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

To decide which relation is wrong we compare the dimensions on the left-hand side (LHS) with those on the right-hand side (RHS). If they differ, the equation is dimensionally incorrect.

First we tabulate the dimensions (in fundamental symbols $$M$$ for mass, $$L$$ for length, $$T$$ for time and $$A$$ for electric current).

$$\begin{aligned} \text{Work }(W)&:&\; \text{force}\times\text{distance}= (MLT^{-2})\,(L)=ML^{2}T^{-2}\\ \text{Torque }(\tau)&:&\; \text{force}\times\text{distance}=ML^{2}T^{-2}\\ \text{Pressure }(p)&:&\; \text{force}/\text{area}=ML^{-1}T^{-2}\\ \text{Surface tension }(s)&:&\; \text{force}/\text{length}=MT^{-2}\\ \text{Density }(\rho)&:&\; ML^{-3}\\ \text{Acceleration }(g)&:&\; LT^{-2}\\ \text{Viscosity }(\eta)&:&\; ML^{-1}T^{-1}\\ \text{Volume }(V)&:&\; L^{3}\\ \text{Current density }(J)&:&\; AL^{-2}\\ \text{Electric field }(E)&:&\; MLT^{-3}A^{-1}\\ \text{Permittivity }(\epsilon)&:&\; M^{-1}L^{-3}T^{4}A^{2} \end{aligned}$$ Length $$(a,r,h,L)$$ has dimension $$L$$, the angle $$\theta$$ is dimensionless.

Option A gives $$W=\tau\theta$$.

LHS $$\rightarrow$$ $$W:ML^{2}T^{-2}$$.

RHS $$\rightarrow$$ $$\tau\theta=(ML^{2}T^{-2})(1)=ML^{2}T^{-2}$$ because $$\theta$$ is dimensionless.

The two sides match, so Option A is dimensionally correct.

Option B gives $$V=\dfrac{\pi p a^{4}}{8\eta L}$$.

LHS $$\rightarrow$$ $$V:L^{3}$$.

RHS step by step:

We have $$p\,a^{4}$$ with dimensions $$\bigl(ML^{-1}T^{-2}\bigr)\,(L^{4}) = ML^{3}T^{-2}.$$

We divide by $$\eta L$$ whose dimensions are $$\bigl(ML^{-1}T^{-1}\bigr)\,(L)=MT^{-1}.$$

Therefore $$ \dfrac{p\,a^{4}}{\eta L}:\; \dfrac{ML^{3}T^{-2}}{MT^{-1}} =\dfrac{ML^{3}T^{-2}}{M}\,T^{1} =L^{3}T^{-1}. $$

So the RHS carries an extra factor $$T^{-1}$$ and has dimension $$L^{3}T^{-1}$$, whereas the LHS is only $$L^{3}$$. They differ, making this relation dimensionally inconsistent.

Option C gives $$h=\dfrac{2s\cos\theta}{\rho r g}$$.

LHS $$\rightarrow$$ $$h:L$$.

RHS numerator $$2s\cos\theta$$ has dimension $$MT^{-2}$$ (since $$\cos\theta$$ is dimensionless).

Denominator $$\rho r g$$ has dimension $$\bigl(ML^{-3}\bigr)\,(L)\,(LT^{-2})=ML^{-1}T^{-2}.$$ Hence $$ \dfrac{MT^{-2}}{ML^{-1}T^{-2}}=L, $$ matching the LHS. Option C is correct.

Option D gives $$J=\epsilon\dfrac{\partial E}{\partial t}$$.

LHS $$\rightarrow$$ $$J:AL^{-2}$$.

First write the time derivative: $$ \dfrac{\partial E}{\partial t} :\; \frac{MLT^{-3}A^{-1}}{T} = MLT^{-4}A^{-1}. $$

Now multiply by $$\epsilon$$: $$ \epsilon\dfrac{\partial E}{\partial t} =\bigl(M^{-1}L^{-3}T^{4}A^{2}\bigr)\bigl(MLT^{-4}A^{-1}\bigr) =L^{-2}A, $$ which is exactly $$AL^{-2}$$, so the dimensions tally. Option D is also correct.

Only Option B fails the dimensional test. Hence, the correct answer is Option B.