Match the physical quantities with their Dimensional formulae a) Coefficient of Viscosity (η) b) Young's Modulus (Y) c) Permittivity (ε) d) Universal Gravitational constant (G) (i) M⁻¹L⁻³T⁴A² (ii) ML⁻¹T⁻¹ (iii) ML⁻¹T⁻² (iv) M⁻¹L³T⁻²

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Physics Q81 – AP-EAMCET 2026

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Home Physics Units & MeasurementsQuestion #541

Match the physical quantities with their Dimensional formulae

\begin{array}{lcl} a) Coefficient of Viscosity (η) & (i) M^{- 1} L^{- 3} T^{4} A^{2} \\ b) Young's Modulus (Y) & (ii) M L^{- 1} T^{- 1} \\ c) Permittivity (ϵ) & (iii) M L^{- 1} T^{- 2} \\ d) Universal Gravitational constant (G) & (iv) M^{- 1} L^{3} T^{- 2} \end{array}

Clue 1 of 3: Dimensional Analysis

Dimensional analysis is a powerful tool to verify physical equations and derive relationships between physical quantities by ensuring dimensional consistency.

🧠 Full Solution Path

Step 1: Determine Dimensional Formula for Coefficient of Viscosity (

η

)✦ Active

The coefficient of viscosity ( $η$ ) is defined by Stokes' law, $F = 6 π η r v$ , where $F$ is force, $r$ is radius, and $v$ is velocity. Rearranging for $η$ : $η = \frac{F}{6 π r v}$ . Since $6 π$ is a dimensionless constant, its dimensions are ignored. The dimensions are:

[η] = \frac{[F]}{[r] [v]} = \frac{[M L T^{- 2}]}{[L] [L T^{- 1}]} = \frac{[M L T^{- 2}]}{[L^{2} T^{- 1}]} = [M L^{- 1} T^{- 1}]

This matches option (ii).

💡 Teacher's Secret Hint

Remember that constants like $2 π$ , $4 π$ , or $6 π$ are dimensionless and do not affect the overall dimensional formula.

Step 2: Determine Dimensional Formula for Young's Modulus (Y)○ Expand

Young's modulus (Y) is defined as stress divided by strain. Strain is dimensionless. Stress is force per unit area ( $S t r e s s = \frac{F}{A}$ ). Therefore, the dimensions of Young's modulus are the same as stress:

[Y] = [S t r e s s] = \frac{[F]}{[A]} = \frac{[M L T^{- 2}]}{[L^{2}]} = [M L^{- 1} T^{- 2}]

This matches option (iii).

💡 Teacher's Secret Hint

Always check if a quantity in a formula is dimensionless (like strain, angles, refractive index). They simplify dimensional analysis.

Step 3: Determine Dimensional Formula for Permittivity (

ϵ

)○ Expand

Permittivity ( $ϵ$ ) appears in Coulomb's Law, $F = \frac{1}{4 π ϵ} \frac{q_{1} q_{2}}{r^{2}}$ , where $F$ is force, $q_{1}, q_{2}$ are charges, and $r$ is distance. Rearranging for $ϵ$ : $ϵ = \frac{1}{4 π F} \frac{q_{1} q_{2}}{r^{2}}$ . The dimension of charge $q$ is $[A T]$ (current $\times$ time).

[ϵ] = \frac{[q_{1}] [q_{2}]}{[F] [r^{2}]} = \frac{[A T] [A T]}{[M L T^{- 2}] [L^{2}]} = \frac{[A^{2} T^{2}]}{[M L^{3} T^{- 2}]} = [M^{- 1} L^{- 3} T^{4} A^{2}]

This matches option (i).

💡 Teacher's Secret Hint

For electromagnetism, current (A) is often taken as a fundamental dimension along with M, L, T.

Step 4: Determine Dimensional Formula for Universal Gravitational Constant (G)○ Expand

The universal gravitational constant (G) is found in Newton's Law of Universal Gravitation, $F = G \frac{m_{1} m_{2}}{r^{2}}$ , where $F$ is gravitational force, $m_{1}, m_{2}$ are masses, and $r$ is distance. Rearranging for G: $G = \frac{F r^{2}}{m_{1} m_{2}}$ .

[G] = \frac{[F] [r^{2}]}{[m_{1}] [m_{2}]} = \frac{[M L T^{- 2}] [L^{2}]}{[M] [M]} = \frac{[M L^{3} T^{- 2}]}{[M^{2}]} = [M^{- 1} L^{3} T^{- 2}]

This matches option (iv).

💡 Teacher's Secret Hint

Be careful with powers of fundamental dimensions; a small error in an exponent can lead to a completely different result.

Step 5: Match the Quantities to their Dimensional Formulae○ Expand

Based on the derivations:

a) Coefficient of Viscosity ( $η$ ) \Rightarrow (ii) $[M L^{- 1} T^{- 1}]$

b) Young's Modulus (Y) \Rightarrow (iii) $[M L^{- 1} T^{- 2}]$

c) Permittivity ( $ϵ$ ) \Rightarrow (i) $[M^{- 1} L^{- 3} T^{4} A^{2}]$

d) Universal Gravitational constant (G) \Rightarrow (iv) $[M^{- 1} L^{3} T^{- 2}]$

This corresponds to option 2.

💡 Teacher's Secret Hint

Double-check each match with the given options to avoid silly mistakes.

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Physics Question 81 – AP-EAMCET 2026

Clue 1 of 3: Dimensional Analysis

🧠 Full Solution Path