The distance covered by a particle in time 't' is given by \(s = at + bt^2\) where 'a' and 'b' are two constants. The dimensional formula of 'b' is:

This question was previously asked in

ALP CBT 2 Physics and Maths Previous Paper 5 (Held On: 22 Jan 2019 Shift 2)

Option 1 : \([M^0 \ L^1 \ T^{-2}]\)

Electric charges and coulomb's law (Basic)

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__Concept__:

Principle of homogeneity of dimensions:

- According to this principle, a physical equation will be dimensionally correct if the dimensions of all the terms occurring on both sides of the equation are the same.
- This principle is based on the fact that only the physical quantities of the same kind can be added, subtracted, or compared.
- Thus, velocity can be added to velocity but not to force.

__Explanation:__

**Given: **

s = at + bt^{2}

From the principle of dimensional homogeneity, the left-hand side of the equation dimensionally equal to the right-hand side of the equation. So, the dimension of (s), (at) and (bt^{2}) is the same.

The dimension formula of distance (s) = [L]

**For the first term:**

∴ [L] = [a] [T]

\(⇒ \left[ a \right] =\frac{[L]}{[T]}=[LT^{-1}]\)

**For the second term:**

⇒ [L] = [b] [T^{2}]

\(⇒ \left[b \right] =\frac{[L]}{[T^2]}=[LT^{-2}]\)

The dimensional formula of 'b' is **[M ^{0 }L^{1} T^{-2}]**.