WAEC SSCE - Physics - 2024 (Essay)

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(a) The uses of polar satellites include:

(i) for location identification

(ii) to collect data on climate change

(iii) observe and predict a natural disaster

(iv) use for communication.

(b) The slope of a graph of tensile stress against tensile strain represents the modulus of elasticity, also known as Young's modulus. Young's modulus quantifies the stiffness of a material. A steeper slope indicates a stiffer material, while a flatter slope indicates a more flexible material.

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(a)i The charged particle must be in motion or have velocity and the particle's velocity must be inclined to the magnetic field at an angle \(\theta\)

(ii) F = Qv\(\beta\)sin\(\theta\)

(iii) for a charged particle moving parallel to the magnetic field,

\(\theta\) = 0 and F = Qv\(\beta\)sin\(\theta\)

                           F = Qv\(\beta\)sin 0 = 0

Thus, F = 0

(b)The main parts of an electrical transformer are the coil, Soft iron core

(ii) An alternating current in the primary coil produces a changing magnetic field in the primary coil. The changing magnetic field around the core gives rise to a changing magnetic flux linking the secondary coil thereby inducing an alternating e.m.f in the secondary coil as stated in Faraday's law.

c(i) Why the speed of the proton remains constant: The speed of the proton remains constant because the magnetic force acts perpendicular to its velocity, providing centripetal acceleration without doing work on it. This means the force changes the direction of the proton's motion but not its speed, resulting in constant kinetic energy and a uniform speed throughout its circular path.

(ii) I. radius of curved path

qv\(\beta\) = \(\frac{mv^2}{r}\)

r = \(\frac{mv}{q\beta}\)

r = \(\frac{1.7 \times 10^{-27}}{1.6 \times 10^{-19}}\) = 0.24 m.

II. time taken by the proton to move through the magnetic field,

t = \(\frac{\pi r}{2v}\) = \(\frac{3.14 \times 0.24}{2 \times 4.5 \times 10^6}\) = 8.37 x 10\(^{-8}\)s

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(a) electrons are only emitted when the photon frequency exceeds the threshold frequency,

- the energy of each photon is proportional to its frequency,

- the emission of photoelectrons depends on the frequency of incident light on the metal,

- the intensity of light does not affect the energy of the electrons emitted/photoelectrons,

- the energy of the photoelectron depends on the frequency of incident radiation.

(b) Given: P = 6 x 10\(^{-3}\) W, no of photons per seconds = 1 x 10\(^{16}\), m\(_e\) = 9.1 x 10\(^{-31}\) kg, e = 1.6 x 10\(^{-19}\) C, W\(_o\) = ?

Photon Energy E = \(\frac{\text{power}}{\text{no of photons per second}}\) = \(\frac{\text{P}}{\text{n}}\)

= \(\frac{6.0 \times 10^{-3}}{1 \times 10^{16}}\)

 =  6 x 10\(^{-19}\)J

But K.E = E +  W\(_o\)

W\(_o\) = E - K.E = E - eV

W\(_o\) = 6 x 10\(^{-19}\) - 2.2 x (1.6 x 10\(^{-19}\)) = 2.48 x 10\(^{-9}\)J

(c) Factors on which radioactive activity depends are: (i) decay constant, (ii) number of unstable nuclei present at a given time (iii) time of the decay process (iv) environmental conditions, (v) type of radioisotope, etc

(d) Given: A = 2.0 x 10\(^6\)Bq, t\(_{\frac{1}{2}}\) = 1.8 x 10\(^8\)s

A = \(\lambda\)N, where \(\lambda\) = decay constant, N = number of radioactive nuclei, A = activity.

But  \(\lambda\) = \(\frac{In(2)}{t_{\frac{1}{2}}}\)

 \(\lambda\) = \(\frac{In(2)}{1.8 \times 10^8}\) ≈ \(\frac{0.693}{1.8 \times 10^8}\) ≈ 3.85 x 10\(^{-9}\)s\(^{-1}\)

Recall, A = \(\lambda\)N

N = \(\frac{A}{\lambda}\) = \(\frac{2.0 \times 10^6}{3.85 \times 10^{-9}}\) = 5.2 x 10\(^{14}\) nuclei

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(a) SURFACE TENSION = \(\frac{\text{force}}{\text{length}}\)

FORCE = mass x accel. due to gravity

Dimension of force = M.LT\(^{-2}\)

Dimension of length = L

Dimension of surface area = \(\frac{MLT^{-2}}{L}\) = MT\(^{-2}\)

(b) The instruments used for measuring the force of gravity at a place are:

(1) Gravimeter, (2) Spring balance, (3) Gravitational accelerometer.

