Question

The frequency of vibration of the string is given by $v = \dfrac{p}{{2l}}{\left[ {\dfrac{F}{m}} \right]^{\dfrac{1}{2}}}$ Here, p is the number of segments in the string and l is the length. The dimensional formula for m will be
A. $[{M^0}L{T^{ - 1}}]$
B. $[M{L^0}{T^{ - 1}}]$
C. $[M{L^{ - 1}}{T^0}]$
D. $[{M^0}{L^0}{T^0}]$

Answer 1

Hint: Dimensions play a key role in physics. Various quantities have various dimensions. Dimensions of the main fundamental quantities are M for mass while L for length T for time K for temperature while A for the current. Dimensional analysis is useful in solving various problems and checking for the validity of the options. We use one of the properties of dimensions here to solve this question.

Complete step by step answer:
There are some rules which we use in case of problems involving dimension usage. Like in the case of any formula, dimensions of the left hand part of the equation must be the same as dimensions of the right hand part of the equation. We can add or subtract quantities of the same dimensions. Logarithm function doesn’t have any dimensions and all trigonometric functions don’t have any dimensions and exponential functions also don’t have any dimensions.
If we have a product of two physical quantities and the dimension of one quantity is inverse of the dimension of the other quantity then that product will be dimensionless.
The equation which we had got is $v = \dfrac{p}{{2l}}{\left[ {\dfrac{F}{m}} \right]^{\dfrac{1}{2}}}$
That equation satisfies dimensionally. Since we need the dimensions of ‘m’ we will take it to the left hand side of the equation. Then we will get
$m = \dfrac{{{p^2}F}}{{4{l^2}{v^2}}}$
‘p’ doesn’t have any dimensions. Dimension of force is $[ML{T^{ - 2}}]$. Dimension of ${l^2}$ is $\left[ {{L^2}} \right]$. Dimension of ${v^2}$ is $\left[ {{T^ - }^2} \right]$. By substituting all these we get dimensions of ‘m’ as
$ \Rightarrow m = \dfrac{{{p^2}F}}{{4{l^2}{v^2}}}$
$\eqalign{
  & \Rightarrow m = \dfrac{{[ML{T^{ - 2}}]}}{{\left[ {{L^2}} \right]\left[ {{T^ - }^2} \right]}} \cr
  & \Rightarrow m = [M{L^{ - 1}}{T^0}] \cr} $

So, the correct answer is “Option C”.

Note: There are some quantities which have units while don't have dimensions. Few instances for that are plane angle which has unit or radian but no dimensions and solid angle which has a unit of steradian and no dimension and angular displacement also has unit of radian but no dimension.