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Q. A projectile is launched at an angle of 30 degrees with an initial speed of 20 m/s. What is the maximum height reached? (g = 10 m/s²)

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

Solution

Maximum height h = (u^2 * sin²θ) / (2g) = (20² * (1/2)²) / (2 * 10) = 5 m.

Correct Answer: B — 10 m

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Q. A projectile is launched at an angle of 30 degrees with an initial speed of 40 m/s. What is the maximum height reached? (g = 10 m/s²)

A. 80 m
B. 60 m
C. 40 m
D. 20 m

Solution

Maximum height = (u^2 * sin²θ) / (2g) = (40^2 * (1/2)) / (2 * 10) = 80 m.

Correct Answer: B — 60 m

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Q. A projectile is launched at an angle of 30 degrees with an initial velocity of 20 m/s. What is the maximum height reached by the projectile?

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

Solution

Maximum height (H) = (u^2 * sin^2(θ)) / (2g) = (20^2 * (1/2)^2) / (2 * 9.8) = 10.2 m.

Correct Answer: B — 10 m

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Q. A projectile is launched at an angle of 30° with an initial speed of 40 m/s. What is the maximum height reached by the projectile?

A. 80 m
B. 60 m
C. 40 m
D. 20 m

Solution

Using the formula H = (u² * sin²θ) / (2g), where u = 40 m/s, θ = 30°, and g = 9.8 m/s², we find H = (40² * (1/4)) / (2*9.8) = 40.82 m, approximately 60 m.

Correct Answer: B — 60 m

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Q. A projectile is launched at an angle of 30° with the horizontal with an initial velocity of 40 m/s. What is the horizontal range of the projectile? (Take g = 10 m/s²)

A. 160 m
B. 200 m
C. 80 m
D. 120 m

Solution

Range R = (u² * sin(2θ))/g = (40² * sin(60°))/10 = (1600 * √3/2)/10 = 80√3 m ≈ 138.56 m.

Correct Answer: B — 200 m

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Q. A projectile is launched with a speed of 30 m/s at an angle of 60 degrees. What is the horizontal component of its velocity?

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

Solution

Horizontal component (v_x) = u * cos(θ) = 30 * 0.5 = 15 m/s.

Correct Answer: B — 20 m/s

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Q. A projectile is launched with a speed of 50 m/s at an angle of 30 degrees. What is the time of flight?

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

Solution

Time of flight (T) = (2 * u * sin(θ)) / g = (2 * 50 * (√3/2)) / 9.8 ≈ 10.2 s.

Correct Answer: B — 10 s

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Q. A projectile is launched with a speed of 50 m/s at an angle of 30 degrees. What is the horizontal component of the velocity?

A. 25 m/s
B. 35 m/s
C. 43.3 m/s
D. 50 m/s

Solution

Horizontal component (u_x) = u * cos(θ) = 50 * (√3/2) = 43.3 m/s.

Correct Answer: C — 43.3 m/s

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Q. A projectile is launched with a speed of 50 m/s at an angle of 60 degrees. What is the horizontal component of its velocity?

A. 25 m/s
B. 50 m/s
C. 43.3 m/s
D. 30 m/s

Solution

Horizontal component (v_x) = v * cos(θ) = 50 * 0.5 = 43.3 m/s.

Correct Answer: C — 43.3 m/s

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Q. A projectile is launched with an initial speed of 30 m/s at an angle of 60 degrees. What is the horizontal component of the velocity?

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

Solution

Horizontal component (vx) = u * cos(θ) = 30 * 0.5 = 15 m/s.

Correct Answer: C — 25 m/s

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Q. A projectile is launched with an initial speed of 30 m/s at an angle of 60 degrees. What is the horizontal component of its velocity?

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

Solution

Horizontal component (vx) = u * cos(θ) = 30 * 0.5 = 15 m/s.

Correct Answer: B — 20 m/s

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Q. A projectile is launched with an initial speed of 40 m/s at an angle of 30 degrees. What is the time of flight?

A. 4 s
B. 5 s
C. 6 s
D. 8 s

Solution

Time of flight (T) = (2u * sin(θ)) / g = (2 * 40 * (√3/2)) / 9.8 ≈ 6.1 s.

Correct Answer: C — 6 s

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Q. A projectile is launched with an initial speed of 40 m/s at an angle of 60 degrees. What is the horizontal component of its velocity?

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

Solution

Horizontal component (vx) = u * cos(θ) = 40 * 0.5 = 20 m/s.

Correct Answer: B — 30 m/s

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Q. A projectile is launched with an initial speed of 50 m/s at an angle of 30 degrees. What is the horizontal component of the velocity?

A. 25 m/s
B. 35 m/s
C. 43.3 m/s
D. 50 m/s

Solution

Horizontal component (vx) = u * cos(θ) = 50 * cos(30) = 50 * (√3/2) = 43.3 m/s.

Correct Answer: C — 43.3 m/s

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Q. A projectile is launched with an initial speed of 50 m/s at an angle of 60 degrees. What is the horizontal component of its velocity?

A. 25 m/s
B. 50 m/s
C. 43.3 m/s
D. 30 m/s

Solution

Horizontal component (v_x) = u * cos(θ) = 50 * 0.5 = 25 m/s.

Correct Answer: C — 43.3 m/s

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Q. A projectile is launched with an initial velocity of 30 m/s at an angle of 30 degrees. What is the horizontal range of the projectile? (Take g = 10 m/s²)

A. 90 m
B. 75 m
C. 100 m
D. 120 m

Solution

Range R = (u² * sin(2θ))/g = (30² * sin(60°))/10 = (900 * √3/2)/10 = 45√3 m ≈ 90 m.

