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Q1.13: NCERT Physics - Part 1 UNITS AND MEASUREMENT 1.13 Solution: A famous relation in physics relates 'moving mass' m to the…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often omit the factor $c$ and treat $1-v^2$ as dimensionless, forgetting that speed must be compared to $c$; always form the Lorentz factor as $\sqrt{1-\frac{v^2}{c^2}}$ so the expression inside the root is dimensionless.

A famous relation in physics relates 'moving mass' m to the 'rest mass' m_ of a particle in terms of its speed v and the speed of light c. A boy recalls the relation almost correctly but forgets where to put the constant c. He writes: m=m0(1−v2)1/2m = \frac{m_0}{(1-v^2)^{1/2}}. Guess where to put the missing c.

Step-by-Step Verified Solution

**Answer:** m=m01−v2/c2m = \frac{m_0}{\sqrt{1 - v^2/c^2}} **Step-by-Step Explanation:** The quantity (1−v2)(1 - v^2) has dimensions of [L2T−2][L^2 T^{-2}], which is not dimensionless. For the equation to be dimensionally correct, we must have (1−v2/c2)(1 - v^2/c^2) in the denominator. Hence, the missing cc goes in the denominator as v2/c2v^2/c^2.

All Questions in this Chapter

Q1.1 Q1.2 Q1.3 Q1.4 Q1.5 Q1.6 Q1.7 Q1.8 Q1.9 Q1.10 Q1.11 Q1.12 Q1.13 Q1.14 Q1.15 Q1.16 Q1.17

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