Q. A 5 kg object is subjected to a net force of 25 N. What is its acceleration?
A.2 m/s²
B.5 m/s²
C.10 m/s²
D.15 m/s²
Solution
Using F = ma, a = F/m = 25 N / 5 kg = 5 m/s².
Correct Answer: C — 10 m/s²
Q. A 5 ohm resistor and a 10 ohm resistor are connected in series. What is the total resistance?
A.15 ohms
B.5 ohms
C.10 ohms
D.2 ohms
Solution
In series, the total resistance is R_total = R1 + R2 = 5 + 10 = 15 ohms.
Correct Answer: A — 15 ohms
Q. A 50 kg object is at rest on a surface. What is the normal force acting on it?
A.0 N
B.50 N
C.100 N
D.500 N
Solution
The normal force equals the weight of the object, which is F = mg = 50 kg * 9.8 m/s² = 490 N.
Correct Answer: C — 100 N
Q. A 50 kg object is moving with a constant velocity. What can be said about the net force acting on it?
A.It is zero
B.It is equal to its weight
C.It is equal to the applied force
D.It is maximum
Solution
If the object is moving with constant velocity, the net force acting on it is zero.
Correct Answer: A — It is zero
Q. A 50 kg object is moving with a velocity of 10 m/s. What is its momentum?
A.500 kg m/s
B.1000 kg m/s
C.1500 kg m/s
D.2000 kg m/s
Solution
Momentum = mass × velocity = 50 kg × 10 m/s = 500 kg m/s.
Correct Answer: B — 1000 kg m/s
Q. A 6 kg object is moving with a velocity of 3 m/s. What is its kinetic energy?
A.27 J
B.36 J
C.54 J
D.18 J
Solution
Kinetic energy KE = 0.5 * m * v² = 0.5 * 6 kg * (3 m/s)² = 27 J.
Correct Answer: B — 36 J
Q. A 6 kg object is subjected to a net force of 12 N. What is its acceleration?
A.2 m/s²
B.3 m/s²
C.4 m/s²
D.5 m/s²
Solution
Using F = ma, acceleration a = F/m = 12 N / 6 kg = 2 m/s².
Correct Answer: B — 3 m/s²
Q. A 60W bulb is connected to a 120V supply. What is the resistance of the bulb?
A.240 ohms
B.120 ohms
C.60 ohms
D.30 ohms
Solution
Using P = V^2 / R, we rearrange to find R = V^2 / P = (120V)^2 / 60W = 240 ohms.
Correct Answer: B — 120 ohms
Q. A bag contains 3 red and 2 blue balls. If one ball is drawn at random, what is the probability that it is red given that it is not blue?
A.1/2
B.3/5
C.2/5
D.3/4
Solution
The total number of balls that are not blue is 3 (red). The probability of drawing a red ball given that it is not blue is 3/5.
Correct Answer: B — 3/5
Q. A bag contains 3 red balls and 2 blue balls. If one ball is drawn at random, what is the probability that it is red?
A.1/5
B.2/5
C.3/5
D.4/5
Solution
The total number of balls is 3 + 2 = 5. The probability of drawing a red ball is 3/5.
Correct Answer: C — 3/5
Q. A ball is dropped from a height of 80 m. How long will it take to reach the ground?
A.4 s
B.5 s
C.6 s
D.8 s
Solution
Using the formula: h = 0.5 * g * t², where h = 80 m and g = 9.8 m/s². Rearranging gives t² = (2 * h) / g = (2 * 80) / 9.8 ≈ 16.33, so t ≈ 4.03 s.
Correct Answer: C — 6 s
Q. A ball is swung in a vertical circle. At the highest point of the circle, what is the condition for the ball to just maintain its circular motion?
A.Weight must be greater than tension
B.Tension must be zero
C.Centripetal force must be zero
D.Weight must be less than tension
Solution
At the highest point, the centripetal force is provided by the weight of the ball. For just maintaining motion, tension can be zero.
Correct Answer: B — Tension must be zero
Q. A ball is thrown at an angle of 45 degrees with an initial speed of 14 m/s. What is the range of the projectile?
A.10 m
B.14 m
C.20 m
D.28 m
Solution
Range (R) = (u² * sin(2θ)) / g = (14² * 1) / 9.8 ≈ 20 m.
Correct Answer: D — 28 m
Q. A ball is thrown downward with an initial speed of 10 m/s from a height of 20 m. How long will it take to hit the ground? (g = 10 m/s²)
A.2 s
B.3 s
C.4 s
D.5 s
Solution
Using the equation of motion: h = ut + 0.5gt². 20 = 10t + 0.5 * 10 * t². Rearranging gives 5t² + 10t - 20 = 0. Solving this quadratic gives t = 2 s.
Correct Answer: B — 3 s
Q. A ball is thrown downward with an initial speed of 10 m/s from a height of 20 m. How long will it take to hit the ground? (Assume g = 10 m/s²)
A.2 s
B.3 s
C.4 s
D.5 s
Solution
Using the equation of motion: h = ut + 0.5gt². 20 = 10t + 0.5 * 10 * t². Rearranging gives 5t² + 10t - 20 = 0. Solving this quadratic gives t = 2 s.
Correct Answer: C — 4 s
Q. A ball is thrown downward with an initial speed of 5 m/s from a height of 20 m. How long will it take to hit the ground? (g = 10 m/s²)
A.2 s
B.3 s
C.4 s
D.5 s
Solution
Using the equation of motion: h = ut + 0.5gt². 20 = 5t + 0.5 * 10 * t². Rearranging gives 5t + 5t² - 20 = 0. Solving the quadratic gives t = 2 s.
