Q. A car is moving on a circular track of radius 100 m. If the maximum speed at which it can move without skidding is 20 m/s, what is the coefficient of friction between the tires and the road?
A.0.1
B.0.2
C.0.3
D.0.4
Solution
The centripetal force required is provided by friction: F = mv^2/r. The frictional force is μmg. Setting them equal gives μ = v^2/(rg). Here, μ = (20^2)/(100*9.8) ≈ 0.4.
Correct Answer: C — 0.3
Q. A car is negotiating a curve of radius 100 m at a speed of 15 m/s. What is the minimum coefficient of friction required to prevent the car from skidding?
A.0.15
B.0.25
C.0.30
D.0.35
Solution
Frictional force = m * a_c; μmg = mv²/r; μ = v²/(rg) = (15 m/s)² / (100 m * 9.8 m/s²) ≈ 0.23.
Correct Answer: B — 0.25
Q. A car moves in a circular path of radius 50 m at a constant speed of 20 m/s. What is the centripetal acceleration?
Q. A car of mass 1000 kg accelerates at 2 m/s². What is the net force acting on the car?
A.2000 N
B.500 N
C.1000 N
D.1500 N
Solution
Using F = ma, the net force is F = 1000 kg * 2 m/s² = 2000 N.
Correct Answer: A — 2000 N
Q. A car of mass 1000 kg accelerates from rest to a speed of 20 m/s. What is the work done on the car?
A.20000 J
B.40000 J
C.80000 J
D.100000 J
Solution
Kinetic energy = 0.5 × m × v^2 = 0.5 × 1000 kg × (20 m/s)^2 = 200000 J.
Correct Answer: B — 40000 J
Q. A car of mass 1000 kg is moving at a speed of 15 m/s. What is the kinetic energy of the car?
A.11250 J
B.22500 J
C.33750 J
D.45000 J
Solution
Kinetic energy KE = (1/2)mv² = (1/2)(1000 kg)(15 m/s)² = 11250 J.
Correct Answer: B — 22500 J
Q. A car of mass 1000 kg is moving at a speed of 20 m/s. What is its kinetic energy?
A.200 J
B.400 J
C.200,000 J
D.400,000 J
Solution
Kinetic energy is given by KE = 0.5mv². Here, KE = 0.5 * 1000 * (20)² = 200,000 J.
Correct Answer: C — 200,000 J
Q. A car of mass 1000 kg is moving with a speed of 20 m/s. What is its kinetic energy?
A.200,000 J
B.400,000 J
C.800,000 J
D.1,000,000 J
Solution
Kinetic Energy = 0.5 × mass × velocity² = 0.5 × 1000 kg × (20 m/s)² = 200,000 J.
Correct Answer: B — 400,000 J
Q. A car of mass 1000 kg is moving with a velocity of 20 m/s. What is the momentum of the car?
A.2000 kg·m/s
B.10000 kg·m/s
C.5000 kg·m/s
D.40000 kg·m/s
Solution
Momentum p = mv = 1000 kg * 20 m/s = 20000 kg·m/s.
Correct Answer: B — 10000 kg·m/s
Q. A car of mass 1000 kg is moving with a velocity of 20 m/s. What is the net force required to bring it to rest in 5 seconds?
A.4000 N
B.2000 N
C.1000 N
D.500 N
Solution
First, find the deceleration: a = (final velocity - initial velocity) / time = (0 - 20 m/s) / 5 s = -4 m/s². Then, F = ma = 1000 kg * 4 m/s² = 4000 N.
Correct Answer: A — 4000 N
Q. A car travels at 90 km/h and a truck at 60 km/h in opposite directions. What is the relative speed of the car with respect to the truck?
A.30 km/h
B.60 km/h
C.150 km/h
D.90 km/h
Solution
Relative speed = Speed of car + Speed of truck = 90 km/h + 60 km/h = 150 km/h.
Correct Answer: C — 150 km/h
Q. A card is drawn from a standard deck of 52 cards. What is the probability that it is a heart given that it is a red card?
A.1/2
B.1/4
C.1/3
D.1/13
Solution
There are 26 red cards (hearts and diamonds). The probability of drawing a heart given that it is red is 13/26 = 1/2.
Correct Answer: A — 1/2
Q. A card is drawn from a standard deck of 52 cards. What is the probability that the card is a heart given that it is a red card?
A.1/2
B.1/4
C.1/3
D.2/3
Solution
There are 26 red cards (hearts and diamonds). The probability of drawing a heart given that the card is red is 13/26 = 1/2.
Correct Answer: A — 1/2
Q. A card is drawn from a standard deck of 52 cards. What is the probability that the card drawn is a heart?
A.1/4
B.1/13
C.1/52
D.3/13
Solution
There are 13 hearts in a deck of 52 cards. The probability of drawing a heart is 13/52 = 1/4.
Correct Answer: A — 1/4
Q. A card is drawn from a standard deck of 52 cards. What is the probability that the card drawn is a queen?
A.1/13
B.1/52
C.1/26
D.3/52
Solution
There are 4 queens in a deck of 52 cards. Probability = 4/52 = 1/13.
