Question:medium

If mass is written as \( m = k c^P G^{-1/2} h^{1/2} \), then the value of \( P \) will be:

Updated On: Jan 13, 2026
  • \( \frac{1}{3} \)

  • \( \frac{1}{2} \)

  • 2

  • \( -\frac{1}{3} \)

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The Correct Option is B

Solution and Explanation

The provided equation is:

\[ m = k c^P G^{-1/2} h^{1/2}, \]

where \( k \) is a dimensionless constant, \( c \) is the speed of light (\([c] = [L][T]^{-1}\)), \( G \) is the gravitational constant (\([G] = [M]^{-1}[L]^3[T]^{-2}\)), and \( h \) is Planck's constant (\([h] = [M][L]^2[T]^{-1}\)).

The dimensions of mass are:

\[ [m] = [M]. \]

The dimensions of each term in the equation are as follows:

\[ [c^P] = ([L][T]^{-1})^P = [L]^P[T]^{-P}, \]

\[ [G^{-1/2}] = ([M]^{-1}[L]^3[T]^{-2})^{-1/2} = [M]^{1/2}[L]^{-3/2}[T], \]

\[ [h^{1/2}] = ([M][L]^2[T]^{-1})^{1/2} = [M]^{1/2}[L][T]^{-1/2}. \]

Substituting these dimensions into the equation yields:

\[ [M] = k \cdot [L]^P[T]^{-P} \cdot [M]^{1/2}[L]^{-3/2}[T]^{-1} \cdot [M]^{1/2}[L][T]^{-1/2}. \]

Combining the dimensions results in:

\[ [M] = [M]^{1/2 + 1/2}[L]^{P - 3/2 + 1}[T]^{-P + 1 - 1/2}. \]

Equating the powers of each dimension:

For mass \([M]\):

\[ 1 = \frac{1}{2} + \frac{1}{2}. \]

For length \([L]\):

\[ 0 = P - \frac{3}{2} + 1. \]

Simplifying this equation gives:

\[ P = \frac{1}{2}. \]

For time \([T]\):

\[ 0 = -P + 1 - \frac{1}{2}. \]

Simplifying this equation yields:

\[ P = \frac{1}{2}. \]

Therefore, the value of \( P \) is:

\[ \frac{1}{2} \]

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