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WAEC SSCE - Physics - 2022 (Objective)

Swali 1 Ripoti

The basic principle of operation of a beam balance is------------------

Hooke's law

Archimedes principle

law of flotation

principle of moments

Maelezo ya Majibu

The basic principle of operation of a beam balance is the principle of moments. This principle states that when an object is placed on one side of a beam balance, it creates a moment (or a turning force) around the pivot point, which is the center of the beam. This moment is counteracted by an equal moment created by a weight placed on the other side of the beam. The weight on the other side is moved along a calibrated scale until the beam is in a horizontal position, indicating that the moments on both sides are equal. This means that the weight of the object can be determined by reading the position of the weight on the calibrated scale.

Soma maelezo ya somo kwenye Equilibrium Of Forces (JAMB)

Soma maelezo ya somo kwenye Equilibrium Of Forces (WAEC)

Equilibrium Of Forces

Equilibrium Of Forces

Tazama Jibu

Swali 8 Ripoti

A galvanometer with a full-scale deflection of 20 mA is Converted to read 8 K by connecting a \(395\,\Omega\) resistor in series with it. Determine the internal resistance of the galvanometer

2.5

Ω

5.0

Ω

8.0

Ω

10.0

Ω

Maelezo ya Majibu

r = resistance of the galvanometer

l = current through galvanometer = 20mA or 0.02A

V1 = p.d across the galvanometer = I X r = 0.02 X r

: V1 = 0.02r

V2 = p.d across the multiplier = 8 - 0.02r

R = resistance of the multiplier = 395Ω

where R = $\frac{V}{I}$

--> 395 = $\frac{8 - 0.02 r}{0.02}$

CROSS MULTIPLY

8−0.02r = 395 * 0.02

8−0.02r = 7.9

--> 0.02r = 8 - 7.9

∴ r = $\frac{0.1}{0.02}$

= 5Ω

Soma maelezo ya somo kwenye Electrostatics (JAMB)

Electrostatics

Tazama Jibu

Swali 12 Ripoti

The linear expansivity, \( \alpha \), and cubic expansivity, \( \gamma \), of a material are related by the equation?

γ

=a

γ

=3

α

γ

=2a

γ

=

\frac{1}{3} α

Maelezo ya Majibu

cubic and linear expansivity are related by this equation. γ=3α.

Soma maelezo ya somo kwenye Thermal Expansion (JAMB)

Thermal Expansion

Tazama Jibu

Swali 20 Ripoti

The engine of a car provides a forward force of \(1240\ \mathrm{N}\) and the total resistive force on the car is \(800\mathrm{N}\). If the mass of the car is \(1220\ \mathrm{kg}\), determine the distance the car has to travel from the rest before acquiring a speed of \(4\ \mathrm{m\ s^{-1}}\).

44.0 m

22.2 m

11.1 m

5.5 m

Maelezo ya Majibu

Acceleration[a] = $\frac{[f o r w a r d - r e s i s t i v e] f o r c e}{m a s s}$

a = $\frac{1240 - 800}{1220}$ → $\frac{440 - 800}{1220}$

a = 0.36m/s $^{2}$ , v = 4m/s, s = ?

Second equation of motion:

s is distance
u is initial velocity
v is final velocity
a is acceleration

v $^{2}$ = u $^{2}$ + 2as

4 $^{2}$ = 0 $^{2}$ + 2 * 0.36 * s

16 = 0.72s

22.2m = s

Tazama Jibu

Swali 30 Ripoti

A lamp is rated 240 V, 60 W. Determine the resistance of the lamp when lit?

120

Ω

240

Ω

540

Ω

960

Ω

Maelezo ya Majibu

P = 60w, V = 240v, R = ?

P = $\frac{V^{2}}{R}$

: 60 = $\frac{240^{2}}{R}$

→ R * 60 = 57,600

R = $\frac{57, 600}{60}$

R = 960 $Ω$

Soma maelezo ya somo kwenye Current Electricity (JAMB)

Current Electricity

Tazama Jibu

Swali 32 Ripoti

The joule is equivalent to?

Kg m

^{2}

s

^{- 1}

Kg

^{2}

m s

^{- 2}

Kg m

^{2}

s

^{- 2}

Kg m s

^{- 1}

Maelezo ya Majibu

Joule as the unit of work done or energy used.

Force = mass × acceleration
Work = Force × Distance
Work = mass x acceleration x distance

Joule = kg × $\frac{m}{s^{2}}$ × m

→ kg m $^{2}$ s $^{- 2}$

Soma maelezo ya somo kwenye Motion (JAMB)

Soma maelezo ya somo kwenye Measurements And Units (JAMB)

Measurements And Units

Tazama Jibu

Swali 33 Ripoti

Three cells each of emf, 1.0 V, and internal resistance, \(2\,\Omega\), are connected in parallel across a \(3\,\Omega\) resistor. Determine the current in the resistor.