Correct Answer: A — 1/13
Q. A charge of +3μC is placed at the origin. What is the electric potential at a point 0.5m away?
A.5400 V
B.1800 V
C.7200 V
D.3600 V
Solution
V = k * q / r = (9 × 10^9) * (3 × 10^-6) / 0.5 = 54000 V.
Correct Answer: D — 3600 V
Q. A charge of +3μC is placed at the origin. What is the potential at a point 0.3m away?
A.9000 V
B.3000 V
C.10000 V
D.15000 V
Solution
V = k * q / r = (9 × 10^9 N m²/C²) * (3 × 10^-6 C) / 0.3 m = 9000 V.
Correct Answer: A — 9000 V
Q. A charge of +3μC is placed in a uniform electric field of strength 1500 N/C. What is the work done in moving the charge 0.2 m in the direction of the field?
A.90 J
B.60 J
C.30 J
D.45 J
Solution
Work done W = F * d = (E * q) * d = (1500 N/C * 3 × 10^-6 C) * 0.2 m = 0.0009 J = 90 J.
Correct Answer: B — 60 J
Q. A charge of 5 μC is placed in an electric field of 2000 N/C. What is the potential energy of the charge?
A.10 mJ
B.1 mJ
C.0.5 mJ
D.2 mJ
Solution
Potential energy (U) = Charge × Electric Field × Distance. Assuming distance = 1 m, U = 5 μC × 2000 N/C = 10 mJ.
Correct Answer: A — 10 mJ
Q. A charged capacitor has a potential difference of 12 V across its plates. If the capacitance is 4 µF, what is the charge stored in the capacitor?
A.48 µC
B.12 µC
C.3 µC
D.24 µC
Solution
Charge Q = C × V = 4 µF × 12 V = 48 µC.
Correct Answer: A — 48 µC
Q. A charged particle moves from a point of higher electric potential to a point of lower electric potential. What happens to its kinetic energy?
A.Increases
B.Decreases
C.Remains constant
D.Cannot be determined
Solution
As the charged particle moves to a lower potential, it loses potential energy, which is converted into kinetic energy, thus increasing its kinetic energy.
Correct Answer: A — Increases
Q. A charged particle moves from a region of high potential to low potential. What happens to its kinetic energy?
A.It increases
B.It decreases
C.It remains constant
D.It becomes zero
Solution
As the charged particle moves from high potential to low potential, it loses potential energy, which is converted into kinetic energy, thus its kinetic energy increases.
Correct Answer: A — It increases
Q. A charged particle moves in a magnetic field. What is the condition for the particle to experience maximum force?
A.Velocity is zero
B.Velocity is parallel to the field
C.Velocity is perpendicular to the field
D.Charge is zero
Solution
The magnetic force on a charged particle is given by F = qvB sin(θ). The force is maximum when the angle θ is 90 degrees, meaning the velocity is perpendicular to the magnetic field.
Correct Answer: C — Velocity is perpendicular to the field
Q. A charged particle moves in a magnetic field. What is the condition for the particle to experience no magnetic force?
A.The particle must be at rest
B.The particle must be moving parallel to the magnetic field
C.The particle must be moving perpendicular to the magnetic field
D.The magnetic field must be zero
Solution
The magnetic force on a charged particle is given by F = q(v × B). If the velocity vector v is parallel to the magnetic field B, the cross product is zero, resulting in no magnetic force.
Correct Answer: B — The particle must be moving parallel to the magnetic field
Q. A charged particle moves in a magnetic field. What is the effect of the magnetic field on the particle's speed?
A.Increases speed
B.Decreases speed
C.No effect on speed
D.Reverses speed
Solution
The magnetic field exerts a force perpendicular to the velocity of the charged particle, which does not change its speed but alters its direction.
Correct Answer: C — No effect on speed
Q. A charged particle moves in a magnetic field. What is the effect of the magnetic field on the particle's motion?
A.It accelerates the particle
B.It changes the particle's speed
C.It changes the particle's direction
D.It has no effect
Solution
A magnetic field exerts a force on a charged particle that is perpendicular to both the velocity of the particle and the magnetic field, changing its direction but not its speed.
Correct Answer: C — It changes the particle's direction
Q. A charged particle moves in a magnetic field. What is the nature of the force acting on it?
A.Always in the direction of motion
B.Always opposite to the direction of motion
C.Perpendicular to the direction of motion
D.Depends on the charge of the particle
Solution
The magnetic force on a charged particle moving in a magnetic field is given by the Lorentz force law, which states that the force is perpendicular to both the velocity of the particle and the magnetic field.
Correct Answer: C — Perpendicular to the direction of motion
Q. A charged particle moves in a magnetic field. What path does it follow?
A.Straight line
B.Circular path
C.Elliptical path
D.Parabolic path
Solution
A charged particle moving in a magnetic field experiences a force perpendicular to its velocity, resulting in circular motion.
