Physics Syllabus (JEE Main)
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 in free fall. What is the force acting on it due to gravity?
A.
6 N
B.
60 N
C.
600 N
D.
0 N
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Solution
Force due to gravity 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 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 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 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 30 degrees with a speed of 30 m/s. What is the time of flight until it hits the ground?
A.
3 s
B.
4 s
C.
5 s
D.
6 s
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Solution
Time of flight (T) = (2u * sin(θ))/g = (2 * 30 * 0.5)/9.8 ≈ 3.06 s.
Correct Answer: C — 5 s
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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
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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: 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
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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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Q. A ball is thrown downwards with an initial velocity 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
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Solution
Using the equation of motion: h = ut + 0.5gt². 20 = 5t + 0.5 * 10 * t². This simplifies to 5t + 5t² - 20 = 0. Solving gives t = 2 s.
Correct Answer: B — 3 s
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Q. A ball is thrown horizontally from the top of a cliff 45 m high. How far from the base of the cliff will it land if it is thrown with a speed of 10 m/s?
A.
20 m
B.
30 m
C.
40 m
D.
50 m
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Solution
Time to fall = sqrt(2h/g) = sqrt(2*45/9.8) ≈ 3.03 s. Horizontal distance = speed * time = 10 * 3.03 ≈ 30.3 m, approximately 30 m.
Correct Answer: B — 30 m
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Q. A ball is thrown horizontally from the top of a cliff 80 m high. How far from the base of the cliff will it land? (Assume g = 10 m/s² and horizontal speed = 20 m/s)
A.
20 m
B.
40 m
C.
60 m
D.
80 m
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Solution
Time to fall = √(2h/g) = √(2*80/10) = 4 s. Horizontal distance = speed * time = 20 m/s * 4 s = 80 m.
Correct Answer: C — 60 m
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Q. A ball is thrown horizontally from the top of a cliff 80 m high. How far from the base of the cliff will it land? (Assume g = 10 m/s² and initial horizontal speed = 20 m/s)
A.
40 m
B.
60 m
C.
80 m
D.
100 m
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Solution
Time to fall = √(2h/g) = √(2*80/10) = 4 s. Horizontal distance = speed * time = 20 * 4 = 80 m.
Correct Answer: C — 80 m
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Q. A ball is thrown horizontally from the top of a cliff with a speed of 15 m/s. If the cliff is 45 m high, how far from the base of the cliff will the ball land?
A.
30 m
B.
45 m
C.
60 m
D.
75 m
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Solution
Time to fall = √(2h/g) = √(2*45/10) = 3 s. Horizontal distance = speed * time = 15 m/s * 3 s = 45 m.
Correct Answer: C — 60 m
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Q. A ball is thrown upwards with a speed of 20 m/s. How high will it go before it starts to fall back?
A.
10 m
B.
20 m
C.
30 m
D.
40 m
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Solution
Maximum height H = (u²)/(2g) = (20²)/(2 * 9.8) ≈ 20.4 m.
Correct Answer: B — 20 m
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Q. A ball is thrown upwards with a speed of 20 m/s. How high will it rise before coming to rest?
A.
10 m
B.
20 m
C.
30 m
D.
40 m
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Solution
Maximum height (H) = (u²)/(2g) = (20²)/(2*9.8) ≈ 20.4 m.
Correct Answer: B — 20 m
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Q. A ball is thrown upwards with a speed of 20 m/s. How high will it rise before coming to a stop?
A.
10 m
B.
20 m
C.
30 m
D.
40 m
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Solution
Maximum height (H) = (u²)/(2g) = (20²)/(2*9.8) ≈ 20.4 m.
Correct Answer: C — 30 m
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Q. A ball is thrown upwards with a speed of 20 m/s. How long will it take to reach the maximum height?
A.
2 s
B.
3 s
C.
4 s
D.
5 s
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Solution
Time to reach maximum height (t) = u/g = 20/9.8 ≈ 2.04 s.
Correct Answer: A — 2 s
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Q. A ball is thrown vertically upward with a speed of 20 m/s. How high will it rise before coming to rest? (g = 10 m/s²)
A.
20 m
B.
30 m
C.
40 m
D.
50 m
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Solution
Using the formula v² = u² + 2as, where v = 0, u = 20 m/s, and a = -g = -10 m/s², we get 0 = (20)² + 2(-10)s, solving gives s = 20 m.
Correct Answer: B — 30 m
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Q. A ball is thrown vertically upward with a speed of 20 m/s. How high will it rise before coming to rest?
A.
20 m
B.
40 m
C.
10 m
D.
80 m
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Solution
Using the formula v² = u² + 2as, where v = 0, u = 20 m/s, a = -9.8 m/s². 0 = (20)² + 2(-9.8)s, solving gives s = 20.4 m, approximately 40 m.
Correct Answer: B — 40 m
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Q. A ball is thrown vertically upward with a speed of 20 m/s. How long will it take to reach the maximum height?
A.
1 s
B.
2 s
C.
3 s
D.
4 s
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Solution
Using v = u - gt, where v = 0, u = 20 m/s, and g = 9.8 m/s², we find t = u/g = 20/9.8 ≈ 2.04 s.
Correct Answer: B — 2 s
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Q. A ball is thrown vertically upward 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 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 ball is thrown vertically upward with an initial velocity of 20 m/s. How high will it rise before coming to a momentary stop? (g = 10 m/s²)
A.
20 m
B.
30 m
C.
40 m
D.
50 m
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Solution
Using the formula v² = u² + 2as, where v = 0, u = 20 m/s, and a = -g = -10 m/s², we get 0 = (20)² + 2(-10)s, solving gives s = 20 m.
Correct Answer: B — 30 m
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