
If the capacitance of a nano capacitor is measured in terms of a unit \['\mu '\] made by combining the electronic charge \['e'\] , bohr radius \['{a_0}'\] , Planck’s constant \['h'\] and speed of light \['c'\] then which of the following relation is possible?
A) \[\mu = \dfrac{{{e^2}c}}{{h{a_0}}}\]
B) \[\mu = \dfrac{{{e^2}h}}{{c{a_0}}}\]
C) \[\mu = \dfrac{{hc}}{{{e^2}{a_0}}}\]
D) \[\mu = \dfrac{{{e^2}{a_0}}}{{hc}}\]
Answer
139.5k+ views
Hint: Check the dimensions of the both sides of the given equations. The one which has dimensions on left hand and right hand equal will be the correct option.
Complete step by step solution:
We will solve this formula with the help of dimensional analysis . If a relation is correct, then the dimensions on the right hand side will be equal to the dimensions on the left hand side . All the physical quantities in physics can be expressed in terms of some sort of combinations of base quantities ( length, mass, time being the most common).
Dimensions of any quantity in physics are the powers to which the fundamental ( base) quantities can be raised to represent that quantity completely .
Now let us assume that given four quantities are dimensionally comparable and are related as follows :
$[\mu ] = {[e]^w}{[{a_0}]^x}{[h]^y}{[c]^z}$ ……….(i)
where $w,x,y,z$ are the powers of the to which these quantities are raised.
$
[{M^{ - 1}}{L^{ - 2}}{T^4}{A^2}] = {[AT]^w}{[L]^x}{[M{L^2}{T^{ - 1}}]^y}{[L{T^{ - 1}}]^z} \\
[{M^{ - 1}}{L^{ - 2}}{T^4}{A^2}] = [{M^y}{L^{x + 2y + z}}{T^{w - y - z}}{A^w}] \\
\\
$
Comparing both sides we get-
$
w = 2 \\
y = - 1 \\
$
$
x + 2y + z = - 2 \\
w - y - z = 4 \\
$
By solving above equations we get: $w = 2,x = 1,y = - 1,z = - 1$
Equation (i) now becomes –
$
[\mu ] = {[e]^2}{[{a_0}]^1}{[h]^{ - 1}}{[c]^{ - 1}} \\
\mu = \dfrac{{{e^2}{a_0}}}{{hc}} \\
$
We have got the answer.
Hence , the correct option is (D).
Note: We have to keep in mind that while writing dimensional formula we need to write it only in terms of fundamental units and not derived units. This will not work if instead we write the derived units.
Complete step by step solution:
We will solve this formula with the help of dimensional analysis . If a relation is correct, then the dimensions on the right hand side will be equal to the dimensions on the left hand side . All the physical quantities in physics can be expressed in terms of some sort of combinations of base quantities ( length, mass, time being the most common).
Dimensions of any quantity in physics are the powers to which the fundamental ( base) quantities can be raised to represent that quantity completely .
Now let us assume that given four quantities are dimensionally comparable and are related as follows :
$[\mu ] = {[e]^w}{[{a_0}]^x}{[h]^y}{[c]^z}$ ……….(i)
where $w,x,y,z$ are the powers of the to which these quantities are raised.
$
[{M^{ - 1}}{L^{ - 2}}{T^4}{A^2}] = {[AT]^w}{[L]^x}{[M{L^2}{T^{ - 1}}]^y}{[L{T^{ - 1}}]^z} \\
[{M^{ - 1}}{L^{ - 2}}{T^4}{A^2}] = [{M^y}{L^{x + 2y + z}}{T^{w - y - z}}{A^w}] \\
\\
$
Comparing both sides we get-
$
w = 2 \\
y = - 1 \\
$
$
x + 2y + z = - 2 \\
w - y - z = 4 \\
$
By solving above equations we get: $w = 2,x = 1,y = - 1,z = - 1$
Equation (i) now becomes –
$
[\mu ] = {[e]^2}{[{a_0}]^1}{[h]^{ - 1}}{[c]^{ - 1}} \\
\mu = \dfrac{{{e^2}{a_0}}}{{hc}} \\
$
We have got the answer.
Hence , the correct option is (D).
Note: We have to keep in mind that while writing dimensional formula we need to write it only in terms of fundamental units and not derived units. This will not work if instead we write the derived units.
Recently Updated Pages
Average fee range for JEE coaching in India- Complete Details

Difference Between Rows and Columns: JEE Main 2024

Difference Between Length and Height: JEE Main 2024

Difference Between Natural and Whole Numbers: JEE Main 2024

Algebraic Formula

Difference Between Constants and Variables: JEE Main 2024

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

Other Pages
Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

Important Questions for CBSE Class 11 Physics Chapter 1 - Units and Measurement

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line
