Q. A ball rolls down a ramp and reaches a speed of 10 m/s at the bottom. If the ramp is 5 m high, what is the ball's moment of inertia if it is a solid sphere?
A.(2/5)m(10^2)
B.(1/2)m(10^2)
C.(1/3)m(10^2)
D.(5/2)m(10^2)
Solution
Using conservation of energy, mgh = (1/2)mv^2 + (1/2)(2/5)mv^2. Solving gives the moment of inertia I = (2/5)m(10^2).
Correct Answer: A — (2/5)m(10^2)
Q. A ball rolls without slipping on a flat surface. If the ball's radius is doubled while keeping its mass constant, how does its moment of inertia change?
A.Increases by a factor of 2
B.Increases by a factor of 4
C.Increases by a factor of 8
D.Remains the same
Solution
The moment of inertia of a solid sphere is (2/5)MR^2. If the radius is doubled, the moment of inertia increases by a factor of 4.
Correct Answer: B — Increases by a factor of 4
Q. A ball rolls without slipping on a flat surface. If the ball's radius is doubled, how does its moment of inertia change?
A.Increases by a factor of 2
B.Increases by a factor of 4
C.Increases by a factor of 8
D.Remains the same
Solution
The moment of inertia for a solid sphere is (2/5)MR^2. If the radius is doubled, the moment of inertia increases by a factor of 4.
Correct Answer: B — Increases by a factor of 4
Q. A beam of light in glass (n=1.5) strikes the glass-air interface at an angle of 60°. Will total internal reflection occur?
A.Yes
B.No
C.Only if the angle is increased
D.Only if the angle is decreased
Solution
To determine if total internal reflection occurs, we first find the critical angle using sin(θc) = 1/n = 1/1.5, which gives θc ≈ 41.8°. Since 60° > 41.8°, total internal reflection will not occur.
Correct Answer: B — No
Q. A beam of light in glass (n=1.5) strikes the glass-air interface at an angle of 60°. What happens to the light?
A.It is refracted into the air.
B.It undergoes total internal reflection.
C.It is absorbed by the glass.
D.It is scattered.
Solution
Since the angle of incidence (60°) is greater than the critical angle (approximately 41.8° for glass to air), total internal reflection occurs.
Correct Answer: B — It undergoes total internal reflection.
Q. A beam of light passes through a thin convex lens with a focal length of 15 cm. If the object is placed 30 cm from the lens, what is the image distance?
A.10 cm
B.15 cm
C.20 cm
D.30 cm
Solution
Using the lens formula, 1/f = 1/v - 1/u; here, f = 15 cm and u = -30 cm. Thus, 1/v = 1/15 + 1/30 = 1/10, giving v = 10 cm.
Correct Answer: C — 20 cm
Q. A block is at rest on a horizontal surface. If the applied force is gradually increased and reaches the maximum static frictional force, what will happen next?
A.The block will remain at rest
B.The block will start moving
C.The block will accelerate
D.The block will slide back
Solution
Once the applied force exceeds the maximum static frictional force, the block will start moving.
Correct Answer: B — The block will start moving
Q. A block is sliding down a frictionless incline of angle θ. If the incline has a coefficient of static friction μs, what is the maximum angle θ for which the block will not slide?
A.tan⁻¹(μs)
B.sin⁻¹(μs)
C.cos⁻¹(μs)
D.μs
Solution
The block will not slide if the component of gravitational force down the incline is less than or equal to the maximum static friction force, leading to θ = tan⁻¹(μs).
Correct Answer: A — tan⁻¹(μs)
Q. A block is sliding down a frictionless incline. If the incline is now covered with a material that has a coefficient of kinetic friction of 0.3, how does this affect the acceleration of the block?
A.Increases acceleration
B.Decreases acceleration
C.No effect on acceleration
D.Acceleration becomes zero
Solution
The presence of kinetic friction opposes the motion, thus decreasing the acceleration of the block compared to a frictionless incline.
