Engineering & Architecture Admissions

JEE Main

Q. A particle with charge q moves with velocity v in a magnetic field B. What is the expression for the magnetic force acting on the particle?

A. F = qvB
B. F = qvB sin(θ)
C. F = qB
D. F = qvB cos(θ)

Solution

The magnetic force acting on a charged particle moving in a magnetic field is given by F = qvB sin(θ), where θ is the angle between the velocity vector and the magnetic field vector.

Correct Answer: B — F = qvB sin(θ)

Q. A pendulum is measured to have a length of 2.0 m with an uncertainty of ±0.1 m. What is the relative uncertainty in the length?

A. 5%
B. 10%
C. 2.5%
D. 1%

Solution

Relative uncertainty = (Uncertainty / Measured value) * 100 = (0.1 / 2.0) * 100 = 5%.

Correct Answer: C — 2.5%

Q. A pendulum swings back and forth with a period of 1 second. If the length of the pendulum is doubled, what will be the new period?

A. 1 s
B. 1.41 s
C. 2 s
D. 4 s

Solution

The period of a simple pendulum is given by T = 2π√(L/g). If L is doubled, T becomes T' = 2π√(2L/g) = √2 * T ≈ 1.41 s.

Correct Answer: B — 1.41 s

Q. A pendulum swings from a height of 2 m. What is the speed at the lowest point of the swing?

A. 2 m/s
B. 4 m/s
C. 6 m/s
D. 8 m/s

Solution

Using conservation of energy, potential energy at the top = kinetic energy at the bottom. mgh = 0.5mv². Solving gives v = √(2gh) = √(2 * 9.8 * 2) = 4 m/s.

Correct Answer: B — 4 m/s

Q. A pendulum swings from a height of 5 m. What is the speed at the lowest point of the swing?

A. 5 m/s
B. 10 m/s
C. 15 m/s
D. 20 m/s

Solution

Using conservation of energy, potential energy at the top = kinetic energy at the bottom. mgh = 0.5mv^2. Solving gives v = sqrt(2gh) = sqrt(2*9.8*5) = 10 m/s.

Correct Answer: B — 10 m/s

Q. A pendulum swings with a period of 1 second. If the length of the pendulum is increased by a factor of 4, what will be the new period?

A. 1 s
B. 2 s
C. 4 s
D. √4 s

Solution

The period T = 2π√(L/g). If L is increased by a factor of 4, T increases by a factor of √4 = 2.

Correct Answer: B — 2 s

Q. A pendulum swings with a period of 1 second. If the length of the pendulum is increased to four times its original length, what will be the new period?

A. 1 s
B. 2 s
C. 4 s
D. √4 s

Solution

The period of a pendulum is given by T = 2π√(L/g). If L is increased to 4L, T becomes 2π√(4L/g) = 2T = 2 seconds.

Correct Answer: B — 2 s

Q. A pendulum swings with a period of 1 second. If the length of the pendulum is tripled, what will be the new period?

A. 1 s
B. 2 s
C. 3 s
D. √3 s

Solution

The period T = 2π√(L/g). If L is tripled, T becomes √3 times longer, so T = 2 s.

Correct Answer: B — 2 s

Q. A pendulum swings with a period of 2 seconds. What is the length of the pendulum?

A. 0.5 m
B. 1 m
C. 2 m
D. 4 m

Solution

The period T of a simple pendulum is given by T = 2π√(L/g). Rearranging gives L = (T²g)/(4π²). Using g = 9.8 m/s², L = (2² * 9.8)/(4π²) ≈ 1 m.

Correct Answer: B — 1 m

Q. A pendulum swings with a small amplitude. What type of motion does it exhibit?

A. Linear motion
B. Rotational motion
C. Simple harmonic motion
D. Circular motion

Solution

A pendulum swinging with a small amplitude exhibits simple harmonic motion.

Correct Answer: C — Simple harmonic motion

Q. A pendulum's period is measured as 2.0 s with an uncertainty of ±0.1 s. What is the percentage uncertainty in the measurement?

A. 5%
B. 10%
C. 2%
D. 1%

Solution

Percentage uncertainty = (Uncertainty / Measured value) * 100 = (0.1 / 2.0) * 100 = 5%.

Correct Answer: A — 5%

Q. A pendulum's period is measured as 2.0 s with an uncertainty of ±0.1 s. What is the relative uncertainty?

A. 5%
B. 10%
C. 2%
D. 1%

Solution

Relative uncertainty = (Uncertainty / Measured value) * 100 = (0.1 / 2.0) * 100 = 5%.

Correct Answer: A — 5%

Q. A person is pushing a heavy box across a floor. If the box does not move, what can be said about the forces acting on it?

A. Net force is zero
B. Net force is positive
C. Net force is negative
D. Only friction is acting

Solution

If the box does not move, the net force acting on it is zero, meaning the applied force equals the frictional force.

Correct Answer: A — Net force is zero

Q. A person is running at 3 m/s on a moving escalator that moves at 2 m/s in the same direction. What is the speed of the person relative to a stationary observer?

A. 1 m/s
B. 3 m/s
C. 5 m/s
D. 2 m/s

Solution

Speed of person relative to observer = Speed of escalator + Speed of person = 2 m/s + 3 m/s = 5 m/s.

Correct Answer: C — 5 m/s

Q. A person is running at 8 m/s in a train moving at 12 m/s in the same direction. What is the speed of the person relative to the ground?

A. 4 m/s
B. 8 m/s
C. 12 m/s
D. 20 m/s

Solution

Speed of person relative to ground = Speed of train + Speed of person = 12 m/s + 8 m/s = 20 m/s.

