Using dimensional analysis, find the correct relation for time period …

Question

Using dimensional analysis, find the correct relation for time period $T$ of a simple pendulum:

Accepted Answer

A. $T \propto \sqrt{\frac{L}{g}}$

Assume $T\propto L^a g^b$.
Then $[T]=[L]^a[LT^{-2}]^b=[L^{a+b}T^{-2b}]$.
Matching powers gives $-2b=1\Rightarrow b=-\frac12$, and $a+b=0\Rightarrow a=\frac12$.
**So $T\propto \sqrt{L/g}$.**

Answer

$T \propto \sqrt{\frac{m}{g}}$

Answer

$T \propto \sqrt{g}\, l$

Answer

$T \propto \frac{L}{g}$

Using dimensional analysis, find the correct relation for time period $T$ of a simple pendulum:

Keep practising Unit 4A1: Types of Chemical Reactions