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.
Correct Answer: A — It increases
Q. A loop of wire is placed in a uniform magnetic field. What happens to the induced EMF if the area of the loop is increased?
A.Increases
B.Decreases
C.Remains the same
D.Depends on the magnetic field strength
Solution
According to Faraday's law of electromagnetic induction, the induced EMF is directly proportional to the rate of change of magnetic flux, which increases with an increase in the area of the loop.
Correct Answer: A — Increases
Q. A machine does 500 J of work in 10 seconds. What is its power output?
A.50 W
B.100 W
C.200 W
D.500 W
Solution
Power is calculated using the formula P = W/t. Here, W = 500 J and t = 10 s, so P = 500 J / 10 s = 50 W.
Correct Answer: B — 100 W
Q. A machine does 500 J of work in 10 seconds. What is its power?
A.50 W
B.100 W
C.200 W
D.500 W
Solution
Power = Work / Time = 500 J / 10 s = 50 W.
Correct Answer: B — 100 W
Q. A man is standing 100 meters away from a building. If the angle of elevation to the top of the building is 45 degrees, what is the height of the building?
A.100 m
B.50 m
C.75 m
D.25 m
Solution
Using tan(45°) = height/distance, we have height = distance * tan(45°) = 100 * 1 = 100 m.
Correct Answer: A — 100 m
Q. A man is standing 50 meters away from a vertical pole. If he looks up at an angle of elevation of 60 degrees to the top of the pole, what is the height of the pole?
A.25 m
B.30 m
C.35 m
D.40 m
Solution
Using tan(60°) = height/50, we have √3 = height/50. Therefore, height = 50√3 ≈ 86.6 m.
Correct Answer: B — 30 m
Q. A man is standing on a hill that is 80 m high. If he looks down at an angle of depression of 30 degrees, how far is he from the base of the hill?
A.40 m
B.60 m
C.80 m
D.100 m
Solution
Using tan(30°) = height/distance, we have 1/√3 = 80/distance. Therefore, distance = 80√3 ≈ 138.56 m.
Correct Answer: A — 40 m
Q. A man is standing on the ground and looking at the top of a 15 m high pole. If he is 20 m away from the base of the pole, what is the angle of elevation?
A.36.87 degrees
B.45 degrees
C.60 degrees
D.30 degrees
Solution
Using tan(θ) = height/distance, we have tan(θ) = 15/20. Therefore, θ = tan⁻¹(0.75) which is approximately 36.87 degrees.
Correct Answer: A — 36.87 degrees
Q. A man is standing on the ground and observes the top of a building at an angle of elevation of 60 degrees. If he is 50 m away from the building, what is the height of the building?
A.25 m
B.43.3 m
C.50 m
D.86.6 m
Solution
Using tan(60°) = height/50, we have √3 = height/50. Therefore, height = 50√3 ≈ 86.6 m.
Correct Answer: B — 43.3 m
Q. A man is standing on the ground and observes the top of a tree at an angle of elevation of 45 degrees. If he is 10 meters away from the tree, what is the height of the tree?
A.5 m
B.10 m
C.15 m
D.20 m
Solution
Using tan(45°) = height/10, we have 1 = height/10. Therefore, height = 10 m.
Correct Answer: B — 10 m
Q. A mass 'm' is lifted to a height 'h' above the Earth's surface. What is the change in gravitational potential energy?
A.mgh
B.mg(h + R)
C.mgR
D.0
Solution
The change in gravitational potential energy when a mass m is lifted to a height h is given by ΔU = mgh.
Correct Answer: A — mgh
Q. A mass attached to a spring oscillates with a frequency of 3 Hz. What is the angular frequency of the motion?
A.3 rad/s
B.6 rad/s
C.9 rad/s
D.12 rad/s
Solution
Angular frequency (ω) = 2πf = 2π(3) = 6 rad/s.
Correct Answer: B — 6 rad/s
Q. A mass attached to a spring oscillates with a frequency of 3 Hz. What is the angular frequency?
A.3 rad/s
B.6 rad/s
C.9 rad/s
D.12 rad/s
Solution
Angular frequency ω = 2πf = 2π(3) = 6 rad/s.
Correct Answer: B — 6 rad/s
Q. A mass attached to a spring oscillates with a period of 2 seconds. What is the frequency of the oscillation?
A.0.25 Hz
B.0.5 Hz
C.1 Hz
D.2 Hz
Solution
Frequency (f) is the reciprocal of the period (T). f = 1/T = 1/2 = 0.5 Hz.
Correct Answer: B — 0.5 Hz
Q. A mass is measured as 5.0 kg with an uncertainty of ±0.1 kg. If this mass is used to calculate weight (W = mg), what is the uncertainty in weight if g = 9.8 m/s²?
A.±0.2 N
B.±0.5 N
C.±0.1 N
D.±0.4 N
Solution
Uncertainty in weight = g * (uncertainty in mass) = 9.8 * 0.1 = ±0.98 N, rounded to ±1 N.
Correct Answer: A — ±0.2 N
Q. A mass m is attached to a string and is whirled in a horizontal circle. If the radius of the circle is halved, what happens to the tension in the string if the speed remains constant?
A.It doubles
B.It remains the same
C.It halves
D.It quadruples
Solution
Tension T = mv²/r. If r is halved, T doubles for constant speed.
Correct Answer: A — It doubles
Q. A mass m is attached to a string and is whirled in a vertical circle. At the highest point of the circle, what is the minimum speed required to keep the mass in circular motion?
A.√(g*r)
B.g*r
C.2g*r
D.g/2
Solution
At the highest point, the centripetal force is provided by the weight. Minimum speed = √(g*r).
Correct Answer: A — √(g*r)
Q. A mass m is attached to a string and is whirled in a vertical circle. At the highest point of the circle, what is the condition for the mass to just complete the circular motion?
A.Tension = 0
B.Tension = mg
C.Tension = 2mg
D.Tension = mg/2
Solution
At the highest point, the centripetal force is provided by the weight of the mass, so T + mg = mv²/r. For T = 0, mg = mv²/r.
Correct Answer: A — Tension = 0
Q. A mass m is attached to a string and is whirled in a vertical circle. At the highest point of the circle, the tension in the string is T. What is the expression for T?
A.T = mg
B.T = mg - mv²/r
C.T = mg + mv²/r
D.T = mv²/r
Solution
At the highest point, T + mg = mv²/r, thus T = mg - mv²/r.
Correct Answer: B — T = mg - mv²/r
Q. A mass m is lifted to a height h in a uniform gravitational field. What is the work done against gravity?
A.mgh
B.gh
C.mg
D.0
Solution
The work done against gravity when lifting a mass m to a height h is given by W = mgh.
