:

• m = mass of sphere
• r = radius of sphere
• ρ = density of sphere
• σ = density of fluid
• η = viscosity of fluid
• v_t = terminal velocity

The forces acting on the sphere are:

Weight downward: \( W = mg = \frac{4}{3}\pi r^3 \rho g \)

Buoyant force upward: \( F_b = \frac{4}{3}\pi r^3 \sigma g \)

Viscous drag upward: \( F_d = 6\pi \eta r v \) (Stokes' law)

Net force: \( F_{net} = W - F_b - F_d = \frac{4}{3}\pi r^3 (\rho - \sigma)g - 6\pi \eta r v \)

At terminal velocity, F_net = 0:

\( \frac{4}{3}\pi r^3 (\rho - \sigma)g = 6\pi \eta r v_t \)

\( v_t = \frac{2}{9} \frac{r^2 (\rho - \sigma)g}{\eta} \)

Initial acceleration (when v=0):

\( a_i = \frac{W - F_b}{m} = \frac{\frac{4}{3}\pi r^3 (\rho - \sigma)g}{\frac{4}{3}\pi r^3 \rho} = \frac{(\rho - \sigma)g}{\rho} \)

Average acceleration: \( a_{avg} = \frac{1}{2}a_i = \frac{(\rho - \sigma)g}{2\rho} \)

Time to reach terminal velocity: \( t = \frac{v_t}{a_{avg}} = \frac{\frac{2}{9} \frac{r^2 (\rho - \sigma)g}{\eta}}{\frac{(\rho - \sigma)g}{2\rho}} = \frac{4}{9} \frac{r^2 \rho}{\eta} \)

This expression for time is independent of fluid density (σ), proving the required result.

(b) (i) Deep water runs slowly. Why? (02 marks)

Answer:

According to the equation of continuity for incompressible fluids (A₁v₁ = A₂v₂), when water flows from a shallow region to a deep region, the cross-sectional area increases. To maintain the same flow rate, the velocity must decrease. Therefore, deep water runs slowly compared to shallow water.

(b) (ii) If the velocity head of stream of water is equal to 10cm, what is the speed of flow? (02 marks)

Answer:

Velocity head is given by \( h = \frac{v^2}{2g} \)

Given h = 10 cm = 0.1 m, g = 9.8 m/s²

\( v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 0.1} = \sqrt{1.96} = 1.4 \, \text{m/s} \)

The speed of flow is 1.4 m/s.

Question 2

20 marks

(a) (i) State Doppler principle concerning with sound waves. (01 marks)

Answer:

The Doppler principle states that the apparent frequency of a sound wave changes when there is relative motion between the source of sound and the observer. The frequency increases when the source and observer are moving toward each other and decreases when they are moving apart.

(a) (ii) We cannot hear echo in a room. Explain. (02 marks)

Answer:

We cannot hear echo in a typical room because:

1. The distance between the sound source and reflecting surfaces is too small. For a distinct echo to be heard, the reflecting surface should be at least 17 meters away (so that the time delay is at least 0.1 seconds).
2. In rooms, multiple reflections (reverberations) occur from various surfaces, causing the sound to persist rather than producing a distinct echo.
3. Sound-absorbing materials in rooms (like furniture, carpets, curtains) reduce reflection intensity.

Question 3

20 marks

(a) (i) Why does not water comes out of a dropper unless the rubber bulb is pressed?

Answer:

Water does not come out of a dropper unless the rubber bulb is pressed due to:

1. Atmospheric pressure: The atmospheric pressure acting on the surface of the liquid in the dropper is greater than the pressure inside the dropper when the bulb is released.
2. Surface tension: The surface tension of water creates a "skin" that prevents water from flowing out freely.
3. Adhesive forces: Water molecules adhere to the glass surface of the dropper, creating a meniscus that resists flow.

When the bulb is pressed, the air pressure inside increases, overcoming these forces and pushing the liquid out.

Question 4

20 marks

(a) (i) What is equipotential surface? (02 marks)

Answer:

An equipotential surface is a surface in a field (electric or gravitational) where every point on the surface has the same potential. In an electric field, it's a surface where the electric potential is constant. No work is done when moving a charge along an equipotential surface.

Question 5

20 marks

(a) (i) State the Biot-Savart law and use it to derive the expression of magnetic field at the centre of current carrying circular coil (04 marks)

Answer:

Biot-Savart Law: The magnetic field dB due to a current element Idl at a point P is given by:

\( d\vec{B} = \frac{\mu_0}{4\pi} \frac{I d\vec{l} \times \hat{r}}{r^2} \)

where:

• μ₀ is the permeability of free space
• I is the current
• dl is the length element of the conductor
• r is the distance from the element to point P
• ŕ is the unit vector in the direction from dl to P

Derivation for magnetic field at center of circular coil:

Consider a circular coil of radius R carrying current I. For an element dl at the circumference:

• The angle between dl and r is 90°
• The distance from dl to center is R
• All dB contributions are in the same direction (perpendicular to the plane of the coil)

\( dB = \frac{\mu_0}{4\pi} \frac{I dl \sin 90^\circ}{R^2} = \frac{\mu_0}{4\pi} \frac{I dl}{R^2} \)

Integrating over the entire circumference (∫dl = 2πR):

\( B = \int dB = \frac{\mu_0}{4\pi} \frac{I}{R^2} \int dl = \frac{\mu_0}{4\pi} \frac{I}{R^2} (2\pi R) \)

\( B = \frac{\mu_0 I}{2R} \)

This is the magnetic field at the center of a circular current-carrying coil.

Question 6

20 marks

(a) (i) Define activity of radioactive decay (01 marks)

Answer:

Activity of radioactive decay is defined as the rate at which radioactive nuclei decay. It is measured in becquerels (Bq) where 1 Bq = 1 decay per second. The activity A is given by A = λN, where λ is the decay constant and N is the number of radioactive nuclei.

End of Examination Paper with Detailed Answers