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Q. A 5 kg object is moving with a velocity of 3 m/s. What is its kinetic energy?
A.
22.5 J
B.
30 J
C.
45 J
D.
60 J
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Solution
Kinetic Energy = 0.5 × mass × velocity² = 0.5 × 5 kg × (3 m/s)² = 22.5 J.
Correct Answer: A — 22.5 J
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Q. A 5 kg object is pulled with a force of 25 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 = 25 N / 5 kg = 5 m/s².
Correct Answer: C — 4 m/s²
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Q. A 5 kg object is pushed with a force of 15 N. If the frictional force is 5 N, what is the acceleration of the object?
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 = 15 N - 5 N = 10 N. Acceleration = F/m = 10 N / 5 kg = 2 m/s².
Correct Answer: B — 2 m/s²
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Q. A 5 kg object is subjected to a net force of 15 N. What is its 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 = 15 N / 5 kg = 3 m/s².
Correct Answer: B — 3 m/s²
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Q. A 5 kg object is subjected to a net force of 15 N. 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 = 15 N / 5 kg = 3 m/s².
Correct Answer: B — 3 m/s²
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Q. A 5 kg object is subjected to a net force of 25 N. What is its acceleration?
A.
2 m/s²
B.
5 m/s²
C.
10 m/s²
D.
15 m/s²
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Solution
Using F = ma, a = F/m = 25 N / 5 kg = 5 m/s².
Correct Answer: C — 10 m/s²
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Q. A 5 kg object is thrown vertically upward with a speed of 20 m/s. What is the maximum height reached by the object? (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) = (20 m/s)² / (2 * 10 m/s²) = 20 m.
Correct Answer: B — 30 m
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Q. A 5 ohm resistor and a 10 ohm resistor are connected in series. What is the total resistance?
A.
15 ohms
B.
5 ohms
C.
10 ohms
D.
2 ohms
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Solution
In series, the total resistance is R_total = R1 + R2 = 5 + 10 = 15 ohms.
Correct Answer: A — 15 ohms
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Q. A 50 kg crate is at rest on a flat surface. What is the normal force acting on the crate?
A.
0 N
B.
50 N
C.
500 N
D.
1000 N
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Solution
The normal force equals the weight of the crate, which is N = mg = 50 kg * 10 m/s² = 500 N.
Correct Answer: C — 500 N
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Q. A 50 kg object is at rest on a surface. What is the normal force acting on it?
A.
0 N
B.
50 N
C.
100 N
D.
500 N
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Solution
The normal force equals the weight of the object, which is F = mg = 50 kg * 9.8 m/s² = 490 N.
Correct Answer: C — 100 N
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Q. A 50 kg object is moving with a constant velocity. What can be said about the net force acting on it?
A.
It is zero
B.
It is equal to its weight
C.
It is equal to the applied force
D.
It is maximum
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Solution
If the object is moving with constant velocity, the net force acting on it is zero.
Correct Answer: A — It is zero
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Q. A 50 kg object is moving with a velocity of 10 m/s. What is its momentum?
A.
500 kg m/s
B.
1000 kg m/s
C.
1500 kg m/s
D.
2000 kg m/s
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Solution
Momentum = mass × velocity = 50 kg × 10 m/s = 500 kg m/s.
Correct Answer: B — 1000 kg m/s
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Q. A 50 kg object is pulled with a force of 200 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, we have a = F/m = 200 N / 50 kg = 4 m/s².
Correct Answer: D — 5 m/s²
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Q. A 50 kg object is pushed with a force of 200 N over a distance of 10 m. What is the work done? (2000)
A.
2000 J
B.
1000 J
C.
500 J
D.
3000 J
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Solution
Work Done = Force * Distance = 200 N * 10 m = 2000 J
Correct Answer: A — 2000 J
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Q. A 6 kg object is at rest on a horizontal surface. If a horizontal force of 12 N is applied, what is the acceleration assuming 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, acceleration a = F/m = 12 N / 6 kg = 2 m/s².
Correct Answer: B — 2 m/s²
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Q. A 6 kg object is dropped from a height. What is the force acting on it just before it hits the ground?
A.
6 N
B.
60 N
C.
12 N
D.
0 N
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Solution
The force acting on it is its weight, F = mg = 6 kg * 10 m/s² = 60 N.
Correct Answer: B — 60 N
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Q. A 6 kg object is moving with a velocity of 3 m/s. What is its kinetic energy?
A.
27 J
B.
36 J
C.
54 J
D.
18 J
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Solution
Kinetic energy KE = 0.5 * m * v² = 0.5 * 6 kg * (3 m/s)² = 27 J.
Correct Answer: B — 36 J
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Q. A 6 kg object is subjected to a net force of 12 N. What is its 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, acceleration a = F/m = 12 N / 6 kg = 2 m/s².
