First slide

Dimensions of physical quantities

Question

The time period T of simple pendulum depends upon length l of the pendulum and gravitational acceleration. The formula for time period of simple pendulum is given by

Moderate

A

$T = 2 π \sqrt{\frac{l}{g}}$
B

$T = 2 π \sqrt{\frac{g}{l}}$
C

$T = \frac{1}{2 π} \sqrt{\lg}$
D

$T = \frac{2 π l}{g}$

Solution

Let $T \propto l^{a} a n d T \propto g^{b}$
where a and b are dimensionless constants
$T = k l^{a} g^{b}$
where, k is dimensionless constant.
$[L H S] [T] = [M^{0} L^{0} T^{- 1}]$
and $[R H S] (l^{a} g^{b}) = {[L]}^{a} {[L T^{- 2}]}^{b}$
$= [L^{a + b} T^{- 2 b}] = [M^{0} L^{a + b} T^{- 2 b}]$
According to homogeneity principle,
[LHS] = [RHS]
or $[M^{0} L^{0} T] = [M^{0} L^{a + b} T^{- 2 b}]$
For dimensional balance, dimensions on both sides should be same.
$∴ a + b = 0$
and -2b =1
$∴ b = - \frac{1}{2}$
and $a = \frac{1}{2}$
$∴ T = 2 π \sqrt{\frac{l}{g}}$
[since, numerical value of k in case of simple pendulum is $2 π$ ]

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