Physics Syllabus (JEE Main)
Q. A 3 kg object is at rest on a frictionless surface. A force of 9 N is applied. What is the final velocity after 3 seconds?
A.
1 m/s
B.
2 m/s
C.
3 m/s
D.
4 m/s
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Solution
First, find acceleration: a = F/m = 9 N / 3 kg = 3 m/s². Then, v = u + at = 0 + 3 m/s² * 3 s = 9 m/s.
Correct Answer: C — 3 m/s
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Q. A 3 kg object is at rest on a frictionless surface. If a force of 9 N is applied, what will be its velocity after 3 seconds?
A.
3 m/s
B.
6 m/s
C.
9 m/s
D.
12 m/s
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Solution
First, find acceleration: a = F/m = 9 N / 3 kg = 3 m/s². Then, v = u + at = 0 + 3 m/s² * 3 s = 9 m/s.
Correct Answer: B — 6 m/s
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Q. A 3 kg object is at rest on a horizontal surface. If a force of 12 N is applied, what is the object's acceleration? (Assume no friction)
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 = 12 N / 3 kg = 4 m/s².
Correct Answer: C — 4 m/s²
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Q. A 3 kg object is at rest on a horizontal surface. If a horizontal force of 12 N is applied, 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 = 12 N / 3 kg = 4 m/s².
Correct Answer: B — 3 m/s²
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Q. A 3 kg object is at rest on a table. If a force of 15 N is applied horizontally, what is the object's acceleration?
A.
5 m/s²
B.
10 m/s²
C.
15 m/s²
D.
20 m/s²
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Solution
Using F = ma, a = F/m = 15 N / 3 kg = 5 m/s².
Correct Answer: A — 5 m/s²
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Q. A 3 kg object is at rest on a table. If a horizontal force of 12 N is applied, what is the acceleration of the object?
A.
4 m/s²
B.
0 m/s²
C.
3 m/s²
D.
12 m/s²
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Solution
Using F = ma, acceleration a = F/m = 12 N / 3 kg = 4 m/s².
Correct Answer: A — 4 m/s²
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Q. A 3 kg object is dropped from a height of 12 m. What is the potential energy at the top?
A.
30 J
B.
36 J
C.
60 J
D.
120 J
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Solution
Potential energy = mgh = 3 kg × 9.8 m/s² × 12 m = 352.8 J.
Correct Answer: D — 120 J
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Q. A 3 kg object is dropped from a height of 15 m. What is its potential energy at the top? (g = 9.8 m/s²)
A.
441 J
B.
294 J
C.
147 J
D.
0 J
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Solution
Potential Energy (PE) = m * g * h = 3 kg * 9.8 m/s² * 15 m = 441 J
Correct Answer: B — 294 J
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Q. A 3 kg object is dropped from a height of 15 m. What is the potential energy at the top?
A.
30 J
B.
45 J
C.
60 J
D.
75 J
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Solution
Potential energy = mass × g × height = 3 kg × 9.8 m/s² × 15 m = 441 J.
Correct Answer: D — 75 J
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Q. A 3 kg object is dropped from a height of 5 m. What is the potential energy at the height?
A.
15 J
B.
30 J
C.
45 J
D.
60 J
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Solution
Potential energy = mass × g × height = 3 kg × 9.8 m/s² × 5 m = 147 J.
Correct Answer: B — 30 J
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Q. A 3 kg object is in free fall. What is the force acting on it due to gravity?
A.
3 N
B.
9 N
C.
30 N
D.
0 N
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Solution
The force due to gravity is F = mg = 3 kg * 9.8 m/s² = 29.4 N, approximately 30 N.
Correct Answer: B — 9 N
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Q. A 3 kg object is lifted to a height of 4 m. What is the work done against gravity?
A.
12 J
B.
24 J
C.
36 J
D.
48 J
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Solution
Work done = mgh = 3 * 9.8 * 4 = 117.6 J.
Correct Answer: B — 24 J
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Q. A 3 kg object is lifted to a height of 5 m. What is the work done against gravity? (g = 9.8 m/s²)
A.
147 J
B.
294 J
C.
441 J
D.
588 J
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Solution
Work done = mgh = 3 kg × 9.8 m/s² × 5 m = 147 J.
Correct Answer: B — 294 J
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Q. A 3 kg object is moving in a circular path of radius 2 m with a speed of 4 m/s. What is the centripetal force acting on it?
A.
6 N
B.
12 N
C.
24 N
D.
36 N
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Solution
Centripetal force F = mv²/r = 3 kg * (4 m/s)² / 2 m = 24 N.
Correct Answer: B — 12 N
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Q. A 3 kg object is moving in a straight line with a constant velocity of 5 m/s. What is the net force acting on the object?
A.
0 N
B.
3 N
C.
15 N
D.
5 N
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Solution
Since the object is moving with constant velocity, the net force acting on it is 0 N.
Correct Answer: A — 0 N
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Q. A 3 kg object is moving with a speed of 10 m/s. What is its kinetic energy?
A.
150 J
B.
300 J
C.
450 J
D.
600 J
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Solution
Kinetic energy = 0.5mv² = 0.5 * 3 * (10)² = 150 J.
Correct Answer: B — 300 J
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Q. A 3 kg object is moving with a speed of 10 m/s. What is its total mechanical energy?
A.
150 J
B.
