A Boy Recalls the Relation Almost Correctly but Forgets Where to Put the Constant C Guess Where to Put the Missing C - Physics


A famous relation in physics relates ‘moving mass’ to the ‘rest mass’ m0 of a particle in terms of its speed 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 = m_0/(1-v^2)^(1/2)`

Guess where to put the missing c.


Solution 1

Given the relation,

m = `m_0/(1-v^2)^(1/2)`

Dimension of m = M1 L0 T0

Dimension of `m_0` = M1 L0 T0

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 only possible when the factor, `(1-v^2)^(1/2)` is dimensionless i.e., (1 – v2) is dimensionless. This is only possible if v2 is divided by c2. Hence, the correct relation is `m = m_0/(1-v^2/c^2)^(1/2)`

Solution 2

From the given equation, m_0/m = `sqrt(1-v^2)`

Left hand side is dimensionless

Therefore, right hand side should also be dimensionless.

It is possible only when `sqrt(1-v^2) should be sqrt(1-v^2/c^2)

Thus the correct formula is m = `m_0 (1 - v^2/c^2)^((-1)/2)`

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Chapter 2: Units and Measurements - Exercises [Page 36]


NCERT Physics Class 11
Chapter 2 Units and Measurements
Exercises | Q 15 | Page 36

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