A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ m0 of a particlein 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:
Answer:
Given the relation,
Dimension of m = M1 L0 T0
Dimension of = 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, is dimensionless
i.e., (1 – v2) is dimensionless. This is only possible if v2 is divided by c2. Hence, the
correct relation is
Answer:
Given the relation,
Dimension of m = M1 L0 T0
Dimension of = 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, is dimensionless
i.e., (1 – v2) is dimensionless. This is only possible if v2 is divided by c2. Hence, the
correct relation is
No comments:
Post a Comment