Question

Write the dimensions of a and b in the relation $P=\dfrac{b-{{x}^{2}}}{at}$. Where P is the power, x is the distance and t is time.
A. $\left[ {{M}^{0}}{{L}^{2}}{{T}^{0}} \right],\left[ {{M}^{-1}}{{L}^{0}}{{T}^{2}} \right]$
B. $\left[ {{M}^{-1}}{{L}^{0}}{{T}^{2}} \right],\left[ {{M}^{0}}{{L}^{2}}{{T}^{0}} \right]$
C. $\left[ {{M}^{-1}}{{L}^{1}}{{T}^{2}} \right],\left[ {{M}^{0}}{{L}^{2}}{{T}^{0}} \right]$
D. $\left[ {{M}^{0}}{{L}^{2}}{{T}^{0}} \right],\left[ {{M}^{-1}}{{L}^{1}}{{T}^{2}} \right]$

Answer 1

In this question we have been asked to find the dimensions of a and b from the given relation. We have been given an equation of power. Therefore, to solve this question, we shall equate the dimensions on the left side of the equation to the right side. We also know that while dealing with dimensions, the dimensions can not be subtracted or added to each other. They can only be multiplied or divided by other dimensions.

Complete answer:
We know that the unit of power is watt. We also know that, the dimensions of power can be given by,
$P=\left[ {{M}^{1}}{{L}^{2}}{{T}^{-3}} \right]$
From the given equation,
$P=\dfrac{b-{{x}^{2}}}{at}$
It is given that x is the distance and t is the time. The dimension of distance is L and that of time is T.
Now, let us equate the dimensions of power with the given equation.
We get,
$\left[ {{M}^{1}}{{L}^{2}}{{T}^{-3}} \right]=\dfrac{b-{{L}^{2}}}{a.T}$
Now, we know that,
$b-{{L}^{2}}=0$
Therefore,
$b={{L}^{2}}$
Therefore, dimensions of b can be given by,
$b=\left[ {{M}^{0}}{{L}^{2}}{{T}^{0}} \right]$ ……………. (1)
Therefore, the equation (A) becomes,
$\left[ {{M}^{1}}{{L}^{2}}{{T}^{-3}} \right]=\dfrac{{{L}^{2}}}{a.T}$
Now, from above equation we can say that,
$\left[ {{M}^{1}}{{L}^{2}}{{T}^{-3}} \right]={{a}^{-1}}{{T}^{-1}}{{L}^{2}}$ ……………………… (2)
Let,
$a={{M}^{x}}{{L}^{y}}{{T}^{z}}$
After substituting the a in equation (2)
We get,
$\left[ {{M}^{1}}{{L}^{2}}{{T}^{-3}} \right]={{[{{M}^{x}}{{L}^{y}}{{T}^{z}}]}^{-1}}[{{T}^{-1}}{{L}^{2}}]$
On comparing both equation and cancelling out the like terms
We get,
${{M}^{1}}{{T}^{-2}}={{M}^{x}}{{L}^{y}}{{T}^{z}}$
From above equation
We get,
$x=1,y=0,z=-2$
Therefore,
$a=\left[ {{M}^{-1}}{{L}^{0}}{{T}^{2}} \right]$ ………….. (3)
From (1) and (3) we can say that,

The correct answer is option B.

Note:
When any equation gives the relation between the fundamental units and derived units in terms of dimensions is called a dimensional formula. The fundamental dimensions are length, time mass, charge and many others. The units derived from fundamental dimensions are called as derived units. In dimensional formula, the addition or subtraction of dimensions is never done. They are only divided or multiplied. The dimensions of unlike units can not be added or subtracted.