Correct Answer: B — Decreases acceleration
Q. A block of mass 10 kg is resting on a horizontal surface. If the coefficient of kinetic friction is 0.3, what is the frictional force acting on the block when it is sliding?
A.30 N
B.20 N
C.10 N
D.15 N
Solution
Frictional force (f_k) = μ_k * N = μ_k * mg = 0.3 * 10 kg * 9.8 m/s² = 29.4 N, approximately 30 N.
Correct Answer: A — 30 N
Q. A block of mass 10 kg is resting on a horizontal surface. If the coefficient of static friction is 0.5, what is the maximum static frictional force acting on the block?
A.25 N
B.50 N
C.75 N
D.100 N
Solution
Maximum static frictional force (Fs) = μs * N = μs * mg = 0.5 * 10 kg * 9.8 m/s² = 49 N, approximately 50 N.
Correct Answer: B — 50 N
Q. A block of mass 2 kg is pushed along a frictionless surface by a constant force of 10 N. What is the work done by the force when the block moves 5 m?
A.10 J
B.20 J
C.30 J
D.50 J
Solution
Work done = Force × Distance = 10 N × 5 m = 50 J.
Correct Answer: B — 20 J
Q. A block of mass 2 kg is pushed along a horizontal surface with a constant force of 10 N. What is the work done by the force after moving the block 5 m?
A.10 J
B.20 J
C.30 J
D.50 J
Solution
Work done = Force × Distance = 10 N × 5 m = 50 J.
Correct Answer: B — 20 J
Q. A block of mass 2 kg is pushed along a horizontal surface with a force of 10 N. If the block moves a distance of 5 m, what is the work done by the force?
A.10 J
B.20 J
C.30 J
D.50 J
Solution
Work done = Force × Distance = 10 N × 5 m = 50 J.
Correct Answer: B — 20 J
Q. A block of mass 2 kg is pushed along a horizontal surface with a force of 10 N. If the block moves a distance of 5 m, what is the work done on the block?
A.10 J
B.20 J
C.30 J
D.50 J
Solution
Work done = Force × Distance = 10 N × 5 m = 50 J.
Correct Answer: B — 20 J
Q. A block of mass 2 kg is released from a height of 10 m. What is its speed just before it hits the ground? (g = 9.8 m/s²)
A.14 m/s
B.20 m/s
C.10 m/s
D.5 m/s
Solution
Using conservation of energy, potential energy at height = kinetic energy just before hitting the ground. mgh = 0.5mv². Solving gives v = sqrt(2gh) = sqrt(2 * 9.8 * 10) = 14 m/s.
Correct Answer: A — 14 m/s
Q. A block of mass 2 kg is released from a height of 5 m. What is its speed just before it hits the ground? (g = 9.8 m/s²)
A.5 m/s
B.10 m/s
C.15 m/s
D.20 m/s
Solution
Using conservation of energy, potential energy at height = kinetic energy just before hitting the ground. mgh = 0.5mv². Solving gives v = √(2gh) = √(2 * 9.8 * 5) = 10 m/s.
Correct Answer: B — 10 m/s
Q. A block of mass 2 kg is sliding down a frictionless incline of angle 30 degrees. What is the acceleration of the block?
A.4.9 m/s²
B.9.8 m/s²
C.3.9 m/s²
D.1.5 m/s²
Solution
The acceleration a = g * sin(θ) = 9.8 * sin(30°) = 9.8 * 0.5 = 4.9 m/s².
Correct Answer: A — 4.9 m/s²
Q. A block of mass 5 kg is resting on a horizontal surface. If a horizontal force of 20 N is applied, what is the acceleration of the block? (Assume no friction)
A.2 m/s²
B.4 m/s²
C.5 m/s²
D.10 m/s²
Solution
Using Newton's second law, F = ma, we have a = F/m = 20 N / 5 kg = 4 m/s².
Correct Answer: A — 2 m/s²
Q. A block of mass 5 kg is resting on a horizontal surface. If the coefficient of static friction is 0.4, what is the maximum static frictional force acting on the block?
