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Question: If \[A = \int\limits_{{x_1}}^{{x_2}} {\dfrac{{dx}}{{B - x}}} \], where x represents position of a bo...

If A=∫x1x2dxB−xA = \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 AB\dfrac{A}{B} are
A. M0LT0{{\rm{M}}^0}{\rm{L}}{{\rm{T}}^0}
B. M0L0T0{{\rm{M}}^0}{{\rm{L}}^0}{{\rm{T}}^0}
C. M0L−1T0{{\rm{M}}^0}{{\rm{L}}^{ - 1}}{{\rm{T}}^0}
D. M0L2T0{{\rm{M}}^0}{{\rm{L}}^2}{{\rm{T}}^0}

Explanation

Solution

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=∫x1x2dxB−xA = \int\limits_{{x_1}}^{{x_2}} {\dfrac{{dx}}{{B - x}}} ……(1)

We can write the above expression after integrating as below:
A=[ln⁡(B−x)]x1x2A = \left[ {\ln \left( {B - x} \right)} \right]_{{x_1}}^{{x_2}}
On substituting the limits of the above expression, we get:

A=[ln⁡(B−x1)−ln⁡(B−x1)] ⇒A=ln⁡[B−x1B−x2]\begin{aligned} 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{aligned}

It is given that x represents a body's position where x1{x_1} and x2{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 x1{x_1} as below:
[x1]=[M0LT0]\left[ {{x_1}} \right] = \left[ {{{\rm{M}}^0}{\rm{L}}{{\rm{T}}^0}} \right]

We know that x2{x_2} is the final position of the given body, so its dimension is also the same as that of x1{x_1} and it can be expressed as below:
[x2]=[M0LT0]\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 x2{x_2}.
[B]=[M0LT0]\left[ B \right] = \left[ {{{\rm{M}}^0}{\rm{L}}{{\rm{T}}^0}} \right]

We know that logarithmic functions are dimensionless, so the term ln⁡[B−x1B−x2]\ln \left[ {\dfrac{{B - {x_1}}}{{B - {x_2}}}} \right] is also dimensionless and can be expressed as:
[ln⁡(B−x1B−x2)]=[M0L0T0]\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⁡[B−x1B−x2]\ln \left[ {\dfrac{{B - {x_1}}}{{B - {x_2}}}} \right].
[A]=[M0L0T0]\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=[A][B]r = \dfrac{{\left[ A \right]}}{{\left[ B \right]}}
Here r represents the dimensional ratio of A and B.

We will substitute [M0L0T0]\left[ {{{\rm{M}}^0}{{\rm{L}}^0}{{\rm{T}}^0}} \right] for [A]\left[ A \right] and [M0LT0]\left[ {{{\rm{M}}^0}{\rm{L}}{{\rm{T}}^0}} \right] for [B]\left[ B \right] in the above expression.

r=[M0L0T0][M0LT0] ⇒=[M0L−1T0]\begin{aligned} 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{aligned}

Therefore, the dimensional ratio of A and B is [M0L−1T0]\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:
∫t1t2dtt=[ln⁡(t)]t1t2\int\limits_{{t_1}}^{{t_2}} {\dfrac{{dt}}{t}} = \left[ {\ln \left( t \right)} \right]_{{t_1}}^{{t_2}}