Q. A plane flies at a speed of 200 m/s at an angle of 30 degrees to the horizontal. What is the vertical component of its velocity?
A.
100 m/s
B.
150 m/s
C.
200 m/s
D.
250 m/s
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Solution
Vertical component (v_y) = v * sin(θ) = 200 * (√3/2) ≈ 173.2 m/s.
Correct Answer: A — 100 m/s
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Q. A plane is flying at 200 m/s in still air. If there is a headwind of 50 m/s, what is the speed of the plane relative to the ground?
A.
150 m/s
B.
200 m/s
C.
250 m/s
D.
300 m/s
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Solution
Speed relative to ground = Speed of plane - Speed of wind = 200 m/s - 50 m/s = 150 m/s.
Correct Answer: A — 150 m/s
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Q. A planet orbits a star in an elliptical path. What remains constant throughout its orbit?
A.
Angular momentum
B.
Kinetic energy
C.
Potential energy
D.
Total energy
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Solution
Angular momentum is conserved in the absence of external torques, even in elliptical orbits.
Correct Answer: A — Angular momentum
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Q. A planet orbits the sun in a circular path. If the radius of the orbit is doubled, what happens to the angular momentum of the planet if its speed remains constant?
A.
Doubles
B.
Halves
C.
Remains the same
D.
Quadruples
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Solution
Angular momentum L = mvr, so if the radius is doubled and speed remains constant, angular momentum doubles.
Correct Answer: A — Doubles
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Q. A planet orbits the sun in a circular path. If the radius of the orbit is halved, what happens to the angular momentum of the planet?
A.
It doubles
B.
It halves
C.
It remains the same
D.
It becomes zero
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Solution
Angular momentum L = mvr; if the radius is halved and speed remains constant, angular momentum halves.
Correct Answer: B — It halves
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Q. A planet orbits the sun in an elliptical path. What remains constant for the planet as it moves in its orbit?
A.
Kinetic energy
B.
Potential energy
C.
Angular momentum
D.
Total energy
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Solution
Angular momentum remains constant for a planet in orbit due to the conservation of angular momentum.
Correct Answer: C — Angular momentum
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Q. A planet orbits the sun in an elliptical path. Which of the following statements is true regarding its angular momentum?
A.
It is constant.
B.
It varies with distance from the sun.
C.
It is zero at perihelion.
D.
It is maximum at aphelion.
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Solution
The angular momentum of a planet about the sun is conserved as there is no external torque acting on it.
Correct Answer: A — It is constant.
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Q. A point charge +Q is placed at the center of a spherical Gaussian surface of radius R. What is the electric flux through the surface?
A.
0
B.
Q/ε₀
C.
Q/(4πε₀R²)
D.
Q/(4πε₀R)
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Solution
The electric flux through the Gaussian surface is given by Φ = Q/ε₀, where Q is the charge enclosed.
Correct Answer: B — Q/ε₀
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Q. A point charge +Q is placed at the center of a spherical Gaussian surface. What is the total electric flux through the surface?
A.
0
B.
Q/ε₀
C.
Q/4πε₀
D.
4πQ/ε₀
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Solution
The total electric flux through the surface is given by Gauss's law as Φ = Q/ε₀, and for a point charge at the center, it results in 4πQ/ε₀.
Correct Answer: D — 4πQ/ε₀
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Q. A point charge of +5 µC is placed at the origin. What is the electric potential at a point 2 m away from the charge?
A.
1125 V
B.
450 V
C.
225 V
D.
0 V
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Solution
The electric potential V at a distance r from a point charge Q is given by V = kQ/r. Here, V = (9 x 10^9 Nm²/C²)(5 x 10^-6 C)/(2 m) = 1125 V.
Correct Answer: A — 1125 V
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Q. A point charge of +Q is placed at the center of a spherical Gaussian surface. What is the total electric flux through the surface?
A.
0
B.
Q/ε₀
C.
Q/2ε₀
D.
4πQ/ε₀
Show solution
Solution
According to Gauss's law, the total electric flux through a closed surface is Φ = Q_enc/ε₀. Here, Q_enc = Q, so Φ = Q/ε₀.
Correct Answer: D — 4πQ/ε₀
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Q. A point charge of +Q is placed at the center of a spherical shell of radius R with surface charge density σ. What is the electric field inside the shell?
A.
0
B.
Q/(4πε₀R²)
C.
σ/ε₀
D.
Q/(4πε₀R)
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Solution
According to Gauss's law, the electric field inside a conductor in electrostatic equilibrium is zero.
