Q. A charged sphere has a radius R and a total charge Q. What is the electric potential at a point outside the sphere at a distance r from the center (r > R)?
A.kQ/R
B.kQ/r
C.kQ/(R+r)
D.0
Solution
For a charged sphere, the electric potential outside the sphere behaves as if all the charge were concentrated at the center, so V = kQ/r.
Correct Answer: B — kQ/r
Q. A child is sitting on a merry-go-round that is rotating. If the child moves towards the center, what happens to the rotational speed of the merry-go-round?
A.Increases
B.Decreases
C.Remains the same
D.Becomes zero
Solution
As the child moves towards the center, the moment of inertia decreases, causing the rotational speed to increase to conserve angular momentum.
Correct Answer: A — Increases
Q. A child is sitting on a merry-go-round that is spinning. If the child moves closer to the center, what happens to the angular velocity of the merry-go-round?
A.Increases
B.Decreases
C.Remains the same
D.Becomes zero
Solution
As the child moves closer to the center, the moment of inertia decreases, causing the angular velocity to increase to conserve angular momentum.
Correct Answer: A — Increases
Q. A child is sitting on a merry-go-round that is spinning. If the child moves towards the center, what happens to the angular velocity of the merry-go-round?
A.Increases
B.Decreases
C.Remains the same
D.Becomes zero
Solution
As the child moves towards the center, the moment of inertia decreases, and to conserve angular momentum, the angular velocity must increase.
Correct Answer: A — Increases
Q. A child sitting at the edge of a merry-go-round throws a ball tangentially. What happens to the angular momentum of the system (merry-go-round + child + ball)?
A.Increases
B.Decreases
C.Remains constant
D.Becomes zero
Solution
Angular momentum of the system remains constant due to conservation of angular momentum.
Correct Answer: C — Remains constant
Q. A circuit contains a 12 V battery and two resistors of 4 ohms and 8 ohms in series. What is the current flowing through the circuit?
A.0.5 A
B.1 A
C.1.5 A
D.2 A
Solution
Total resistance R = 4 + 8 = 12 ohms. Current I = V/R = 12 V / 12 ohms = 1 A.
Correct Answer: B — 1 A
Q. A circuit contains a 12V battery and two resistors of 4 ohms and 8 ohms in series. What is the total current in the circuit?
A.1 A
B.0.5 A
C.2 A
D.3 A
Solution
Total resistance R = R1 + R2 = 4 + 8 = 12 ohms. Current I = V/R = 12V / 12 ohms = 1 A.
Correct Answer: B — 0.5 A
Q. A circuit contains a 9V battery and two resistors in series: 3Ω and 6Ω. What is the voltage across the 6Ω resistor?
A.6V
B.3V
C.9V
D.0V
Solution
The total resistance R_total = 3Ω + 6Ω = 9Ω. The current I = V/R_total = 9V / 9Ω = 1A. Voltage across 6Ω, V = I * R = 1A * 6Ω = 6V.
Correct Answer: A — 6V
Q. A circuit contains a 9V battery and two resistors of 3 ohms and 6 ohms in series. What is the voltage across the 6 ohm resistor?
A.6V
B.3V
C.9V
D.4.5V
Solution
Total resistance R_total = 3 + 6 = 9 ohms. Current I = V/R_total = 9V / 9Ω = 1 A. Voltage across 6 ohm resistor = I * R = 1 A * 6Ω = 6V.
Correct Answer: A — 6V
Q. A circuit contains a 9V battery and two resistors of 3 ohms and 6 ohms in series. What is the voltage drop across the 6 ohm resistor?
A.3 V
B.6 V
C.9 V
D.4.5 V
Solution
Total resistance R_total = 3 + 6 = 9 ohms. Current I = V/R_total = 9V / 9Ω = 1 A. Voltage drop across 6 ohm resistor = I * R = 1 A * 6Ω = 6 V.
Correct Answer: B — 6 V
Q. A circuit contains a 9V battery and two resistors of 3Ω and 6Ω in series. What is the voltage across the 6Ω resistor?
A.6V
B.3V
C.9V
D.0V
Solution
The total resistance R_total = 3Ω + 6Ω = 9Ω. The current I = V/R_total = 9V / 9Ω = 1A. Voltage across 6Ω, V = I * R = 1A * 6Ω = 6V.
Correct Answer: A — 6V
Q. A circuit has a voltage of 12V and a resistance of 4Ω. What is the current flowing through the circuit?
A.3A
B.4A
C.12A
D.48A
Solution
Using Ohm's Law, I = V/R = 12V / 4Ω = 3A.
Correct Answer: A — 3A
Q. A circular loop of radius R carries a current I. What is the magnetic field at the center of the loop?
A.μ₀I/(2R)
B.μ₀I/R
C.μ₀I/(4R)
D.μ₀I/(8R)
Solution
The magnetic field at the center of a circular loop carrying current I is given by the formula B = (μ₀I)/(2R), where μ₀ is the permeability of free space.
Correct Answer: B — μ₀I/R
Q. A circular loop of wire carries a current. What is the direction of the magnetic field at the center of the loop?
A.Out of the plane
B.Into the plane
C.Clockwise
D.Counterclockwise
Solution
Using the right-hand rule, the magnetic field at the center of a current-carrying circular loop is directed out of the plane of the loop.
Correct Answer: A — Out of the plane
Q. A circular loop of wire is placed in a uniform magnetic field. If the magnetic field is increased, what happens to the induced EMF in the loop?
A.Increases
B.Decreases
C.Remains constant
D.Becomes zero
Solution
According to Faraday's law of electromagnetic induction, an increase in magnetic field through the loop induces an EMF in the loop.
