Physics Syllabus (JEE Main)
Q. A car accelerates from rest to a speed of 30 m/s in 10 seconds. What is the distance covered by the car during this time?
A.
150 m
B.
300 m
C.
400 m
D.
600 m
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Solution
Using the formula d = ut + 0.5at², where u = 0, a = (30 m/s) / 10 s = 3 m/s², we get d = 0 + 0.5 * 3 * (10)² = 150 m.
Correct Answer: B — 300 m
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Q. A car accelerates from rest to a speed of 30 m/s. If the mass of the car is 800 kg, what is the work done on the car?
A.
360,000 J
B.
480,000 J
C.
600,000 J
D.
720,000 J
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Solution
Work done = Change in Kinetic Energy = 0.5 × mass × (final speed² - initial speed²) = 0.5 × 800 kg × (30 m/s)² = 360,000 J.
Correct Answer: B — 480,000 J
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Q. A car accelerates uniformly from rest to a speed of 20 m/s in 10 seconds. What is the distance covered by the car during this time?
A.
100 m
B.
200 m
C.
300 m
D.
400 m
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Solution
Using the formula: distance = initial velocity * time + 0.5 * acceleration * time^2. Here, initial velocity = 0, final velocity = 20 m/s, time = 10 s. Acceleration = (final velocity - initial velocity) / time = 2 m/s². Distance = 0 + 0.5 * 2 * 10² = 100 m.
Correct Answer: B — 200 m
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Q. A car accelerates uniformly from rest to a speed of 25 m/s in 10 seconds. What is the distance covered by the car during this time?
A.
100 m
B.
125 m
C.
150 m
D.
200 m
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Solution
Distance (s) = ut + (1/2)at^2. Here, a = (v-u)/t = (25-0)/10 = 2.5 m/s². So, s = 0 + (1/2)*2.5*10^2 = 125 m.
Correct Answer: B — 125 m
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Q. A car engine does 3000 J of work in 5 seconds. What is the average power output of the engine?
A.
600 W
B.
800 W
C.
1000 W
D.
1200 W
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Solution
Using the formula P = W/t, we find P = 3000 J / 5 s = 600 W.
Correct Answer: C — 1000 W
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Q. A car engine does 3000 J of work in 5 seconds. What is the power of the engine?
A.
600 W
B.
800 W
C.
1000 W
D.
1200 W
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Solution
Using the formula P = W/t, we have P = 3000 J / 5 s = 600 W.
Correct Answer: A — 600 W
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Q. A car is moving at 80 km/h and a motorcycle at 60 km/h in the same direction. What is the relative speed of the motorcycle with respect to the car?
A.
20 km/h
B.
60 km/h
C.
80 km/h
D.
140 km/h
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Solution
Relative speed = Speed of motorcycle - Speed of car = 60 km/h - 80 km/h = -20 km/h (20 km/h behind).
Correct Answer: A — 20 km/h
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Q. A car is moving at 80 km/h and a motorcycle is moving at 100 km/h in the same direction. What is the speed of the motorcycle relative to the car?
A.
20 km/h
B.
80 km/h
C.
100 km/h
D.
180 km/h
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Solution
Relative speed = Speed of motorcycle - Speed of car = 100 km/h - 80 km/h = 20 km/h.
Correct Answer: A — 20 km/h
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Q. A car is moving at 80 km/h and a motorcycle is moving at 100 km/h in the same direction. What is the relative speed of the motorcycle with respect to the car?
A.
20 km/h
B.
180 km/h
C.
100 km/h
D.
80 km/h
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Solution
Relative speed = Speed of motorcycle - Speed of car = 100 km/h - 80 km/h = 20 km/h.
Correct Answer: A — 20 km/h
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Q. A car is moving at 80 km/h and a motorcycle is moving at 60 km/h in the same direction. What is the relative speed of the motorcycle with respect to the car?
A.
20 km/h
B.
60 km/h
C.
80 km/h
D.
140 km/h
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Solution
Relative speed = Speed of motorcycle - Speed of car = 60 km/h - 80 km/h = -20 km/h (20 km/h behind).