Correct Answer: D — 20 m/s

Q. A person is standing 25 meters away from a vertical cliff. If the angle of elevation to the top of the cliff is 60 degrees, what is the height of the cliff?

A. 10 m
B. 15 m
C. 20 m
D. 25 m

Solution

Using tan(60°) = height/25, we have √3 = height/25. Therefore, height = 25√3 ≈ 43.3 m.

Correct Answer: C — 20 m

Q. A person is standing 30 m away from the base of a tree. If the angle of elevation to the top of the tree is 60 degrees, what is the height of the tree?

A. 15 m
B. 20 m
C. 25 m
D. 30 m

Solution

Using tan(60°) = height/30, we have √3 = height/30. Therefore, height = 30√3 = 25 m.

Correct Answer: C — 25 m

Q. A person is standing 30 m away from the foot of a tree. If the angle of elevation to the top of the tree is 60 degrees, what is the height of the tree?

A. 15 m
B. 20 m
C. 25 m
D. 30 m

Solution

Using tan(60°) = height/30, we have √3 = height/30. Therefore, height = 30√3 = 25 m.

Correct Answer: C — 25 m

Q. A person is standing 40 m away from a building. If the angle of elevation to the top of the building is 60 degrees, what is the height of the building?

A. 20√3 m
B. 40 m
C. 30 m
D. 50 m

Solution

Using tan(60°) = height/40, we have √3 = height/40. Therefore, height = 40√3 m.

Correct Answer: A — 20√3 m

Q. A person is standing 50 m away from a building. If the angle of elevation to the top of the building is 30 degrees, what is the height of the building?

A. 25√3 m
B. 50 m
C. 30 m
D. 40 m

Solution

Using tan(30°) = height/50, we have 1/√3 = height/50. Therefore, height = 50/√3 m.

Correct Answer: A — 25√3 m

Q. A person is standing 50 meters away from a vertical pole. If the angle of elevation of the top of the pole is 60 degrees, what is the height of the pole?

A. 25 m
B. 30 m
C. 35 m
D. 40 m

Solution

Using tan(60°) = height/50, we have √3 = height/50. Therefore, height = 50√3 ≈ 86.6 m.

Correct Answer: B — 30 m

Q. A person is standing at a distance of 20 m from a vertical pole. If the angle of elevation to the top of the pole is 45 degrees, what is the height of the pole?

A. 20 m
B. 10 m
C. 30 m
D. 15 m

Solution

Using tan(45°) = height/distance, we have height = distance * tan(45°) = 20 * 1 = 20 m.

Correct Answer: A — 20 m

Q. A person is standing at a distance of 40 m from a tree. If the angle of elevation to the top of the tree is 60 degrees, what is the height of the tree?

A. 20√3 m
B. 40 m
C. 30 m
D. 10√3 m

Solution

Using tan(60°) = height/distance, we have height = distance * tan(60°) = 40√3 m.

Correct Answer: A — 20√3 m

Q. A person is standing at a distance of 40 meters from the base of a building. If the angle of elevation to the top of the building is 60 degrees, what is the height of the building?

A. 20 m
B. 30 m
C. 40 m
D. 50 m

Solution

Using tan(60°) = height/40, we have √3 = height/40. Therefore, height = 40√3 ≈ 69.28 m.

Correct Answer: C — 40 m

Q. A person is walking at 2 m/s on a train moving at 10 m/s. What is the speed of the person relative to the ground?

A. 2 m/s
B. 8 m/s
C. 10 m/s
D. 12 m/s

Solution

Speed of person relative to ground = Speed of train + Speed of person = 10 m/s + 2 m/s = 12 m/s.

Correct Answer: D — 12 m/s

Q. A person standing 20 meters away from a vertical cliff observes the top of the cliff at an angle of elevation of 75 degrees. What is the height of the cliff?

A. 10 m
B. 15 m
C. 20 m
D. 25 m

Solution

Using tan(75°) = height/20, we have height = 20 * tan(75°) ≈ 20 * 3.732 = 74.64 m.

Correct Answer: D — 25 m

Q. A person standing 40 m away from a building observes the top of the building at an angle of elevation of 30 degrees. What is the height of the building?

A. 10 m
B. 20 m
C. 30 m
D. 40 m

Solution

Using tan(30°) = height/40, we have 1/√3 = height/40. Therefore, height = 40/√3 ≈ 23.1 m.

Correct Answer: B — 20 m

Q. A person standing 50 m away from a building observes the top of the building at an angle of elevation of 60 degrees. What is the height of the building?

A. 25 m
B. 30 m
C. 35 m
D. 40 m

Solution

Using tan(60°) = height/50, we have √3 = height/50. Therefore, height = 50√3 ≈ 86.6 m.

Correct Answer: B — 30 m

Q. A person standing on the ground observes the top of a tree at an angle of elevation of 45 degrees. If the person is 10 meters away from the tree, what is the height of the tree?

A. 5 m
B. 10 m
C. 15 m
D. 20 m

Solution

Using tan(45°) = height/10, we have 1 = height/10. Therefore, height = 10 m.

Correct Answer: B — 10 m

Q. A person walks at 4 km/h in a train moving at 36 km/h. What is the speed of the person relative to the ground?

A. 32 km/h
B. 36 km/h
C. 40 km/h
D. 4 km/h

Solution

Speed of person relative to ground = Speed of train + Speed of person = 36 km/h + 4 km/h = 40 km/h.

Correct Answer: C — 40 km/h

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