Physics Syllabus (JEE Main)
Q. A 2 kg object is pulled with a force of 8 N. What is the acceleration of the object?
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A.
2 m/s²
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B.
4 m/s²
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C.
6 m/s²
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D.
8 m/s²
Solution
Using F = ma, we find a = F/m = 8 N / 2 kg = 4 m/s².
Correct Answer: B — 4 m/s²
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Q. A 2 kg object is raised to a height of 5 m. What is the potential energy gained?
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A.
10 J
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B.
20 J
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C.
30 J
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D.
40 J
Solution
Potential Energy (PE) = m * g * h = 2 kg * 9.8 m/s^2 * 5 m = 98 J
Correct Answer: B — 20 J
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Q. A 2 kg object is sliding on a frictionless surface with a velocity of 4 m/s. What is the momentum of the object?
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A.
8 kg·m/s
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B.
2 kg·m/s
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C.
4 kg·m/s
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D.
16 kg·m/s
Solution
Momentum p = mv = 2 kg * 4 m/s = 8 kg·m/s.
Correct Answer: A — 8 kg·m/s
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Q. A 2 kg object is thrown upwards with a speed of 10 m/s. What is the maximum height it reaches?
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A.
5 m
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B.
10 m
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C.
15 m
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D.
20 m
Solution
Using energy conservation: KE_initial = PE_max; 0.5 × 2 kg × (10 m/s)² = 2 kg × 9.8 m/s² × h; h = 5.1 m.
Correct Answer: A — 5 m
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Q. A 2 kg object is thrown vertically upwards with a speed of 10 m/s. What is the maximum height it reaches? (g = 10 m/s²)
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A.
5 m
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B.
10 m
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C.
15 m
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D.
20 m
Solution
Using energy conservation: KE at bottom = PE at top; 1/2 mv^2 = mgh; h = v^2/(2g) = (10^2)/(2*10) = 5 m
Correct Answer: B — 10 m
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Q. A 20 kg object is dropped from a height. What is the force acting on it just before it hits the ground?
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A.
0 N
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B.
20 N
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C.
200 N
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D.
10 N
Solution
The force acting on the object is its weight: F = mg = 20 kg * 10 m/s² = 200 N.
Correct Answer: C — 200 N
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Q. A 20 kg object is pulled with a force of 100 N. If the frictional force is 40 N, what is the acceleration of the object?
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A.
2 m/s²
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B.
3 m/s²
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C.
4 m/s²
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D.
5 m/s²
Solution
Net force = applied force - friction = 100 N - 40 N = 60 N. Acceleration a = F/m = 60 N / 20 kg = 3 m/s².
Correct Answer: B — 3 m/s²
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Q. A 2000 W heater operates for 3 hours. How much energy does it consume in kilowatt-hours?
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A.
6 kWh
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B.
5 kWh
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C.
4 kWh
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D.
3 kWh
Solution
Energy consumed in kilowatt-hours is calculated as E = P * t. Here, P = 2000 W = 2 kW and t = 3 hours. Thus, E = 2 kW * 3 h = 6 kWh.
Correct Answer: A — 6 kWh
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Q. A 25 kg object is pulled with a force of 100 N. What is the acceleration of the object?
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A.
2 m/s²
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B.
3 m/s²
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C.
4 m/s²
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D.
5 m/s²
Solution
Using F = ma, we have a = F/m = 100 N / 25 kg = 4 m/s².
Correct Answer: D — 5 m/s²
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Q. A 25 kg object is subjected to a force of 50 N. What is its acceleration?
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A.
1 m/s²
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B.
2 m/s²
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C.
3 m/s²
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D.
4 m/s²
Solution
Using F = ma, we find a = F/m = 50 N / 25 kg = 2 m/s².
Correct Answer: B — 2 m/s²
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Q. A 3 kg ball is thrown vertically upwards with a speed of 10 m/s. What is the maximum height it reaches? (g = 9.8 m/s²)
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A.
5.1 m
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B.
10.2 m
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C.
15.3 m
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D.
20.0 m
Solution
Using conservation of energy, KE at the bottom = PE at the top. 0.5mv² = mgh. Solving gives h = v²/(2g) = (10)²/(2*9.8) = 5.1 m.
Correct Answer: A — 5.1 m
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Q. A 3 kg block is at rest on a horizontal surface. If a horizontal force of 12 N is applied, what is the block's acceleration assuming no friction?
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A.
2 m/s²
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B.
3 m/s²
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C.
4 m/s²
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D.
5 m/s²
Solution
Acceleration a = F/m = 12 N / 3 kg = 4 m/s².
Correct Answer: A — 2 m/s²
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Q. A 3 kg block is on a table and a horizontal force of 15 N is applied. If the frictional force is 5 N, what is the acceleration of the block?
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A.
2 m/s²
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B.
3 m/s²
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C.
5 m/s²
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D.
10 m/s²
Solution
Net force = applied force - friction = 15 N - 5 N = 10 N. Acceleration a = F/m = 10 N / 3 kg = 3.33 m/s².
Correct Answer: A — 2 m/s²
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Q. A 3 kg block is pulled with a force of 12 N. If the frictional force is 3 N, what is the acceleration of the block?
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A.
3 m/s²
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B.
4 m/s²
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C.
2 m/s²
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D.
1 m/s²
Solution
Net force = applied force - friction = 12 N - 3 N = 9 N. Acceleration a = F/m = 9 N / 3 kg = 3 m/s².
Correct Answer: B — 4 m/s²
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Q. A 3 kg block is sliding down a frictionless incline of 30 degrees. What is the acceleration of the block?
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A.
3.9 m/s²
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B.
4.9 m/s²
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C.
9.8 m/s²
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D.
1.5 m/s²
Solution
The acceleration down the incline is given by a = g * sin(θ) = 9.8 m/s² * sin(30°) = 4.9 m/s².
Correct Answer: A — 3.9 m/s²
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Q. A 3 kg block is sliding on a frictionless surface with a velocity of 4 m/s. What is the momentum of the block?
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A.
12 kg·m/s
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B.
8 kg·m/s
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C.
6 kg·m/s
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D.
4 kg·m/s
Solution
Momentum p = mv = 3 kg * 4 m/s = 12 kg·m/s.
Correct Answer: A — 12 kg·m/s
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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?
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A.
1 m/s
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B.
2 m/s
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C.
3 m/s
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D.
4 m/s
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?
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A.
3 m/s
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B.
6 m/s
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C.
9 m/s
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D.
12 m/s
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)
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A.
2 m/s²
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B.
3 m/s²
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C.
4 m/s²
-
D.
5 m/s²
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
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C.
4 m/s²
-
D.
5 m/s²
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?
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A.
5 m/s²
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B.
10 m/s²
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C.
15 m/s²
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D.
20 m/s²
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
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B.
0 m/s²
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C.
3 m/s²
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D.
12 m/s²
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?
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A.
30 J
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B.
36 J
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C.
60 J
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D.
120 J
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 the potential energy at the top?
-
A.
30 J
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B.
45 J
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C.
60 J
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D.
75 J
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?
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A.
15 J
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B.
30 J
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C.
45 J
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D.
60 J
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
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B.
9 N
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C.
30 N
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D.
0 N
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?
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A.
12 J
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B.
24 J
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C.
36 J
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D.
48 J
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)
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A.
147 J
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B.
294 J
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C.
441 J
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D.
588 J
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?
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A.
6 N
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B.
12 N
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C.
24 N
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D.
36 N
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?
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A.
0 N
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B.
3 N
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C.
15 N
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D.
5 N
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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