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2023 NABTEB Physics Questions and Answers Expo

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2023 NABTEB Physics Questions and Answers Expo

2023 NABTEB Physics Questions and Answers

The National Business and Technical Examinations Board (NABTEB) Has Scheduled The 2023 Nabteb Physics Questions and Answers Paper To Kick of on Friday 7th July, 2023.

This brings the attention of candidates writing the exam in to searching for 2023 NABTEB Physics Questions and Answers, NABTEB Physics Expo 2023, and etc.

2023 NABTEB Physics Questions and Answers

In this section, you will read the steps and requirements needed for you to get Nabteb Physics 2023 Questions And Answers before exam.

NABTEB Physics 2023 Paper is Categorized in to 2 parts:

  • NABTEB Physics Theory 2023
  • NABTEB Physics Objective 2023

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NABTEB Physics Expo 2023

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NABTEB Physics Answer 2023

INSTRUCTION: ANSWER FOUR QUESTIONS ONLY

(1ai)
The law of flotation states that an object will float in a fluid if the buoyant force acting on it is equal to or greater than the gravitational force acting on it.

(1aii)
(i) Shipbuilding: The law of flotation is crucial in designing ships and boats to ensure they stay afloat. By calculating the weight of the ship and the density of water, shipbuilders can determine the shape and size needed to keep the vessel buoyant.

(ii) Life Jackets: Life jackets are designed based on the principle of flotation. They are filled with materials that displace enough water to support the weight of a person, allowing them to float in water and prevent drowning.

(iii) Submarine Design: Understanding the law of flotation is essential for designing submarines. By controlling the amount of water they displace, submarines can control their buoyancy and adjust their depth in the water.

(1bi)
The “centre of gravity” of a body refers to the point through which the entire weight of the body appears to act. It is the average position of the weight of an object or a system of particles.

(1bii)
The higher the centre of gravity of a body, the less stable it becomes. Conversely, the lower the centre of gravity of a body, the more stable it is

(1c)
To find the center of gravity of the wooden timber such that when pivoted, it remains horizontal, you can use the following formula:

x = (W1 * d1 + W2 * d2) / (W1 + W2)

Where:
x is the distance from the thin end to the center of gravity.
W1 is the weight at the fat end.
W2 is the weight at the thin end.
d1 is the distance from the fat end to the pivot point.
d2 is the distance from the thin end to the pivot point.

In this case, the values are:
W1 = 550 N (weight at the fat end)
W2 = 350 N (weight at the thin end)
d1 = 5.5 m (distance from the fat end to the pivot point)
d2 = 0 m (distance from the thin end to the pivot point)

Using these values in the formula:

x = (550 N * 5.5 m + 350 N * 0 m) / (550 N + 350 N)
x = (3025 Nm + 0 Nm) / 900 N
x = 3025 Nm / 900 N
x ≈ 3.36 m

Therefore, the center of gravity of the timber, such that when pivoted, it remains horizontal, is approximately 3.36 meters from the thin end.
=============================

(2ai)
(i) Expansion: When a body is heated, its particles gain energy and move more vigorously.
(ii) Change in state: Heat can cause a substance to change its state from solid to liquid (melting) or from liquid to gas (vaporization).
(iii) Change in chemical properties: Heat can initiate chemical reactions by providing the necessary activation energy
(iv) Change in temperature: Heat energy increases the kinetic energy of particles, causing them to move faster and collide more frequently.

(2aii)
TABULATE
=BOILING=
(i) Boiling occurs at a specific temperature called the boiling point, which is unique to each substance.
(ii) It takes place throughout the entire liquid, including the bulk and surface.
(iii) It is a rapid process and results in the formation of bubbles within the liquid due to the formation of vapor.
(iv) Boiling occurs when the vapor pressure of the liquid equals or exceeds the atmospheric pressure.

=EVAPORATION=
(i) Evaporation occurs at any temperature below the boiling point of a substance.
(ii) It takes place only at the surface of the liquid.
(iii) It is a relatively slow process and does not result in the formation of bubbles.
(iv) Evaporation occurs when the kinetic energy of some liquid particles exceeds the surface tension and escapes as vapor.

(2b)
The statement “The specific heat capacity of water is 4200 J/kg·K” means that it requires 4200 joules of energy to raise the temperature of one kilogram (1 kg) of water by one degree Kelvin (1 K). Specific heat capacity is a measure of the amount of heat energy required to change the temperature of a substance.

