__Question 2.15__:

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 relativity due to Albert Einstein). A boy recalls the relation almost correctly but forgets where to put the constant c. He writes :

m = m

_{0 }

**/**(1-v

^{2})

^{1/2}

Guess where to put the missing c.

__Solution__:

Given the relation,

m = m

_{0 }

**/**(1-v

^{2})

^{1/2}

Dimension of

*m*= M

^{1}L

^{0}T

^{0}

Dimension of m

_{0}= M

^{1}L

^{0}T

^{0}

Dimension of

*v*= M

^{0}L

^{1}T

^{–1}

Dimension of

*v*

^{2}= M

^{0}L

^{2}T

^{–2}

Dimension of

*c*= M

^{0}L

^{1}T

^{–1}

The given formula will be dimensionally correct only when the dimension of L.H.S is the same as that of R.H.S. This is only possible when the factor, (1-v

^{2})

^{1/2}is dimensionless i.e., (1 –

*v*

^{2}) is dimensionless. This is only possible if

*v*

^{2}is divided by c

^{2}. Hence, the correct relation is

^{}m = m

_{0}

**/**(1 - v

^{2}/c

^{2})

^{1/2}

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