A.10 N
B.20 N
C.15 N
D.25 N
Solution
Maximum static frictional force (Fs) = μs * N = μs * mg = 0.4 * (5 kg * 10 m/s²) = 20 N.
Correct Answer: B — 20 N
Q. A block on a frictionless surface is attached to a spring and undergoes simple harmonic motion. If the spring constant is 200 N/m and the mass is 2 kg, what is the period of oscillation?
A.0.5 s
B.1 s
C.2 s
D.4 s
Solution
The period T is given by T = 2π√(m/k). Here, T = 2π√(2/200) = 2π√(0.01) = 2π(0.1) = 0.2π ≈ 0.63 s.
Correct Answer: B — 1 s
Q. A block on 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 block on a spring oscillates with a period of 1.5 seconds. If the mass of the block is halved, what will be the new period?
A.1.5 s
B.1.22 s
C.1.73 s
D.1.0 s
Solution
The period of a mass-spring system is T = 2π√(m/k). Halving the mass does not change the period since it is independent of mass in this case.
Correct Answer: A — 1.5 s
Q. A block slides down a frictionless incline of angle 30 degrees. If the incline has a coefficient of kinetic friction of 0.2, what is the acceleration of the block?
A.4.9 m/s²
B.3.9 m/s²
C.2.9 m/s²
D.1.9 m/s²
Solution
Net force = mg sin(30) - μmg cos(30). Acceleration a = (mg sin(30) - μmg cos(30))/m = g(sin(30) - μ cos(30)). Substituting g = 10 m/s² gives a = 10(0.5 - 0.2 * √3/2) = 4.9 m/s².
Correct Answer: B — 3.9 m/s²
Q. A boat can travel at 10 km/h in still water. If it is moving downstream in a river flowing at 5 km/h, what is the speed of the boat relative to the riverbank?
A.5 km/h
B.10 km/h
C.15 km/h
D.20 km/h
Solution
Speed downstream = Speed of boat + Speed of river = 10 km/h + 5 km/h = 15 km/h.
Correct Answer: C — 15 km/h
Q. A boat can travel at 12 km/h in still water. If it is going downstream in a river flowing at 4 km/h, what is the speed of the boat relative to the riverbank?
A.8 km/h
B.12 km/h
C.16 km/h
D.20 km/h
Solution
Speed of boat downstream = Speed of boat + Speed of river = 12 km/h + 4 km/h = 16 km/h.
Correct Answer: C — 16 km/h
Q. A boat can travel at 12 km/h in still water. If it is moving downstream in a river flowing at 4 km/h, what is the speed of the boat relative to the riverbank?
A.8 km/h
B.12 km/h
C.16 km/h
D.4 km/h
Solution
Speed of boat relative to riverbank = Speed of boat + Speed of river = 12 km/h + 4 km/h = 16 km/h.
Correct Answer: C — 16 km/h
Q. A boat can travel at 12 km/h in still water. If it travels downstream in a river flowing at 3 km/h, what is its speed relative to the riverbank?
A.9 km/h
B.12 km/h
C.15 km/h
D.18 km/h
Solution
Speed downstream = Speed of boat + Speed of river = 12 km/h + 3 km/h = 15 km/h.
Correct Answer: C — 15 km/h
Q. A boat is moving upstream at 10 km/h in a river that flows downstream at 5 km/h. What is the speed of the boat relative to the ground?
A.5 km/h
B.10 km/h
C.15 km/h
D.20 km/h
Solution
Speed of boat relative to ground = Speed of boat - Speed of river = 10 km/h - 5 km/h = 5 km/h.
Correct Answer: A — 5 km/h
Q. A boat is moving upstream at 10 km/h in a river that flows downstream at 5 km/h. What is the speed of the boat relative to the riverbank?
A.5 km/h
B.10 km/h
C.15 km/h
D.20 km/h
Solution
Speed of boat relative to riverbank = Speed of boat - Speed of river = 10 km/h - 5 km/h = 5 km/h.