Correct Answer: B — 3 m/s²
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Q. A 60W bulb is connected to a 120V supply. What is the resistance of the bulb?
A.
240 ohms
B.
120 ohms
C.
60 ohms
D.
30 ohms
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Solution
Using P = V^2 / R, we rearrange to find R = V^2 / P = (120V)^2 / 60W = 240 ohms.
Correct Answer: B — 120 ohms
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Q. A bag contains 3 red and 2 blue balls. If one ball is drawn at random, what is the probability that it is red given that it is not blue?
A.
1/2
B.
3/5
C.
2/5
D.
3/4
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Solution
The total number of balls that are not blue is 3 (red). The probability of drawing a red ball given that it is not blue is 3/5.
Correct Answer: B — 3/5
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Q. A bag contains 3 red balls and 2 blue balls. If one ball is drawn at random, what is the probability that it is red?
A.
1/5
B.
2/5
C.
3/5
D.
4/5
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Solution
The total number of balls is 3 + 2 = 5. The probability of drawing a red ball is 3/5.
Correct Answer: C — 3/5
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Q. A ball is dropped from a height of 80 m. How long will it take to reach the ground?
A.
4 s
B.
5 s
C.
6 s
D.
8 s
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Solution
Using the formula: h = 0.5 * g * t², where h = 80 m and g = 9.8 m/s². Rearranging gives t² = (2 * h) / g = (2 * 80) / 9.8 ≈ 16.33, so t ≈ 4.03 s.
Correct Answer: C — 6 s
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Q. A ball is dropped from a height of 80 m. How long will it take to reach the ground? (Assume g = 10 m/s²)
A.
4 s
B.
5 s
C.
6 s
D.
8 s
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Solution
Using the formula: h = 0.5 * g * t^2. 80 = 0.5 * 10 * t^2. Solving gives t^2 = 16, so t = 4 s.
Correct Answer: B — 5 s
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Q. A ball is swung in a vertical circle. At the highest point of the circle, what is the condition for the ball to just maintain its circular motion?
A.
Weight must be greater than tension
B.
Tension must be zero
C.
Centripetal force must be zero
D.
Weight must be less than tension
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Solution
At the highest point, the centripetal force is provided by the weight of the ball. For just maintaining motion, tension can be zero.
Correct Answer: B — Tension must be zero
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Q. A ball is thrown at an angle of 45 degrees with an initial speed of 14 m/s. What is the range of the projectile?
A.
10 m
B.
14 m
C.
20 m
D.
28 m
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Solution
Range (R) = (u² * sin(2θ)) / g = (14² * 1) / 9.8 ≈ 20 m.
Correct Answer: D — 28 m
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Q. A ball is thrown at an angle of 45 degrees with an initial speed of 28 m/s. What is the vertical component of the velocity at the peak of its trajectory?
A.
0 m/s
B.
14 m/s
C.
20 m/s
D.
28 m/s
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Solution
At the peak, the vertical component of velocity is 0 m/s.
Correct Answer: A — 0 m/s
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Q. A ball is thrown at an angle of 60 degrees with a speed of 15 m/s. What is the horizontal range of the ball?
A.
20 m
B.
30 m
C.
40 m
D.
50 m
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Solution
Range (R) = (u^2 * sin(2θ)) / g = (15^2 * sin(120)) / 9.8 = 38.5 m.
Correct Answer: C — 40 m
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Q. A ball is thrown downward with an initial speed of 10 m/s from a height of 20 m. How long will it take to hit the ground? (Assume g = 10 m/s²)
A.
2 s
B.
3 s
C.
4 s
D.
5 s
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Solution
Using the equation of motion: h = ut + 0.5gt². 20 = 10t + 0.5 * 10 * t². Rearranging gives 5t² + 10t - 20 = 0. Solving this quadratic gives t = 2 s.
Correct Answer: C — 4 s
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Q. A ball is thrown downward with an initial speed of 10 m/s from a height of 20 m. How long will it take to hit the ground? (g = 10 m/s²)
A.
2 s
B.
3 s
C.
4 s
D.
5 s
Show solution
Solution
Using the equation of motion: h = ut + 0.5gt². 20 = 10t + 0.5 * 10 * t². Rearranging gives 5t² + 10t - 20 = 0. Solving this quadratic gives t = 2 s.
Correct Answer: B — 3 s
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Q. A ball is thrown downward with an initial speed of 5 m/s from a height of 20 m. How long will it take to hit the ground? (g = 10 m/s²)
A.
2 s
B.
3 s
C.
4 s
D.
5 s
Show solution
Solution
Using the equation of motion: h = ut + 0.5gt². 20 = 5t + 0.5 * 10 * t². Rearranging gives 5t + 5t² - 20 = 0. Solving the quadratic gives t = 2 s.
Correct Answer: B — 3 s
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