The Question is from neet-physics units-and-measurement conversion-of-units-for-one-system-to-another, and it was created by the author Mr. ANIL KUMAR.G

Question

The dimensions of emf in MKS is

Mr. ANIL KUMAR.G · Accepted Answer

## CORE INFORMATION
✓ **Answer**: D
## CONCEPT FRAMEWORK
**Primary Concept**: Dimensional analysis of Electromotive Force (EMF).
**Related Topics**: Work, Charge, Potential Difference, Energy
**Prerequisites**: Understanding of dimensional analysis, definition of EMF, relationship between work, charge, and potential difference.
## SOLUTION PATHWAY
**Given Information**:
    *   We need to find the dimensions of EMF in MKS system.
    *   EMF is the work done per unit charge.
**Method & Steps**:
    1.  **Relate EMF to Work and Charge:** EMF ($\epsilon$) is defined as the work done (W) per unit charge (Q). So, $\epsilon = \frac{W}{Q}$.
    Reasoning: This is the fundamental definition of EMF.
    Formula/Rule: $\epsilon = \frac{W}{Q}$
    2.  **Determine the Dimensions of Work:** Work is force times distance, $W = F \cdot d$.  Force (F) has dimensions of mass times acceleration, which is $\left[ {{m{MLT}}^{ - 2}} ight]$. Distance (d) has dimensions of length $\left[ {{m{L}}} ight]$. Therefore, the dimensions of work are $\left[ {{m{MLT}}^{ - 2}} ight] 	imes \left[ {{m{L}}} ight] = \left[ {{m{ML}}^2{m{T}}^{ - 2}} ight]$.
    Working: $W = Fd  \Rightarrow [W]=[MLT^{-2}][L] = [ML^2T^{-2}]$
    Key Point: Work is a form of energy.
    3.  **Determine the Dimensions of Charge:** Charge (Q) is a fundamental quantity, so its dimension is $\left[ {{m{Q}}} ight]$.
    Working: [Q] is a fundamental dimension.
    Key Point: We are using charge as a fundamental dimension in this problem.
    4.  **Calculate the Dimensions of EMF:** Now we can divide the dimensions of work by the dimensions of charge: $\left[ {\epsilon} ight] = \frac{{\left[ {{m{ML}}^2{m{T}}^{ - 2}} ight]}}{{\left[ {{m{Q}}} ight]}} = \left[ {{m{ML}}^2{m{T}}^{ - 2}{m{Q}}^{ - 1}} ight]$
    Result: The dimensions of EMF are $\left[ {{m{ML}}^2{m{T}}^{ - 2}{m{Q}}^{ - 1}} ight]$.
    Verification: The derived dimensions match the correct answer.
## OPTION ANALYSIS
[A] :x: This option incorrectly has length raised to the power of -1 and charge to the power of -2.
[B] :x: This option incorrectly has length raised to the power of -2 and charge to the power of -2.
[C] :x: This option incorrectly has length raised to the power of 1 and charge to the power of -1.
[D] ✓: This option correctly represents the dimensions of EMF as  $\left[ {{m{ML}}^2{m{T}}^{ - 2}{m{Q}}^{ - 1}} ight]$
## LEARNING AIDS
**Common Mistakes**:
    1. **Confusing Dimensions:**  Forgetting to use the correct dimensions for force, work, and charge. Always break down the quantities to their fundamental dimensions (M, L, T, Q).
    2. **Incorrect Formula:** Using the wrong relationship between EMF, work, and charge. Remember the basic definition of EMF.
**Quick Tips**:
    1.  Remember the relationship: EMF = Work / Charge.
    2.  Start with basic definitions and work your way up.

Mr. ANIL KUMAR.G · Answer

$\left[ {{\rm{M}}{{\rm{L}}^{ - 1}}{{\rm{T}}^{ - 2}}{{\rm{Q}}^{ - 2}}} \right]$

Mr. ANIL KUMAR.G · Answer

$\left[ {{\rm{M}}{{\rm{L}}^{ - 2}}{{\rm{T}}^{ - 2}}{{\rm{Q}}^{ - 2}}} \right]$

Mr. ANIL KUMAR.G · Answer

$\left[ {{\rm{ML}}{{\rm{T}}^{ - 2}}{{\rm{Q}}^{ - 1}}} \right]$

Mr. ANIL KUMAR.G · Answer

The correct Option is "$\left[ {{\rm{M}}{{\rm{L}}^2}{{\rm{T}}^{ - 2}}{{\rm{Q}}^{ - 1}}} \right]$"

NEET Physics Units And Measurement Conversion Of Units For One System To Another MCQs

Select Topic

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

NEET Subjects

Our Bots

Company