Q. A circular loop of wire carries a current. What is the shape of the magnetic field lines inside the loop?
A.
Straight lines
B.
Concentric circles
C.
Uniform field
D.
Radial lines
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Solution
Inside a circular loop of wire carrying current, the magnetic field lines are uniform and parallel, indicating a uniform magnetic field.
Correct Answer: C — Uniform field
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Q. A circular loop of wire carrying current produces a magnetic field. What is the shape of the magnetic field lines? (2020)
A.
Straight lines
B.
Concentric circles
C.
Ellipses
D.
Spirals
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Solution
The magnetic field lines around a circular loop of wire are concentric circles centered on the axis of the loop.
Correct Answer: B — Concentric circles
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Q. A circular loop of wire carrying current produces a magnetic field. What is the shape of the magnetic field lines inside the loop? (2020)
A.
Straight lines
B.
Concentric circles
C.
Parallel lines
D.
Uniform field
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Solution
Inside the loop, the magnetic field lines are nearly uniform and parallel, indicating a strong and consistent magnetic field.
Correct Answer: D — Uniform field
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Q. A circular loop of wire is placed in a uniform magnetic field. If the magnetic field is increased, what happens to the induced EMF in the loop?
A.
Increases
B.
Decreases
C.
Remains constant
D.
Becomes zero
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Solution
According to Faraday's law of electromagnetic induction, an increase in magnetic field through the loop induces an EMF in the loop.
Correct Answer: A — Increases
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Q. A circular loop of wire is placed in a uniform magnetic field. What happens to the induced EMF if the magnetic field strength is doubled?
A.
Induced EMF is halved
B.
Induced EMF remains the same
C.
Induced EMF is doubled
D.
Induced EMF is quadrupled
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Solution
According to Faraday's law of electromagnetic induction, the induced EMF is directly proportional to the rate of change of magnetic flux. If the magnetic field strength is doubled, the induced EMF will also double.
Correct Answer: C — Induced EMF is doubled
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Q. A circular loop of wire is placed in a uniform magnetic field. What happens to the induced EMF if the area of the loop is increased?
A.
Increases
B.
Decreases
C.
Remains the same
D.
Depends on the magnetic field strength
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Solution
According to Faraday's law of electromagnetic induction, the induced EMF is proportional to the rate of change of magnetic flux. Increasing the area increases the flux, thus increasing the induced EMF.
Correct Answer: A — Increases
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Q. A class has an average score of 65. If a new student with a score of 85 joins, how does the average change if the class size increases by one?
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Solution
Total score = 65 * n + 85. New average = (65n + 85) / (n + 1).
Correct Answer: B — 67
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Q. A class of 30 students has an average score of 75. If one student scores 90, what will be the new average? (2022)
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Solution
Total score = 75 * 30 = 2250. New total = 2250 - 75 + 90 = 2265. New average = 2265 / 30 = 75.5.
Correct Answer: B — 77
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Q. A clock shows 3:15. What is the angle between the hour and the minute hand? (2023)
A.
45 degrees
B.
90 degrees
C.
112.5 degrees
D.
135 degrees
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Solution
At 3:15, the hour hand is at 97.5 degrees and the minute hand is at 90 degrees. The angle is 97.5 - 90 = 7.5 degrees.
Correct Answer: C — 112.5 degrees
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Q. A coil has an inductance of 0.5 H and is connected to a 50 Hz AC supply. What is the inductive reactance? (2020)
A.
25.13 Ω
B.
31.42 Ω
C.
50 Ω
D.
100 Ω
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Solution
Inductive reactance Xl = 2πfL = 2π(50)(0.5) = 31.42 Ω.
Correct Answer: B — 31.42 Ω
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Q. A coil of wire has 100 turns and is placed in a magnetic field of strength 0.5 T. If the area of the coil is 0.1 m² and the magnetic field is perpendicular to the coil, what is the magnetic flux through the coil? (2022)
A.
5 Wb
B.