0.90A

0.39A

0.30A

0.01A

Maelezo ya Majibu

Given Data: Emf = 1v, r = 2 ohms, R = 3 ohms, I = ?

3 resistance in parallel = $\frac{1}{r_{1}}$ $\frac{1}{r_{2}}$ + $\frac{1}{r_{3}}$

$\frac{1}{r_{T}}$ = $\frac{1}{2}$ + $\frac{1}{2}$ + $\frac{1}{2}$

$\frac{1}{r_{T}}$ = $\frac{1 + 1 + 1}{2}$

$\frac{1}{r_{T}}$ = $\frac{3}{2}$

cross multiply

r $_{T}$ = $\frac{2}{3}$ or 0.67

E = I(R+r)

1 = I(3+0.67)

1 = I(3.67)

1 ÷ 3.67 = I

0.27 = I

: I ≈ 0.30A

Tazama Jibu

Swali 36 Ripoti

If the kinetic energy of an electron is 100 eV, what is the wavelength of the de-Broglie wave associated with it? [h = 6.6 x \(10^{-34}\) J s, e = 1.6 x \(10^{19}\) C, \(m_e\), = 9.1 x \(10^{-31}\) kg]

1.22 x 10

^{- 10}

m

4.10 x 10

^{- 10}

m

3.90 x 10

^{- 10}

m

5.50 x 10

^{- 14}

m

Maelezo ya Majibu

de broglie wavelength formula in terms of kinetic energy:

λ = $\frac{h}{\sqrt{2 m E}}$

λ = $\frac{6.6 * 10^{- 34}}{\sqrt{2 * 9.1 * 10^{- 31} * 100}}$

λ = 5.50 x 10 $^{- 14}$ m

Soma maelezo ya somo kwenye Thermal Expansion (JAMB)

Thermal Expansion

Tazama Jibu

Swali 40 Ripoti

The correct relationship between the displacement, s, of a particle initially at rest in a linear motion and the time, t, is?

s $α$ t $^{- 2}$

s

α

t

^{- 1}

s

α

t

s

α

t

^{2}

Maelezo ya Majibu

The correct relationship between the displacement, s, of a particle initially at rest in a linear motion and the time, t, is s α t2. This means that the displacement of a particle is directly proportional to the square of the time elapsed when the particle is moving in a straight line with a constant acceleration. In other words, if the time is doubled, the displacement will be four times greater; if the time is tripled, the displacement will be nine times greater, and so on. This relationship is also known as the second equation of motion, which is commonly used in physics to describe the motion of objects under constant acceleration.

Soma maelezo ya somo kwenye Rectilinear Acceleration (WAEC)

Soma maelezo ya somo kwenye Speed And Velocity (WAEC)

Rectilinear Acceleration

Speed And Velocity

Tazama Jibu

Swali 42 Ripoti

An inductor of inductance 10 H is connected across an a.c circuit source of 50 V, 100 Hz. What is the current in the circuit? [\( \pi = 3.14 \)]

0.200A

0.070A

0.050A

0.008A

Maelezo ya Majibu

Given Data: L = 10, V = 50, F = 100, I = ?

Inductive Reactance [X $_{L}$ ] = 2 $π$ FL → 2 * 3.14 * 100 * 10

X $_{L}$ = 6,280

Current[I] = $\frac{V}{X_{L}}$ → $\frac{50}{6280}$

I = 0.0079

≈ 0.008A

Soma maelezo ya somo kwenye Light (JAMB)

Tazama Jibu

Swali 44 Ripoti

The period of a 10 kHz radio wave traveling at \(10^8\) ms\(^{-1}\) is--

3.0 x 10

^{- 5}

s

1.0 x 10

^{- 4}

s

1.0 x 10

^{4}

s

3.0 x 10

^{4}

s

Maelezo ya Majibu

The formula for the period of a wave is given by T = 1/f, where T is the period and f is the frequency. In this case, the frequency of the radio wave is 10 kHz, which means that f = 10,000 Hz. We can use the formula v = λf, where v is the velocity of the wave, λ is the wavelength, and f is the frequency. Solving for λ, we get λ = v/f. Substituting the given values, we get λ = 10 m/s / 10,000 Hz = 0.001 m. Now, we can use the formula T = λ/v to find the period. Substituting the given values, we get T = 0.001 m / 10 m/s = 0.0001 s. Therefore, the period of the 10 kHz radio wave traveling at 10 m/s is 0.0001 seconds or 1.0 x 10^-4 s. The answer is the second option.

Soma maelezo ya somo kwenye Electromagnetic Waves: Types Of Radiation In Electromagnetic Spectrum (WAEC)

Electromagnetic Waves: Types Of Radiation In Electromagnetic Spectrum

Tazama Jibu

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