Physics
Q. A 4 kg block is pulled with a force of 20 N. What is the acceleration of the block? (2020)
A.
5 m/s²
B.
4 m/s²
C.
3 m/s²
D.
2 m/s²
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Solution
Using F = ma, acceleration (a) = F/m = 20 N / 4 kg = 5 m/s².
Correct Answer: A — 5 m/s²
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Q. A 4 kg object is lifted to a height of 2 m. What is the work done against gravity? (2022)
A.
40 J
B.
80 J
C.
20 J
D.
60 J
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Solution
Work done = mass × g × height = 4 kg × 10 m/s² × 2 m = 80 J
Correct Answer: B — 80 J
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Q. A 5 kg block of ice at 0°C is converted to water at 0°C. If the latent heat of fusion of ice is 334 kJ/kg, how much heat is absorbed?
A.
1670 kJ
B.
334 kJ
C.
167 kJ
D.
3340 kJ
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Solution
Heat absorbed (Q) = m * Lf = 5 kg * 334 kJ/kg = 1670 kJ.
Correct Answer: B — 334 kJ
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Q. A 5 kg object is dropped from a height of 10 m. What is its potential energy at the top? (g = 9.8 m/s²)
A.
490 J
B.
98 J
C.
50 J
D.
100 J
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Solution
Potential Energy = mgh = 5 kg * 9.8 m/s² * 10 m = 490 J.
Correct Answer: A — 490 J
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Q. A 5 kg object is dropped 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.
15 m/s
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Solution
Using energy conservation, Potential Energy = Kinetic Energy at the ground: mgh = 0.5mv²; v = √(2gh) = √(2 × 9.8 m/s² × 10 m) = 14 m/s
Correct Answer: A — 14 m/s
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Q. A 50 kg crate is pushed with a force of 200 N. If the frictional force is 50 N, what is the acceleration of the crate? (2022)
A.
2 m/s²
B.
3 m/s²
C.
4 m/s²
D.
5 m/s²
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Solution
Net force = 200 N - 50 N = 150 N. Acceleration a = F/m = 150 N / 50 kg = 3 m/s².
Correct Answer: B — 3 m/s²
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Q. A 50 kg crate is pushed with a force of 200 N. What is the acceleration of the crate? (Assume no friction) (2021)
A.
2 m/s²
B.
4 m/s²
C.
5 m/s²
D.
10 m/s²
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Solution
Using F = ma, acceleration a = F/m = 200 N / 50 kg = 4 m/s².
Correct Answer: B — 4 m/s²
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Q. A 50 kg object is moving with a velocity of 5 m/s. What is its momentum? (2023)
A.
100 kg·m/s
B.
150 kg·m/s
C.
200 kg·m/s
D.
250 kg·m/s
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Solution
Momentum p = mv = 50 kg * 5 m/s = 250 kg·m/s.
Correct Answer: A — 100 kg·m/s
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Q. A 50 kg object is pulled with a force of 200 N. What is its acceleration? (2023)
A.
2 m/s²
B.
3 m/s²
C.
4 m/s²
D.
5 m/s²
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Solution
Using F = ma, we find a = F/m = 200 N / 50 kg = 4 m/s².
Correct Answer: D — 5 m/s²
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Q. A 50 kg object is pushed with a force of 200 N. What is the frictional force if the object moves with constant velocity? (2023)
A.
0 N
B.
50 N
C.
200 N
D.
100 N
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Solution
If the object moves with constant velocity, the frictional force equals the applied force, which is 200 N.
Correct Answer: C — 200 N
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Q. A ball is thrown horizontally from a height of 20 m. How long does it take to hit the ground? (2019)
A.
1 s
B.
2 s
C.
3 s
D.
4 s
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Solution
Using the equation of motion for free fall: h = (1/2)gt². 20 m = (1/2)(9.8)t². Solving gives t ≈ 2 s.
Correct Answer: B — 2 s
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Q. A ball is thrown upwards with a speed of 20 m/s. How long will it take to reach the maximum height? (Take g = 10 m/s²) (2023)
A.
1 s
B.
2 s
C.
3 s
D.
4 s
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Solution
Time to reach maximum height = initial velocity / g = 20 / 10 = 2 s.
Correct Answer: B — 2 s
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Q. A ball is thrown upwards with a speed of 25 m/s. How long will it take to reach the maximum height? (Assume g = 10 m/s²)
A.
2.5 s
B.
3 s
C.
4 s
D.
5 s
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Solution
Time to reach maximum height = initial velocity / g = 25 / 10 = 2.5 s.
Correct Answer: A — 2.5 s
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Q. A ball is thrown upwards with a velocity of 20 m/s. How high will it go before coming to rest? (2023)
A.
10 m
B.
20 m
C.
30 m
D.
40 m
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Solution
Using the formula h = v²/(2g), where g = 10 m/s², h = (20 m/s)² / (2 * 10 m/s²) = 20 m.
Correct Answer: C — 30 m
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Q. A ball is thrown upwards with an initial velocity of 20 m/s. How long will it take to reach the maximum height? (g = 10 m/s²)
A.
1 s
B.
2 s
C.
3 s
D.
4 s
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Solution
Using the formula: t = v/g = 20/10 = 2 s.
Correct Answer: B — 2 s
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Q. A ball is thrown vertically upwards with a speed of 15 m/s. How high will it rise before coming to a stop? (2023)
A.
11.25 m
B.
15 m
C.
22.5 m
D.
30 m
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Solution
Using the formula: height = (initial velocity^2) / (2 * g), where g = 9.8 m/s². Height = (15^2) / (2 * 9.8) = 11.25 m.
