Physics Syllabus (JEE Main)
Q. A 5 kg block is placed on a surface with a coefficient of static friction of 0.6. What is the maximum static friction force before the block starts to move?
A.
30 N
B.
20 N
C.
15 N
D.
25 N
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Solution
Maximum static friction force = μs * N = 0.6 * (5 kg * 9.8 m/s²) = 29.4 N, approximately 30 N.
Correct Answer: A — 30 N
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Q. A 5 kg block is resting on a frictionless surface. A force of 10 N is applied to it. What is the acceleration of the block?
A.
1 m/s²
B.
2 m/s²
C.
3 m/s²
D.
4 m/s²
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Solution
Using Newton's second law, F = ma, we have a = F/m = 10 N / 5 kg = 2 m/s².
Correct Answer: B — 2 m/s²
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Q. A 5 kg block is resting on a frictionless surface. If a force of 10 N is applied to it, what will be its acceleration?
A.
1 m/s²
B.
2 m/s²
C.
5 m/s²
D.
10 m/s²
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Solution
Using Newton's second law, F = ma, we have a = F/m = 10 N / 5 kg = 2 m/s².
Correct Answer: B — 2 m/s²
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Q. A 5 kg block is resting on a horizontal surface. What is the normal force acting on the block?
A.
5 N
B.
10 N
C.
15 N
D.
20 N
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Solution
The normal force equals the weight of the block, which is mass times gravity: 5 kg * 9.8 m/s² = 49 N.
Correct Answer: B — 10 N
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Q. A 5 kg object is at rest on a surface. If a horizontal force of 15 N is applied, what is the object's acceleration? (Assume friction is negligible)
A.
2 m/s²
B.
3 m/s²
C.
4 m/s²
D.
5 m/s²
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Solution
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 5 kg object is dropped from a height of 10 m. What is the total mechanical energy just before it hits the ground?
A.
0 J
B.
50 J
C.
100 J
D.
200 J
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Solution
Total mechanical energy is conserved. At height, PE = mgh = 5 * 9.8 * 10 = 490 J. Just before hitting the ground, total energy = 490 J.
Correct Answer: C — 100 J
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Q. A 5 kg object is dropped from a height of 20 m. What is the potential energy at the top? (g = 9.8 m/s²)
A.
980 J
B.
490 J
C.
196 J
D.
9800 J
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Solution
Potential energy is given by PE = mgh. Here, PE = 5 * 9.8 * 20 = 980 J.
Correct Answer: A — 980 J
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Q. A 5 kg object is dropped from a height of 20 m. What is the total mechanical energy just before it hits the ground?
A.
0 J
B.
100 J
C.
200 J
D.
500 J
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Solution
Total mechanical energy is conserved. At height, PE = mgh = 5 * 9.8 * 20 = 980 J. Just before hitting the ground, total energy = 980 J.
Correct Answer: C — 200 J
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Q. A 5 kg object is dropped from a height. What is the force acting on it just before it hits the ground?
A.
5 N
B.
10 N
C.
15 N
D.
20 N
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Solution
The only force acting on it is its weight, W = mg = 5 kg * 10 m/s² = 50 N.
Correct Answer: B — 10 N
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Q. A 5 kg object is lifted to a height of 10 m. What is the gravitational potential energy gained?
A.
50 J
B.
100 J
C.
200 J
D.
500 J
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Solution
Potential Energy (PE) = m * g * h = 5 kg * 10 m/s^2 * 10 m = 500 J
Correct Answer: B — 100 J
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Q. A 5 kg object is lifted to a height of 10 m. What is the potential energy gained by the object?
A.
50 J
B.
100 J
C.
150 J
D.
200 J
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Solution
Potential energy (PE) = mgh = 5 kg * 9.8 m/s² * 10 m = 490 J.
Correct Answer: B — 100 J
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Q. A 5 kg object is lifted to a height of 10 m. What is the potential energy gained?
A.
50 J
B.
100 J
C.
200 J
D.
500 J
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Solution
Potential Energy (PE) = m * g * h = 5 kg * 9.8 m/s^2 * 10 m = 490 J (approx. 500 J)
Correct Answer: B — 100 J
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Q. A 5 kg object is lifted to a height of 10 m. What is the work done against gravity?
