### Q.15:- A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ m0 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.

**Answer:-**

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}