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
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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
Show solution
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
Show solution
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
Show solution
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
Show solution
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
Show solution
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
Show solution
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
Show solution
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
Show solution
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
Show solution
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
Show solution
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
Show solution
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
Show solution
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
Show solution
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
Show solution
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
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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
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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
Show solution
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 at a speed of 20 km/h, how long will it take to complete a full scan? (2022)
A.
8 hours
B.
10 hours
C.
12 hours
D.
15 hours
Show solution
Solution
Time = Distance / Speed = 200 km / 20 km/h = 10 hours
Correct Answer: B — 10 hours
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Q. A radar system can detect objects up to a distance of 50 km. If it scans an area of 20 km radius, what is the area it can cover? (2023)
A.
1256 km²
B.
314 km²
C.
400 km²
D.
500 km²
Show solution
Solution
Area = π × r² = π × (20 km)² ≈ 1256 km²
Correct Answer: A — 1256 km²
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Q. A ray of light in a medium with n=1.33 strikes the boundary with air at an angle of 45°. What is the behavior of the ray?
A.
Total internal reflection
B.
Partial reflection and refraction
C.
Complete absorption
D.
Total refraction
Show solution
Solution
Since 45° < θc ≈ 48.6°, the ray will partially reflect and refract.
Correct Answer: B — Partial reflection and refraction
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Q. A ray of light in glass (n=1.5) strikes the glass-air interface at an angle of 30°. What will be the angle of refraction in air?
A.
60°
B.
30°
C.
45°
D.
90°
Show solution
Solution
Using Snell's law, n1 * sin(θ1) = n2 * sin(θ2). Here, n1 = 1.5, θ1 = 30°, n2 = 1.0. Thus, sin(θ2) = (1.5 * sin(30°))/1.0 = 0.75, giving θ2 ≈ 60°.
Correct Answer: A — 60°
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Q. A ray of light passes from air into glass at an angle of 45 degrees. What is the angle of refraction if the refractive index of glass is 1.5? (2022)
A.
30 degrees
B.
45 degrees
C.
60 degrees
D.
90 degrees
Show solution
Solution
Using Snell's law, n1 * sin(theta1) = n2 * sin(theta2). Here, n1 = 1 (air), theta1 = 45 degrees, n2 = 1.5 (glass). Thus, sin(theta2) = (1 * sin(45))/1.5 = 0.471, giving theta2 ≈ 30 degrees.
Correct Answer: A — 30 degrees
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Q. A ray of light passes from air into glass at an angle of incidence of 30 degrees. If the refractive index of glass is 1.5, what is the angle of refraction?
A.
20 degrees
B.
30 degrees
C.
18.4 degrees
D.
22 degrees
Show solution
Solution
Using Snell's law, n1 * sin(i) = n2 * sin(r). Here, n1 = 1 (air), i = 30 degrees, n2 = 1.5 (glass). Thus, sin(r) = (1 * sin(30))/1.5 = 0.333, giving r ≈ 18.4 degrees.
Correct Answer: C — 18.4 degrees
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Q. A ray of light passes from air into glass at an angle of incidence of 45 degrees. If the refractive index of glass is 1.5, what is the angle of refraction?
A.
30 degrees
B.
45 degrees
C.
60 degrees
D.
90 degrees
Show solution
Solution
Using Snell's law, n1 * sin(i) = n2 * sin(r). Here, n1 = 1 (air), n2 = 1.5 (glass), i = 45 degrees. Solving gives r = 30 degrees.
Correct Answer: A — 30 degrees
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Q. A ray of light passes from air into glass with a refractive index of 1.5. If the angle of incidence is 30 degrees, what is the angle of refraction?
A.
20 degrees
B.
30 degrees
C.
18.4 degrees
D.
22 degrees
Show solution
Solution
Using Snell's law, n1 * sin(i) = n2 * sin(r). Here, n1 = 1, n2 = 1.5, i = 30 degrees. Solving gives r ≈ 18.4 degrees.
Correct Answer: C — 18.4 degrees
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Q. A ray of light passes from air into water at an angle of incidence of 30 degrees. What is the angle of refraction?
A.
22 degrees
B.
30 degrees
C.
45 degrees
D.
18 degrees
Show solution
Solution
Using Snell's law, n1 * sin(θ1) = n2 * sin(θ2). Here, n1 = 1 (air), n2 = 1.33 (water), θ1 = 30 degrees. Thus, sin(θ2) = (1 * sin(30))/1.33 = 0.375. Therefore, θ2 = sin^(-1)(0.375) which is approximately 22 degrees.
Correct Answer: A — 22 degrees
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Q. A ray of light passes from air into water at an angle of incidence of 30 degrees. What is the angle of refraction in water (n = 1.33)?
A.
22.5 degrees
B.
30 degrees
C.
40 degrees
D.
45 degrees
Show solution
Solution
Using Snell's law, n1 * sin(θ1) = n2 * sin(θ2). Here, sin(30) = 0.5, so 1 * 0.5 = 1.33 * sin(θ2). Solving gives θ2 ≈ 22.5 degrees.
Correct Answer: A — 22.5 degrees
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Q. A ray of light passes from air into water at an angle of incidence of 30 degrees. What is the angle of refraction in water? (Refractive index of water = 1.33)
A.
22.5 degrees
B.
30 degrees
C.
40 degrees
D.
20 degrees
Show solution
Solution
Using Snell's law, n1 * sin(θ1) = n2 * sin(θ2), we find θ2 = sin^(-1)(sin(30 degrees)/1.33) = 22.5 degrees.
Correct Answer: A — 22.5 degrees
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Q. A ray of light passes from air into water. If the angle of incidence is 30 degrees, what is the angle of refraction?
A.
22 degrees
B.
30 degrees
C.
45 degrees
D.
18 degrees
Show solution
Solution
Using Snell's law, n1 * sin(θ1) = n2 * sin(θ2). Here, n1 = 1 (air), n2 = 1.33 (water). Solving gives θ2 ≈ 22 degrees.
Correct Answer: A — 22 degrees
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