Engineering & Architecture Admissions
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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Q. A car travels 100 m in 4 seconds. What is its average speed?
A.
20 m/s
B.
25 m/s
C.
30 m/s
D.
35 m/s
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Solution
Average speed = total distance / total time = 100 m / 4 s = 25 m/s.
Correct Answer: B — 25 m/s
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Q. A car travels 100 m in 5 seconds. What is its average speed?
A.
20 m/s
B.
25 m/s
C.
15 m/s
D.
30 m/s
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Solution
Average speed = total distance / total time = 100 m / 5 s = 20 m/s.
Correct Answer: B — 25 m/s
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Q. A car travels 100 m north and then 100 m east. What is the magnitude of the displacement from the starting point? (2000)
A.
100 m
B.
141.42 m
C.
200 m
D.
50 m
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Solution
Displacement = √(100^2 + 100^2) = √20000 = 141.42 m.
Correct Answer: B — 141.42 m
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Q. A car travels around a circular track of radius 50 m at a speed of 15 m/s. What is the centripetal force acting on the car if its mass is 1000 kg?
A.
450 N
B.
225 N
C.
150 N
D.
75 N
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Solution
Centripetal force (F_c) = mv²/r = 1000 kg * (15 m/s)² / 50 m = 4500 N.
Correct Answer: A — 450 N
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Q. A car travels at 90 km/h and a truck at 60 km/h in opposite directions. What is the relative speed of the car with respect to the truck?
A.
30 km/h
B.
60 km/h
C.
150 km/h
D.
90 km/h
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Solution
Relative speed = Speed of car + Speed of truck = 90 km/h + 60 km/h = 150 km/h.
Correct Answer: C — 150 km/h
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Q. A car travels at a speed of 80 km/h and a bike travels at 60 km/h. If they start from the same point and travel in the same direction, how far apart will they be after 1 hour?
A.
20 km
B.
10 km
C.
30 km
D.
40 km
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Solution
Relative speed = 80 - 60 = 20 km/h. Distance apart after 1 hour = 20 km.
Correct Answer: A — 20 km
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Q. A card is drawn from a standard deck of 52 cards. What is the probability that it is a heart given that it is a red card?
A.
1/2
B.
1/4
C.
1/3
D.
1/13
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Solution
There are 26 red cards (hearts and diamonds). The probability of drawing a heart given that it is red is 13/26 = 1/2.
Correct Answer: A — 1/2
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Q. A card is drawn from a standard deck of 52 cards. What is the probability that the card drawn is a queen?
A.
1/13
B.
1/52
C.
1/26
D.
3/52
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Solution
There are 4 queens in a deck of 52 cards. Probability = 4/52 = 1/13.
Correct Answer: A — 1/13
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Q. A card is drawn from a standard deck of 52 cards. What is the probability that the card is a heart given that it is a red card?
A.
1/2
B.
1/4
C.
1/3
D.
2/3
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Solution
There are 26 red cards (hearts and diamonds). The probability of drawing a heart given that the card is red is 13/26 = 1/2.
Correct Answer: A — 1/2
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Q. A card is drawn from a standard deck of 52 cards. What is the probability that the card drawn is a heart?
A.
1/4
B.
1/13
C.
1/52
D.
3/13
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Solution
There are 13 hearts in a deck of 52 cards. The probability of drawing a heart is 13/52 = 1/4.
Correct Answer: A — 1/4
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Q. A changing magnetic field induces a current in a closed loop. What is this phenomenon called?
A.
Electromagnetic induction
B.
Magnetic resonance
C.
Electrolysis
D.
Magnetism
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Solution
This phenomenon is known as electromagnetic induction, where a changing magnetic field induces an electromotive force (EMF) in a conductor.
Correct Answer: A — Electromagnetic induction
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Q. A changing magnetic field induces an electric field. This phenomenon is known as?
A.
Electromagnetic induction
B.
Electrostatics
C.
Magnetostatics
D.
Electrodynamics
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Solution
The phenomenon of a changing magnetic field inducing an electric field is known as electromagnetic induction.
Correct Answer: A — Electromagnetic induction
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Q. A charge of +10μC is placed at the origin. What is the electric potential at a point 2m away from the charge?
A.
4500 V
B.
2250 V
C.
5000 V
D.
1000 V
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Solution
V = k * q / r = (9 × 10^9) * (10 × 10^-6) / 2 = 4500 V.
Correct Answer: B — 2250 V
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Q. A charge of +10μC is placed in a uniform electric field of strength 500 N/C. What is the work done in moving the charge 2m in the direction of the field?
A.
10 J
B.
1 J
C.
100 J
D.
0.5 J
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Solution
Work done W = F * d = (E * q) * d = (500 N/C * 10 × 10^-6 C) * 2m = 0.01 J.
Correct Answer: A — 10 J
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