Question

The frequency of vibration of a 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) $\left[ {{M}^{0}}L{{T}^{-1}} \right]$
b) $\left[ M{{L}^{0}}{{T}^{-1}} \right]$
c) $\left[ M{{L}^{-1}}{{T}^{0}} \right]$
d) $\left[ {{M}^{0}}{{L}^{0}}{{T}^{0}} \right]$

Answer 1

Hint : To obtain the formula of any quantity in the derived physical quantity, express the derived quantity in the form of its dimensions. All the quantities should be expressed in the form of fundamental physical quantities for simplicity.

Complete solution:
Some of the fundamental physical quantities are mentioned below:
Dimension of length=$\left[ L \right]$
Dimension of mass=$\left[ M \right]$
Dimension of time=$\left[ T \right]$
Dimension of electric current=$\left[ A \right]$
Dimension of thermodynamic temperature=$\left[ K \right]$
Dimension of luminous intensity=$\left[ cd \right]$
Dimension of the amount of substance=$\left[ mol \right]$
These are the seven fundamental quantities of the world.

Now let us define the dimension of a physical quantity in order to make the understanding of the concept more simpler,
Dimensions of a physical quantity are the powers (or exponents) to which the fundamental quantity must be raised to represent that quantity completely.

Now let us obtain the dimensional formula of m
$v=\dfrac{p}{2l}{{\left[ \dfrac{F}{m} \right]}^{\dfrac{1}{2}}}$

First let us substitute the dimensional formula of all the quantities known to us.

$v=\left[ {{M}^{0}}{{L}^{0}}{{T}^{-1}} \right]$
$F=\left[ ML{{T}^{-2}} \right]$
$l=\left[ {{M}^{0}}L{{T}^{0}} \right]$ ,
p is the number of segments in the string. Since p being constant it does not have a dimensional formula.

After substituting the dimensional formula of the quantities in the equation of frequency of string, the equation reduce to,

$\left[ {{M}^{0}}{{L}^{0}}{{T}^{-1}} \right]=\dfrac{p}{2\left[ {{M}^{0}}L{{T}^{0}} \right]}{{\left[ \dfrac{\left[ ML{{T}^{-2}} \right]}{m} \right]}^{\dfrac{1}{2}}}$
Since p and $2$are constants we can write the above equation as
$\left[ {{M}^{0}}{{L}^{0}}{{T}^{-1}} \right]=\dfrac{k}{\left[ {{M}^{0}}L{{T}^{0}} \right]}{{\left[ \dfrac{\left[ ML{{T}^{-2}} \right]}{m} \right]}^{\dfrac{1}{2}}}$
Cross multiplying and squaring on both the sides squaring on both the side we get,
$\left[ {{M}^{0}}{{L}^{-2}}{{T}^{-2}} \right]m=\left[ ML{{T}^{-2}} \right]$
And hence m=$\left[ M{{L}^{-1}}{{T}^{0}} \right]$

Note: The derived quantities should be expressed in the fundamental physical quantities. The laws of exponent should be used properly. The power fundamental of a physical quantity adds up on multiplication and reduces on division