Physics Syllabus (JEE Main)
Q. A 1 kg ball is thrown upwards with a force of 10 N. What is the net force acting on the ball at the peak of its motion?
A.
0 N
B.
10 N
C.
5 N
D.
20 N
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Solution
At the peak, the only force acting on the ball is its weight (10 N down), so the net force is 0 N.
Correct Answer: A — 0 N
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Q. A 1 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 = 0.5 mv², 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 1 kg ball is thrown vertically upwards with a speed of 10 m/s. What is the maximum height it reaches? (g = 10 m/s²)
A.
5 m
B.
10 m
C.
15 m
D.
20 m
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Solution
Using energy conservation, initial kinetic energy = mgh. 0.5 × 1 kg × (10 m/s)² = 1 kg × 10 m/s² × h. h = 5 m.
Correct Answer: B — 10 m
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Q. A 1 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 energy conservation: KE_initial = PE_max; 0.5 × m × v² = mgh; h = v²/(2g) = (20 m/s)²/(2 × 9.8 m/s²) = 20.4 m.
Correct Answer: B — 20 m
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Q. A 1 kg ball is thrown vertically upwards with a speed of 20 m/s. What is the maximum height it reaches? (g = 10 m/s²)
A.
20 m
B.
30 m
C.
40 m
D.
50 m
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Solution
Using energy conservation: Initial K.E. = Potential Energy at max height. 0.5mv² = mgh. h = v²/(2g) = (20 m/s)² / (2 × 10 m/s²) = 20 m.
Correct Answer: B — 30 m
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Q. A 1 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.
40.8 m
D.
50.0 m
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Solution
Using energy conservation, initial kinetic energy = mgh; 0.5 × 1 kg × (20 m/s)² = 9.8 m × h; h = 20.4 m.
Correct Answer: A — 20.4 m
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Q. A 1 kg ball is thrown vertically upwards with a speed of 20 m/s. What is the maximum height it reaches? (Assume 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), h = (20 m/s)² / (2 * 10 m/s²) = 400 / 20 = 20 m.
Correct Answer: B — 30 m
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Q. A 1 kg mass is attached to a spring and compressed by 0.2 m. If the spring constant is 100 N/m, what is the potential energy stored in the spring?
A.
2 J
B.
4 J
C.
6 J
D.
8 J
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Solution
Potential energy in a spring is given by PE = 0.5kx². Thus, PE = 0.5 * 100 * (0.2)² = 2 J.
Correct Answer: B — 4 J
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Q. A 1 kg mass is attached to a spring with a spring constant of 200 N/m. What is the maximum extension of the spring when the mass is released from rest?
A.
0.5 m
B.
1 m
C.
2 m
D.
0.25 m
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Solution
Using Hooke's law, F = kx, where F = mg = 1 kg * 9.8 m/s² = 9.8 N. Thus, x = F/k = 9.8 N / 200 N/m = 0.049 m.
Correct Answer: A — 0.5 m
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Q. A 1 kg mass is attached to a spring with a spring constant of 200 N/m. What is the maximum force exerted by the spring when it is compressed by 0.1 m?
A.
2 N
B.
5 N
C.
20 N
D.
10 N
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Solution
Using Hooke's law, F = kx = 200 N/m * 0.1 m = 20 N.
Correct Answer: C — 20 N
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Q. A 1 kg mass is attached to a spring with a spring constant of 200 N/m. What is the force exerted by the spring when it is compressed by 0.1 m?
A.
2 N
B.
5 N
C.
10 N
D.
20 N
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Solution
Using Hooke's Law, F = kx = 200 N/m * 0.1 m = 20 N.
Correct Answer: C — 10 N
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Q. A 1 kg mass is dropped from a height of 1 m. What is its speed just before it hits the ground?
A.
1 m/s
B.
2 m/s
C.
3 m/s
D.
4 m/s
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Solution
Using conservation of energy, v = √(2gh) = √(2 * 9.8 * 1) = 4.43 m/s.
Correct Answer: B — 2 m/s
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Q. A 1 kg mass is dropped from a height of 1 m. What is the speed just before it hits the ground?
A.
1 m/s
B.
2 m/s
C.
3 m/s
D.
4 m/s
Show solution
Solution
Using conservation of energy, v = sqrt(2gh) = sqrt(2 * 9.8 * 1) = 4.43 m/s.
Correct Answer: B — 2 m/s
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Q. A 1 kg mass is dropped from a height of 10 m. What is its speed just before it hits the ground?
A.
5 m/s
B.
10 m/s
C.
14 m/s
D.
20 m/s
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Solution
Using conservation of energy, v = √(2gh) = √(2 * 9.8 * 10) = 14 m/s.
Correct Answer: C — 14 m/s
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Q. A 1 kg mass is dropped from a height of 10 m. What is the speed just before it hits the ground?
A.
5 m/s
B.
10 m/s
C.
15 m/s
D.
