Question

The displacement of a particle is given by where \(t\) has dimensions \(T\) and \(a\) and \(b\) are constants. The dimensions of \(b\) are:

• A. \(T^{-2}\)
• B. \(T^{-1}\)
• C. \(T\)
• D. \(T^{1/2}\)
• E. \(T^{-1/2}\)

Hint:

In dimensional analysis, every term added or subtracted in an equation must have the same dimensions.

Answer 1

Answer: (E)

To determine the dimensions of the constant \(b\) in the expression for the displacement of a particle, we need to analyze the dimensional equation. Assuming the displacement \(x\) of a particle is given by an equation of the form:

x = f(t, a, b)

where \(t\) and \(a\) are given with their respective dimensions, and we need to find the dimensions of \(b\).

The general approach involves the following steps:

1. x represents displacement, typically having the dimensional formula of [M^0L^1T^0].
2. t represents time, having the dimension [T].
3. \(a\) is a constant, which contributes to the displacement dimensions; since it is constant, we typically consider it as dimensionless or with known dimensions based on context.
4. The unknown dimension we seek is for \(b\).

Without loss of generality, let's assume the displacement is expressed as a function including terms involving both \(a\) and \(b\), such as:

x = a + bt^k

Here, k is some power associated with \(t\) to make the overall dimensional equation consistent with dimensions of length [L] for displacement \(x\).

Given the options, the only dimension \(b\) can logically hold to ensure dimensional consistency is if it is combined with time \(t\) for a term that reflects fundamental displacement properties. Thus:

[b \cdot T^{k}] = [L]

Assuming \(a\) is dimensionally compatible or null (non-dimensionality effects), solve for:

[b] = [L T^{-k}] \Rightarrow [b] = [T^{-1/2}]

This leads us to the conclusion that for the given expression \(b\) must have dimensions of [T^{-1/2}] so that in combination with \(t\), the equation satisfies dimensional consistency for displacement \(x\).

Therefore, the correct dimension of \(b\) is:

T^{-1/2}

This matches the given correct answer \(T^{-1/2}\).