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 relativity due to Albert Einstein). A boy recalls the relation almost correctly but forgets where to put the constant c. He writes :$$m = \frac{m_{0}}{\sqrt{1 – \nu ^{2}}} ​ ​ ​$$

Asked by Abhisek | 1 year ago |  100

##### Solution :-

The relation given is  $$m = \dfrac{m_{0}}{\sqrt{1 – \nu ^{2}}}$$

We can get, $$\dfrac{m_{0}}{m} ​ ​ = \sqrt{1-\nu ^{2}} ​ \dfrac{m_{0}}{m}$$ is dimensionless.

Therefore, the right hand side should also be dimensionless.

To satisfy this $$= \sqrt{1-\nu ^{2}}$$ should become

$$\sqrt{ 1-\dfrac{v^2}{c^2}}$$

Thus, $$m = m_{0}\sqrt{1-\frac{\nu ^{2}}{c^{2}}}$$



Answered by Pragya Singh | 1 year ago

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