Dimension of force $F = [MLT^{-2}] = 100 N $ ... (i)

Length $(L) = [L] = 10 m $ ... (ii)

Time $(t) = [T] = 100 s $ ... (iii)

Substituting values of $L$ and $T$ from Eqs. (ii) and (iii) in Eq. (i), we get

$\begin{aligned}& \Rightarrow \frac{M \times 10}{100 \times 100} =100 \\\\ & \Rightarrow M =100 \times 1000 \mathrm{~kg} \\\\ & M =10^5 \mathrm{~kg} \end{aligned}$

(a) Plane angle $ \theta = \frac{L}{r} \text{radian}$

its unit is radian but has no dimensions

(b) Strain $ = \frac{\Delta L}{L} = \frac{\text{Change in length}}{\text{length}}$

It has neither unit nor dimensions

(c) Gravitational constant (G) = $6.67 \times 10^{-11} \text{N} m^{2} / \text{kg}^{2}$

(d) Reynold’s number is a constant which has no unit.

We know that angle $ \theta = \frac{l}{r} \text{radian}$

Given, $ \theta = \frac{ \pi}{6} = \frac{l}{31 \text{ cm}}$

Hence, length $l = 31 \times \frac{ \pi}{6} \text{ cm} = \frac{31 \times 3.14}{6} \text{ cm} = 16.22 \text{ cm}$

Rounding off to three significant figures it would be $16.2 \text{ cm}$.

We know that solid angle $ \Omega = \frac{\text{Area}}{(\text{Distance})^{2}}$

$= \frac{1 \text{cm}^{2}}{(5 \text{cm})^{2}}= \frac{1}{25} = 4 \times 10^{-2} \text{ steradian}$

(∵ Area = $1 \text{cm}^{2}$, distance = $5 \text{ cm}$)

Note We should not confuse, solid angle with plane angle $ \theta = \frac{l}{r} \text{ radian}$.

Now, by the principle of homogeneity, i.e., dimensions of LHS and RHS should be equal, hence

$[LHS] = [RHS]$

$$ \Rightarrow $$$[L] = [A] = L$

As $\omega t - kx$ should be dimensionless, $[\omega t] = [kx] = 1$

$$ \Rightarrow $$$[\omega] T = [k] L = 1$

$$ \Rightarrow $$$[\omega] = T^{-1}$ and $[k] = L^{-1}$