| ⇦ | 
| ⇨ |
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: - If α (current gain) of transistor is 0.98, then what is the value of β
- A bicyclist comes to a skidding stop in 10 m.During this process, the force
- The potential differences across the resistance, capacitance and inductance
- In Young’s double slit interference experiment, using two coherent waves
- The force of repulsion between two electrons at a certain distance is F.
Topics: Electrostatics
(146)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- If α (current gain) of transistor is 0.98, then what is the value of β
- A bicyclist comes to a skidding stop in 10 m.During this process, the force
- The potential differences across the resistance, capacitance and inductance
- In Young’s double slit interference experiment, using two coherent waves
- The force of repulsion between two electrons at a certain distance is F.
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