Q. A family has 2 children. What is the probability that both children are boys?
A.1/4
B.1/2
C.1/3
D.1/5
Solution
The possible combinations of children are BB, BG, GB, GG. Out of these, only 1 combination is both boys (BB). Thus, the probability is 1/4.
Correct Answer: A — 1/4
Q. A family has 3 children. What is the probability that at least one child is a girl given that at least one child is a boy?
A.1/2
B.2/3
C.3/4
D.1/4
Solution
The only combinations with at least one boy are: BBB, BBG, BGB, GBB, BGG, GBG, GGB. Out of these, all combinations except BBB have at least one girl. Thus, P(At least one girl | At least one boy) = 6/7.
Correct Answer: B — 2/3
Q. A fiber optic cable uses total internal reflection to transmit light. What is the primary requirement for this to work effectively?
A.The core must have a higher refractive index than the cladding
B.The cladding must have a higher refractive index than the core
C.The light must be monochromatic
D.The cable must be straight
Solution
For total internal reflection to occur in a fiber optic cable, the core must have a higher refractive index than the cladding.
Correct Answer: A — The core must have a higher refractive index than the cladding
Q. A fiber optic cable uses total internal reflection. What is the role of the cladding?
A.To increase the refractive index.
B.To decrease the refractive index.
C.To prevent light loss.
D.To enhance light absorption.
Solution
The cladding has a lower refractive index than the core, ensuring that light is kept within the core through total internal reflection.
Correct Answer: C — To prevent light loss.
Q. A figure skater spins with arms extended. When she pulls her arms in, what happens to her angular momentum?
A.Increases
B.Decreases
C.Remains the same
D.Becomes zero
Solution
Angular momentum remains the same; however, her angular velocity increases due to a decrease in moment of inertia.
Correct Answer: C — Remains the same
Q. A figure skater spins with arms extended. When she pulls her arms in, what happens to her rotational speed?
A.Increases
B.Decreases
C.Remains the same
D.Becomes zero
Solution
Pulling her arms in decreases her moment of inertia, causing her rotational speed to increase to conserve angular momentum.
Correct Answer: A — Increases
Q. A figure skater spins with arms extended. When she pulls her arms in, what happens to her angular velocity?
A.Increases
B.Decreases
C.Remains the same
D.Becomes zero
Solution
By conservation of angular momentum, pulling arms in decreases moment of inertia, thus increasing angular velocity.
Correct Answer: A — Increases
Q. A fluid with a viscosity of 0.1 Pa·s flows through a pipe of radius 0.05 m. If the pressure difference across the pipe is 1000 Pa, what is the flow rate?
A.0.01 m³/s
B.0.02 m³/s
C.0.03 m³/s
D.0.04 m³/s
Solution
Using Poiseuille's law, the flow rate Q = (π * r^4 * ΔP) / (8 * η * L). Assuming L = 1 m, Q = (π * (0.05)^4 * 1000) / (8 * 0.1 * 1) = 0.01 m³/s.
Correct Answer: A — 0.01 m³/s
Q. A flywheel is rotating with an angular speed of 20 rad/s. If it comes to rest in 5 seconds, what is the angular deceleration?
Q. A force of 10 N is applied at a distance of 0.5 m from the pivot point. What is the torque about the pivot?
A.2.0 Nm
B.5.0 Nm
C.10.0 Nm
D.20.0 Nm
Solution
Torque (τ) = Force (F) × Distance (d) = 10 N × 0.5 m = 5 Nm.
Correct Answer: A — 2.0 Nm
Q. A force of 10 N is applied to a 2 kg object. What is the net force acting on the object if it is also experiencing a frictional force of 4 N?
A.6 N
B.10 N
C.14 N
D.4 N
Solution
Net force = applied force - frictional force = 10 N - 4 N = 6 N.
Correct Answer: A — 6 N
Q. A force of 10 N is applied to a 2 kg object. What is the object's acceleration?
A.5 m/s²
B.10 m/s²
C.2 m/s²
D.20 m/s²
Solution
Using F = ma, acceleration a = F/m = 10 N / 2 kg = 5 m/s².
Correct Answer: A — 5 m/s²
Q. A force of 15 N is applied to a 3 kg object. What is the resulting acceleration?
A.3 m/s²
B.5 m/s²
C.7 m/s²
D.10 m/s²
Solution
Using F = ma, we find a = F/m = 15 N / 3 kg = 5 m/s².
