| ⇦ | 
| ⇨ |
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: - An electric motor runs on DC source of emf 200 V and draws a current of 10 A.
- An interference pattern was made by using red light. If the red light
- A bar magnet of magnetic moment M is placed at right angles to a magnetic
- Planck’s constant has same dimensions as those of
- A tyre filled with air (27° C and 2 atm) bursts. The temperature of air is (ˠ=1.5)
Topics: Electrostatics
(146)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- An electric motor runs on DC source of emf 200 V and draws a current of 10 A.
- An interference pattern was made by using red light. If the red light
- A bar magnet of magnetic moment M is placed at right angles to a magnetic
- Planck’s constant has same dimensions as those of
- A tyre filled with air (27° C and 2 atm) bursts. The temperature of air is (ˠ=1.5)
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