Question

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

• A. MLT\(^{-1}\)
• B. M\(^{-1}\)T\(^{-1}\)
• C. ML\(^{-1}\)T\(^{-1}\)
• D. Dimensionless

Hint:

For equations involving exponentials, the argument of the exponential must be dimensionless.

Answer 1

The Correct Option is C

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).