0.5 Wb
C.
10 Wb
D.
0.05 Wb
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Solution
Magnetic flux (Φ) = B × A × cos(θ). Here, B = 0.5 T, A = 0.1 m², and θ = 0° (cos(0) = 1). Thus, Φ = 0.5 T × 0.1 m² × 1 = 0.05 Wb.
Correct Answer: A — 5 Wb
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Q. A coil of wire has 100 turns and is placed in a magnetic field of strength 0.5 T. If the area of the coil is 0.1 m², what is the maximum magnetic flux through the coil? (2022)
A.
5 Wb
B.
0.5 Wb
C.
10 Wb
D.
1 Wb
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Solution
Magnetic flux (Φ) = B × A × N = 0.5 T × 0.1 m² × 100 = 5 Wb.
Correct Answer: A — 5 Wb
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Q. A coil of wire is placed in a changing magnetic field. What happens to the induced current if the resistance of the coil is increased?
A.
Induced current increases
B.
Induced current decreases
C.
Induced current remains the same
D.
Induced current becomes zero
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Solution
According to Ohm's law, if the resistance increases while the induced EMF remains constant, the induced current will decrease.
Correct Answer: B — Induced current decreases
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Q. A coil of wire is placed in a changing magnetic field. What phenomenon is observed?
A.
Electromagnetic induction
B.
Magnetic resonance
C.
Electrolysis
D.
Thermal conduction
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Solution
According to Faraday's law of electromagnetic induction, a changing magnetic field induces an electromotive force (EMF) in the coil.
Correct Answer: A — Electromagnetic induction
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Q. A coil of wire is placed in a magnetic field. If the magnetic field strength is increased, what happens to the induced EMF?
A.
Increases
B.
Decreases
C.
Remains constant
D.
Becomes zero
Show solution
Solution
According to Faraday's law of electromagnetic induction, an increase in magnetic field strength induces a greater EMF.
Correct Answer: A — Increases
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Q. A coil of wire is placed in a magnetic field. If the magnetic field strength is doubled, what happens to the induced EMF?
A.
It doubles
B.
It remains the same
C.
It halves
D.
It quadruples
Show solution
Solution
Doubling the magnetic field strength will double the induced EMF, as it is directly proportional to the magnetic field strength.
Correct Answer: A — It doubles
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Q. A coil of wire is placed in a magnetic field. If the magnetic field strength is increased, what happens to the induced EMF in the coil?
A.
It increases
B.
It decreases
C.
It remains the same
D.
It becomes zero
Show solution
Solution
According to Faraday's law of electromagnetic induction, the induced EMF in a coil is directly proportional to the rate of change of magnetic flux. Increasing the magnetic field strength increases the magnetic flux, thus increasing the induced EMF.
Correct Answer: A — It increases
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Q. A coil with 100 turns and an area of 0.01 m² is placed in a magnetic field of 0.5 T. What is the magnetic flux through the coil?
A.
0.5 Wb
B.
0.1 Wb
C.
0.05 Wb
D.
0.01 Wb
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Solution
Magnetic flux Φ = B * A * N = 0.5 T * 0.01 m² * 100 = 0.5 Wb.
Correct Answer: A — 0.5 Wb
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Q. A coil with 100 turns is placed in a magnetic field that changes at a rate of 0.5 T/s. What is the induced EMF in the coil?
A.
50 V
B.
100 V
C.
200 V
D.
25 V
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Solution
Using Faraday's law, EMF = -N * (dΦ/dt) = -100 * 0.5 = -50 V. The induced EMF is 50 V.
Correct Answer: B — 100 V
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Q. A coil with 100 turns is placed in a magnetic field that changes from 0.2 T to 0.5 T in 2 seconds. What is the induced EMF?
A.
15 V
B.
30 V
C.
5 V
D.
10 V
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Solution
Induced EMF (ε) = -N(dB/dt) = -100 * (0.5 - 0.2)/2 = -15 V.
