| ⇦ |
| ⇨ |
An electric dipole of length 1 cm is placed with the axis making an angle of 30⁰ to an electric field of strength 10⁴ NC⁻¹. If it experiences a torque of Nm, the potential energy of the dipole is
Options
(a) 2.45 J
(b) 0.0245 J
(c) 245.0 J
(d) 24.5 J
Correct Answer:
24.5 J
Explanation:
No explanation available. Be the first to write the explanation for this question by commenting below.
Related Questions: - Four identical cells of emf Ԑ and internal resistance r are to be connected in series.
- A 1m long wire of diameter 0.31 mm has a resistance of 4.2Ω. If it is replaced by
- If the angle of minimum deviation is 60° for an equilateral prism, then the refractive
- The mean radius of the earth is R, its angular speed on its own axis is ?
- Imagine a new planet having the same density as that of earth but it is 3 times bigger
Topics: Electrostatics
(146)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Four identical cells of emf Ԑ and internal resistance r are to be connected in series.
- A 1m long wire of diameter 0.31 mm has a resistance of 4.2Ω. If it is replaced by
- If the angle of minimum deviation is 60° for an equilateral prism, then the refractive
- The mean radius of the earth is R, its angular speed on its own axis is ?
- Imagine a new planet having the same density as that of earth but it is 3 times bigger
Topics: Electrostatics (146)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends

Torque=PE sin¢
10√2= P×10^4 sin 30°
P=2√2×10^-3
U=PE cos 30°
U=2√2×10^-3× 10^4 cos 30°
=24.5 J
Torque=PE sin¢
10√2= P×10^4 sin 30°
P=2√2×10^-3
U=PE cos 30°
U=2√2×10^-3× 10^4 cos 30°
=24.5 J
Torque=PE sin¢
10√2= P×10^4 sin 30°
P=2√2×10^-3
U=PE cos 30°
U=2√2×10^-3× 10^4 cos 30°
=24.5 J