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:

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Solution

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

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A famous relation in physics relates moving mass m to the rest mass $m_{0}$ of a particle in terms of its

speed v and the speed of light c. (This relation first arose as a consequence of special theory of relatively

given by Albert Einstein). A boy recalls the relation almost correctly but forgets where to put the constant

c. He writes $m = \frac{m_{0}}{(1 - v^{2})^{\frac{1}{2}}} .$ Guess where to put the missing c?

A famous relation in physics relates moving mass m to the rest mass $m_{0}$ of a particle in terms of its

speed v and the speed of light c. (This relation first arose as a consequence of special theory of relatively

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c. He writes $m = \frac{m_{0}}{(1 - v^{2})^{\frac{1}{2}}} .$ Guess where to put the missing c?

m_{0}

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