Question

A famous relation in physics relates moving mass m to the rest mass $m_0$ of a particle in terms of its

speed v and the speed of light c. (This relation first arose as a consequence of special theory of relatively

given by Albert Einstein). A boy recalls the relation almost correctly but forgets where to put the constant

c. He writes $m=\frac{m_0}{(1-v^2{)}^{\frac{1}{2}}}.$ Guess where to put the missing c?

• A. $m=\frac{m_{0}c}{(1-v^2{)}^{\frac{1}{2}}}$
• B. $m=\frac{m_0}{(c^2-v^2{)}^{\frac{1}{2}}}$
• C. $m=\frac{m_0}{(1-\frac{v^2}{c^2}{)}^{\frac{1}{2}}}$
• D. None of these

Answer 1

The correct option is C

$m=\frac{m_0}{(1-\frac{v^2}{c^2}{)}^{\frac{1}{2}}}$

Dimension of L.H.S. should be equal to R.H.S.

$m=\frac{m_0}{(1-v^2{)}^{\frac{1}{2}}}$

L.H.S. = [M]

R.H.S. we already have $m_0$ whose dimension will be [M]

So denominator should be dimensionless

Denominator is $(1-v^2)$

1 is dimensionless

But v has dimension $\left[\frac{L}{T}\right]$

Also c has same dimension

$\Rightarrow \frac{v^2}{c^2}$ will be dimensionless

$m=\frac{m_0}{[1-(\frac{V}{C}{)}^2{]}^{\frac{1}{2}}}$