A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ m0 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 = m0 / (1-v2)1/2 Guess where to put the missing c. | Learn NCERT solution | Education portal Class 11 Physics | Study online Unit-2 Units and Measurements



Q.15:- A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ m0 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 = m0 / (1-v2)1/2
Guess where to put the missing c.

 

Answer:-



Given the relation,
m = m/ (1-v2)1/2
Dimension of m = M1 L0 T0
Dimension of m0 = 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-v2)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 = m0 / (1 – v2/c2)1/2