20 m/s
Show solution
Solution
Using conservation of energy, potential energy at height = kinetic energy just before hitting the ground. mgh = 0.5mv^2. Solving gives v = sqrt(2gh) = sqrt(2*9.8*10) = 14 m/s.
Correct Answer: B — 10 m/s
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Q. A 1 kg mass is lifted to a height of 5 m. How much work is done against gravity?
A.
5 J
B.
10 J
C.
15 J
D.
20 J
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Solution
Work done against gravity = mgh = 1 * 9.8 * 5 = 49 J.
Correct Answer: B — 10 J
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Q. A 1 kg mass is lifted vertically 10 m. How much work is done against gravity? (g = 9.8 m/s²)
A.
9.8 J
B.
19.6 J
C.
29.4 J
D.
39.2 J
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Solution
Work done = mgh = 1 kg × 9.8 m/s² × 10 m = 98 J.
Correct Answer: B — 19.6 J
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Q. A 1 kg object is pushed with a force of 10 N over a distance of 3 m. What is the work done?
A.
10 J
B.
20 J
C.
30 J
D.
40 J
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Solution
Work done = Force × Distance = 10 N × 3 m = 30 J.
Correct Answer: C — 30 J
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Q. A 1 kg object is pushed with a force of 10 N. If the frictional force is 4 N, what is the net force acting on the object?
A.
6 N
B.
10 N
C.
4 N
D.
0 N
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Solution
Net force = applied force - frictional force = 10 N - 4 N = 6 N.
Correct Answer: A — 6 N
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Q. A 10 kg block is at rest on a horizontal surface. If a horizontal force of 30 N is applied, what will be 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 = 30 N / 10 kg = 3 m/s².
Correct Answer: C — 3 m/s²
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Q. A 10 kg block is at rest on a horizontal surface. If a horizontal force of 30 N is applied, what will be its acceleration? (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, a = F/m = 30 N / 10 kg = 3 m/s².
Correct Answer: C — 3 m/s²
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Q. A 10 kg block is at rest on a horizontal surface. If a horizontal force of 30 N is applied, what will be 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 F = ma, acceleration a = F/m = 30 N / 10 kg = 3 m/s².
Correct Answer: C — 3 m/s²
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Q. A 10 kg object falls freely from a height of 20 m. What is its potential energy at the top?
A.
100 J
B.
200 J
C.
300 J
D.
400 J
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Solution
Potential energy is given by PE = mgh = 10 * 9.8 * 20 = 1960 J.
Correct Answer: B — 200 J
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Q. A 10 kg object falls freely from a height of 20 m. What is its speed just before hitting the ground? (g = 9.8 m/s²)
A.
10 m/s
B.
14 m/s
C.
20 m/s
D.
28 m/s
Show solution
Solution
Using conservation of energy, v = √(2gh) = √(2 * 9.8 * 20) = 28 m/s.
Correct Answer: D — 28 m/s
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Q. A 10 kg object falls from a height of 20 m. What is its potential energy at the top?
A.
100 J
B.
200 J
C.
300 J
D.
400 J
Show solution
Solution
Potential energy is given by PE = mgh = 10 * 9.8 * 20 = 1960 J.
Correct Answer: D — 400 J
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Q. A 10 kg object falls from a height of 20 m. What is its speed just before it hits the ground?
A.
10 m/s
B.
14 m/s
C.
20 m/s
D.
28 m/s
Show solution
Solution
Using conservation of energy, v = √(2gh) = √(2 * 9.8 * 20) = 20 m/s.
Correct Answer: C — 20 m/s
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Q. A 10 kg object falls from a height of 20 m. What is its speed just before it hits the ground? (g = 9.8 m/s²)
A.
10 m/s
B.
14 m/s
C.
20 m/s
D.
28 m/s
Show solution
Solution
Using conservation of energy, v = √(2gh) = √(2 * 9.8 * 20) = 19.8 m/s.
Correct Answer: B — 14 m/s
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Q. A 10 kg object is at rest on a horizontal surface. If a force of 30 N is applied, what is the object's acceleration assuming a frictional force of 10 N?
A.
2 m/s²
B.
3 m/s²
C.
4 m/s²
D.
5 m/s²
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Solution
Net force = 30 N - 10 N = 20 N. Using F = ma, a = F/m = 20 N / 10 kg = 2 m/s².
Correct Answer: B — 3 m/s²
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Q. A 10 kg object is at rest on a surface. If a force of 30 N is applied, what is the object's acceleration assuming the friction is negligible?
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 = 30 N / 10 kg = 3 m/s².
Correct Answer: C — 3 m/s²
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Q. A 10 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
Show solution
Solution
Using conservation of energy: Potential Energy at height = Kinetic Energy just before hitting ground. mgh = 0.5mv². v = √(2gh) = √(2 × 9.8 m/s² × 20 m) = 19.8 m/s.
Correct Answer: B — 19.8 m/s
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