Correct Answer: B — 3 s
Q. A ball is thrown horizontally from the top of a cliff 80 m high. How far from the base of the cliff will it land? (Assume g = 10 m/s² and initial horizontal speed = 20 m/s)
A.40 m
B.60 m
C.80 m
D.100 m
Solution
Time to fall = √(2h/g) = √(2*80/10) = 4 s. Horizontal distance = speed * time = 20 * 4 = 80 m.
Correct Answer: C — 80 m
Q. A ball is thrown horizontally from the top of a cliff 80 m high. How far from the base of the cliff will it land? (Assume g = 10 m/s² and horizontal speed = 20 m/s)
A.20 m
B.40 m
C.60 m
D.80 m
Solution
Time to fall = √(2h/g) = √(2*80/10) = 4 s. Horizontal distance = speed * time = 20 m/s * 4 s = 80 m.
Correct Answer: C — 60 m
Q. A ball is thrown upwards with a speed of 20 m/s. How high will it rise before coming to a stop?
A.10 m
B.20 m
C.30 m
D.40 m
Solution
Maximum height (H) = (u²)/(2g) = (20²)/(2*9.8) ≈ 20.4 m.
Correct Answer: C — 30 m
Q. A ball is thrown upwards with a speed of 20 m/s. How long will it take to reach the maximum height?
A.2 s
B.3 s
C.4 s
D.5 s
Solution
Time to reach maximum height (t) = u/g = 20/9.8 ≈ 2.04 s.
Correct Answer: A — 2 s
Q. A ball is thrown vertically upward with a speed of 20 m/s. How high will it rise before coming to rest?
A.20 m
B.40 m
C.10 m
D.80 m
Solution
Using the formula v² = u² + 2as, where v = 0, u = 20 m/s, a = -9.8 m/s². 0 = (20)² + 2(-9.8)s, solving gives s = 20.4 m, approximately 40 m.
Correct Answer: B — 40 m
Q. A ball is thrown vertically upward with a speed of 20 m/s. How high will it rise before coming to rest? (g = 10 m/s²)
A.20 m
B.30 m
C.40 m
D.50 m
Solution
Using the formula v² = u² + 2as, where v = 0, u = 20 m/s, and a = -g = -10 m/s², we get 0 = (20)² + 2(-10)s, solving gives s = 20 m.
Correct Answer: B — 30 m
Q. A ball is thrown vertically upward with a speed of 20 m/s. What is the maximum height it reaches? (g = 10 m/s²)
A.20 m
B.30 m
C.40 m
D.50 m
Solution
Using the formula h = v²/(2g) = (20 m/s)² / (2 * 10 m/s²) = 20 m.
Correct Answer: B — 30 m
Q. A ball is thrown vertically upward with an initial velocity of 20 m/s. How high will it rise before coming to a momentary stop? (g = 10 m/s²)
A.20 m
B.30 m
C.40 m
D.50 m
Solution
Using the formula v² = u² + 2as, where v = 0, u = 20 m/s, and a = -g = -10 m/s², we get 0 = (20)² + 2(-10)s, solving gives s = 20 m.
Correct Answer: B — 30 m
Q. A ball is thrown vertically upwards with a speed of 15 m/s. What is the maximum height it reaches? (g = 9.8 m/s²)
A.11.5 m
B.15.0 m
C.22.5 m
D.30.0 m
Solution
Using the formula h = v² / (2g), h = (15 m/s)² / (2 × 9.8 m/s²) = 11.5 m.
Correct Answer: A — 11.5 m
Q. A ball is thrown vertically upwards with a speed of 20 m/s. How high will it rise before coming to rest momentarily? (g = 10 m/s²)
A.20 m
B.40 m
C.10 m
D.30 m
Solution
Using the formula v² = u² + 2as, where v = 0, u = 20 m/s, and a = -g = -10 m/s². 0 = (20)² + 2(-10)h, solving gives h = 20 m.
Correct Answer: B — 40 m
Q. A ball is thrown vertically upwards with a speed of 20 m/s. What is the maximum height it reaches? (g = 10 m/s²)
A.20 m
B.40 m
C.30 m
D.10 m
Solution
Using the formula h = v²/(2g), we have h = (20 m/s)² / (2 * 10 m/s²) = 20 m.
Correct Answer: B — 40 m
Q. A ball is thrown vertically upwards with a speed of 30 m/s. What is the maximum height it reaches? (g = 9.8 m/s²)
A.45.9 m
B.46.0 m
C.46.1 m
D.46.2 m
Solution
Using conservation of energy, initial kinetic energy = potential energy at maximum height. 0.5mv² = mgh. Solving gives h = v²/(2g) = (30)²/(2 * 9.8) = 45.9 m.
Correct Answer: B — 46.0 m
Q. A ball is tied to a string and swung in a vertical circle. At the highest point of the circle, what is the condition for the ball to remain in circular motion?
A.Tension must be zero
B.Tension must be maximum
C.Weight must be zero
D.Centripetal force must be zero
Solution
At the highest point, the tension can be zero if the centripetal force is provided entirely by the weight.
Correct Answer: A — Tension must be zero
Q. A ball is tied to a string and swung in a vertical circle. At the highest point of the circle, what is the condition for the ball to just maintain circular motion?
A.Tension = 0
B.Tension = mg
C.Tension > mg
D.Tension < mg
Solution
At the highest point, the centripetal force is provided by the weight, so T + mg = mv²/r, T = 0.