Correct Answer: A — Increases
Q. A circular 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 proportional to the rate of change of magnetic flux. Increasing the area increases the flux, thus increasing the induced EMF.
Correct Answer: A — Increases
Q. A circular loop of wire is placed in a uniform magnetic field. What happens to the induced EMF if the magnetic field strength is doubled?
A.Induced EMF is halved
B.Induced EMF remains the same
C.Induced EMF is doubled
D.Induced EMF is quadrupled
Solution
According to Faraday's law of electromagnetic induction, the induced EMF is directly proportional to the rate of change of magnetic flux. If the magnetic field strength is doubled, the induced EMF will also double.
Correct Answer: C — Induced EMF is doubled
Q. A coil of wire is placed in a changing magnetic field. What phenomenon is observed?
A.Electromagnetic induction
B.Magnetic resonance
C.Electrolysis
D.Thermal conduction
Solution
According to Faraday's law of electromagnetic induction, a changing magnetic field induces an electromotive force (EMF) in the coil.
Correct Answer: A — Electromagnetic induction
Q. A coil of wire is placed in a magnetic field. If the magnetic field strength is doubled, what happens to the induced EMF?
A.It doubles
B.It remains the same
C.It halves
D.It quadruples
Solution
Doubling the magnetic field strength will double the induced EMF, as it is directly proportional to the magnetic field strength.
Correct Answer: A — It doubles
Q. A coil with 100 turns is placed in a magnetic field that changes at a rate of 0.5 T/s. What is the induced EMF in the coil?
A.50 V
B.100 V
C.200 V
D.25 V
Solution
Using Faraday's law, EMF = -N * (dΦ/dt) = -100 * 0.5 = -50 V. The induced EMF is 50 V.
Correct Answer: B — 100 V
Q. A coil with 100 turns is placed in a magnetic field that changes from 0.5 T to 1.5 T in 2 seconds. What is the induced EMF?
A.50 V
B.100 V
C.200 V
D.400 V
Solution
Induced EMF = -N * (ΔB/Δt) = -100 * ((1.5 - 0.5)/2) = -100 * (1/2) = -50 V. The magnitude is 50 V.
Correct Answer: B — 100 V
Q. A conical pendulum consists of a mass attached to a string that swings in a horizontal circle. If the angle of the string with the vertical is θ, what is the expression for the tension in the string?
A.mg/cos(θ)
B.mg/sin(θ)
C.mg/tan(θ)
D.mg
Solution
Tension T = mg/cos(θ) to balance the vertical component of weight.
Correct Answer: A — mg/cos(θ)
Q. A conical pendulum consists of a mass m attached to a string of length L, swinging in a horizontal circle. What is the expression for the tension in the string?
A.T = mg
B.T = mg/cos(θ)
C.T = mg/sin(θ)
D.T = m(v²/r)
Solution
In a conical pendulum, T = mg/cos(θ) where θ is the angle with the vertical.
Correct Answer: B — T = mg/cos(θ)
Q. A conical pendulum swings in a horizontal circle. If the angle of the string with the vertical is θ, what is the expression for the tension in the string?
A.T = mg
B.T = mg/cos(θ)
C.T = mg/sin(θ)
D.T = mg tan(θ)
Solution
The vertical component of tension balances the weight: T cos(θ) = mg, thus T = mg/cos(θ).
Correct Answer: B — T = mg/cos(θ)
Q. A conical pendulum swings in a horizontal circle. If the angle of the string with the vertical is θ, what is the relationship between the tension and the gravitational force acting on the pendulum bob?
A.T = mg
B.T = mg cos(θ)
C.T = mg sin(θ)
D.T = mg tan(θ)
Solution
The vertical component of tension balances the weight: T cos(θ) = mg.
Correct Answer: B — T = mg cos(θ)
Q. A conical pendulum swings in a horizontal circle. If the angle of the string with the vertical is θ, what is the relationship between the tension in the string and the gravitational force?
A.T = mg
B.T = mg/cos(θ)
C.T = mg/sin(θ)
D.T = mg/tan(θ)
Solution
Tension T provides the centripetal force and balances the weight, T = mg/cos(θ).
Correct Answer: B — T = mg/cos(θ)
Q. A conical pendulum swings with a constant speed. If the angle of the string with the vertical is θ, what is the expression for the tension in the string?
A.mg/cos(θ)
B.mg/sin(θ)
C.mg/tan(θ)
D.mg
Solution
Tension T = mg/cos(θ) to balance the vertical component of weight.
Correct Answer: A — mg/cos(θ)
Q. A convex lens has a focal length of 15 cm. What is the power of the lens?
A.+6.67 D
B.+7.5 D
C.+10 D
D.+15 D
Solution
Power (P) = 1/f (in meters). f = 0.15 m, so P = 1/0.15 = +6.67 D.
Correct Answer: B — +7.5 D
Q. A copper wire has a resistivity of 1.68 x 10^-8 Ω·m. What is the resistance of a 100 m long wire with a cross-sectional area of 1 mm²?
A.1.68 Ω
B.0.168 Ω
C.0.0168 Ω
D.16.8 Ω
Solution
Resistance (R) = ρ * (L / A) = 1.68 x 10^-8 * (100 / 1 x 10^-6) = 1.68 Ω.
Correct Answer: A — 1.68 Ω
Q. A cube of side length a has a charge Q at one of its corners. What is the total electric flux through the cube?
A.Q/ε₀
B.Q/(6ε₀)
C.Q/(12ε₀)
D.0
Solution
The charge at one corner contributes 1/8 of its flux to the cube. Thus, total flux = (1/8)Q/ε₀ = Q/(12ε₀).
