| ⇦ | 
| ⇨ |
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: - Two identical capacitors are first connected in series and then in parallel.
- A wheel of mass 10 kg has a moment of inertia of 10 kgm² about its own axis
- A nucleus at rest splits into two nuclear parts having radii in the ratio 1:2.
- Two forces of magnitude 8N and 15N respectively act at a point so as to make the resultant force
- Four cells, each of emf E and internal resistance r, are connected in series across
Topics: Electrostatics
(146)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Two identical capacitors are first connected in series and then in parallel.
- A wheel of mass 10 kg has a moment of inertia of 10 kgm² about its own axis
- A nucleus at rest splits into two nuclear parts having radii in the ratio 1:2.
- Two forces of magnitude 8N and 15N respectively act at a point so as to make the resultant force
- Four cells, each of emf E and internal resistance r, are connected in series across
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