Ncert Exercises & Solutions

A famous relation in physics relates ‘moving mass m to the ‘rest mass’ $m_{0}$ $m_{0}$ 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 :

$m = \frac{m_{0}}{{(1 - v^{2})}^{1 / 2}}$ $m = \frac{m_{0}}{{(1 - v^{2})}^{1 / 2}}$ .

Guess where to put the missing c.

Given the relation,

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

Dimension of $m = [M^{1} L^{0} T^{0}]$ $m = [M^{1} L^{0} T^{0}]$

Dimension of $m_{0} = [M^{1} L^{0} T^{0}]$ $m_{0} = [M^{1} L^{0} T^{0}]$

Dimension of $v = [M^{0} L^{1} T^{- 1}]$ $v = [M^{0} L^{1} T^{- 1}]$

Dimension of

v^{2} = [M^{0} L^{2} T^{- 2}]

$v^{2} = [M^{0} L^{2} T^{- 2}]$

Dimension of $c = [M^{0} L^{1} T^{- 1}]$ $c = [M^{0} L^{1} T^{- 1}]$

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, ${(1 - v^{2})}^{1 / 2}$ ${(1 - v^{2})}^{1 / 2}$ is dimensionless i.e., (1 – $v^{2}$ $v^{2}$ ) is dimensionless. This is only possible if $v^{2}$ $v^{2}$ is divided by $c^{2}$ $c^{2}$ . Hence, the correct relation is-

$m = \frac{m_{0}}{{(1 - \frac{v^{2}}{c^{2}})}^{1 / 2}}$ $m = \frac{m_{0}}{{(1 - \frac{v^{2}}{c^{2}})}^{1 / 2}}$

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