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Question: The period of a body SHM is represented by \(T \propto {P^a}{D^b}{S^c}\), where \(P\) is the pressur...

The period of a body SHM is represented by TPaDbScT \propto {P^a}{D^b}{S^c}, where PP is the pressure, DD the density and SS is the surface tension then the values of aa, bb and cc is
A) 1,3,131,3,\dfrac{1}{3}
B) 32,12,1 - \dfrac{3}{2},\dfrac{1}{2},1
C) 1,2,3 - 1, - 2,3
D) 12,32,12 - \dfrac{1}{2}, - \dfrac{3}{2}, - \dfrac{1}{2}

Explanation

Solution

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

Complete step-by-step answer
Use the dimensional analysis to solve this problem. Pressure is given by force per unit area,
P=FAP = \dfrac{F}{A}
Dimensional formula:
[P]=[ML1T2][P] = [M{L^{ - 1}}{T^{ - 2}}]
Density is the ratio of mass to volume,
D=MVD = \dfrac{M}{V}
Dimensional formula:
[D]=[ML3T0][D] = [M{L^{ - 3}}{T^0}]
Surface tension is force per unit length,
S=FLS = \dfrac{F}{L}
Dimensional formula:
[S]=[ML0T2][S] = [M{L^0}{T^{ - 2}}]
Let TT be a quantity with dimensions,
[T]=[M0L0T1][T] = [{M^0}{L^0}{T^1}].
Given that,
TPaDbScT \propto {P^a}{D^b}{S^c}
[M0L0T1]=[ML1T2]a[ML3T0]b[ML0T2]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}
[M0L0T1]=[Ma+b+cLa3bT2a2c][{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=0a + b + c = 0,
a3b=0- a - 3b = 0,
2a2c=0- 2a - 2c = 0,
Solve the equations simultaneously.
a=32,b=12,c=1a = - \dfrac{3}{2},b = \dfrac{1}{2},c = 1
Hence, the values of a, b and c is a=32,b=12,c=1a = - \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.