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 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.
Correct Answer: C — 600 J
Q. A gas has an RMS speed of 500 m/s. If the molar mass of the gas is 0.02 kg/mol, what is the temperature of the gas?
A.250 K
B.500 K
C.1000 K
D.2000 K
Solution
Using the formula v_rms = sqrt((3RT)/M), we can rearrange to find T = (v_rms^2 * M) / (3R). Substituting v_rms = 500 m/s and M = 0.02 kg/mol gives T = 500 K.
Correct Answer: B — 500 K
Q. A hollow sphere has a charge +Q distributed uniformly on its surface. What is the electric field at a point inside the sphere?
A.Q/(4πε₀r²)
B.0
C.Q/(4πε₀R²)
D.Q/(4πε₀R)
Solution
The electric field inside a hollow charged sphere is zero due to symmetry.
Correct Answer: B — 0
Q. A hollow sphere with charge Q has a radius R. What is the electric field at a point inside the sphere?
A.Q/(4πε₀R²)
B.0
C.Q/(4πε₀R)
D.Q/(2πε₀R²)
Solution
Inside a hollow charged sphere, the electric field is zero due to symmetry, as per Gauss's law.
Correct Answer: B — 0
Q. A hollow spherical conductor carries a charge Q. What is the electric field inside the cavity?
A.Q/(4πε₀r²)
B.0
C.Q/(4πε₀R²)
D.Q/(4πε₀R)
Solution
Inside the cavity of a hollow conductor, the electric field is zero due to the shielding effect of the conductor.
Correct Answer: B — 0
Q. A jar contains 4 red, 5 green, and 6 blue marbles. If one marble is drawn at random, what is the probability that it is either red or green?
A.4/15
B.3/5
C.9/15
D.1/3
Solution
Total marbles = 4 + 5 + 6 = 15. Probability of red or green = (4 + 5)/15 = 9/15 = 3/5.
Correct Answer: B — 3/5
Q. A jar contains 4 red, 5 green, and 6 blue marbles. If one marble is drawn at random, what is the probability that it is not blue?
Q. A jar contains 4 red, 5 green, and 6 blue marbles. What is the probability of picking a green marble?
A.5/15
B.1/3
C.1/5
D.1/2
Solution
The total number of marbles is 4 + 5 + 6 = 15. The probability of picking a green marble is 5/15 = 1/3.
Correct Answer: B — 1/3
Q. A jar contains 4 red, 5 green, and 6 blue marbles. What is the probability of randomly selecting a green marble?
A.1/3
B.5/15
C.5/15
D.1/5
Solution
The total number of marbles is 4 + 5 + 6 = 15. The probability of selecting a green marble is 5/15 = 1/3.
Correct Answer: A — 1/3
Q. A jar contains 5 red, 3 green, and 2 yellow marbles. If one marble is drawn at random, what is the probability that it is either red or green?
A.1/2
B.4/5
C.2/5
D.3/5
Solution
The total number of marbles is 5 + 3 + 2 = 10. The number of favorable outcomes (red or green) is 5 + 3 = 8. Thus, the probability is 8/10 = 4/5.
Correct Answer: B — 4/5
Q. A kite is flying at a height of 100 m. If the angle of elevation from a point on the ground to the kite is 30 degrees, how far is the point from the base of the kite?
A.100 m
B.200 m
C.300 m
D.400 m
Solution
Using tan(30°) = height/distance, we have 1/√3 = 100/distance. Therefore, distance = 100√3 ≈ 173.2 m.
Correct Answer: B — 200 m
Q. A kite is flying at a height of 100 meters. If the angle of depression from the kite to a point on the ground is 30 degrees, how far is the point from the point directly below the kite?
A.50 m
B.60 m
C.70 m
D.80 m
Solution
Using tan(30°) = 100/distance, we have 1/√3 = 100/distance. Therefore, distance = 100√3 ≈ 173.21 m.
