Physics Syllabus (JEE Main)
Q. A 1000 W heater operates for 3 hours. How much energy does it consume?
A.
3000 J
B.
1080000 J
C.
3600000 J
D.
1000 J
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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
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Q. A 12 kg box is pulled with a force of 60 N at an angle of 30 degrees to the horizontal. What is the horizontal component of the force?
A.
30 N
B.
60 N
C.
51.96 N
D.
20 N
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Solution
The horizontal component is F_horizontal = F * cos(θ) = 60 N * cos(30°) = 60 N * (√3/2) = 51.96 N.
Correct Answer: C — 51.96 N
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Q. A 12 kg object is at rest on a horizontal surface. What is the normal force acting on it?
A.
0 N
B.
12 N
C.
120 N
D.
100 N
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Solution
The normal force equals the weight of the object: N = mg = 12 kg * 10 m/s² = 120 N.
Correct Answer: C — 120 N
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Q. A 12 kg object is in free fall. What is the force acting on it due to gravity?
A.
12 N
B.
24 N
C.
36 N
D.
48 N
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Solution
The force due to gravity is F = mg = 12 kg * 9.8 m/s² = 117.6 N (approximately 120 N).
Correct Answer: B — 24 N
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Q. A 12 kg object is moving with a velocity of 5 m/s. What is its kinetic energy?
A.
150 J
B.
120 J
C.
100 J
D.
75 J
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Solution
Kinetic energy KE = 0.5 * m * v² = 0.5 * 12 kg * (5 m/s)² = 150 J.
Correct Answer: B — 120 J
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Q. A 12 kg object is moving with a velocity of 5 m/s. What is its momentum?
A.
30 kg·m/s
B.
60 kg·m/s
C.
90 kg·m/s
D.
120 kg·m/s
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Solution
Momentum p = mv = 12 kg * 5 m/s = 60 kg·m/s.
Correct Answer: B — 60 kg·m/s
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Q. A 12 kg object is pulled with a force of 24 N. If the object experiences a frictional force of 4 N, what is its acceleration?
A.
1.67 m/s²
B.
2 m/s²
C.
2.5 m/s²
D.
3 m/s²
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Solution
Net force = 24 N - 4 N = 20 N. Acceleration a = F/m = 20 N / 12 kg = 1.67 m/s².
Correct Answer: B — 2 m/s²
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Q. A 15 kg box is pushed with a force of 30 N. If the frictional force opposing the motion is 10 N, what is the acceleration of the box?
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 = 30 N - 10 N = 20 N. Acceleration a = F/m = 20 N / 15 kg = 1.33 m/s².
Correct Answer: B — 2 m/s²
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Q. A 15 kg box is pushed with a force of 45 N. What is the acceleration of the box? (Assume no friction)
A.
1 m/s²
B.
2 m/s²
C.
3 m/s²
D.
4 m/s²
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Solution
Using F = ma, we have a = F/m = 45 N / 15 kg = 3 m/s².
Correct Answer: C — 3 m/s²
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Q. A 15 kg box is pushed with a force of 60 N on a rough surface with a frictional force of 15 N. What is the box's acceleration?
A.
3 m/s²
B.
4 m/s²
C.
5 m/s²
D.
6 m/s²
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Solution
Net force = 60 N - 15 N = 45 N. Acceleration a = F/m = 45 N / 15 kg = 3 m/s².
Correct Answer: B — 4 m/s²
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Q. A 15 kg box is pushed with a force of 60 N. If the frictional force is 20 N, what is the acceleration of the box?
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 - friction = 60 N - 20 N = 40 N. Acceleration a = F/m = 40 N / 15 kg = 2.67 m/s².
Correct Answer: B — 3 m/s²
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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
B.
3 m/s²
C.
4 m/s²
D.
5 m/s²
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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
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Q. A 15 kg object is at rest on a horizontal surface. What is the normal force acting on it?
A.
0 N
B.
15 N
C.
150 N
D.
75 N
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Solution
Normal force = weight = mg = 15 kg * 10 m/s² = 150 N.
Correct Answer: C — 150 N
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Q. A 15 kg object is at rest on a table. What is the force exerted by the table on the object?
A.
0 N
B.
15 N
C.
150 N
D.
10 N
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Solution
The normal force equals the weight of the object: 15 kg * 10 m/s² = 150 N.
Correct Answer: C — 150 N
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Q. A 15 kg object is in equilibrium. What is the net force acting on it?
