WAEC SSCE - Physics - 2022 (Objective)

Joule as the unit of work done or energy used.

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

Joule = kg × ms2 × m

→ kg m2 ${}^2$ s−2 ${}^{-2}$ 

Objects at a distance greater than two times of the focal length produces a real image, inverted and diminished in size.

→ 2 * 15 = 30cm

: At >30cm this is applicable  

The kinetic-molecular theory of gases assumes that ideal gas molecules:

(1) are constantly moving;

(2) have negligible volume;

(3) have negligible intermolecular forces;

(4) undergo perfectly elastic collisions; and

(5) have an average kinetic energy proportional to the ideal gas's absolute temperature.

Since the current flowing through a parallel resonance circuit is the product of voltage divided by impedance, at resonance the impedance, Z is at its maximum value.

 (Z=R )

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 = VI $\frac{\mathrm{<br>}}{}$ 

--> 395 = 8−0.02r0.02 $\frac{\mathrm{<br>}}{}$

CROSS MULTIPLY

8−0.02r = 395 * 0.02

8−0.02r = 7.9

--> 0.02r = 8 - 7.9

∴ r = 0.10.02 $\frac{0.1}{0.02}$

 = 5Ω

Refraction occurs when light travels from one medium to another which changes the speed at which the light travels.

This causes light to bend upon incidence with the interface of a new material.

 Angle of refraction is smaller than angle of incidence when light ray travels from rarer medium to denser medium.

I.e  light can either refract towards the normal(when slowing down while crossing the boundary) or away from the normal (when speeding up while crossing the boundary).

When light undergoes refraction, its frequency remains the same.

de broglie wavelength formula in terms of kinetic energy:

 λ = h2mE√ $\frac{\mathrm{<br>}}{}$

 λ = 6.6∗10−342∗9.1∗10−31∗100√ $\frac{\mathrm{<br>}}{}$

 λ = 5.50 x 10−14 ${}^{<br>}$m

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

3 resistance in parallel = 1r1 + 1r2 $\frac{\mathrm{<br>}}{}$ + 1r3 $\frac{\mathrm{<span\; style="word-spacing:\; normal;\; font-weight:\; var(--bs-body-font-weight);">3</span>}}{}$

1rT= 12 $\frac{1}{2}$ + 12 $\frac{1}{2}$ + 12 $\frac{1}{2}$

1rT = 1+1+12 $\frac{1+1+1}{2}$

1rT $\frac{\mathrm{<br>}}{}$ = 32 $\frac{3}{2}$

cross multiply

rT ${}_{\mathrm{<br>}}$ = 23 $\frac{\mathrm{<br>}}{}$ 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

The greatest force is experienced when the conductor is at right angle to the field.

I.e at perpendicular or Ø = 90°

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

Inductive Reactance [XL ${}_{\mathrm{<br>}}$] = 2π $\mathrm{<br>}$FL → 2 * 3.14 * 100 * 10

XL
 = 6,280

Current[I] = VXL $\frac{\mathrm{<br>}}{}$ → 506280 $\frac{50}{6280}$

I = 0.0079 

≈ 0.008A 

Acceleration[a] = [forward−resistive]forcemass

a = 1240−8001220 $\frac{1240-800}{1220}$ → 440−8001220 $\frac{440-800}{1220}$

a = 0.36m/s2 ${}^2$,  v = 4m/s, s = ?

Second equation of motion:

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

v2 ${}^2$ = u2 ${}^2$ + 2as

42 ${}^2$ = 02 ${}^2$ + 2 * 0.36 * s

16 =  0.72s 

22.2m = s

 friction, force that resists the sliding or rolling of one solid object over another.

Examples of friction in our daily life

• Driving of a a vehicle on a surface.
• Applying brakes to stop a moving vehicle.
• Skating.
• Walking on the road.

Vernier calipers commonly used in industry provide a precision to 0.01 mm (10 micrometres), or 0.01cm or one thousandth of an inch.

 Usually, it is the total length of the main scale. Vernier calipers, generally, have a range of 300 mm

300 millimeters equal 30.0 centimeters (300mm = 30.0cm).

It is a graph that represents the force that acts on a body in a certain amount of time.
Force on the object is represented by the y-axis and time taken by the object is represented by the y axis.
The area under the curve indicates the impulse, the change in momentum that the object undergoes.

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

P = V2R $\frac{{\mathrm{<br>}}^{}}{}$

: 60 = 2402R $\frac{{\mathrm{<br>}}^{}}{}$

→ R * 60 = 57,600

R = 5 $\frac{57,600}{60}$

R = 960Ω