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Question : 15 of 33
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A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ mo of a particle in terms of its speed v and the speed of light, c. (This relation first arose as a consequence of special theory of relativity due to Albert Einstein). A boy recalls the relation almost correctly but forgets where to put the constant c. He writes :
m=mo(1−v2)1/2m = \frac{m_o}{(1-v^2)^{1/2}}.
Guess where to put the missing c.
Solution:  
From principle of homogenetity of dimensions both sides ofabove formula must be same dimensions. For this, (1−v2)1/2(1 - v^2)^{1/2} must bedimensionless.
Therefore, instead of (1−v2)1/2,(1-v^2)^{1/2}, it will be (1−v2c2)1/2\left(1 - \frac{v^2}{c^2}\right)^{1/2}
Hence relation should be mo(1−v2c2)1/2\frac{m_o}{\left(1 - \frac{v^2}{c^2}\right)^{1/2}}.
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