Question

If density (D) acceleration (a) and force (F) are taken as basic quantities, then time period has dimensions
A. $\dfrac{1}{6}$ in F
B. $ - \dfrac{1}{6}$ in F
C. $ - \dfrac{2}{3}$ in F
D. All the above are true

Answer 1

Hint: Dimensional formula of a given physical quantity is defined as the expression which shows how and which of the fundamental quantities represent the dimensions of a physical quantity. To find the solution we will write the dimensional formulas of the given quantities and equate it with the dimensional formula of time period.

Complete step-by-step answer:
Time period of a wave is defined as the time taken by any element of the string to complete one oscillation. The dimensional formula of the period of time is given by,
$T = \left[ {{M^0}{L^0}{T^1}} \right]$
Density of a substance is defined as its mass per unit volume. Its SI unit is $kg/{m^3}$. Its dimensional formula is given by,
D$ = \left[ {M{L^{ - 3}}} \right]$
The rate of change of velocity of an object with time is known as its acceleration. It is a vector quantity. Its SI unit is $m/{s^2}$. Its dimensional formula is given by,
a$ = \left[ {L{T^{ - 2}}} \right]$
Force is defined as an external agent which is capable of changing the state of rest or motion. Its SI unit is Newton. Its dimensional formula is given by,
F$ = \left[ {ML{T^{ - 2}}} \right]$
$\eqalign{
  & {M^0}{L^0}{T^1} = {\left[ D \right]^x}{\left[ a \right]^y}{\left[ F \right]^z} \cr
  & {M^0}{L^0}{T^1} = {\left[ {M{L^{ - 3}}} \right]^x}{\left[ {L{T^{ - 2}}} \right]^y}{\left[ {ML{T^{ - 2}}} \right]^z} \cr
  & {M^0}{L^0}{T^1} = {M^{x + z}}{L^{ - 3x + y + z}}{T^{ - 2y - 2z}} \cr} $
Comparing the quantities, we will get,
$\eqalign{
  & \Rightarrow x + z = 0 \cr
  & \Rightarrow - 3x + y + z = 0 \cr
  & \Rightarrow - 2y - 2z = 1 \cr} $
After solving these equations, we will get,
$x = - \dfrac{1}{6},y = - \dfrac{2}{3},z = \dfrac{1}{6}$
So, we were required to find the dimension of $F$.
Therefore, ${\left[ F \right]^z} = \dfrac{1}{6}$
Thus, $\dfrac{1}{6}$ in F is the required dimension of time period.

So, the correct answer is “Option A”.

Note: The equation obtained by equating a physical quantity with its dimensional formula is called the dimensional equation of the given physical quantity. The physical constant quantities which have no dimensions are called dimensionless constants.