Physics Syllabus (JEE Main)
Q. A 1000 kg car accelerates from rest to a speed of 20 m/s in 10 seconds. What is the average power exerted by the engine?
-
A.
2000 W
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B.
4000 W
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C.
5000 W
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D.
6000 W
Solution
First, calculate the work done: W = (1/2)mv^2 = (1/2)(1000 kg)(20 m/s)^2 = 200000 J. Then, power is P = W/t = 200000 J / 10 s = 20000 W.
Correct Answer: C — 5000 W
Q. A 1000 W heater operates for 1 hour. How much energy does it consume?
-
A.
3600 J
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B.
1000 J
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C.
3600000 J
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D.
100000 J
Solution
Energy consumed is E = P * t. Here, P = 1000 W and t = 1 hour = 3600 seconds. Thus, E = 1000 W * 3600 s = 3600000 J.
Correct Answer: C — 3600000 J
Q. A 1000 W heater operates for 3 hours. How much energy does it consume?
-
A.
3000 J
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B.
1080000 J
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C.
3600000 J
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D.
1000 J
Solution
Energy consumed is E = P * t. Here, P = 1000 W and t = 3 hours = 10800 seconds. Thus, E = 1000 W * 10800 s = 10800000 J.
Correct Answer: C — 3600000 J
Q. A 12 kg object is at rest on a horizontal surface. What is the normal force acting on it?
-
A.
0 N
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B.
12 N
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C.
120 N
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D.
100 N
Solution
The normal force equals the weight of the object: N = mg = 12 kg * 10 m/s² = 120 N.
Correct Answer: C — 120 N
Q. A 15 kg cart is pushed with a force of 60 N. If the frictional force opposing the motion is 15 N, what is the acceleration of the cart?
-
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
Net force = 60 N - 15 N = 45 N. Using F = ma, a = F/m = 45 N / 15 kg = 3 m/s².
Correct Answer: B — 3 m/s²
Q. A 15 kg object is in equilibrium. What is the net force acting on it?
-
A.
0 N
-
B.
15 N
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C.
30 N
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D.
45 N
Solution
In equilibrium, the net force acting on the object is 0 N.
Correct Answer: A — 0 N
Q. A 15 kg object is pushed with a force of 45 N. What is the acceleration of the object?
-
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, acceleration a = F/m = 45 N / 15 kg = 3 m/s².
Correct Answer: B — 3 m/s²
Q. A 2 kg ball is thrown vertically upwards with a speed of 20 m/s. What is the maximum height it reaches?
-
A.
10 m
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B.
20 m
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C.
30 m
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D.
40 m
Solution
Using conservation of energy, KE at launch = PE at max height. 0.5mv² = mgh. Solving gives h = v²/(2g) = (20)²/(2*9.8) = 20.41 m.
Correct Answer: B — 20 m
Q. A 2 kg ball is thrown vertically upwards with a speed of 20 m/s. What is the maximum height it reaches? (g = 9.8 m/s²)
-
A.
20.4 m
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B.
30.4 m
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C.
40.4 m
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D.
50.4 m
Solution
Using conservation of energy, KE at the bottom = PE at the maximum height. 0.5mv² = mgh. Solving gives h = v²/(2g) = (20)²/(2 * 9.8) = 20.4 m.
Correct Answer: B — 30.4 m
Q. A 2 kg block is sliding down a frictionless incline of 30 degrees. What is the acceleration of the block?
-
A.
4.9 m/s²
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B.
9.8 m/s²
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C.
3.9 m/s²
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D.
1.96 m/s²
Solution
The acceleration a = g sin(θ) = 9.8 m/s² * sin(30°) = 9.8 m/s² * 0.5 = 4.9 m/s².
Correct Answer: A — 4.9 m/s²
Q. A 2 kg object is dropped from a height of 10 m. What is its potential energy at the top?
-
A.
20 J
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B.
40 J
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C.
60 J
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D.
80 J
Solution
Potential Energy = mass × g × height = 2 kg × 9.8 m/s² × 10 m = 196 J.
Correct Answer: B — 40 J
Q. A 2 kg object is dropped from a height of 10 m. What is the speed of the object just before it hits the ground? (g = 9.8 m/s²)
-
A.
10 m/s
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B.
14 m/s
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C.
20 m/s
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D.
