Units and Measurements Part 1

As we know that, Dimensional formula of volume $=[M^{0} L^{3} T^{0}]$ ...(i) Since, $F=6 \pi \eta r v$ $\therefore \eta \propto \frac{F}{r v}$ where, η is viscosity and F is force. $\therefore [\eta] = \frac{[ML T^{-2}]}{[L \cdot L T^{-1}]} = [M L^{-1} T^{-1}]$ So,$\left[\frac{\pi p a^{4}}{8 \eta L}\right] = \frac{[ML^{-1} T^{-2}][L^{4}]}{[ML^{-1} T^{-1}][L^{1}]}$ $=[M^{0} L^{3} T^{-1}]$ ...(ii) Since, Eq. (i) is not equal to Eq (ii), so option (a) is wrong. Now, since formula of capillary rise in tube, $h = \frac{2 s \cos \theta}{\rho g r}$ Dimensional formula of LHS part, ∴ [h] = [L] Dimensional formula of RHS part $= \frac{[s]}{[\rho][g][r]} = \frac{[MT^{-2}]}{[ML^{-3}][LT^{-2}][L]} = [L]$ Hence, $h = \frac{2 s \cos \theta}{\rho r g}$ is dimensionally correct. So, option (b) will also be dimensionally correct. In option (c), $J = \varepsilon \frac{dE}{\partial t}$ ...(iii) $\Rightarrow J = \varepsilon \frac{E}{t}$ Dimension of current density J is calculated as Since, $J = \frac{l}{A}$ $\therefore [J] = \frac{[l]}{[A]} = \frac{[A]}{[L^{2}]}$ $\Rightarrow [J] = [AL^{-2}]$ ...(iv) Again, we know that $E = \frac{1}{4 \pi \varepsilon} \cdot \frac{q}{r^{2}}$ $\Rightarrow \varepsilon E = \frac{1}{4 \pi} \cdot \frac{q}{r^{2}}$ $\frac{\Rightarrow \varepsilon E}{t} = \frac{1}{4 \pi} \cdot \frac{q}{t r^{2}}$ $\Rightarrow \left[ \frac{\varepsilon E}{t} \right] = \frac{[q]}{[t][r^{2}]} = \frac{[AT]}{[T][L^{2}]}$ $\left[ \frac{\varepsilon E}{t} \right] = [AL^{-2}]$ ...(v) Form Eqs. (iv) and (v), we see that Eq. (iii) is dimensionally correct. In option (d) W = τθ and $\tau = r \times F$ So, dimensional formula of $[\tau][\theta] = [r][F] = [L][MLT^{-2}] = [ML^{2} T^{-2}]$ $= [W]$