Correct Answer: B — 30 V
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Q. A coil with 100 turns is placed in a magnetic field that changes from 0.5 T to 1.5 T in 2 seconds. What is the induced EMF in the coil?
A.
50 V
B.
100 V
C.
200 V
D.
400 V
Show solution
Solution
Using Faraday's law, EMF = -N * (ΔB/Δt) = -100 * ((1.5 - 0.5) / 2) = -100 * (1 / 2) = -50 V. The magnitude is 50 V.
Correct Answer: B — 100 V
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Q. A coil with 100 turns is placed in a magnetic field that changes from 0.5 T to 1.5 T in 2 seconds. What is the induced EMF?
A.
50 V
B.
100 V
C.
200 V
D.
400 V
Show solution
Solution
Induced EMF = -N * (ΔB/Δt) = -100 * ((1.5 - 0.5)/2) = -100 * (1/2) = -50 V. The magnitude is 50 V.
Correct Answer: B — 100 V
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Q. A coin is tossed three times. What is the probability of getting at least one head?
A.
1/8
B.
1/2
C.
7/8
D.
3/8
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Solution
Probability of getting no heads (all tails) = (1/2)^3 = 1/8. Therefore, probability of at least one head = 1 - 1/8 = 7/8.
Correct Answer: C — 7/8
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Q. A composite body consists of a solid cylinder and a solid sphere, both of mass M and radius R. What is the total moment of inertia about the same axis?
A.
(7/10) MR^2
B.
(9/10) MR^2
C.
(11/10) MR^2
D.
(13/10) MR^2
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Solution
The total moment of inertia is I_cylinder + I_sphere = (1/2 MR^2) + (2/5 MR^2) = (7/10) MR^2.
Correct Answer: A — (7/10) MR^2
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Q. A concave lens has a focal length of -10 cm. What is the nature of the image formed when an object is placed at 15 cm from the lens?
A.
Real and inverted
B.
Virtual and upright
C.
Real and upright
D.
Virtual and inverted
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Solution
Concave lenses always produce virtual and upright images regardless of the object distance.
Correct Answer: B — Virtual and upright
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Q. A concave lens has a focal length of -12 cm. What is the image distance when the object is placed at 24 cm?
A.
8 cm
B.
12 cm
C.
16 cm
D.
20 cm
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Solution
Using the lens formula, 1/f = 1/v - 1/u, we find v = 8 cm.
Correct Answer: A — 8 cm
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Q. A concave lens has a focal length of -15 cm. What is the nature of the image formed by the lens when an object is placed at 30 cm from the lens?
A.
Real and inverted
B.
Virtual and erect
C.
Real and erect
D.
Virtual and inverted
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Solution
For a concave lens, the image formed is virtual and erect when the object is placed beyond the focal length.
Correct Answer: B — Virtual and erect
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Q. A concave lens has a focal length of -20 cm. What is the nature of the image formed when an object is placed at 30 cm from the lens?
A.
Real and inverted
B.
Virtual and erect
C.
Real and erect
D.
Virtual and inverted
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Solution
For a concave lens, the image formed is virtual and erect when the object is placed at any distance.
Correct Answer: B — Virtual and erect
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Q. A concave lens has a focal length of 25 cm. What is the image distance when the object is placed at 50 cm?
A.
-16.67 cm
B.
-25 cm
C.
-50 cm
D.
-75 cm
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Solution
Using the lens formula, the image distance is calculated to be -16.67 cm.
Correct Answer: A — -16.67 cm
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Q. A concave mirror has a focal length of 10 cm. An object is placed 30 cm in front of the mirror. Where will the image be formed?
A.
10 cm
B.
15 cm
C.
20 cm
D.
30 cm
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Solution
Using the mirror formula, 1/f = 1/v + 1/u, where f = -10 cm (concave mirror), u = -30 cm. Solving gives v = -15 cm, which means the image is formed 15 cm in front of the mirror.
Correct Answer: C — 20 cm
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