Which of the following equations is dimensionally incorrect? Where, t = time, h = height, s = surfac

Q: How do I quickly find the dimension of viscosity $\eta$?

You can easily derive it using Stokes' Law: $F = 6\pi\eta r v$. Rearranging for $\eta$, we get $[\eta] = \frac{[F]}{[r][v]} = \frac{[MLT^{-2}]}{[L][LT^{-1}]} = [ML^{-1}T^{-1}]$.

Q: What is the dimension of permittivity $\varepsilon$?

From Coulomb's Law, $F = \frac{1}{4\pi\varepsilon} \frac{q_1 q_2}{r^2}$. So, $[\varepsilon] = \frac{[q^2]}{[F][r^2]} = \frac{[A^2 T^2]}{[MLT^{-2}][L^2]} = [M^{-1}L^{-3}T^4 A^2]$. However, in option (c), grouping $\varepsilon E$ together makes the calculation much faster!

Q: Why is angle $\theta$ considered dimensionless?

Angle is defined as the ratio of arc length to radius ($\theta = l/r$). Since both are lengths, their dimensions cancel out ($[L]/[L]$), making angle a dimensionless quantity $[M^0 L^0 T^0]$.

Q: What does the expression in option (a) actually represent?

The expression $\frac{\pi p a^4}{8 \eta L}$ is Poiseuille's equation for the *volume flow rate* (volume flowing per unit time, $dV/dt$), not just volume. That's why its dimension is $[L^3 T^{-1}]$ instead of $[L^3]$.

Q: Is current density $J$ a vector or scalar, and what are its dimensions?

Current density is a vector quantity. Its magnitude is current per unit area ($I/A$), and its direction is the direction of current flow. Its dimension is simply $[AL^{-2}]$.

Question

Which of the following equations is dimensionally incorrect ?
Where, $t = 	ext{time}$, $h = 	ext{height}$, $s = 	ext{surface tension}$, $	heta = 	ext{angle}$, $ho = 	ext{density}$, $a, r = 	ext{radius}$, $g = 	ext{acceleration due to gravity}$, $V = 	ext{volume}$, $p = 	ext{pressure}$, $W = 	ext{work done}$, $	au = 	ext{torque}$, $\varepsilon = 	ext{permittivity}$, $E = 	ext{electric field}$, $J = 	ext{current density}$, $L = 	ext{length}$.

Options:

(A) $V = \frac{\pi p a^4}{8 \eta L}$

(B) $h = \frac{2s \cos	heta}{ho r g}$

(C) $J = \varepsilon \frac{\partial E}{\partial t}$

(D) $W = 	au 	heta$

Sigma Percentile · Accepted Answer

### The Power of Dimensional Analysis

Imagine you are a detective, and your magnifying glass is the **Principle of Dimensional Homogeneity**. This powerful principle states that you cannot add apples to oranges, nor can you equate them. In the language of physics, for any equation to be valid, the dimensions on the left-hand side (LHS) must perfectly mirror the dimensions on the right-hand side (RHS).

In this problem, we are presented with four suspects—four physical equations. Our mission is to interrogate each one and find the imposter that is dimensionally incorrect.

### Interrogating Option (a): The Flow Rate Trap

Let's start with the first equation:
$$V = \frac{\pi p a^4}{8 \eta L}$$

On the left, we have $V$, which represents volume. The dimension of volume is straightforward:
$$[	ext{LHS}] = [V] = [L^3]$$

Now, let's break down the right-hand side. We need the dimensions of pressure $p$ and the coefficient of viscosity $\eta$. 
Pressure is force distributed over an area:
$$[p] = \frac{[MLT^{-2}]}{[L^2]} = [ML^{-1}T^{-2}]$$

Viscosity can be deduced from Stokes' Law ($F = 6\pi\eta r v$):
$$[\eta] = \frac{[MLT^{-2}]}{[L][LT^{-1}]} = [ML^{-1}T^{-1}]$$

Let's substitute these into the RHS, keeping in mind that $\pi$ and $8$ are dimensionless constants:
$$[	ext{RHS}] = \frac{[ML^{-1}T^{-2}] [L^4]}{[ML^{-1}T^{-1}] [L]}$$

Simplifying the numerator and denominator:
$$[	ext{RHS}] = \frac{[ML^3T^{-2}]}{[ML^0T^{-1}]} = [L^3T^{-1}]$$

Aha! The LHS is $[L^3]$, but the RHS is $[L^3T^{-1}]$. They do not match! This equation is dimensionally incorrect.

**Fun Fact:** The expression on the right is actually Poiseuille's equation for **volume flow rate** (volume per unit time, $\frac{dV}{dt}$), not just volume. That extra $T^{-1}$ was the smoking gun!

### Verifying the Innocent: Option (b)

Even though we caught the culprit, let's quickly verify the others. Option (b) is the formula for capillary rise:
$$h = \frac{2s \cos	heta}{ho r g}$$

The LHS is height, so $[	ext{LHS}] = [L]$.
For the RHS, surface tension $s$ is force per unit length $[MT^{-2}]$, density $ho$ is $[ML^{-3}]$, radius $r$ is $[L]$, and gravity $g$ is $[LT^{-2}]$.
$$[	ext{RHS}] = \frac{[MT^{-2}]}{[ML^{-3}] [L] [LT^{-2}]}$$

Notice how the $M$ and $T^{-2}$ terms beautifully cancel out:
$$[	ext{RHS}] = \frac{1}{[L^{-1}]} = [L]$$

The dimensions match perfectly. This equation is innocent.

### A Smart Trick for Option (c)

Option (c) relates current density to the rate of change of the electric field (Displacement Current):
$$J = \varepsilon \frac{\partial E}{\partial t}$$

Current density $J$ is current per unit area, so $[	ext{LHS}] = [AL^{-2}]$.
For the RHS, finding the dimensions of permittivity $\varepsilon$ and electric field $E$ separately can be tedious. Let's use a smart trick! From Coulomb's Law, we know:
$$E = \frac{q}{4\pi\varepsilon r^2} \implies \varepsilon E = \frac{q}{4\pi r^2}$$

This means the product $\varepsilon E$ has the same dimensions as charge over distance squared!
$$[	ext{RHS}] = \frac{[\varepsilon E]}{[t]} = \frac{[q]}{[r^2][t]}$$

Since charge $q$ is current times time ($[AT]$):
$$[	ext{RHS}] = \frac{[AT]}{[L^2][T]} = [AL^{-2}]$$

A perfect match once again!

### Conclusion

By systematically applying the principle of dimensional homogeneity, we confidently identified that option (a) is the only dimensionally incorrect equation. Always remember to check your dimensions—it is the ultimate lie detector in physics!

Select Answer:

Visualized Solution

Principle of Homogeneity

Dimensions for Option (a)

Substituting Dimensions

Evaluating RHS

Comparing LHS and RHS

Verifying Option (b)

Verifying Option (c)

The Sigma Insight: Dimensional Analysis

The Power of Dimensional Analysis

Interrogating Option (a)

Verifying the Innocent

A Smart Trick for Option (c)

Conclusion

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