A.
50 J
B.
100 J
C.
150 J
D.
200 J
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Solution
Work done against gravity is W = mgh = 5 * 9.8 * 10 = 490 J.
Correct Answer: D — 200 J
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Q. A 5 kg object is lifted to a height of 3 m. What is the work done against gravity?
A.
15 J
B.
30 J
C.
45 J
D.
60 J
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Solution
Work done against gravity is W = mgh = 5 * 9.8 * 3 = 147 J.
Correct Answer: B — 30 J
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Q. A 5 kg object is moving with a velocity of 10 m/s. What is its kinetic energy?
A.
25 J
B.
50 J
C.
100 J
D.
250 J
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Solution
Kinetic energy (KE) = 0.5 * mass * velocity² = 0.5 * 5 * (10)² = 250 J.
Correct Answer: C — 100 J
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Q. A 5 kg object is moving with a velocity of 3 m/s. What is its kinetic energy?
A.
22.5 J
B.
30 J
C.
45 J
D.
60 J
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Solution
Kinetic Energy = 0.5 × mass × velocity² = 0.5 × 5 kg × (3 m/s)² = 22.5 J.
Correct Answer: A — 22.5 J
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Q. A 5 kg object is pulled with a force of 25 N. What is the acceleration of the object?
A.
2 m/s²
B.
3 m/s²
C.
4 m/s²
D.
5 m/s²
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Solution
Using F = ma, acceleration a = F/m = 25 N / 5 kg = 5 m/s².
Correct Answer: C — 4 m/s²
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Q. A 5 kg object is pushed with a force of 15 N. If the frictional force is 5 N, what is the acceleration of the object?
A.
1 m/s²
B.
2 m/s²
C.
3 m/s²
D.
4 m/s²
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Solution
Net force = applied force - friction = 15 N - 5 N = 10 N. Acceleration = F/m = 10 N / 5 kg = 2 m/s².
Correct Answer: B — 2 m/s²
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Q. A 5 kg object is subjected to a net force of 15 N. What is its acceleration?
A.
2 m/s²
B.
3 m/s²
C.
4 m/s²
D.
5 m/s²
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Solution
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 5 kg object is subjected to a net force of 15 N. What is the object's acceleration?
A.
2 m/s²
B.
3 m/s²
C.
4 m/s²
D.
5 m/s²
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Solution
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 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
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Solution
Using F = ma, a = F/m = 25 N / 5 kg = 5 m/s².
Correct Answer: C — 10 m/s²
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Q. A 5 kg object is thrown vertically upward with a speed of 20 m/s. What is the maximum height reached by the object? (g = 10 m/s²)
A.
20 m
B.
30 m
C.
40 m
D.
50 m
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Solution
Using the formula h = v²/(2g) = (20 m/s)² / (2 * 10 m/s²) = 20 m.
Correct Answer: B — 30 m
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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
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Solution
In series, the total resistance is R_total = R1 + R2 = 5 + 10 = 15 ohms.
Correct Answer: A — 15 ohms
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Q. A 50 kg crate is at rest on a flat surface. What is the normal force acting on the crate?
A.
0 N
B.
50 N
C.
500 N
D.
1000 N
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Solution
The normal force equals the weight of the crate, which is N = mg = 50 kg * 10 m/s² = 500 N.
Correct Answer: C — 500 N
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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
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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
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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
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Solution
If the object is moving with constant velocity, the net force acting on it is zero.
Correct Answer: A — It is zero
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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
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Solution
Momentum = mass × velocity = 50 kg × 10 m/s = 500 kg m/s.
Correct Answer: B — 1000 kg m/s
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Q. A 50 kg object is pulled with a force of 200 N. What is the acceleration of the object?
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 have 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 over a distance of 10 m. What is the work done? (2000)
A.
2000 J
B.
1000 J
C.
500 J
D.
3000 J
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Solution
Work Done = Force * Distance = 200 N * 10 m = 2000 J
Correct Answer: A — 2000 J
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Q. A 50 kg person jumps off a diving board with a speed of 5 m/s. What is the momentum of the person just before hitting the water?
A.
100 kg·m/s
B.
200 kg·m/s
C.
250 kg·m/s
D.
300 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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