Question

A boy recalls the relation almost correctly but forgets where to put the constant $c$ i.e. speed of light. He writes $m=\frac{m_0}{\sqrt{1-v^2}}$, where $m$ and $m_0$ stand for masses and $v$ for speed. Right place of $c$ is

• A. $m=\frac{c{m}_0}{\sqrt{1-v^2}}$
• B. $m=\frac{m_0}{c\sqrt{1-v^2}}$
• C. $m=\frac{m_0}{\sqrt{c^2-v^2}}$
• D. $m=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$

Answer 1

The correct option is D $m=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$

Given relation is $m=\frac{m_0}{\sqrt{1-v^2}}$, so, both L.H.S and R.H.S should have dimension of mass.
Therefore, in $1-v^2$, $v^2$ should be dimensionless.
Hence, it should be $1-\frac{v^2}{c^2}$
So, the relation should be
$m=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$