(2c)
Mass of block, Mb = 150g
Specific heat capacity of metal block, Cb= ?
Change in temperature of metal block = 100-43 = 57°C
∆θb = 57K

Heat supplied by block = Mb Cb ∆θb
= 150*Cb*57 = 8550Cb Joules

Heat gained by water = Mw Cw ∆θw
= 250*4.2*(43-30)
= 13650 Joules

Heat gained by calorimeter= Mc Cc ∆θc
= 80*0.4*(43-30)
= 416 Joules

Heat supplied by block = Heat gained by water + heat gained by calorimeter

8550Cb = 13650 + 416
8550Cb = 14066

Cb= 14066/8550
Cb= 1.645 J/gk
=============================

(3ai)
Dispersion of light refers to the phenomenon where light waves of different wavelengths are separated or spread out as they pass through a medium, such as a prism or a drop of water. This separation occurs because different wavelengths of light have different speeds or indices of refraction in the medium. As a result, the light waves bend or refract at different angles, causing them to separate and create a spectrum of colors.

(3aii)
– Light Source: A stable and intense light source that emits white light is needed.
– Collimator: To ensure parallel rays of light enter the prism or the dispersing medium
– Dispersing Medium: A medium that can effectively disperse light, such as a glass prism or a diffraction grating, is required
– Prism or Grating: A glass prism or a diffraction grating is used to disperse the light into its component colors.

(3bi)
=EXPERIMENT TO DETERMINE THE REFRACTIVE INDEX OF A RECTANGULAR GLASS BLOCK=

Materials needed:
– Rectangular glass block
– Ray box or a bright light source
– Protractor
– White paper or a screen
– Ruler
– Pencil

Procedure:
– Place the rectangular glass block on a table or flat surface.

– Position the ray box or bright light source such that it emits a narrow beam of light towards the glass block. The light beam should be incident on one of the large faces of the glass block.

– Allow the light to pass through the glass block, and it will emerge from the opposite face. Adjust the setup so that the emergent ray is clearly visible.

– On the white paper or screen, trace the incident ray (the one entering the glass block) and the emergent ray (the one leaving the glass block). Ensure that the two rays are straight and clearly marked.

– Measure the angle of incidence (θ₁) between the incident ray and the normal (an imaginary line perpendicular to the surface) of the glass block.

– Measure the angle of refraction (θ₂) between the emergent ray and the normal of the glass block.

– Repeat the experiment for different angles of incidence, ensuring a wide range of values.

– Use the measured values of θ₁ and θ₂ to calculate the refractive index (n) of the glass block using Snell’s law:
n = sin(θ₁) / sin(θ₂)

Where n is the refractive index of the glass block.

(3bii)
– Ensure that the surfaces of the glass block are clean and free from any dirt or smudges.
– When measuring angles using a protractor, avoid parallax errors.

(3c)
Given data:
Focal length (f) of the concave mirror = 10cm
Object height (h) = 2cm
Object distance (u) = 6cm
Image distance (v) = ?

(i) Position of the image (v):

Using the formula:
1/f = 1/v + 1/u

Substitute the given values:

1/10 = 1/v + 1/6

To solve for v, find the reciprocal of both sides of the equation:
v = 1/(1/10 – 1/6)
v = 1/(3/30 – 5/30)
v = 1/(-2/30)
v = -30/2
v = -15 cm

The negative sign in the result indicates that the image formed is a real image, and it is formed on the same side as the object, which means it is on the left side of the mirror.
(ii) Size of the image (height):
The magnification (M) is given by:
M = -v/u
Substitute the given values:
M = -(-15)/6
M = 15/6
M = 2.5

The positive value of the magnification indicates that the image is upright (erect).
Now, to find the size of the image (height), multiply the magnification by the object height:
Size of the image = Magnification × Object height
Size of the image = 2.5 × 2cm
Size of the image = 5cm

(iii) Nature of the image:
If the magnification is positive, as we found above, the image is upright (erect). Since the object is placed at a distance greater than the focal length (u > f), the image is also real.

:. The image is real and upright (erect).
=============================

(4ai)
Ohm’s law states that the current passing through a conductor is directly proportional to the voltage across the conductor, provided its temperature and other physical conditions remain constant.

