Question

Given m is the mass per unit length, T is the tension & L is the length of the wire the dimensional formula for $L{\left( {\dfrac{m}{T}} \right)^{1/2}}$ is same as that for:-
A. Wavelength
B. Velocity
C. Time period
D. Frequency

Answer 1

Hint: Here we will use the concept of the dimensional analysis. It is used to determine the relationships between the numbers of variables. Also, use the basic rules of power and exponents to simplify. The mathematical expression which shows the powers to which the fundamental units are to be raised to get one unit of the derived quantity is called the dimensional formula of that quantity. If “x” is the unit of the derived quantity represented by $x = {M^a}{L^b}{T^c}$ where ${M^a}{L^b}{T^c}$is called the dimensional formula and the exponents a, b and c are called the dimensions.

Complete step by step answer:
Let us suppose that –
$x = L{\left( {\dfrac{m}{T}} \right)^{1/2}}$
In the above equation, Place $x = k{M^a}{L^b}{T^c}$
K is constant
Dimensions of mass per unit length, $ = [{M^1}{L^{ - 1}}{T^0}]$
Length $ = [L]$
Tension, $T = [{M^1}{L^1}{T^{ - 2}}]$
$ \Rightarrow k{M^a}{L^b}{T^c} = [{L^1}]\left( {\dfrac{{{{[{M^1}{L^{ - 1}}{T^0}]}^{1/2}}}}{{{{[{M^1}{L^1}{T^{ - 2}}]}^{1/2}}}}} \right)$
Simplify using the laws of power and exponent –
$ \Rightarrow k{M^a}{L^b}{T^c} = [{L^1}]\left( {\dfrac{{[{M^{1/2}}{L^{ - 1/2}}{T^0}]}}{{[{M^{1/2}}{L^{1/2}}{T^{ - 2/2}}]}}} \right)$
Like and equal terms in the division form cancel each other. Therefore, ${M^{1/2}}$ is removed from the denominator and the numerator as they both cancel each other.
$ \Rightarrow k{M^a}{L^b}{T^c} = [{L^1}]\left( {\dfrac{{[{L^{ - 1/2}}{T^0}]}}{{[{L^{1/2}}{T^{ - 1}}]}}} \right)$
By the law of the power and exponent – the sign of the power changes when they are moved from the denominator to the numerator and vice- versa.
\[
   \Rightarrow k{M^a}{L^b}{T^c} = [{L^1}][{L^{ - \dfrac{1}{2} - \dfrac{1}{2}}}{T^1}] \\
   \Rightarrow k{M^a}{L^b}{T^c} = [{L^1}][{L^{ - 1}}{T^1}] \\
   \Rightarrow k{M^a}{L^b}{T^c} = [{L^{1 - 1}}][{T^1}] \\
  k{M^a}{L^b}{T^c} = [{T^1}] \\
 \]
By comparing we get, $a = 0,b = 0\,{\text{and c = 1}}$
Also, we know that the dimension of time period is $[T]$
Therefore, the required answer is – the dimensional formula for $L{\left( {\dfrac{m}{T}} \right)^{1/2}}$ is the same as that for time period.

So, the correct answer is “Option C”.

Note:
The dimensional analysis is the mathematical tool to calculate the physical parameters and its analysis involves the fundamental units of the dimensions M (mass), Length (L) and T (Time). Remember the basic power and exponent laws to solve these types of problems.