Class 11 NCERT PHYSICS • Chapter 1: Units and Measurement • Question 13

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We need to determine the correct placement of the constant $c$ in the mass-velocity relation using dimensional analysis.

Concept: Principle of Homogeneity of Dimensions

This principle states that the dimensions of each term in a physical equation must be identical. Furthermore, any quantity subtracted from or added to a dimensionless number (like $1$ ) must itself be dimensionless.

Given:

Moving mass: $m$ with dimension $[M]$
Rest mass: $m_0$ with dimension $[M]$
Velocity of the particle: $v$ with dimension $[L T^{-1}]$
Speed of light: $c$ with dimension $[L T^{-1}]$
Incorrect formula:

m = \frac{m_0}{(1 - v^2)^{1/2}}

Solving:

Analyze the dimensions of the mass ratio:

\frac{m}{m_0} = \frac{1}{(1 - v^2)^{1/2}}

Since both $m$ and $m_0$ represent mass, the ratio $\frac{m}{m_0}$ is dimensionless ( $M^0 L^0 T^0$ ).

For the equation to be dimensionally consistent, the denominator $(1 - v^2)^{1/2}$ must also be dimensionless.
In the expression $1 - v^2$ , the number $1$ is dimensionless. According to the Principle of Homogeneity of Dimensions, the term $v^2$ must also be dimensionless to be subtracted from $1$ .
Calculate the dimensions of $v^2$ :

[v^2] = [L T^{-1}]^2 = L^2 T^{-2}

To make this term dimensionless, we must divide $v^2$ by a quantity with the same dimensions. Since the speed of light $c$ has dimensions $[L T^{-1}]$ , its square $c^2$ has dimensions:

[c^2] = [L T^{-1}]^2 = L^2 T^{-2}

By dividing $v^2$ by $c^2$ , we obtain a dimensionless ratio:

\left[ \frac{v^2}{c^2} \right] = \frac{L^2 T^{-2}}{L^2 T^{-2}} = 1

Replacing $v^2$ with $\frac{v^2}{c^2}$ in the original formula restores dimensional consistency.

Answer

m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}

NCERT Solutions

Units and Measurement

Guess where to put the missing $c$ .

SOLUTION

We need to determine the correct placement of the constant $c$ in the mass-velocity relation using dimensional analysis.

Concept: Principle of Homogeneity of Dimensions

Given:

Solving:

Answer

Units and Measurement

Guess where to put the missing ccc.

SOLUTION

We need to determine the correct placement of the constant ccc in the mass-velocity relation using dimensional analysis.

Concept: Principle of Homogeneity of Dimensions

Given:

Solving:

Answer

Guess where to put the missing $c$ .

We need to determine the correct placement of the constant $c$ in the mass-velocity relation using dimensional analysis.