Question

The acceleration of a point particle is given by the equation \[ \frac{d^2\mathbf{x}}{dt^2} = \alpha \frac{\mathbf{x}}{|\mathbf{x}|^7} + \beta \frac{d\mathbf{x}}{dt} \] where $\mathbf{x}$ denotes position and $t$ denotes time. Which of the following relations show the correct dimensions for $\alpha$ and $\beta$?

• A. $[\alpha] = [\text{M}^0\text{L}^7\text{T}^{-2}]$, $[\beta] = [\text{M}^0\text{L}^0\text{T}^{-1}]$
• B. $[\alpha] = [\text{M}^1\text{L}^6\text{T}^{-2}]$, $[\beta] = [\text{M}^0\text{L}^0\text{T}^{-3}]$
• C. $[\alpha] = [\text{M}^0\text{L}^6\text{T}^{-1}]$, $[\beta] = [\text{M}^0\text{L}^1\text{T}^{-2}]$
• D. $[\alpha] = [\text{M}^0\text{L}^7\text{T}^{-2}]$, $[\beta] = [\text{M}^0\text{L}^0\text{T}^0]$

Hint:

The principle of dimensional homogeneity requires that every term separated by a plus or minus sign in an equation must have the exact same dimensions as the term on the other side of the equal sign.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are given a differential equation describing the acceleration of a point particle.
Using the principle of dimensional homogeneity, all terms in a physical equation must have the exact same dimensions.
We will find the dimensions of $\alpha$ and $\beta$ by equating the dimensions of the terms on the right-hand side to those of acceleration on the left-hand side.

Step 2: Key Formula or Approach:
The dimensions of the basic quantities are:
- Position: $[\mathbf{x}] = [\text{L}]$
- Time: $[t] = [\text{T}]$
- Acceleration (LHS): $\left[ \frac{d^2\mathbf{x}}{dt^2} \right] = [\text{L}\text{T}^{-2}]$

Step 3: Detailed Explanation:

• Let us analyze the first term on the right-hand side:
\[ \left[ \alpha \frac{\mathbf{x}}{|\mathbf{x}|^7} \right] = [\text{L}\text{T}^{-2}] \] \[ [\alpha] \frac{[\text{L}]}{[\text{L}]^7} = [\text{L}\text{T}^{-2}] \] \[ [\alpha] [\text{L}]^{-6} = [\text{L}\text{T}^{-2}] \] \[ [\alpha] = [\text{L}]^7 [\text{T}]^{-2} = [\text{M}^0\text{L}^7\text{T}^{-2}] \]
• Now, let us analyze the second term on the right-hand side:
\[ \left[ \beta \frac{d\mathbf{x}}{dt} \right] = [\text{L}\text{T}^{-2}] \] \[ [\beta] [\text{L}\text{T}^{-1}] = [\text{L}\text{T}^{-2}] \] \[ [\beta] = [\text{T}]^{-1} = [\text{M}^0\text{L}^0\text{T}^{-1}] \]

Step 4: Final Answer:
The dimensions are $[\alpha] = [\text{M}^0\text{L}^7\text{T}^{-2}]$ and $[\beta] = [\text{M}^0\text{L}^0\text{T}^{-1}]$.