Correct Answer: A — 11.25 m
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Q. A ball is thrown vertically upwards with a speed of 20 m/s. How high will it go? (g = 10 m/s²) (2023)
A.
10 m
B.
20 m
C.
30 m
D.
40 m
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Solution
Using the formula h = v²/(2g), we have h = (20 m/s)² / (2 * 10 m/s²) = 20 m.
Correct Answer: B — 20 m
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Q. A beam of light passes through a narrow slit and produces a diffraction pattern. What is the angle for the first minimum? (2022)
A.
λ/a
B.
2λ/a
C.
λ/2a
D.
a/λ
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Solution
The angle for the first minimum in single-slit diffraction is given by a sin θ = λ, hence θ = λ/a.
Correct Answer: A — λ/a
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Q. A beam of light passes through a narrow slit and produces a diffraction pattern. What happens to the width of the central maximum if the slit width is decreased? (2019)
A.
It increases
B.
It decreases
C.
It remains the same
D.
It becomes zero
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Solution
According to the diffraction principle, as the slit width decreases, the width of the central maximum increases.
Correct Answer: A — It increases
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Q. A beam of light passes through a narrow slit and produces a diffraction pattern. What is the angle of the first minimum? (2019)
A.
λ/a
B.
λ/2a
C.
2λ/a
D.
3λ/a
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Solution
The angle of the first minimum in single-slit diffraction is given by sin(θ) = λ/a, where a is the width of the slit.
Correct Answer: A — λ/a
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Q. A block of mass 2 kg is placed on a frictionless surface and is pushed with a force of 10 N. What is its acceleration?
A.
2 m/s²
B.
5 m/s²
C.
10 m/s²
D.
20 m/s²
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Solution
Using Newton's second law, F = ma. Here, F = 10 N and m = 2 kg. Therefore, a = F/m = 10/2 = 5 m/s².
Correct Answer: B — 5 m/s²
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Q. A block of mass 2 kg is sliding down a frictionless incline of height 5 m. What is its speed at the bottom?
A.
5 m/s
B.
10 m/s
C.
15 m/s
D.
20 m/s
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Solution
Using conservation of energy, potential energy at the top (mgh) converts to kinetic energy at the bottom (1/2 mv²). Thus, v = sqrt(2gh) = sqrt(2 * 9.8 * 5) ≈ 10 m/s.
Correct Answer: B — 10 m/s
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Q. A block of mass 2 kg is sliding down a frictionless incline of height 5 m. What is its speed at the bottom of the incline?
A.
10 m/s
B.
5 m/s
C.
20 m/s
D.
15 m/s
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Solution
Using conservation of energy, potential energy at the top = kinetic energy at the bottom. mgh = 0.5mv². Thus, v = sqrt(2gh) = sqrt(2 * 9.81 * 5) = 10 m/s.
Correct Answer: A — 10 m/s
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Q. A block of mass 5 kg is pulled on a horizontal surface with a force of 20 N. If the frictional force is 5 N, what is the acceleration of the block? (2020)
A.
2 m/s²
B.
3 m/s²
C.
4 m/s²
D.
5 m/s²
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Solution
Net force = Applied force - Frictional force = 20 N - 5 N = 15 N. Using F = ma, a = F/m = 15 N / 5 kg = 3 m/s².
Correct Answer: B — 3 m/s²
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Q. A block of mass 5 kg is resting on a frictionless surface. A force of 10 N is applied to it. What will be its acceleration? (2020)
A.
2 m/s²
B.
5 m/s²
C.
10 m/s²
D.
20 m/s²
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Solution
Using Newton's second law, F = ma, acceleration (a) = F/m = 10 N / 5 kg = 2 m/s².
Correct Answer: B — 5 m/s²
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Q. A car accelerates from rest at a constant rate of 2 m/s². How far does it travel in 5 seconds? (2021)
A.
10 m
B.
20 m
C.
25 m
D.
50 m
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Solution
Using the equation of motion: s = ut + (1/2)at². Here, u = 0, a = 2 m/s², t = 5 s. So, s = 0 + (1/2)(2)(5²) = 25 m.
Correct Answer: C — 25 m
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Q. A car accelerates from rest at a rate of 2 m/s². What is its speed after 5 seconds? (2023)
A.
5 m/s
B.
10 m/s
C.
15 m/s
D.
20 m/s
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Solution
Using the formula v = u + at, where u = 0, a = 2 m/s², and t = 5 s, we get v = 0 + 2 * 5 = 10 m/s.
Correct Answer: B — 10 m/s
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Q. A car accelerates from rest to a speed of 20 m/s in 5 seconds. What is its acceleration?
A.
2 m/s²
B.
4 m/s²
C.
5 m/s²
D.
10 m/s²
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Solution
Acceleration is calculated using the formula a = (final velocity - initial velocity) / time. Here, a = (20 m/s - 0 m/s) / 5 s = 4 m/s².
Correct Answer: B — 4 m/s²
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Q. A car accelerates from rest to a speed of 20 m/s. If its mass is 1000 kg, what is the kinetic energy of the car at this speed? (2000)
A.
200 J
B.
1000 J
C.
2000 J
D.
4000 J
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Solution
Kinetic Energy (KE) = 0.5 × m × v² = 0.5 × 1000 kg × (20 m/s)² = 0.5 × 1000 × 400 = 200000 J.
Correct Answer: C — 2000 J
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Q. A car accelerates from rest to a speed of 30 m/s in 10 seconds. What is the acceleration of the car? (2022)
A.
3 m/s²
B.
2 m/s²
C.
1 m/s²
D.
4 m/s²
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Solution
Acceleration (a) = (final velocity - initial velocity) / time = (30 m/s - 0) / 10 s = 3 m/s².
Correct Answer: A — 3 m/s²
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