Question

If \[A = \int\limits_{{x_1}}^{{x_2}} {\dfrac{{dx}}{{B - x}}} \], where x represents position of a body, A and B are unknown quantities, then the dimensions of \[\dfrac{A}{B}\] are
A. \[{{\rm{M}}^0}{\rm{L}}{{\rm{T}}^0}\]
B. \[{{\rm{M}}^0}{{\rm{L}}^0}{{\rm{T}}^0}\]
C. \[{{\rm{M}}^0}{{\rm{L}}^{ - 1}}{{\rm{T}}^0}\]
D. \[{{\rm{M}}^0}{{\rm{L}}^2}{{\rm{T}}^0}\]

Answer 1

Hint: We will use the concept of homogeneity of dimensions that say that the dimension of two adding quantities is the same; also, the dimensions of both sides of an equation should be the same.

Complete step by step answer:
Let us write the given value of A.
\[A = \int\limits_{{x_1}}^{{x_2}} {\dfrac{{dx}}{{B - x}}} \]……(1)

We can write the above expression after integrating as below:
\[A = \left[ {\ln \left( {B - x} \right)} \right]_{{x_1}}^{{x_2}}\]
On substituting the limits of the above expression, we get:
\begin{align*}
A &= \left[ {\ln \left( {B - {x_1}} \right) - \ln \left( {B - {x_1}} \right)} \right]\\
\Rightarrow A &= \ln \left[ {\dfrac{{B - {x_1}}}{{B - {x_2}}}} \right]
\end{align*}
It is given that x represents a body's position where \[{x_1}\] and \[{x_2}\] represents its initial and final positions. We know that the position of a body is measured in meters because it is a fundamental unit length. Therefore we can say write the dimension of position \[{x_1}\] as below:
\[\left[ {{x_1}} \right] = \left[ {{{\rm{M}}^0}{\rm{L}}{{\rm{T}}^0}} \right]\]

We know that \[{x_2}\] is the final position of the given body, so its dimension is also the same as that of \[{x_1}\] and it can be expressed as below:
\[\left[ {{x_2}} \right] = \left[ {{{\rm{M}}^0}{\rm{L}}{{\rm{T}}^0}} \right]\]

From the concept of homogeneity of dimensions, we can say that B's dimension should be the same as \[{x_2}\].
\[\left[ B \right] = \left[ {{{\rm{M}}^0}{\rm{L}}{{\rm{T}}^0}} \right]\]

We know that logarithmic functions are dimensionless, so the term \[\ln \left[ {\dfrac{{B - {x_1}}}{{B - {x_2}}}} \right]\] is also dimensionless and can be expressed as:
\[\left[ {\ln \left( {\dfrac{{B - {x_1}}}{{B - {x_2}}}} \right)} \right] = \left[ {{{\rm{M}}^0}{{\rm{L}}^0}{{\rm{T}}^0}} \right]\]

Again, using the concept of homogeneity of dimensions, we can say that the dimension of A is the same as that of the term's dimension \[\ln \left[ {\dfrac{{B - {x_1}}}{{B - {x_2}}}} \right]\].
\[\left[ A \right] = \left[ {{{\rm{M}}^0}{{\rm{L}}^0}{{\rm{T}}^0}} \right]\]

Let us write the expression for the ratio of the dimension of A and dimension of B.
\[r = \dfrac{{\left[ A \right]}}{{\left[ B \right]}}\]
Here r represents the dimensional ratio of A and B.

We will substitute \[\left[ {{{\rm{M}}^0}{{\rm{L}}^0}{{\rm{T}}^0}} \right]\] for \[\left[ A \right]\] and \[\left[ {{{\rm{M}}^0}{\rm{L}}{{\rm{T}}^0}} \right]\] for \[\left[ B \right]\] in the above expression.
\begin{align*}
r &= \dfrac{{\left[ {{{\rm{M}}^0}{{\rm{L}}^0}{{\rm{T}}^0}} \right]}}{{\left[ {{{\rm{M}}^0}{\rm{L}}{{\rm{T}}^0}} \right]}}\\
\Rightarrow &= \left[ {{{\rm{M}}^0}{{\rm{L}}^{ - 1}}{{\rm{T}}^0}} \right]
\end{align*}
Therefore, the dimensional ratio of A and B is \[\left[{{{\rm{M}}^0}{{\rm{L}}^{ - 1}}{{\rm{T}}^0}} \right]\], and option (C) is correct.

Note:Using the identity for integration, we will integrate the given expression using the formula as written below:
\[\int\limits_{{t_1}}^{{t_2}} {\dfrac{{dt}}{t}} = \left[ {\ln \left( t \right)} \right]_{{t_1}}^{{t_2}}\]