Welcome to “EXAMRUNZ.COM” For Waec 2025 Expo
 Questions & Answers

---

Wednesday, 21st May_2025

Physics 2 (Essay), 2:00pm -3:30pm
Physics 1 (Objective), 3:30pm – 4:45pm

---

PHYSICS OBJ
01-10: BBCDACBCBB
11-21: ADBBCCDCBC
21-30: CDABAAADBA
31-40: ABABCCBCDA
41-50: BBACBBCABC

COMPLETED ✅

---

1. CLICK TO VIEW WAEC PHYSICS  THEORY QUESTION PAPER
2. CLICK TO VIEW WAEC PHYSICS  THEORY QUESTION PAPER

THEORY  ANSWERS 

---

 

---

 

---

---

 

---

 

---

---

PART II: ANSWER THREE(3) QUESTIONS ONLY

(8ai)
A semiconductor is said to be doped when a small amount of impurity is intentionally added to a pure semiconductor to increase its electrical conductivity.

(8aii)
I. Intrinsic semiconductor:
An intrinsic semiconductor is a pure semiconductor material without any significant impurities added.
II. Extrinsic semiconductor:
An extrinsic semiconductor is a semiconductor that has been doped with impurity atoms (dopants) to increase its conductivity.

(8b)

 

(8c)

(8d)
Stopping potential is defined as the minimum negative potential (voltage) applied to the collector electrode in a photoelectric experiment that just stops the most energetic photoelectrons emitted from reaching the collector.
===========================

(9a)
Deformation refers to the change in shape or size of an object under an external force, while elasticity refers to the ability of an object to return to its original shape or size after the external force is removed.

(9b)

(9bii)
Slope = ∆F/∆e
Point 1: (Extension = 10.0 cm, Force = 2.0 N)
Point 2: (Extension = 30.0 cm, Force = 6.0 N)
S = (6.0-2.0)/(30.0-10.0)
S = 4.0/20.0
S = 0.2N/cm

(9biii)
The slope represents the spring constant
k, indicating the stiffness of the spring.

(9c)
Given:
Mass (m) = 40 g = 0.04 kg
Elastic constant (k) = 350 N/m
Extension (x) = 4 cm = 0.04 m
Elastic potential energy stored in the rubber(E) = (1/2)kx²
= (1/2) × 35 × (0.04)²
(1/2) × 35 × 0.0016
= 0.28 Jgkj
This energy is converted to kinetic energy of the stone:
K = (1/2)mv² = E
(1/2)mv² = 0.28 J
(1/2) × 0.04 × v² = 0.28
v² = (2 × 0.28)/0.04
v² = 14
v = √14
v = 3.74 m/s

(9d)
(i) Bulk modulus
Bulk modulus is a measure of a material’s resistance to compression under uniform pressure.
(ii) Energy gained by the wire
Given:
Mass (m) = 2.5g = 0.0025kg
Extension (x) = 2cm = 0.02m
g = 10m/s²
Force (F) = mg = 0.0025 × 10 = 0.025N
Energy gained (E) = (1/2)Fx
= (1/2) × 0.025 × 0.02
= 2.5 × 10⁻⁴ J

---

(10a)
(i) First overtone;
The first overtone is the second harmonic of a vibrating system, having a frequency twice that of the fundamental frequency.
(ii) End correction;
End correction accounts for the fact that the antinode of a standing wave in a pipe doesn’t exactly coincide with the physical end.

(10b)
Given:
Length (L) = 40 cm = 0.4 m
End correction (e) = 3.2 cm = 0.032 m
Velocity of sound (v) = 330 m/s
Effective length = L + e = 0.4 + 0.032 = 0.432m
Wave length (λ)= 4Leff
λ = 4×0.432
= 1.728m
Frequency (f) = v/λ=
= 330/1.728
= 191Hz

(10ci)
Given: n = 8
n = 360° / β – 1
360° / β = n + 1
360° / β = 8 + 1
360° / β = 9
β = 360° / 9
β = 40°

(10cii)
(i) Mirage involves refraction and total internal reflection in air, while total internal reflection occurs at a medium boundary.
(ii) Mirage produces an inverted image, while total internal reflection produces a mirror-like reflection.

(10ciii)
Given: f₀ = 3 cm, fe = 6 cm, L = 20 cm
v₀ = L – fe = 20 – 6 = 14 cm
1/f₀ = 1/v₀ – 1/u₀
1/3 = 1/14 – 1/u₀
1/u₀ = 1/14 – 1/3
1/u₀ = (3 – 14) / 42
1/u₀ = -11 / 42
u₀ = -42 / 11
u₀ = -3.82 cm

---

 

NUMBER11

---

NUMBER 12

HOW WE SEND ANSWERS:

Do you  Wish To Subscribe For Your Nine (9) Subjects or  All Subjects  at Once? Click Here 

 

We all Also Available in NABTEB AND NECO Examination

Thank you.