Question

The period of a body SHM is represented by $T \propto {P^a}{D^b}{S^c}$, where $P$ is the pressure, $D$ the density and $S$ is the surface tension then the values of $a$, $b$ and $c$ is
A) $1,3,\dfrac{1}{3}$
B) $ - \dfrac{3}{2},\dfrac{1}{2},1$
C) $ - 1, - 2,3$
D) $ - \dfrac{1}{2}, - \dfrac{3}{2}, - \dfrac{1}{2}$

Answer 1

Hint
Pressure, density and surface tension are related as $T \propto {P^a}{D^b}{S^c}$. The dimensions of these quantities will be substituted and equated with $T$. Apply principle of homogeneity and solve the equations of powers to find the values of $a$, $b$ and $c$.

Complete step-by-step answer
Use the dimensional analysis to solve this problem. Pressure is given by force per unit area,
$P = \dfrac{F}{A}$
Dimensional formula:
$[P] = [M{L^{ - 1}}{T^{ - 2}}]$
Density is the ratio of mass to volume,
$D = \dfrac{M}{V}$
Dimensional formula:
$[D] = [M{L^{ - 3}}{T^0}]$
Surface tension is force per unit length,
$S = \dfrac{F}{L}$
Dimensional formula:
$[S] = [M{L^0}{T^{ - 2}}]$
Let $T$ be a quantity with dimensions,
$[T] = [{M^0}{L^0}{T^1}]$.
Given that,
$T \propto {P^a}{D^b}{S^c}$
$[{M^0}{L^0}{T^1}] = {[M{L^{ - 1}}{T^{ - 2}}]^a}{[M{L^{ - 3}}{T^0}]^b}{[M{L^0}{T^{ - 2}}]^c}$
$[{M^0}{L^0}{T^1}] = [{M^{a + b + c}}{L^{ - a - 3b}}{T^{ - 2a - 2c}}]$
Equate the corresponding powers of mass, length and time on both sides. k is the constant of proportionality.
$a + b + c = 0$,
$ - a - 3b = 0$,
$ - 2a - 2c = 0$,
Solve the equations simultaneously.
$a = - \dfrac{3}{2},b = \dfrac{1}{2},c = 1$
Hence, the values of a, b and c is $a = - \dfrac{3}{2},b = \dfrac{1}{2},c = 1$
The correct option is (B).

Note
Applications of dimensional analysis:
(i) To find the unit of physical quantity.
(ii) To find dimensions of physical constant or coefficients.
(iii) To convert physical quantities to other systems.
(iv) To check the correctness of dimensions.