### Question 2.15

NCERT Class XI Physics

A famous relation in physics relates 'moving mass' m to the 'rest mass' mo of a particle in terms of its speed v and the speed of light, c. (This relation first arose as a consequence of special relativity due to Albert Einstein). A boy recalls the relation almost correctly but forgets where to put the constant c. He writes : m=m0/√(1-v2). Guess where to put the missing c.

### Video Explanation

(detailed solution given after this video)

Please watch this youtube video for a quick explanation of the solution:

### Solution in Detail

$ \begin{aligned} &\text{given } m = \frac{m_0}{\sqrt{1 - v^2}}\\\\ &\implies \frac{m}{m_0} = \frac{1}{\sqrt{1 - v^2}}\\\\ &\text{must be dimensionally correct}\\\\ &\therefore \bigg[\frac{m}{m_0}\bigg] = \bigg[\frac{1}{\sqrt{1 - v^2}}\bigg]\\\\ &\implies \text{LHS is dimensionless}\\\\ &\text{but RHS has an unbalanced }\\ &\text{dimension of }v^2\text{ that needs to} \\ &\text{cancelled by a square of speed } c^2\\\\ &\therefore \text{either RHS } = \frac{1}{\sqrt{1 - \displaystyle \frac{v^2}{c^2}}}\text{. . .(1)}\\\\ &\text{or RHS } = \frac{1}{\sqrt{1 - \displaystyle \frac{c^2}{v^2}}}\text{. . .(2)}\\\\ &\text{(2) is rejected as it fails at } v = 0\\\\ &\therefore m = \frac{m_0}{\sqrt{1 - \displaystyle \frac{v^2}{c^2}}}\\\\ &\therefore \text{c must be in denominator of v}\:\underline{Ans} \end{aligned} $

### Physical Concept and Solution in Brief

A relation has to be dimensionally correct.

We observe that the numerators on both sides are masses of dimension [M]. The denominator on RHS must, therefore, be dimension-less, which is possible if c forms a fraction with v either as (i)v/c or (ii)c/v. The latter can be rejected because v cannot be zero. Hence c must be in the denominator of v.

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