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The uses of ferromagnetic materials are: 

(i) Electromagnets
(ii) Magnetic Storage
(iii) Magnetic Sensors
(iv) Magnetic Shielding
(v) Permanent Magnets
(vi) Inductive Components, others are transformers, electric bells, electric generators, etc.

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(ai) Increasing the temperature of a liquid decreases the viscosity of the liquid. (aii) Increasing the temperature of a gas increases the viscosity of the gas

(b) Factors that determine the magnitude of the moment of a force about a point are: 1. Angle between the distance from the turning point and the line of action of the force. 2. magnitude of the applied force. 3. perpendicular distance from the pivot to the line of action of the applied force.

(c) SEE THE DIAGRAM ABOVE.

(ii) Taking a moment about the support: clockwise moment = anti-clockwise moment

400 x 10 = m x (\(\frac{L}{2}\) - 10) -- ------- from figure 1 above

4000 = m x (\(\frac{L}{2}\) - 10) - - - -- - - (i)

400 x 6 = m x (\(\frac{L}{2}\) - 15)  - - - - -- - - -(ii)

divide eqn i by ii

\(\frac{4000}{2400}\) = \(\frac{m \times \frac{L}{2} - 10)}{m \times  \frac{L}{2} - 15)}\)

\(\frac{5}{3}\) =  \(\frac{\frac{L}{2} - 10)}{\frac{L}{2} - 15)}\)

3 x (\(\frac{L}{2}\) - 10)  = 5 x (\(\frac{L}{2}\) - 15)

\(\frac{3L}{2}\) - 30) = \(\frac{5L}{2}\) - 75)

\(\frac{5L}{2}\) - \(\frac{3L}{2}\) = 75 - 30 = 45

L = \(\frac{2L}{2}\) = 45 cm

From equation (ii) 400 x 6 = m x (\(\frac{L}{2}\) - 15) 

2400 =  m x (\(\frac{45}{2}\) - 15) 

2400 = m x 7.5

m = \(\frac{2400}{7.5}\) = 320 g

Therefore, I. m = 320g and II. L = 45 cm

(d) Reasons for less pressure at the top of a mountain: 

The lower pressure at the top of a mountain is due to the reduced number of air molecules in the atmosphere above that elevation, resulting in a lighter air column and fewer molecular collisions, which together create lower atmospheric pressure compared to sea level.

(e)(i) Density = \(\rho\) = \(\frac{m}{v}\)

 m = v x \(\rho\), but v = A x h 

m = A x h x \(\rho\) = 13.6 x 10\(^3\) x 40 x 20 x 10\(^{-6}\) = 10.88kg

(e)(ii) pressure exerted by the surface of the mercury

P\(_t\) = P\(_{atm}\) + pressure of mercury

P\(_{atm}\) = P\(_t\) - \(\rho\) x h x g = 1.42 x 10\(^5\) - 13.6 x 10\(^3\) x 0.2 x 10 = 1.148 x 10\(^5\)Pa

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(a) The incident light rays must be parallel and close to the principal axis

(b) Spherical aberration as a defect of curved mirrors: When a wide parallel beam is incident on a spherical mirror, it is not brought to the same focus. This makes the image distorted.

(c) SEE THE DIAGRAM ABOVE

(d) Accommodation is the ability of the eyes to alter its focal length to form a clear image 

(e) Quantitatively, (i) far point = infinity, Near point = 25 cm

P\(_{near point}\) = \(\frac{1}{f}\) = \(\frac{1}{u}\) + \(\frac{1}{v}\)

(ii) for near point, u = 0.25 m

P\(_{near point}\) =  \(\frac{1}{f}\) = \(\frac{1}{0.25}\) + \(\frac{1}{v}\)

similarly, for far point, u = infinity(∞)

P\(_{far point}\) = \(\frac{1}{f}\) = \(\frac{1}{u}\) + \(\frac{1}{v}\)

P\(_{far point}\)  = \(\frac{1}{∞}\) + \(\frac{1}{v}\)

                        = 0 + \(\frac{1}{v}\)

So, change in optical power

ΔP = P\(_{near point}\) - P\(_{far point}\)

ΔP = 4 + \(\frac{1}{v}\) - \(\frac{1}{v}\)

ΔP = 4 dioptres.

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(i) Stress: The base unit of stress is the Pascal (Pa), which is equivalent to kg⋅m\(^{−1}\)⋅s\(^{−2}\)

(ii) Force Constant (Spring Constant,k): The base unit of the spring constant is also in Newtons per meter (N/m), which is equivalent to kg⋅s\(^{−2}\)

(iii) Planck's Constant (h): The base unit of Planck's constant is Joule seconds (J·s), which is equivalent to kg⋅m\(^2\)s\(^{−1}\)
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