Q. If the magnetic field through a loop is increased uniformly, what happens to the induced current in the loop?
A.
It flows in the direction of the magnetic field
B.
It flows in the opposite direction to the magnetic field
C.
It remains constant
D.
It stops flowing
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Solution
According to Lenz's law, the induced current will flow in the opposite direction to oppose the increase in magnetic flux.
Correct Answer: B — It flows in the opposite direction to the magnetic field
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Q. If the magnetic field through a loop is increasing at a constant rate, what can be said about the induced current?
A.
It is constant
B.
It is increasing
C.
It is decreasing
D.
It is zero
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Solution
If the magnetic field is increasing at a constant rate, the induced EMF is also increasing, which means the induced current is increasing.
Correct Answer: B — It is increasing
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Q. If the magnetic flux through a coil changes from 0.2 Wb to 0.5 Wb in 2 seconds, what is the average induced EMF? (2021)
A.
150 V
B.
100 V
C.
75 V
D.
50 V
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Solution
Average induced EMF = -ΔΦ/Δt = -(0.5 Wb - 0.2 Wb) / 2 s = -0.15 Wb/s = 150 V.
Correct Answer: B — 100 V
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Q. If the magnetic flux through a coil changes from 0.5 Wb to 0.2 Wb in 2 seconds, what is the average induced emf? (2023)
A.
150 V
B.
100 V
C.
75 V
D.
50 V
Show solution
Solution
Average induced emf = -ΔΦ/Δt = -(0.2 - 0.5)/2 = 0.15 V = 150 V.
Correct Answer: B — 100 V
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Q. If the magnetic flux through a coil changes, what happens to the induced EMF? (2023)
A.
It remains constant
B.
It increases
C.
It decreases
D.
It is induced
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Solution
An induced EMF is generated when there is a change in magnetic flux through a coil.
Correct Answer: D — It is induced
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Q. If the mass of a satellite is doubled while keeping its orbital radius constant, what happens to the gravitational force acting on it?
A.
It doubles.
B.
It remains the same.
C.
It halves.
D.
It quadruples.
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Solution
The gravitational force acting on the satellite will double if its mass is doubled, as gravitational force is directly proportional to mass.
Correct Answer: A — It doubles.
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Q. If the mass of a simple harmonic oscillator is doubled while keeping the spring constant the same, how does the period change?
A.
Increases
B.
Decreases
C.
Remains the same
D.
Cannot be determined
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Solution
The period T = 2π√(m/k). If m is doubled, T increases.
Correct Answer: A — Increases
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Q. If the mass of an object is 10 kg and it is located at a point where the gravitational potential is -30 J/kg, what is the gravitational potential energy of the object?
A.
-300 J
B.
-30 J
C.
300 J
D.
30 J
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Solution
Gravitational potential energy U = m * V = 10 * (-30) = -300 J.
Correct Answer: A — -300 J
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Q. If the mass of an object is 5 kg and it is at a height of 10 m above the ground, what is its gravitational potential energy?
A.
50 J
B.
100 J
C.
500 J
D.
0 J
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Solution
Gravitational potential energy U = mgh = 5 * 10 * 10 = 500 J.
Correct Answer: B — 100 J
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Q. If the mass of an object is doubled, how does its gravitational potential energy change at a constant height?
A.
It remains the same
B.
It doubles
C.
It halves
D.
It quadruples
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Solution
The gravitational potential energy U = mgh will double if the mass m is doubled at constant height h.
Correct Answer: B — It doubles
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Q. If the mass of an object is halved and the distance from the center of the Earth is doubled, what happens to the gravitational potential energy?
A.
It remains the same.
B.
It doubles.
C.
It halves.
D.
It quadruples.
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Solution
Gravitational potential energy is directly proportional to mass and inversely proportional to distance, so it halves.
Correct Answer: C — It halves.
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Q. If the mass of an object is halved, how does its gravitational potential energy change at a constant height?
A.
It doubles
B.
It halves
C.
It remains the same
D.
It becomes zero
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Solution
The gravitational potential energy U = mgh is directly proportional to the mass, so if the mass is halved, the potential energy also halves.
Correct Answer: B — It halves
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Q. If the mass of an object is measured as 10.0 kg with an uncertainty of ±0.2 kg, what is the percentage error?
A.
2%
B.
0.2%
C.
0.02%
D.
0.5%
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Solution
Percentage error = (absolute error / measured value) * 100 = (0.2 / 10.0) * 100 = 2%.
Correct Answer: A — 2%
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Q. If the mass of an object is measured as 10.0 kg with an uncertainty of ±0.2 kg, what is the percentage error in the mass measurement?
A.
2%
B.
0.2%
C.
0.5%
D.
1%
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Solution
Percentage error = (absolute error / measured value) * 100 = (0.2 / 10.0) * 100 = 2%.
Correct Answer: D — 1%
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Q. If the mass of an object is measured as 50 kg with an uncertainty of ±0.5 kg, what is the percentage error in the mass measurement?
A.
1%
B.
0.1%
C.
0.5%
D.
2%
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Solution
Percentage error = (absolute error / measured value) * 100 = (0.5 / 50) * 100 = 1%.
Correct Answer: A — 1%
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Q. If the mass of an object is measured in kilograms and its volume in cubic meters, what is the unit of density? (2022)
A.
kg/m
B.
kg/m^3
C.
m^3/kg
D.
m/kg
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Solution
Density is defined as mass per unit volume, so the unit of density is kg/m³.
