| ⇦ |
| ⇨ |
The dimension of 1/2ε₀E², where ε₀ is permittivity of free space and E is electric field, is:
Options
(a) ML²T⁻²
(b) ML⁻¹T⁻²
(c) ML²T⁻¹
(d) MLT⁻¹
Correct Answer:
ML⁻¹T⁻²
Explanation:
1/2 ?¬ツタE² represents energy density i.e., energy per unit volume.
⇒ [ 1/2 ?¬ツタE² ] = ML²T⁻² / L³
= ML⁻¹T⁻²
Related Questions: - Three resistances P,Q,R each of 2Ω and an unknown resistances S form the four arms
- Calculate the focal length of a reading glass of a person, if the distance
- If a wire is stretched to four times its length, then the specific resistance
- In depletion layer of unbiased P-N junction
- If emf induced in a coil is 2 V by changing the current in it from 8 A to 6 A
Question Type: Memory
(964)
Difficulty Level: Easy
(1008)
Topics: Physical World and Measurement
(103)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Three resistances P,Q,R each of 2Ω and an unknown resistances S form the four arms
- Calculate the focal length of a reading glass of a person, if the distance
- If a wire is stretched to four times its length, then the specific resistance
- In depletion layer of unbiased P-N junction
- If emf induced in a coil is 2 V by changing the current in it from 8 A to 6 A
Question Type: Memory (964)
Difficulty Level: Easy (1008)
Topics: Physical World and Measurement (103)
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

Leave a Reply