Q. A kite is flying at a height of 100 meters. If the angle of depression from the kite to a point on the ground is 30 degrees, how far is the point from the point directly below the kite?
A.50 m
B.60 m
C.70 m
D.80 m
Solution
Using tan(30°) = 100/distance, we have 1/√3 = 100/distance. Therefore, distance = 100√3 ≈ 173.21 m.
Correct Answer: A — 50 m
Q. A kite is flying at a height of 30 m. If the angle of elevation from a point on the ground to the kite is 60 degrees, how far is the point from the base of the kite?
A.15√3 m
B.30 m
C.10√3 m
D.20 m
Solution
Using tan(60°) = height/distance, we have distance = height/tan(60°) = 30/√3 = 15√3 m.
Correct Answer: A — 15√3 m
Q. A ladder 10 m long reaches a window 8 m above the ground. What is the angle of elevation of the ladder from the ground?
A.30 degrees
B.45 degrees
C.60 degrees
D.75 degrees
Solution
Using sin(θ) = opposite/hypotenuse, we have sin(θ) = 8/10. Therefore, θ = sin⁻¹(0.8) ≈ 53.13 degrees.
Correct Answer: C — 60 degrees
Q. A ladder 15 meters long reaches a window 12 meters above the ground. What is the angle of elevation of the ladder from the ground?
A.30 degrees
B.45 degrees
C.60 degrees
D.75 degrees
Solution
Using sin(θ) = opposite/hypotenuse, we have sin(θ) = 12/15. Therefore, θ = sin⁻¹(0.8) ≈ 53.13 degrees.
Correct Answer: C — 60 degrees
Q. A length is measured as 100.0 m with an uncertainty of ±0.5 m. What is the significant figure of the measurement?
A.2
B.3
C.4
D.5
Solution
The measurement has 4 significant figures (100.0).
Correct Answer: C — 4
Q. A length is measured as 15.0 m with an uncertainty of ±0.2 m. What is the total uncertainty if this length is used in a calculation involving addition with another length of 10.0 m (±0.1 m)?
A.0.3 m
B.0.2 m
C.0.1 m
D.0.4 m
Solution
Total uncertainty = √((0.2)² + (0.1)²) = √(0.04 + 0.01) = √0.05 ≈ 0.224 m.
Correct Answer: A — 0.3 m
Q. A length is measured as 15.0 m with an uncertainty of ±0.3 m. What is the fractional error in this measurement?
Q. A light ray traveling in a medium with a refractive index of 1.6 strikes a boundary with air at an angle of 50°. What will be the outcome?
A.Total internal reflection occurs.
B.Light is refracted into the air.
C.Light is absorbed.
D.Light is scattered.
Solution
The critical angle for this scenario is approximately 38.7°. Since 50° is greater than the critical angle, total internal reflection occurs.
Correct Answer: A — Total internal reflection occurs.
Q. A light ray traveling in a medium with n=2.0 strikes a boundary with air at an angle of incidence of 45°. What will be the angle of refraction in air?
A.22.5°
B.45°
C.60°
D.90°
Solution
Using Snell's law, n1 * sin(θ1) = n2 * sin(θ2). Here, n1 = 2.0, θ1 = 45°, and n2 = 1.0 (air). Thus, 2.0 * sin(45°) = 1.0 * sin(θ2) leads to sin(θ2) = √2, which gives θ2 = 90°.
Correct Answer: D — 90°
Q. A light ray traveling in diamond (n=2.42) strikes the diamond-air interface. What is the critical angle?
A.24.4°
B.36.9°
C.42.5°
D.49.5°
Solution
Using sin(θc) = n2/n1, where n1 = 2.42 (diamond) and n2 = 1.0 (air), we find sin(θc) = 1.0/2.42, leading to θc ≈ 24.4°.
Correct Answer: A — 24.4°
Q. A long straight wire carries a uniform linear charge density λ. What is the electric field at a distance r from the wire?
A.λ/(2πε₀r)
B.λ/(4πε₀r²)
C.λ/(2πε₀r²)
D.0
Solution
Using Gauss's law for a cylindrical surface around the wire, the electric field E at a distance r is E = λ/(2πε₀r).
Correct Answer: A — λ/(2πε₀r)
Q. A loop of wire is moved into a magnetic field at a constant speed. What happens to the induced EMF as it enters the field?
A.It increases
B.It decreases
C.It remains constant
D.It becomes zero
Solution
As the loop enters the magnetic field, the rate of change of magnetic flux increases, leading to an increase in induced EMF.
Correct Answer: A — It increases
Q. A loop of wire is moved into a uniform magnetic field at a constant speed. What happens to the induced EMF as it enters the field?
A.It increases
B.It decreases
C.It remains constant
D.It becomes zero
Solution
As the loop enters the magnetic field, the area exposed to the magnetic field increases, leading to an increase in the induced EMF.
Correct Answer: A — It increases
Q. A loop of wire is placed in a changing magnetic field. What phenomenon is this an example of?
A.Electromagnetic induction
B.Magnetic resonance
C.Electrostatics
D.Magnetic hysteresis
Solution
This is an example of electromagnetic induction, where a changing magnetic field induces an electromotive force (EMF) in the loop of wire.
Correct Answer: A — Electromagnetic induction
Q. A loop of wire is placed in a uniform magnetic field. If the field strength is increased, what happens to the induced EMF?
A.It increases
B.It decreases
C.It remains constant
D.It becomes zero
Solution
According to Faraday's law, an increase in magnetic field strength leads to an increase in the induced EMF.