Correct Answer: A — 90 m

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Q. A projectile is launched with an initial velocity of 30 m/s at an angle of 30 degrees to the horizontal. What is the horizontal range of the projectile? (Take g = 10 m/s²)

A. 90 m
B. 75 m
C. 100 m
D. 120 m

Solution

Range R = (u² * sin(2θ))/g = (30² * sin(60°))/10 = (900 * √3/2)/10 = 45√3 m ≈ 90 m.

Correct Answer: A — 90 m

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Q. A projectile is launched with an initial velocity of 30 m/s at an angle of 30° to the horizontal. What is the horizontal range of the projectile? (Take g = 10 m/s²)

A. 90 m
B. 75 m
C. 100 m
D. 120 m

Solution

Range R = (u² * sin(2θ))/g = (30² * sin(60°))/10 = (900 * √3/2)/10 = 45√3 m ≈ 90 m.

Correct Answer: A — 90 m

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Q. A projectile is launched with an initial velocity of 30 m/s at an angle of 45 degrees. What is the maximum height reached?

A. 22.5 m
B. 30 m
C. 45 m
D. 15 m

Solution

Maximum height (H) = (u² * sin²θ) / (2g) = (30² * (1/2)) / (2 * 9.8) = 22.5 m.

Correct Answer: A — 22.5 m

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Q. A projectile is launched with an initial velocity of 30 m/s at an angle of 45 degrees. What is the maximum height reached by the projectile? (2023)

A. 22.5 m
B. 30 m
C. 45 m
D. 15 m

Solution

Maximum height (H) = (u² * sin²θ) / (2g). Here, u = 30 m/s, θ = 45°, g = 9.81 m/s². H = (30² * (1/2)) / (2 * 9.81) = 22.5 m.

Correct Answer: A — 22.5 m

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Q. A projectile is launched with an initial velocity of 30 m/s at an angle of 60 degrees. What is the horizontal component of the velocity?

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

Solution

Horizontal component (u_x) = u * cos(θ) = 30 * 0.5 = 15 m/s.

Correct Answer: B — 20 m/s

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Q. A projectile is launched with an initial velocity of 40 m/s at an angle of 30 degrees. What is the horizontal range of the projectile? (Take g = 10 m/s²)

A. 160 m
B. 200 m
C. 80 m
D. 120 m

Solution

Range R = (u² * sin(2θ))/g = (40² * sin(60°))/10 = (1600 * √3/2)/10 = 80√3 m ≈ 138.56 m.

Correct Answer: B — 200 m

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Q. A projectile is launched with an initial velocity of 40 m/s at an angle of 45 degrees. What is the time of flight?

A. 4 s
B. 5 s
C. 6 s
D. 7 s

Solution

Time of flight (T) = (2u * sin(θ)) / g = (2*40*sin(45))/9.8 = 5.14 s.

Correct Answer: B — 5 s

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Q. A projectile is launched with an initial velocity of 40 m/s at an angle of 45 degrees. What is the vertical component of its velocity?

A. 20 m/s
B. 28.28 m/s
C. 30 m/s
D. 40 m/s

Solution

Vertical component Vy = u * sin(θ) = 40 * (√2/2) = 28.28 m/s.

Correct Answer: B — 28.28 m/s

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Q. A projectile is launched with an initial velocity of 40 m/s at an angle of 45 degrees. What is the vertical component of the velocity?

A. 20 m/s
B. 28.28 m/s
C. 30 m/s
D. 40 m/s

Solution

Vertical component Vy = u * sin(θ) = 40 * (√2/2) = 28.28 m/s.

Correct Answer: B — 28.28 m/s

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Q. A projectile is thrown with a speed of 20 m/s at an angle of 45 degrees. What is the vertical component of the velocity?

A. 10 m/s
B. 14.14 m/s
C. 20 m/s
D. 28.28 m/s

Solution

Vertical component (u_y) = u * sin(θ) = 20 * (√2/2) = 14.14 m/s.

Correct Answer: B — 14.14 m/s

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Q. A proton moves in a magnetic field and experiences a force. If the velocity of the proton is doubled, what happens to the magnetic force?

A. It doubles
B. It halves
C. It remains the same
D. It quadruples

Solution

The magnetic force is proportional to the velocity of the charge, so if the velocity is doubled, the magnetic force also doubles.

Correct Answer: A — It doubles

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Q. A radar system can detect objects up to 100 km away. If an object is detected at 75 km, how far is it from the radar? (2022)

A. 25 km
B. 50 km
C. 75 km
D. 100 km

Solution

Distance from Radar = Maximum Range - Detected Distance = 100 km - 75 km = 25 km

Correct Answer: A — 25 km

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Q. A radar system can detect objects up to 100 km away. If it scans an area of 50 km radius, what is the area it can cover? (2023)

A. 7850 km²
B. 1500 km²
C. 2000 km²
D. 2500 km²

Solution

Area = π × r² = π × (50 km)² ≈ 7850 km²

Correct Answer: A — 7850 km²

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Q. A radar system can detect objects up to 200 km away. If it scans an area in a circular pattern, what is the area it can cover? (2023)

A. 12560 km²
B. 31415 km²
C. 40000 km²
D. 50000 km²

Solution

Area = πr² = π(200 km)² = 125600 km² (approx 31415 km²)

Correct Answer: B — 31415 km²

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Major Competitive Exams

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