Question

Check the correctness of equation $v = \dfrac{1}{{2l}}\sqrt {\dfrac{T}{M}} $.

Answer 1

Hint:To check the accuracy of the equation, we have to show that the left hand side of the equation is dimensionally equal to the right hand side of the equation.Dimensional Equations are the equations that result from equating a physical quantity by its dimensional formulae.

Complete step by step answer:
 The dimensional equation helps in the expression of physical quantities in terms of fundamental or base quantities.
$v = \dfrac{1}{{2l}}\sqrt {\dfrac{T}{M}} $
Where $T$ represents the Tension, $M$ represents the Mass per unit Length, $l$ represents the Length and $v$ represents the Frequency.

Dimensional formula is based on the base quantities which are mass, length and time which are raised to their powers.
-$M$ represents the base quantity of mass
-$L$ represents the base quantity of length
-$T$ represents the base quantity of time

Now we shall write the dimensional formula for all the terms in the equation,
The Dimension of Tension is represented as $T = \left[ {ML{T^{ - 2}}} \right]$
The Dimension of mass per unit length is represented as $M = \left[ {M{L^{ - 1}}} \right]$
The Dimension of length is represented as $l = \left[ L \right]$
The Dimension of frequency is represented as $v = \left[ {{T^{ - 1}}} \right]$

To check the accuracy or correctness of an equation, we have to ensure that the Left Hand Side of the equation is equal to its Right Hand Side. First, we shall find the Left hand side of the equation
$L.H.S = \left[ {{T^{ - 1}}} \right]$
Now, we shall find the Right hand side of the equation
$R.H.S = \dfrac{{{M^{\dfrac{1}{2}}}{L^{\dfrac{1}{2}}}{T^{ - 1}}}}{{{M^{\dfrac{1}{2}}}{L^{\dfrac{{ - 1}}{2}}}L}}$
On solving further, we get,
\[R.H.S = \dfrac{{{L^{\dfrac{1}{2}}}{T^{ - 1}}}}{{{L^{\dfrac{{ - 1}}{2}}}L}} = \dfrac{{{L^{\dfrac{1}{2}}}{T^{ - 1}}}}{{{L^{\dfrac{1}{2}}}}} = {T^{ - 1}}\]
$\therefore L.H.S = R.H.S$
Since the Left Hand Side of the equation is equal to the Right Hand Side of the equation.

Therefore, the equation $v = \dfrac{1}{{2l}}\sqrt {\dfrac{T}{M}} $ is correct.

Note: In the case of trigonometric, logarithmic, and exponential equations, dimensional formulae are undefined, which means we cannot predict the existence of quantities using these functions. The physical quantities that can be used in Dimensional Formulae are limited. Dimensional Formula cannot be used to figure out what proportionality constants we need to use. Only addition and subtraction are used in dimensional formulas.