Correct Answer: A — 20 km/h
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Q. A car is moving at 80 km/h and a motorcycle is moving at 60 km/h in the same direction. What is the speed of the motorcycle relative to the car?
A.
20 km/h
B.
60 km/h
C.
80 km/h
D.
140 km/h
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Solution
Relative speed = Speed of motorcycle - Speed of car = 60 km/h - 80 km/h = -20 km/h (20 km/h behind the car).
Correct Answer: A — 20 km/h
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Q. A car is moving in a circular path of radius 50 m with a constant speed of 20 m/s. What is the centripetal acceleration of the car?
A.
2 m/s²
B.
4 m/s²
C.
8 m/s²
D.
10 m/s²
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Solution
Centripetal acceleration (a_c) = v²/r = (20 m/s)² / (50 m) = 8 m/s².
Correct Answer: B — 4 m/s²
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Q. A car is moving in a circular path of radius 50 m with a speed of 15 m/s. What is the angular displacement after 10 seconds?
A.
1 rad
B.
2 rad
C.
3 rad
D.
4 rad
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Solution
Angular displacement θ = (v/r)t = (15/50)(10) = 3 rad.
Correct Answer: D — 4 rad
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Q. A car is moving in a circular track of radius 100 m at a speed of 20 m/s. What is the time period of one complete revolution?
A.
10 s
B.
20 s
C.
30 s
D.
40 s
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Solution
Circumference = 2πr = 2π(100) = 200π m. Time period (T) = Circumference / Speed = 200π / 20 = 10π s.
Correct Answer: B — 20 s
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Q. A car is moving in a circular track of radius 50 m with a speed of 15 m/s. What is the angular momentum of the car if its mass is 1000 kg? (2000)
A.
7500 kg m²/s
B.
10000 kg m²/s
C.
15000 kg m²/s
D.
20000 kg m²/s
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Solution
Angular momentum L = mvr = 1000 * 15 * 50 = 750000 kg m²/s.
Correct Answer: A — 7500 kg m²/s
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Q. A car is moving on a circular track of radius 100 m. If the maximum speed at which it can move without skidding is 20 m/s, what is the coefficient of friction between the tires and the road?
A.
0.1
B.
0.2
C.
0.3
D.
0.4
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Solution
The centripetal force required is provided by friction: F = mv^2/r. The frictional force is μmg. Setting them equal gives μ = v^2/(rg). Here, μ = (20^2)/(100*9.8) ≈ 0.4.
Correct Answer: C — 0.3
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Q. A car is negotiating a curve of radius 100 m at a speed of 15 m/s. What is the minimum coefficient of friction required to prevent the car from skidding?
A.
0.15
B.
0.25
C.
0.30
D.
0.35
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Solution
Frictional force = m * a_c; μmg = mv²/r; μ = v²/(rg) = (15 m/s)² / (100 m * 9.8 m/s²) ≈ 0.23.
Correct Answer: B — 0.25
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Q. A car moves in a circular path of radius 50 m at a constant speed of 20 m/s. What is the centripetal acceleration?
A.
4 m/s²
B.
8 m/s²
C.
10 m/s²
D.
16 m/s²
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Solution
Centripetal acceleration (a_c) = v²/r = (20²)/50 = 8 m/s².
Correct Answer: C — 10 m/s²
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Q. A car moves in a circular path of radius 50 m with a constant speed of 20 m/s. What is the centripetal acceleration?
A.
4 m/s²
B.
8 m/s²
C.
10 m/s²
D.
16 m/s²
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Solution
Centripetal acceleration (a_c) = v²/r = (20²)/50 = 8 m/s².
Correct Answer: C — 10 m/s²
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Q. A car moving with a speed of 30 m/s applies brakes and comes to a stop in 5 seconds. What is the deceleration of the car?
A.
3 m/s²
B.
4 m/s²
C.
5 m/s²
D.
6 m/s²
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Solution
Deceleration = (final velocity - initial velocity) / time = (0 - 30) / 5 = -6 m/s².
