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
Answers
Answered by
6
Final Answer : Dim (m(o)) = M (Mass)
Steps and Understanding :
1)This is Einstein's Reduced Mass Equation,

2) dim(LHS) = dim(RHS)

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.
Steps and Understanding :
1)This is Einstein's Reduced Mass Equation,
2) dim(LHS) = dim(RHS)
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.
Answered by
12
Given equation,
m = m₀(1 - V₂/c₂)¹⁾²
∴ m₀ =
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.
m = m₀(1 - V₂/c₂)¹⁾²
∴ m₀ =
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.
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