| ⇦ |
| ⇨ |
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 coherent monochromatic beams of intensities I and 4I respectively, are superposed.
- The concentric spheres of radii R and r have positive charges q₁ and q₂ with equal
- The distance of the closest approach of an alpha particle fired at a nucleus
- The respective speeds of the molecules are 1,2,3,4 and 5 km/s.
- An air core coil and an electric bulb are connected in series with an AC source.
Topics: Electrostatics
(146)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Two coherent monochromatic beams of intensities I and 4I respectively, are superposed.
- The concentric spheres of radii R and r have positive charges q₁ and q₂ with equal
- The distance of the closest approach of an alpha particle fired at a nucleus
- The respective speeds of the molecules are 1,2,3,4 and 5 km/s.
- An air core coil and an electric bulb are connected in series with an AC source.
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