Correct Answer: B — 5 m/s²
Q. A force of 15 N is applied to a 5 kg object. What is the object's acceleration?
A.3 m/s²
B.2 m/s²
C.1 m/s²
D.4 m/s²
Solution
Using F = ma, acceleration a = F/m = 15 N / 5 kg = 3 m/s².
Correct Answer: A — 3 m/s²
Q. A force of 15 N is applied to move an object 3 m at an angle of 60 degrees to the horizontal. What is the work done by the force?
A.15 J
B.20 J
C.30 J
D.45 J
Solution
Work done = F × d × cos(θ) = 15 N × 3 m × cos(60°) = 15 N × 3 m × 0.5 = 22.5 J.
Correct Answer: C — 30 J
Q. A force of 15 N is applied to move an object 4 m in the direction of the force. What is the work done?
A.30 J
B.60 J
C.75 J
D.90 J
Solution
Work done = Force × Distance = 15 N × 4 m = 60 J.
Correct Answer: D — 90 J
Q. A force of 20 N is applied at an angle of 60 degrees to the lever arm of length 2 m. What is the torque about the pivot?
A.10 Nm
B.20 Nm
C.17.32 Nm
D.34.64 Nm
Solution
Torque (τ) = F × r × sin(θ) = 20 N × 2 m × sin(60°) = 20 × 2 × (√3/2) = 20√3 Nm ≈ 34.64 Nm.
Correct Answer: C — 17.32 Nm
Q. A force of 30 N is applied to a 5 kg object. What is the object's acceleration?
A.3 m/s²
B.6 m/s²
C.9 m/s²
D.12 m/s²
Solution
Using F = ma, a = F/m = 30 N / 5 kg = 6 m/s².
Correct Answer: B — 6 m/s²
Q. A force of 40 N is applied at an angle of 60 degrees to the lever arm of length 2 m. What is the torque about the pivot?
A.20 Nm
B.40 Nm
C.34.64 Nm
D.69.28 Nm
Solution
Torque (τ) = F × r × sin(θ) = 40 N × 2 m × sin(60°) = 40 × 2 × (√3/2) = 40√3 Nm ≈ 34.64 Nm.
Correct Answer: C — 34.64 Nm
Q. A force of 50 N is applied at an angle of 30 degrees to the lever arm of length 1 m. What is the torque about the pivot?
A.25 Nm
B.43.3 Nm
C.50 Nm
D.86.6 Nm
Solution
Torque (τ) = F × d × sin(θ) = 50 N × 1 m × sin(30°) = 50 N × 1 m × 0.5 = 25 Nm.
Correct Answer: B — 43.3 Nm
Q. A force of 50 N is applied at an angle of 60 degrees to the lever arm of length 1 m. What is the torque about the pivot?
A.25 Nm
B.43.3 Nm
C.50 Nm
D.0 Nm
Solution
Torque (τ) = F × r × sin(θ) = 50 N × 1 m × sin(60°) = 50 N × 1 m × (√3/2) ≈ 43.3 Nm.
Correct Answer: B — 43.3 Nm
Q. A force of 50 N is applied at an angle of 60 degrees to the lever arm of length 2 m. What is the torque about the pivot?
A.25 Nm
B.50 Nm
C.43.3 Nm
D.100 Nm
Solution
Torque (τ) = F × r × sin(θ) = 50 N × 2 m × sin(60°) = 50 × 2 × (√3/2) = 43.3 Nm.
Correct Answer: C — 43.3 Nm
Q. A gas at 300 K has an RMS speed of 400 m/s. What will be its RMS speed at 600 K?
A.400 m/s
B.400 sqrt(2) m/s
C.800 m/s
D.200 m/s
Solution
The RMS speed is proportional to the square root of the temperature. Therefore, at 600 K, the RMS speed will be 400 * sqrt(600/300) = 400 * sqrt(2) m/s.
Correct Answer: B — 400 sqrt(2) m/s
Q. A gas expands isothermally at 300 K from a volume of 1 m³ to 2 m³. What is the work done by the gas?
A.0 J
B.300 J
C.600 J
D.150 J
Solution
The work done in an isothermal expansion is W = nRT ln(Vf/Vi). Assuming 1 mole of gas, W = 1 * 8.31 * 300 * ln(2) ≈ 600 J.