300 J
C.
450 J
D.
600 J
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Solution
Total mechanical energy = KE + PE. Assuming PE = 0, KE = 0.5mv² = 0.5 * 3 * (10)² = 150 J.
Correct Answer: B — 300 J
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Q. A 3 kg object is moving with a speed of 4 m/s. If a net work of 24 J is done on the object, what will be its final speed?
A.
4 m/s
B.
6 m/s
C.
8 m/s
D.
10 m/s
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Solution
Initial kinetic energy = 0.5 × m × v^2 = 0.5 × 3 kg × (4 m/s)^2 = 24 J. Final kinetic energy = Initial + Work done = 24 J + 24 J = 48 J. Final speed = √(2 × KE/m) = √(2 × 48 J / 3 kg) = 6.93 m/s.
Correct Answer: C — 8 m/s
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Q. A 3 kg object is moving with a speed of 4 m/s. What is its kinetic energy?
A.
12 J
B.
24 J
C.
36 J
D.
48 J
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Solution
Kinetic energy = 0.5mv² = 0.5 * 3 * (4)² = 24 J.
Correct Answer: B — 24 J
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Q. A 3 kg object is moving with a speed of 4 m/s. What is its total mechanical energy?
A.
24 J
B.
32 J
C.
48 J
D.
60 J
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Solution
Total mechanical energy = KE + PE. Assuming it is at ground level, PE = 0. KE = 0.5mv² = 0.5 * 3 * (4)² = 24 J.
Correct Answer: B — 32 J
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Q. A 3 kg object is moving with a speed of 5 m/s. What is its kinetic energy?
A.
15 J
B.
25 J
C.
35 J
D.
45 J
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Solution
Kinetic energy is given by KE = 0.5mv² = 0.5 * 3 * (5)² = 37.5 J, approximately 35 J.
Correct Answer: B — 25 J
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Q. A 3 kg object is moving with a velocity of 4 m/s. If a force of 12 N is applied in the opposite direction, what will be its velocity after 2 seconds?
A.
0 m/s
B.
2 m/s
C.
4 m/s
D.
6 m/s
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Solution
Acceleration = F/m = -12 N / 3 kg = -4 m/s². Change in velocity = a * t = -4 m/s² * 2 s = -8 m/s. Final velocity = 4 m/s - 8 m/s = -4 m/s (object stops).
Correct Answer: B — 2 m/s
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Q. A 3 kg object is moving with a velocity of 4 m/s. If it comes to a stop, what is the work done by the friction force?
A.
-24 J
B.
-48 J
C.
-12 J
D.
-36 J
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Solution
Work done by friction = - Kinetic Energy = - (0.5 × 3 kg × (4 m/s)²) = -24 J.
Correct Answer: B — -48 J
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Q. A 3 kg object is moving with a velocity of 4 m/s. What is its kinetic energy?
A.
12 J
B.
24 J
C.
36 J
D.
48 J
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Solution
Kinetic Energy = 0.5 × mass × velocity² = 0.5 × 3 kg × (4 m/s)² = 24 J.
Correct Answer: B — 24 J
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Q. A 3 kg object is moving with a velocity of 4 m/s. What is its momentum?
A.
6 kg·m/s
B.
12 kg·m/s
C.
9 kg·m/s
D.
15 kg·m/s
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Solution
Momentum = mass × velocity = 3 kg × 4 m/s = 12 kg·m/s.
Correct Answer: B — 12 kg·m/s
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Q. A 3 kg object is moving with a velocity of 6 m/s. What is its kinetic energy?
A.
54 J
B.
36 J
C.
18 J
D.
72 J
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Solution
Kinetic Energy (KE) = 1/2 * m * v^2 = 1/2 * 3 * (6^2) = 54 J
Correct Answer: B — 36 J
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Q. A 3 kg object is pushed with a force of 12 N over a distance of 4 m. If the object starts from rest, what is its final speed? (Assume no friction)
A.
2 m/s
B.
3 m/s
C.
4 m/s
D.
5 m/s
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Solution
Work done = Force × Distance = 12 N × 4 m = 48 J. Kinetic energy = 0.5 × mass × v²; 48 J = 0.5 × 3 kg × v²; v² = 32; v = 4 m/s.
Correct Answer: C — 4 m/s
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Q. A 3 kg object is pushed with a force of 12 N over a distance of 4 m. What is the net work done if friction does 8 J of work?
A.
28 J
B.
32 J
C.
36 J
D.
40 J
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Solution
Net work done = Work done by force - Work done against friction = (12 N × 4 m) - 8 J = 48 J - 8 J = 40 J.
Correct Answer: B — 32 J
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Q. A 3 kg object is pushed with a force of 15 N over a distance of 4 m. If the object experiences a frictional force of 3 N, what is the net work done on the object?
A.
48 J
B.
60 J
C.
72 J
D.
84 J
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Solution
Net force = Applied force - Friction = 15 N - 3 N = 12 N. Work done = Net force × Distance = 12 N × 4 m = 48 J.
Correct Answer: B — 60 J
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Q. A 3 kg object is pushed with a force of 15 N over a distance of 4 m. What is the work done on the object?
A.
30 J
B.
45 J
C.
60 J
D.
75 J
Show solution
Solution
Work done = Force × Distance = 15 N × 4 m = 60 J.
Correct Answer: C — 60 J
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