Numerical Problems Based on Dimensional Correctness of Physical Relations for Class 11 Physics

By the Principle of Homogeneity, the dimensions of all terms on both sides of an equation must be identical.

(i) $v = u + at$
LHS: $[v] = [\text{LT}^{-1}]$
RHS: $[u] + [a][t] = [\text{LT}^{-1}] + [\text{LT}^{-2}][\text{T}] = [\text{LT}^{-1}] + [\text{LT}^{-1}]$
Since LHS = RHS, it is dimensionally correct.

(ii) $s = ut + \frac{1}{2}at^2$
LHS: $[s] = [\text{L}]$
RHS: $[u][t] + [a][t^2] = ([\text{LT}^{-1}][\text{T}]) + ([\text{LT}^{-2}][\text{T}^2]) = [\text{L}] + [\text{L}]$ (Constants like 1/2 are dimensionless).
Since LHS = RHS, it is dimensionally correct.

(iii) $v^2 – u^2 = 2as$
LHS: $[v^2] – [u^2] = [\text{LT}^{-1}]^2 – [\text{LT}^{-1}]^2 = [\text{L}^2\text{T}^{-2}] – [\text{L}^2\text{T}^{-2}]$
RHS: $[a][s] = [\text{LT}^{-2}][\text{L}] = [\text{L}^2\text{T}^{-2}]$ (Constant 2 is dimensionless).
Since LHS = RHS, it is dimensionally correct.

Rearranging Stokes’ Law formula for $\eta$:

Writing the dimensions of each physical quantity (Note that $6\pi$ is a dimensionless constant):

• Force $[F] = [\text{MLT}^{-2}]$
• Radius $[r] = [\text{L}]$
• Velocity $[v] = [\text{LT}^{-1}]$

Substituting the dimensions:

According to the Principle of Homogeneity, quantities can only be added or equated if they have the same dimensions. Therefore, the dimension of each term on the right side must equal the dimension of distance $x$ ($[\text{L}]$).

Equating each term to $[\text{L}]$:

• $[a] = [\text{L}]$
• $[b][\text{T}] = [\text{L}] \implies [b] = \frac{[\text{L}]}{[\text{T}]} = [\text{LT}^{-1}]$
• $[c][\text{T}^2] = [\text{L}] \implies [c] = \frac{[\text{L}]}{[\text{T}^2]} = [\text{LT}^{-2}]$
• $[d][\text{T}^3] = [\text{L}] \implies [d] = \frac{[\text{L}]}{[\text{T}^3]} = [\text{LT}^{-3}]$

We need to check if the dimensions on the LHS equal the dimensions on the RHS.

LHS: Critical velocity $[v_c] = [\text{LT}^{-1}]$

RHS: $K$ is a dimensionless constant (Reynolds number), so it has no dimensions. The other quantities are:

• Coefficient of viscosity $[\eta] = [\text{ML}^{-1}\text{T}^{-1}]$
• Radius $[r] = [\text{L}]$
• Density $[\rho] = [\text{ML}^{-3}]$

Since $[\text{LHS}] = [\text{RHS}]$, the relation is dimensionally correct.

Here, $V$ is the rate of flow, which means volume flowing per unit time. Its dimensional formula is:

Now evaluate the RHS. The constant $\frac{\pi}{8}$ is dimensionless.

• Pressure $[P] = \frac{\text{Force}}{\text{Area}} = [\text{ML}^{-1}\text{T}^{-2}]$
• Radius to power 4 $[r^4] = [\text{L}^4]$
• Coefficient of viscosity $[\eta] = [\text{ML}^{-1}\text{T}^{-1}]$
• Length $[l] = [\text{L}]$

Combine terms in the numerator and denominator:

Divide the two:

Since $[\text{LHS}] = [\text{RHS}]$, Poiseuille’s equation is dimensionally correct.