| ⇦ | 
| ⇨ |
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: - 3 persons are initially at the 3 corners of an equilateral triangle
- A thin rod of length L and mass M is held vertically with one end on the floor
- The cadmium rods are used in nuclear reactor to
- If the minimum energy of photons needed to produce photoelectrons is 3 eV,
- Two coils of self inductance 2 mH and 8 mH are placed so close together
Topics: Electrostatics
(146)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- 3 persons are initially at the 3 corners of an equilateral triangle
- A thin rod of length L and mass M is held vertically with one end on the floor
- The cadmium rods are used in nuclear reactor to
- If the minimum energy of photons needed to produce photoelectrons is 3 eV,
- Two coils of self inductance 2 mH and 8 mH are placed so close together
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