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Units And Measurement

Question
CBSEENPH11019883

A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ moof 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 :

straight m space equals space fraction numerator straight m subscript straight o over denominator left parenthesis square root of 1 minus straight v squared end root right parenthesis to the power of bevelled 1 half end exponent end fraction

Guess where to put the missing c.

Solution

Given the relation, 
straight m space equals space fraction numerator straight m subscript straight o over denominator left parenthesis square root of 1 minus straight v squared end root right parenthesis to the power of bevelled 1 half end exponent end fraction 
Dimension of m = M1 L0 T0
Dimension of mo = M1 L0 T
Dimension of v = M0 L1 T–1
Dimension of v2 = M0 L2 T–2
Dimension of c = M0 L1 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 possible when the factor in the denominator is dimensionless. This is only possible if straight nu squared space is space divided space by space straight c squared. 
Hence the correct relation is 
straight m space equals space straight m subscript straight o over open parentheses 1 minus begin display style straight v squared over straight c squared end style close parentheses to the power of bevelled 1 half end exponent