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 1

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)`

#### Solution 2

From the given equation, m_0/m = `sqrt(1-v^2)`

Left hand side is dimensionless

Therefore, right hand side should also be dimensionless.

It is possible only when `sqrt(1-v^2) should be sqrt(1-v^2/c^2)

Thus the correct formula is m = `m_0 (1 - v^2/c^2)^((-1)/2)`