Physics And Measurement Question 258
Question: A boy recalls the relation almost correctly but for gets where to put the constant c (speed of light). He writes; $ m=\frac{m _{0}}{\sqrt{1-v^{2}}} $ , where m and $ m _{0} $ stand masses and v for speed. Right place of c is
Options:
A) $ m=\frac{cm _{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}}}} $
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Answer:
Correct Answer: D
Solution:
[d] In, $ 1-v^{2},v^{2} $ should be dimensionless, so it should be $ 1-\frac{v^{2}}{c^{2}} $
BETA
