| ⇦ | 
| ⇨ |
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: - A uniform rope of length 12 m and mass 6 kg hangs vertically from a rigid support.
- In Young’s double slit experiment, for which colour the fringe width is the least
- A cell can be balanced against 110 cm and 100 cm of potentiometer wire, respectively
- The heart of a man pumps 5 litres of blood through the arteries per minute at
- A particle with total energy E is moving in a potential energy region U(x)
Topics: Electrostatics
(146)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A uniform rope of length 12 m and mass 6 kg hangs vertically from a rigid support.
- In Young’s double slit experiment, for which colour the fringe width is the least
- A cell can be balanced against 110 cm and 100 cm of potentiometer wire, respectively
- The heart of a man pumps 5 litres of blood through the arteries per minute at
- A particle with total energy E is moving in a potential energy region U(x)
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