WAEC Physics Past Questions & Answers - Page 7

Earthing of an electrical appliance meant the outer metallic case of appliance is connected with earth, wire which is welded to the end of a copper rod and it is burned in the ground. It save the appliance during short-circuiting by passing Excessive current to earth

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\(\Omega\)

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 )

Moderation is the process of the reduction of the initial high speed (high kinetic energy) of the free neutron.

Since energy is conserved, this reduction of the neutron speed takes place by transfer of energy to a material called a moderator.

There are several different types of moderating materials, and each have places where they are used more effectively.

Typically-used moderator materials include heavy water, light water, and graphite.

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

Inductive Reactance [X\(_L\)] = 2\(\pi\)FL → 2 * 3.14 * 100 * 10

X\(_L\) = 6,280

Current[I] = \(\frac{V}{X_L}\) → \(\frac{50}{6280}\)

I = 0.0079 

≈ 0.008A 

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 = \(\frac{8−0.02r}{0.02}\)

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

The potential energy for a positive charge increases when it moves against an electric field

And decreases when it moves with the electric field;

The opposite is true for a negative charge. Unless the unit charge crosses a changing magnetic field, its potential at any given point does not depend on the path taken.

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

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

I.e at perpendicular or Ø = 90°

The photoelectric effect is the emission of electrons when electromagnetic radiation, such as light, hits a material