Correct Answer: B — kg/m^3
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Q. If the mass of an object is tripled, how does its gravitational potential energy change at a constant height?
A.
It triples
B.
It doubles
C.
It remains the same
D.
It increases by a factor of nine
Show solution
Solution
Gravitational potential energy U = mgh, so if mass is tripled, U also triples.
Correct Answer: A — It triples
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Q. If the mass of an object is tripled, how does the gravitational potential energy change for a constant height?
A.
It triples
B.
It doubles
C.
It halves
D.
It remains the same
Show solution
Solution
Gravitational potential energy is directly proportional to mass, so it triples.
Correct Answer: A — It triples
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Q. If the mass of the Earth is 6 x 10^24 kg and the radius is 6.4 x 10^6 m, what is the gravitational acceleration at the surface of the Earth?
A.
9.8 m/s²
B.
10 m/s²
C.
9.81 m/s²
D.
11 m/s²
Show solution
Solution
Using g = G * M / R², where G = 6.67 x 10^-11 N m²/kg², M = 6 x 10^24 kg, and R = 6.4 x 10^6 m, we find g ≈ 9.8 m/s².
Correct Answer: A — 9.8 m/s²
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Q. If the mass of the Earth is 6 x 10^24 kg and the radius is 6.4 x 10^6 m, what is the gravitational force on a 1 kg mass at the surface?
A.
9.8 N
B.
6.67 N
C.
10 N
D.
8.9 N
Show solution
Solution
Using F = G * (m1 * m2) / r^2, F = (6.67 x 10^-11) * (6 x 10^24 * 1) / (6.4 x 10^6)^2 = 9.8 N.
Correct Answer: A — 9.8 N
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Q. If the mass of the Earth is 6 × 10^24 kg and its radius is 6.4 × 10^6 m, what is the acceleration due to gravity at its surface?
A.
9.8 m/s²
B.
10 m/s²
C.
9.81 m/s²
D.
8.9 m/s²
Show solution
Solution
g = G * M / R² = (6.67 × 10^-11) * (6 × 10^24) / (6.4 × 10^6)² = 9.81 m/s²
Correct Answer: C — 9.81 m/s²
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Q. If the mass of the Earth is M and its radius is R, what is the gravitational acceleration at the surface of the Earth?
A.
g = GM/R^2
B.
g = GMR^2
C.
g = G/MR^2
D.
g = MR/G
Show solution
Solution
The gravitational acceleration at the surface of the Earth is given by g = GM/R^2.
Correct Answer: A — g = GM/R^2
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Q. If the mass of the Earth is M and the radius is R, what is the gravitational acceleration at the surface of the Earth?
A.
g = GM/R^2
B.
g = GMR^2
C.
g = G/R
D.
g = M/R^2
Show solution
Solution
The gravitational acceleration at the surface of the Earth is given by g = GM/R^2.
Correct Answer: A — g = GM/R^2
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Q. If the mass of the Earth is M and the radius is R, what is the gravitational force acting on a satellite of mass m at a height h?
A.
GmM/R^2
B.
GmM/(R+h)^2
C.
GmM/(R-h)^2
D.
GmM/h^2
Show solution
Solution
The gravitational force F acting on a satellite at height h is given by F = GmM/(R+h)^2, where G is the gravitational constant.
Correct Answer: B — GmM/(R+h)^2
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Q. If the mass of the Earth were to double while its radius remained the same, what would happen to the gravitational acceleration at its surface?
A.
It would double
B.
It would remain the same
C.
It would halve
D.
It would increase by a factor of four
Show solution
Solution
Gravitational acceleration is directly proportional to mass, so if the mass doubles, g also doubles.
Correct Answer: A — It would double
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Q. If the mass of the Earth were to double while its radius remains the same, what would happen to the gravitational acceleration at its surface?
A.
It would double
B.
It would remain the same
C.
It would halve
D.
It would increase by a factor of four
Show solution
Solution
Gravitational acceleration is directly proportional to mass; thus, if the mass doubles, g also doubles.
Correct Answer: A — It would double
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Q. If the mass of the Earth were to double while its radius remains the same, what would happen to the acceleration due to gravity? (2022)
A.
It would double
B.
It would remain the same
C.
It would halve
D.
It would increase by a factor of four
Show solution
Solution
If the mass of the Earth doubles, the acceleration due to gravity also doubles, as it is directly proportional to mass.
Correct Answer: A — It would double
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Q. If the mass of the Earth were to double while the radius remains the same, what would happen to the weight of an object on its surface?
A.
It would double
B.
It would remain the same
C.
It would increase by a factor of four
D.
It would decrease by half
Show solution
Solution
Weight is proportional to mass; thus, if the Earth's mass doubles, the weight of an object on its surface also doubles.
Correct Answer: A — It would double
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Q. If the mass of the Earth were to double, what would happen to the gravitational force between the Earth and an object on its surface?
A.
It would double
B.
It would remain the same
C.
It would halve
D.
It would quadruple
Show solution
Solution
F ∝ M; if M doubles, F also doubles.
Correct Answer: A — It would double
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Q. If the mass of the Earth were to double, what would happen to the gravitational force it exerts on an object at its surface?
A.
It would double
B.
It would remain the same
C.
It would increase by a factor of four
D.
It would decrease by half
Show solution
Solution
If the mass of the Earth doubles, the gravitational force on an object at its surface would also double.
Correct Answer: A — It would double
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