Consider a spherical particle of mass \( m \) and radius \( r \) moving in a medium. Its velocity at any time is given by \( v = v_0 \exp\left( -\frac{6 \pi X r t}{m} \right) \), where \( v_0 \) is the initial velocity of the particle. The dimensions of \( X \) are
Show Hint
- MLT\(^{-1}\)
- M\(^{-1}\)T\(^{-1}\)
- ML\(^{-1}\)T\(^{-1}\)
- Dimensionless
The Correct Option is C
Solution and Explanation
Step 1: Understanding the equation.
The given equation for velocity \( v \) has the exponential form where \( X \) has dimensions that we need to find. From the equation, we see that the argument of the exponential must be dimensionless. This gives the equation:
\[
\left( \frac{6 \pi X r t}{m} \right) \text{ must be dimensionless}.
\]
Step 2: Analyze the dimensions of the variables.
The dimensions of the variables are as follows:
- \( r \) (radius) has dimensions of length \([L]\).
- \( t \) (time) has dimensions of time \([T]\).
- \( m \) (mass) has dimensions of mass \([M]\).
Thus, the dimensions of the term \( \frac{r t}{m} \) are:
\[
\left[ \frac{r t}{m} \right] = \frac{L T}{M}.
\]
For the entire expression to be dimensionless, the dimensions of \( X \) must cancel the dimensions of \( \frac{r t}{m} \), which means \( X \) must have the dimensions:
\[
[X] = \frac{M}{L T}.
\]
Step 3: Conclusion.
The dimensions of \( X \) are \( \text{ML}^{-1}\text{T}^{-1} \), so the correct answer is (C).
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