30 m/s
Solution
Using energy conservation, mgh = 0.5mv²; v = sqrt(2gh) = sqrt(2 × 9.8 m/s² × 10 m) = 14 m/s.
Correct Answer: B — 14 m/s
Q. A 2 kg object is dropped from a height of 20 m. What is its potential energy at the top? (g = 9.8 m/s²)
-
A.
39.2 J
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B.
196 J
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C.
78.4 J
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D.
98.0 J
Solution
Potential Energy (PE) = m × g × h = 2 kg × 9.8 m/s² × 20 m = 392 J.
Correct Answer: B — 196 J
Q. A 2 kg object is dropped from a height of 5 m. What is its potential energy at the top?
-
A.
10 J
-
B.
20 J
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C.
30 J
-
D.
40 J
Solution
Potential Energy = mass × g × height = 2 kg × 9.8 m/s² × 5 m = 98 J.
Correct Answer: B — 20 J
Q. A 2 kg object is dropped from a height of 5 m. What is the potential energy at the height?
-
A.
10 J
-
B.
20 J
-
C.
30 J
-
D.
40 J
Solution
Potential energy = mgh = 2 kg × 9.8 m/s² × 5 m = 98 J.
Correct Answer: B — 20 J
Q. A 2 kg object is dropped from a height of 5 m. What is the potential energy at the top? (g = 9.8 m/s²)
-
A.
19.6 J
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B.
39.2 J
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C.
49 J
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D.
98 J
Solution
Potential energy = mgh = 2 kg × 9.8 m/s² × 5 m = 98 J.
Correct Answer: B — 39.2 J
Q. A 2 kg object is in free fall. What is the net force acting on it?
-
A.
0 N
-
B.
2 N
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C.
20 N
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D.
19.6 N
Solution
The net force is equal to the weight of the object: F = mg = 2 kg * 9.8 m/s² = 19.6 N.
Correct Answer: C — 20 N
Q. A 2 kg object is moving with a speed of 3 m/s. What is its kinetic energy?
-
A.
6 J
-
B.
9 J
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C.
12 J
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D.
18 J
Solution
Kinetic energy = 0.5 × m × v² = 0.5 × 2 kg × (3 m/s)² = 9 J.
Correct Answer: B — 9 J
Q. A 2 kg object is moving with a speed of 3 m/s. What is its total mechanical energy?
-
A.
9 J
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B.
12 J
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C.
15 J
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D.
18 J
Solution
Total mechanical energy = kinetic energy + potential energy. KE = 0.5mv² = 0.5 * 2 * (3)² = 9 J.
Correct Answer: B — 12 J
Q. A 2 kg object is moving with a velocity of 3 m/s. What is its momentum?
-
A.
3 kg·m/s
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B.
6 kg·m/s
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C.
9 kg·m/s
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D.
12 kg·m/s
Solution
Momentum = mass × velocity = 2 kg × 3 m/s = 6 kg·m/s.
Correct Answer: B — 6 kg·m/s
Q. A 2 kg object is pulled with a force of 8 N. What is the acceleration of the object?
-
A.
2 m/s²
-
B.
4 m/s²
-
C.
6 m/s²
-
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
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?
-
A.
8 kg·m/s
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B.
2 kg·m/s
-
C.
4 kg·m/s
-
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
Q. A 2000 W heater operates for 3 hours. How much energy does it consume in kilowatt-hours?
-
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
Q. A 25 kg object is subjected to a force of 50 N. What is its acceleration?
-
A.
1 m/s²
-
B.
2 m/s²
-
C.
3 m/s²
-
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
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?
-
A.
2 m/s²
-
B.
3 m/s²
-
C.
5 m/s²
-
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
Q. A 3 kg block is sliding down a frictionless incline of 30 degrees. What is the acceleration of the block?
-
A.
3.9 m/s²
-
B.
4.9 m/s²
-
C.
9.8 m/s²
-
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
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?
-
A.
12 kg·m/s
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B.
8 kg·m/s
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C.
6 kg·m/s
-
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
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
Solution
Using F = ma, a = F/m = 12 N / 3 kg = 4 m/s².
Correct Answer: C — 4 m/s²
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
Solution
Using F = ma, a = F/m = 15 N / 3 kg = 5 m/s².
Correct Answer: A — 5 m/s²
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
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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