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 player has a batting average of 45 runs per match. If he plays 10 matches, what is his total expected score? (2020)
A.
400
B.
450
C.
500
D.
550
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Solution
Total expected score = batting average * number of matches = 45 runs/match * 10 matches = 450 runs.
Correct Answer: B — 450
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Q. A player has a batting average of 50 runs after 10 innings. If he scores 60 runs in his next innings, what will be his new average? (2022)
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Solution
Total runs after 10 innings = 50 * 10 = 500. New total runs = 500 + 60 = 560. New average = 560 / 11 = 50.91, which rounds to 55.
Correct Answer: B — 55
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Q. A player has a batting average of 50 runs per match. If he has played 20 matches, how many runs has he scored in total? (2020)
A.
800
B.
1000
C.
900
D.
950
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Solution
Total runs = Batting average * Number of matches = 50 * 20 = 1000 runs.
Correct Answer: B — 1000
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Q. A player has a batting average of 50 runs. If he has played 20 matches, how many runs has he scored in total? (2019)
A.
800
B.
1000
C.
900
D.
700
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Solution
Total runs scored = Batting average * Number of matches = 50 * 20 = 1000 runs.
Correct Answer: A — 800
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Q. A player hits a ball at an angle of 30 degrees with the horizontal. If the initial speed is 20 m/s, what is the vertical component of the velocity? (2022)
A.
10 m/s
B.
15 m/s
C.
20 m/s
D.
5 m/s
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Solution
Vertical component = Initial speed × sin(angle) = 20 m/s × sin(30°) = 20 m/s × 0.5 = 10 m/s.
Correct Answer: A — 10 m/s
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Q. A player hits a baseball at an angle of 30 degrees with an initial speed of 40 m/s. What is the horizontal component of the velocity? (2020)
A.
20
B.
34.64
C.
30
D.
35
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Solution
Horizontal component = initial speed × cos(angle) = 40 m/s × cos(30°) = 40 × √3/2 ≈ 34.64 m/s.
Correct Answer: B — 34.64
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Q. A player runs 100 meters in 12 seconds. What is his average speed in m/s? (2019)
A.
8.33
B.
9.00
C.
7.50
D.
8.00
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Solution
Average speed = Distance / Time = 100 meters / 12 seconds = 8.33 m/s.
Correct Answer: A — 8.33
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Q. A player runs 5 km in 25 minutes. What is his speed in km/h? (2023)
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Solution
Total time in hours = 25 minutes / 60 = 25/60 hours. Speed = distance/time = 5 km / (25/60) h = 12 km/h.
Correct Answer: A — 10
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Q. A player scores 10 goals in 5 matches. What is the average number of goals scored per match? (2019)
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Solution
Average goals per match = total goals / total matches = 10 / 5 = 2.
Correct Answer: B — 2
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Q. A player scores 15 points in a game. If he plays 10 games, what is the total number of points scored? (2023)
A.
150
B.
120
C.
130
D.
140
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Solution
Total points scored = 15 points/game * 10 games = 150 points.
Correct Answer: A — 150
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Q. A player scores 60% of the total points in a game. If the total points are 150, how many points did the player score? (2022)
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Solution
Points scored = 60% of 150 = 0.6 × 150 = 90 points.
Correct Answer: A — 90
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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
Show solution
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
Show solution
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
Show solution
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
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 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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