A.
0 N
B.
15 N
C.
30 N
D.
45 N
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Solution
In equilibrium, the net force acting on the object is 0 N.
Correct Answer: A — 0 N
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Q. A 15 kg object is pushed with a force of 45 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 = 45 N / 15 kg = 3 m/s².
Correct Answer: B — 3 m/s²
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Q. A 15 kg object is subjected to a net force of 45 N. What is its acceleration?
A.
1 m/s²
B.
2 m/s²
C.
3 m/s²
D.
4 m/s²
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Solution
Using F = ma, acceleration a = F/m = 45 N / 15 kg = 3 m/s².
Correct Answer: C — 3 m/s²
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Q. A 2 kg ball is thrown vertically upwards with a speed of 10 m/s. What is the maximum height it reaches?
A.
5 m
B.
10 m
C.
15 m
D.
20 m
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Solution
Using energy conservation: mgh = 0.5mv², h = v²/(2g) = (10 m/s)²/(2 × 9.8 m/s²) = 5.1 m.
Correct Answer: A — 5 m
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Q. A 2 kg ball is thrown vertically upwards with a speed of 15 m/s. What is the maximum height it reaches?
A.
5 m
B.
10 m
C.
15 m
D.
20 m
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Solution
Using conservation of energy, KE at the bottom = PE at the top. 0.5mv^2 = mgh. Solving gives h = v^2/(2g) = (15^2)/(2*9.8) = 11.47 m.
Correct Answer: B — 10 m
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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
B.
30.4 m
C.
40.4 m
D.
50.4 m
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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
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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
B.
20 m
C.
30 m
D.
40 m
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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
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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
B.
9.8 m/s²
C.
3.9 m/s²
D.
1.96 m/s²
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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
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Q. A 2 kg block is sliding on a frictionless surface. If a force of 6 N is applied, what is its acceleration?
A.
2 m/s²
B.
3 m/s²
C.
4 m/s²
D.
6 m/s²
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Solution
Using F = ma, acceleration a = F/m = 6 N / 2 kg = 3 m/s².
Correct Answer: B — 3 m/s²
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Q. A 2 kg object is dropped from a height of 10 m. What is its potential energy at the top?
A.
20 J
B.
40 J
C.
60 J
D.
80 J
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Solution
Potential Energy = mass × g × height = 2 kg × 9.8 m/s² × 10 m = 196 J.
Correct Answer: B — 40 J
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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
B.
14 m/s
C.
20 m/s
D.
30 m/s
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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
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Q. A 2 kg object is dropped from a height of 15 m. What is its speed just before it hits the ground? (g = 9.8 m/s²)
A.
17.15 m/s
B.
12.25 m/s
C.
14.14 m/s
D.
10.0 m/s
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Solution
Using conservation of energy, potential energy at height = kinetic energy just before hitting the ground. mgh = 0.5mv². Solving gives v = sqrt(2gh) = sqrt(2 * 9.8 * 15) = 17.15 m/s.
Correct Answer: A — 17.15 m/s
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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
B.
196 J
C.
78.4 J
D.
98.0 J
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Solution
Potential Energy (PE) = m × g × h = 2 kg × 9.8 m/s² × 20 m = 392 J.
Correct Answer: B — 196 J
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Q. A 2 kg object is dropped from a height of 20 m. What is its speed just before it hits the ground? (Assume g = 10 m/s²)
A.
10 m/s
B.
14 m/s
C.
20 m/s
D.
40 m/s
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Solution
Using conservation of energy, Potential Energy at height = Kinetic Energy just before hitting ground. mgh = 0.5 mv². v = √(2gh) = √(2 × 10 m/s² × 20 m) = 20 m/s.
Correct Answer: C — 20 m/s
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Q. A 2 kg object is dropped from a height of 20 m. What is its speed just before it hits the ground? (g = 9.8 m/s²)
A.
14 m/s
B.
19.8 m/s
C.
20 m/s
D.
28 m/s
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Solution
Using conservation of energy: Potential Energy at height = Kinetic Energy just before hitting ground. mgh = 0.5 mv². v = √(2gh) = √(2 × 9.8 m/s² × 20 m) = 19.8 m/s.
Correct Answer: B — 19.8 m/s
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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
C.
30 J
D.
40 J
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Solution
Potential Energy = mass × g × height = 2 kg × 9.8 m/s² × 5 m = 98 J.
Correct Answer: B — 20 J
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