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Step-by-Step Dimensional Analysis
Step 1: Examine Option (1):
$v = \frac{\pi\,p\,a^4}{8\,\eta\,L}$
• According to the question statement, the symbol $v$ represents volume (or, in some contexts, volumetric flow rate), both of which have dimensions different from a simple velocity.
• Let us check the dimensions on the right-hand side:
Pressure, $p$, has dimensions: $[M L^{-1} T^{-2}]$.
Radius (or length), $a$, raised to the 4th power: $[L^4]$.
Viscosity, $\eta$, has dimensions: $[M L^{-1} T^{-1}]$.
Length, $L$, in the denominator: $[L]$.
• Multiplying and dividing these:
$
\text{Dimension of RHS} \;=\; \frac{[M L^{-1} T^{-2}]\,[L^4]}{[M L^{-1} T^{-1}]\,[L]}
\;=\; \frac{[M L^3 T^{-2}]}{[M L^2 T^{-1}]}
\;=\; [L T^{-1}],
$
which is the dimension of velocity, not volume ($[L^3]$) or volumetric flow rate ($[L^3 T^{-1}]$). Hence, if $v$ is taken to be volume or volumetric flow rate, this equation is dimensionally inconsistent.
Step 2: Examine Option (2):
$h = \frac{2\,s\,\cos \theta}{\rho\,r\,g}$
• Dimension of $h$ is $[L]$.
• Check the right-hand side:
Surface Tension $s$: $[M\,T^{-2}]$ (force per unit length).
cos θ is dimensionless.
Density $\rho$: $[M\,L^{-3}]$.
Radius (length) $r$: $[L]$.
Acceleration due to gravity $g$: $[L\,T^{-2}]$.
• So the denominator has dimension $[M\,L^{-3}]\times [L]\times [L\,T^{-2}] = [M\,L^{-1}\,T^{-2}]$.
• Hence,
$
\text{Dimension of RHS}
= \frac{[M\,T^{-2}]}{[M\,L^{-1}\,T^{-2}]}
= [L].
$
• This matches the dimension of $h$, so Option (2) is dimensionally correct.
Step 3: Examine Option (3):
$J = \varepsilon\, \frac{\partial E}{\partial t}$
• Current density $J$ has dimension $[I\,L^{-2}]$, where $I$ is electric current.
• Permittivity $\varepsilon$ has dimension $[M^{-1}L^{-3}T^2\,I^2]$.
• Electric field $E$ has dimension $[M\,L\,T^{-3}\,I^{-1}]$. Therefore,
$
\frac{\partial E}{\partial t}
$
has dimension $[M\,L\,T^{-4}\,I^{-1}]$.
• Multiplying by $\varepsilon$:
$
\varepsilon \times \frac{\partial E}{\partial t}
=
\bigl[M^{-1}L^{-3}T^2\,I^2\bigr]
\times
\bigl[M\,L\,T^{-4}\,I^{-1}\bigr]
=
[I^{2-1}\,T^{2-4}\,M^{-1+1}\,L^{-3+1}]
=
[I\,T^{-2}\,L^{-2}].
$
• In some formulations of Maxwell’s equations, additional constants (such as $\mu_0$) may appear, ensuring the final expression indeed yields a correct current density dimension. As used in many textbooks, $J = \varepsilon\, \partial E/\partial t$ can be consistent in the context of displacement current (with the correct system of units). Hence, within typical SI-based definitions, Option (3) is not usually taken as dimensionally incorrect in that context.
Step 4: Examine Option (4):
$W = \Gamma\,\theta$
• Work $W$ has dimension $[M\,L^2\,T^{-2}]$.
• Torque $\Gamma$ also has dimension of work or energy, i.e., $[M\,L^2\,T^{-2}]$.
• $\theta$ (angle) is dimensionless.
• Thus the product $\Gamma \theta$ has dimension $[M\,L^2\,T^{-2}]$, matching that of work, making Option (4) dimensionally correct.
Conclusion
Since Option (1) yields a dimension of velocity on the right-hand side instead of volume or volumetric flow rate, it is the dimensionally incorrect equation.
