Question

m c and V respectively denote the mass, speed and velocity of light then in the equation:
m= m0 (1-v2/c2)1/2
find the dimension of m0

Answer 1

Final Answer : Dim (m(o)) = M (Mass)

Steps and Understanding :
1)This is Einstein's Reduced Mass Equation,

$m = \frac{m(o)}{ \sqrt{1 - {( \frac{v}{c}) }^{2} } }$

2) dim(LHS) = dim(RHS)
$= > dim(m) = dim( \frac{m(o)}{ \sqrt{1 - \frac{ {v}^{2} }{ {c}^{2} } } } )$

Since, Denominator in RHS is constant as 1 is constant and (v/c) is also constant by constant law of addition units.

3)=> dim(m(o)) = dim(m)
=> dim(m(0)) = M

Therefore, dimension of m(o) is mass M.

Answer 2

Given equation,

m = m₀(1 - V₂/c₂)¹⁾²
∴ m₀ = $\frac{m}{ 1 - \sqrt{ \frac{V^{2} }{ c^{2} } } }$

In this, Dimension of Mass = M

Now, 
 Dimension of V = Dimension of Length/Dimension of Time
   = L/T
   = LT⁻¹
 Dimension of c (speed of the light) = Dimension of Length/Time
  = L/T
  = LT⁻¹

From the Equation,
 √ (1 - V²/c²) is 1, i.e. constant.

∴ [m₀] = M/1
          = M


Hence, the Dimension of m₀ is M.


Hope it helps.