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
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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 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 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 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 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 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 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
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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 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 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?
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? (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 in water? (Refractive index of water = 1.33)
A.
22.1 degrees
B.
30 degrees
C.
36.9 degrees
D.
45 degrees
Show solution
Solution
Using Snell's law, n1*sin(θ1) = n2*sin(θ2), we find θ2 = 22.1 degrees.
Correct Answer: A — 22.1 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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Q. A ray of light passes from air into water. If the angle of incidence is 30 degrees, what is the angle of refraction in water (n = 1.33)?
A.
22.5 degrees
B.
30 degrees
C.
22 degrees
D.
40 degrees
Show solution
Solution
Using Snell's law: n1 * sin(θ1) = n2 * sin(θ2). Here, sin(30) = 0.5, so sin(θ2) = (1 * 0.5) / 1.33 = 0.375. Thus, θ2 = sin^(-1)(0.375) ≈ 22.5 degrees.
Correct Answer: A — 22.5 degrees
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Q. A ray of light passes through the center of curvature of a concave mirror. What will be the angle of reflection?
A.
0 degrees
B.
30 degrees
C.
45 degrees
D.
90 degrees
Show solution
Solution
When a ray of light passes through the center of curvature, it strikes the mirror perpendicularly, resulting in an angle of reflection of 0 degrees.
Correct Answer: A — 0 degrees
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Q. A ray of light passes through the center of curvature of a concave mirror. What is the angle of reflection?
A.
0 degrees
B.
30 degrees
C.
45 degrees
D.
90 degrees
Show solution
Solution
When a ray passes through the center of curvature, it strikes the mirror perpendicularly, resulting in an angle of reflection of 0 degrees.
Correct Answer: A — 0 degrees
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Q. A ray of light passes through the optical center of a lens. What happens to the ray?
A.
It bends towards the normal
B.
It bends away from the normal
C.
It continues in a straight line
D.
It reflects back
Show solution
Solution
A ray of light passing through the optical center of a lens continues in a straight line without bending.
Correct Answer: C — It continues in a straight line
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Q. A ray of light strikes a plane mirror at an angle of 30 degrees. What is the angle of reflection?
A.
30 degrees
B.
60 degrees
C.
90 degrees
D.
45 degrees
Show solution
Solution
According to the law of reflection, the angle of reflection is equal to the angle of incidence, which is 30 degrees.
Correct Answer: A — 30 degrees
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Q. A ray of light strikes a plane mirror at an angle of incidence of 30 degrees. What is the angle of reflection?
A.
30 degrees
B.
45 degrees
C.
60 degrees
D.
90 degrees
Show solution
Solution
According to the law of reflection, the angle of reflection is equal to the angle of incidence. Therefore, the angle of reflection is 30 degrees.
Correct Answer: A — 30 degrees
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Q. A ray of light traveling in air strikes the surface of water at an angle of incidence of 45 degrees. 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). For air (n1=1) and water (n2=1.33), we find r = 45 degrees.
Correct Answer: B — 45 degrees
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Q. A rectangular plate of mass M and dimensions a x b is rotated about an axis along one of its edges. What is its moment of inertia?
A.
1/3 Ma^2
B.
1/12 Ma^2 + 1/3 Mb^2
C.
1/2 Ma^2
D.
1/4 Ma^2 + 1/3 Mb^2
Show solution
Solution
The moment of inertia of a rectangular plate about an edge is I = 1/12 Ma^2 + 1/3 Mb^2.
Correct Answer: B — 1/12 Ma^2 + 1/3 Mb^2
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Q. A resistor of 10 ohms is connected to a 20V battery. What is the current flowing through the resistor?
A.
0.5 A
B.
1 A
C.
2 A
D.
5 A
Show solution
Solution
Using Ohm's Law, I = V/R = 20V / 10Ω = 2 A.
Correct Answer: C — 2 A
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