Correct Answer: A — 0
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Q. A point charge Q is placed at the center of a cube. What is the electric flux through one face of the cube?
A.
Q/ε₀
B.
Q/6ε₀
C.
Q/4ε₀
D.
0
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Solution
The total flux through the cube is Q/ε₀. Since there are 6 faces, the flux through one face is Q/(6ε₀).
Correct Answer: B — Q/6ε₀
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Q. A point charge Q is placed at the center of a spherical Gaussian surface. What is the electric flux through the surface?
A.
0
B.
Q/ε₀
C.
Q/4πε₀
D.
Q²/ε₀
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Solution
According to Gauss's law, the electric flux Φ = Q/ε₀ when a point charge Q is at the center of a spherical surface.
Correct Answer: B — Q/ε₀
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Q. A potentiometer is used to compare two emf sources. If the first source gives a balance length of 60cm and the second gives 90cm, what is the ratio of their emfs?
A.
2:3
B.
3:2
C.
1:1
D.
4:5
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Solution
The ratio of emfs is equal to the ratio of the balance lengths, so it is 60cm:90cm = 2:3.
Correct Answer: B — 3:2
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Q. A potentiometer is used to compare two EMFs. If the known EMF is 6V and the length of the wire is 120 cm, what is the potential gradient if the length of the wire is used to balance an unknown EMF of 4V?
A.
0.05 V/cm
B.
0.03 V/cm
C.
0.04 V/cm
D.
0.02 V/cm
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Solution
The potential gradient is calculated as (6V / 120 cm) = 0.05 V/cm. For the unknown EMF of 4V, the length used would be (4V / 0.05 V/cm) = 80 cm.
Correct Answer: C — 0.04 V/cm
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Q. A potentiometer wire has a length of 10 m and a potential difference of 5 V across it. What is the potential gradient?
A.
0.5 V/m
B.
1 V/m
C.
2 V/m
D.
5 V/m
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Solution
The potential gradient is calculated as the potential difference divided by the length of the wire: 5 V / 10 m = 0.5 V/m.
Correct Answer: A — 0.5 V/m
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Q. A potentiometer wire has a resistance of 10 ohms and is connected to a 5 V battery. What is the current flowing through the wire?
A.
0.5 A
B.
1 A
C.
2 A
D.
5 A
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Solution
Using Ohm's law (V = IR), the current can be calculated as I = V/R = 5 V / 10 ohms = 0.5 A.
Correct Answer: B — 1 A
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Q. A potentiometer wire has a uniform cross-section and a length of 10 m. If a potential difference of 5 V is applied, what is the potential gradient?
A.
0.5 V/m
B.
1 V/m
C.
2 V/m
D.
5 V/m
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Solution
The potential gradient is calculated as V/L = 5 V / 10 m = 0.5 V/m.
Correct Answer: B — 1 V/m
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Q. A potentiometer wire has a uniform cross-section and a potential difference of 12V across it. If the length of the wire is 6m, what is the potential gradient?
A.
2 V/m
B.
4 V/m
C.
6 V/m
D.
8 V/m
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Solution
Potential gradient = Voltage / Length = 12V / 6m = 2 V/m.
Correct Answer: B — 4 V/m
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Q. A potentiometer wire has a uniform cross-section and a total length of 10 m. If a potential difference of 5 V is applied across it, what is the potential gradient?
A.
0.5 V/m
B.
1 V/m
C.
2 V/m
D.
5 V/m
Show solution
Solution
The potential gradient is calculated as V/L = 5 V / 10 m = 0.5 V/m.
Correct Answer: B — 1 V/m
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Q. A potentiometer wire is made of constantan. What is the advantage of using this material?
A.
High conductivity
B.
Low temperature coefficient
C.
High resistance
D.
Low cost
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Solution
Constantan has a low temperature coefficient, making it stable for accurate measurements.
Correct Answer: B — Low temperature coefficient
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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 by the projectile?
A.
5 m
B.
10 m
C.
15 m
D.
20 m
Show solution
Solution
Using the formula: h = (u² * sin²θ) / (2g), where u = 20 m/s, θ = 30°, g = 9.8 m/s². h = (20² * (1/4)) / (2 * 9.8) = 5.1 m, approximately 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 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
Show solution
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
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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 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
Show solution
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
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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
Show solution
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
Show solution
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 its velocity?
A.
15 m/s
B.
20 m/s
C.
25 m/s
D.
30 m/s
Show solution
Solution
Horizontal component (vx) = u * cos(θ) = 30 * 0.5 = 15 m/s.
Correct Answer: B — 20 m/s
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