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WAEC SSCE - Physics - 2022 (Objective)

Tambaya 1 Rahoto

The power of the lens is +2.5D. What is the radius of the curvature?

2.5 cm

25.0 cm

40.0 cm

80.0 cm

Bayanin Amsa

The power of a lens, P is defined as the reciprocal of the focal length of the lens, f in meters, expressed in diopters (D). Therefore, we can use the formula P = 1/f to find the focal length of the lens. For a thin lens, the radius of curvature, R of each surface of the lens is related to the focal length, f, by the formula: 1/f = (n - 1) × (1/R1 - 1/R2) Where n is the refractive index of the lens medium, R1 is the radius of curvature of one surface and R2 is the radius of curvature of the other surface. For a single curved surface of a thin lens, the formula reduces to: 1/f = (n - 1) × (1/R) where R is the radius of curvature of the curved surface. Rearranging the formula, we get: R = 1 / ((n-1) x f) Substituting P = 2.5D = 2.5m⁻¹ in place of f, and assuming the refractive index of the lens medium is 1, we get: R = 1 / ((1-1) x 2.5) = undefined Therefore, it is not possible to determine the radius of curvature of the lens from the given information. The answer is undefined.

Karanta bayanin darasi akan Structure Of The Atom (Nigeria Only) (WAEC)

Structure Of The Atom (Nigeria Only)

Tambaya 12 Rahoto

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

Bayanin Amsa

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

Karanta bayanin darasi akan Thermal Expansion (JAMB)

Thermal Expansion

Tambaya 15 Rahoto

The joule is equivalent to?

Kg m

^{2}

s

^{- 1}

Kg

^{2}

m s

^{- 2}

Kg m

^{2}

s

^{- 2}

Kg m s

^{- 1}

Bayanin Amsa

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 $^{- 2}$

Karanta bayanin darasi akan Motion (JAMB)

Karanta bayanin darasi akan Measurements And Units (JAMB)

Measurements And Units

Tambaya 22 Rahoto

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

Bayanin Amsa

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

Karanta bayanin darasi akan Thermal Expansion (JAMB)

Thermal Expansion

Tambaya 23 Rahoto

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}

Bayanin Amsa

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.

Karanta bayanin darasi akan Rectilinear Acceleration (WAEC)

Karanta bayanin darasi akan Speed And Velocity (WAEC)

Rectilinear Acceleration

Speed And Velocity

Tambaya 29 Rahoto

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

120

Ω

240

Ω

540

Ω

960

Ω

Bayanin Amsa

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 $Ω$

Karanta bayanin darasi akan Current Electricity (JAMB)

Current Electricity

Tambaya 30 Rahoto

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

Bayanin Amsa

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

Karanta bayanin darasi akan Electrostatics (JAMB)

Electrostatics

Tambaya 35 Rahoto

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

Bayanin Amsa

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

Karanta bayanin darasi akan Light (JAMB)

Tambaya 43 Rahoto

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

Bayanin Amsa

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

Tambaya 44 Rahoto

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

Bayanin Amsa

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.

Karanta bayanin darasi akan Electromagnetic Waves: Types Of Radiation In Electromagnetic Spectrum (WAEC)

Electromagnetic Waves: Types Of Radiation In Electromagnetic Spectrum

Tambaya 45 Rahoto

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

Bayanin Amsa

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

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