(4aii)
– Semiconductors
– Diodes

(4aiii)
– Length
– Material
– Cross sectional area
– Temperature

(4b)
To compare the electromotive force (EMF) of two cells using a potentiometer, follow these steps:

(i) Set up the potentiometer: Connect the potentiometer to a known and stable voltage source, such as a standard cell or a battery with a well-known EMF.

(ii) Balance the potentiometer: Adjust the position of the sliding contact (the jockey) on the potentiometer wire until there is no current flowing through the galvanometer connected in series with the potentiometer.

(iii) Note the length: Mark the length of the wire on the potentiometer where the jockey is balanced. This provides a reference point for future measurements.

(iv) Swap the cells: Now, disconnect the standard cell and replace it with one of the cells whose EMF you want to compare.

(v) Re-balance the potentiometer: Move the jockey along the potentiometer wire until the galvanometer again shows no deflection (null point).

(vi) Compare lengths: Measure the length of the wire on the potentiometer where the jockey is balanced for the new cell.

(vii) Calculate EMF: The ratio of the two lengths represents the ratio of EMFs of the two cells. Assuming the resistance of the potentiometer wire is constant, the EMF of the unknown cell can be calculated using the EMF of the standard cell:
EMF of unknown cell = (Length for standard cell) * (EMF of standard cell) / (Length for unknown cell)

(4ci)
Diagram

(4cii)
E = I(R+r)
I.e r = E/I – R= 6/0.3 -16 = 20 – 16
r = 4
Internal resistance (r) = 4Ω

(4ciii)
Let the value of the resistance be R¹
Equivalent external resistance = RR¹/R+R¹

R¹ = (16R¹/16+R¹)
E = I(Rt + r)
6 = 0.5[(16R¹/16+R¹) + 4]

12= [16R¹/16+R¹] + 4
16R¹/16+R¹ = 8
16R¹ = 8R¹ + 128
8R¹ = 128
R¹ = 128/8
R¹ = 16Ω
=============================

(5ai)
Nuclear fusion is the process in which two or more atomic nuclei combine to form a heavier nucleus, releasing a large amount of energy in the process. This process occurs at extremely high temperatures and pressures, such as those found in the core of the Sun. In nuclear fusion, lighter elements, typically isotopes of hydrogen, combine to form heavier elements WHILE Nuclear fission on the other hand, is the process in which the nucleus of a heavy atom (such as uranium-235 or plutonium-239) is split into two or more smaller nuclei, along with the release of energy. This process occurs spontaneously or can be induced by bombarding the nucleus with neutrons. Nuclear fission is the basis for energy production in nuclear power plants and atomic bombs.

(5aii)
=CHARACTERISTICS OF NUCLEAR ACTIVITY=
(i) Radioactivity: Nuclear reactions often result in the emission of radiation due to the unstable nature of some isotopes involved
(ii) High Energy Density: Nuclear reactions have an incredibly high energy density compared to conventional chemical reactions

=APPLICATIONS OF NUCLEAR ENERGY=
(i) Electricity generation: Nuclear power plants use controlled nuclear fission reactions to produce heat, which is then used to generate electricity.
(ii) Medical Applications: Nuclear energy plays a vital role in medicine, particularly in diagnostic imaging and cancer treatment

(5bi)
Radiation hazard refers to the potential harmful effects of exposure to ionizing radiation. Ionizing radiation consists of particles or electromagnetic waves with enough energy to remove electrons from atoms or molecules, leading to the formation of ions. The ionization process can damage biological tissues and DNA, potentially causing health issues such as radiation sickness, cancer, and genetic mutations.

(5bii)
(i) Shielding: Using appropriate shielding materials, such as lead or concrete, can block or absorb radiation, reducing exposure to harmful levels. The choice of shielding depends on the type and energy of radiation involved.
(ii) Time limitation: Minimizing the time spent in areas with radiation sources helps reduce the overall exposure. Reducing the duration of exposure lowers the potential risk.
(iii) Distance: Increasing the distance between oneself and the radiation source can significantly reduce the exposure. Radiation intensity decreases with distance, so staying farther away from the source decreases the dose received.

(5c)
Equation:
²²⁶₈₈Ra —> ²²²₈₆Rn + ⁴₂He + Energy

Mass defect = Mass of products – Mass of reactants
∆m = (221.938490u + 4.001504u) – 225.947540u
∆m = – 0.007546u
|∆m| = 0.007546u

Energy released = 0.007546 x 931 MeV
= 7.025326 MeV
=============================

Completed!.

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