1.Check the dimensional correctness of $v^2 - u^2 = 2as$.
Sol: LHS = $[L T^{-1}]^2 = [L^2 T^{-2}]$. RHS = $[L T^{-2}] [L] = [L^2 T^{-2}]$. Dimensionally correct!
2.Find the percentage error in density if mass has $1\%$ error and length of cube has $2\%$ error.
Sol: $\rho = \frac{m}{L^3} \implies \frac{\Delta\rho}{\rho}\times 100\% = \frac{\Delta m}{m}\times 100\% + 3\left(\frac{\Delta L}{L}\times 100\%\right) = 1\% + 3(2\%) = 7\%$.
3.Find dimensions of $a$ and $b$ in van der Waals equation $\left(P + \frac{a}{V^2}\right)(V - b) = RT$.
Sol: $[a] = [P][V^2] = [M L^{-1} T^{-2}][L^6] = [M L^5 T^{-2}]$. $[b] = [V] = [L^3]$.
4.Convert $1 \text{ Joule}$ into CGS unit (Ergs) using dimensional analysis.
Sol: $[Energy] = [M L^2 T^{-2}]$. $n_2 = 1 \times \left(\frac{1\text{ kg}}{1\text{ g}}\right)^1 \left(\frac{1\text{ m}}{1\text{ cm}}\right)^2 \left(\frac{1\text{ s}}{1\text{ s}}\right)^{-2} = 10^3 \times 10^4 = 10^7\text{ Ergs}$.
5.The period of oscillation of a simple pendulum is $T = 2\pi \sqrt{\frac{L}{g}}$. Derive this relation using dimensional analysis.
Sol: Let $T \propto L^a g^b \implies [T] = [L]^a [L T^{-2}]^b = [L]^{a+b} [T]^{-2b}$. $a+b=0, -2b=1 \implies b=-1/2, a=1/2 \implies T = k \sqrt{L/g}$.
