| ⇦ |
| ⇨ |
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: - At 10⁰C the value of the density of a fixed mass of an ideal gas divided by its pressure
- A circular disc of radius 0.2 meter is placed in a uniform magnetic field of induction
- A thin circular ring of mass M and radius R is rotating in a horizontal plane about
- The ratio of magnetic dipole moment of an electron of charge e and mass m
- A 220 volt and 1000 watt bulb is connected across a 110 volt mains supply.
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
- At 10⁰C the value of the density of a fixed mass of an ideal gas divided by its pressure
- A circular disc of radius 0.2 meter is placed in a uniform magnetic field of induction
- A thin circular ring of mass M and radius R is rotating in a horizontal plane about
- The ratio of magnetic dipole moment of an electron of charge e and mass m
- A 220 volt and 1000 watt bulb is connected across a 110 volt mains supply.
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