WAEC SSCE - Physics - 2024 (Essay)

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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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(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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Given: time of flight T = 4s, g = 10ms\(^{-2}\), H\(_{max}\)

\(\theta\) = ?

T = \(\frac{2u sin\theta}{g}\)

4 = \(\frac{2 \times u sin \theta}{10}\)

40 = 2u sin\(\theta\)

sin\(\theta\) = \(\frac{20}{u}\)

But, H\(_{max}\) = \(\frac{u^2 sin^2 \theta}{2g}\)

H\(_{max}\) = \(\frac{u^2 \times (\frac{20}{u})^2}{2 \times 10}\)

H\(_{max}\) = \(\frac{20 \times 20}{20}\) = 20 metres.(u\(^2\)s cancel out)

Thus H\(_{max}\) attained = 20metres.

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From Hooke's law, F = kx where x = extension, k = force constant

K =  680Nm\(^{-1}\), F = 306N

x = \(\frac{\text{F}}{\text{x}}\) = \(\frac{306}{680}\) ≈ 0.450m

Fractional Change in Length: Given the fractional change is 1.5:

1.5 = \(\frac{\text{x}}{\text{L}}\) 

L = \(\frac{0.450}{1.5}\) = 0.30m.

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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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(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)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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To show that \(\lambda\) = \(\frac{\text{h}}{\sqrt{2meV_o}}\)

The de broglie wavelength \(\lambda\), is given by 

\(\lambda\) = \(\frac{\text{h}}{\text{p}}\)

where p = mv

\(\lambda\) = \(\frac{\text{h}}{\text{mv}}\)

When an electron is accelerated through a potential difference V\(_0\), it gains kinetic energy given K.E.= eV\(_0\)

The kinetic energy can also be expressed in terms of the mass m of the electron and its speed v: K.E = \(\frac{1}{2}\)mv\(^2\)

K.E = eV\(_0\) = \(\frac{1}{2}\)mv\(^2\)

v = \(\sqrt{\frac{2eV_o}{m}}\)

Substituting v =  \(\sqrt{\frac{2eV_o}{m}}\) into the de Broglie equation

\(\lambda\) = \(\frac{\text{h}}{m \times \sqrt{\frac{2eV_o}{m}}}\) = \(\frac{\text{h}}{\sqrt{\frac{2eV_o m^2}{m}}}\)

\(\lambda\) =  \(\frac{\text{h}}{\sqrt{2meV_o}}\)

Note: use the fact that \(\sqrt{m}\) x \(\sqrt{m}\) = m