Correct Answer: A — 3 m/s²
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Q. A car of mass 1000 kg accelerates at 2 m/s². What is the net force acting on the car?
A.
2000 N
B.
500 N
C.
1000 N
D.
1500 N
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Solution
Using F = ma, the net force is F = 1000 kg * 2 m/s² = 2000 N.
Correct Answer: A — 2000 N
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Q. A car of mass 1000 kg accelerates from rest to a speed of 20 m/s in 10 seconds. What is the net force acting on the car? (2000)
A.
100 N
B.
200 N
C.
300 N
D.
400 N
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Solution
Acceleration a = (final speed - initial speed) / time = (20 m/s - 0) / 10 s = 2 m/s². Net force F = ma = 1000 kg * 2 m/s² = 2000 N.
Correct Answer: B — 200 N
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Q. A car of mass 1000 kg accelerates from rest to a speed of 20 m/s. What is the work done on the car?
A.
20000 J
B.
40000 J
C.
80000 J
D.
100000 J
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Solution
Kinetic energy = 0.5 × m × v^2 = 0.5 × 1000 kg × (20 m/s)^2 = 200000 J.
Correct Answer: B — 40000 J
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Q. A car of mass 1000 kg is moving at a speed of 15 m/s. What is the kinetic energy of the car?
A.
11250 J
B.
22500 J
C.
33750 J
D.
45000 J
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Solution
Kinetic energy KE = (1/2)mv² = (1/2)(1000 kg)(15 m/s)² = 11250 J.
Correct Answer: B — 22500 J
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Q. A car of mass 1000 kg is moving at a speed of 20 m/s. What is its kinetic energy?
A.
200 J
B.
400 J
C.
200,000 J
D.
400,000 J
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Solution
Kinetic energy is given by KE = 0.5mv². Here, KE = 0.5 * 1000 * (20)² = 200,000 J.
Correct Answer: C — 200,000 J
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Q. A car of mass 1000 kg is moving with a speed of 20 m/s. What is its kinetic energy?
A.
200,000 J
B.
400,000 J
C.
800,000 J
D.
1,000,000 J
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Solution
Kinetic Energy = 0.5 × mass × velocity² = 0.5 × 1000 kg × (20 m/s)² = 200,000 J.
Correct Answer: B — 400,000 J
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Q. A car of mass 1000 kg is moving with a velocity of 15 m/s. What is the kinetic energy of the car?
A.
11250 J
B.
15000 J
C.
22500 J
D.
30000 J
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Solution
Kinetic energy KE = (1/2)mv² = (1/2)(1000 kg)(15 m/s)² = 11250 J.
Correct Answer: A — 11250 J
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Q. A car of mass 1000 kg is moving with a velocity of 20 m/s. If the brakes are applied and the car comes to a stop in 5 seconds, what is the average force applied by the brakes?
A.
2000 N
B.
4000 N
C.
5000 N
D.
6000 N
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Solution
The change in momentum is 1000 kg * 20 m/s = 20000 kg·m/s. The average force is F = Δp/Δt = 20000 kg·m/s / 5 s = 4000 N.
Correct Answer: B — 4000 N
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Q. A car of mass 1000 kg is moving with a velocity of 20 m/s. What is the momentum of the car?
A.
2000 kg·m/s
B.
10000 kg·m/s
C.
5000 kg·m/s
D.
40000 kg·m/s
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Solution
Momentum p = mv = 1000 kg * 20 m/s = 20000 kg·m/s.
Correct Answer: B — 10000 kg·m/s
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Q. A car of mass 1000 kg is moving with a velocity of 20 m/s. What is the net force required to bring it to rest in 5 seconds?
A.
4000 N
B.
2000 N
C.
1000 N
D.
500 N
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Solution
First, find the deceleration: a = (final velocity - initial velocity) / time = (0 - 20 m/s) / 5 s = -4 m/s². Then, F = ma = 1000 kg * 4 m/s² = 4000 N.
Correct Answer: A — 4000 N
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