Question

Find the expression for the time period of a pendulum using dimensional analysis.

• A. $k\sqrt{\frac{g}{l}}$
• B. $k\sqrt{\frac{l}{g}}$
• C. kl
• D. $\frac{kl}{g}$

Answer 1

The correct option is B

$k\sqrt{\frac{l}{g}}$

$T=k{l}^a{g}^b\left[M^{\circ }{L}^{\circ }{T}^1\right]=L^a[L{T}^{-2}{]}^b\text{Comparing the exponents of fundamental quantities on both sides}a+b=0-2b=1\Rightarrow b=-\frac{1}{2}anda=\frac{1}{2}\therefore T=k{\left[\frac{l}{g}\right]}^{\frac{1}{2}}$