Units & Dimensions

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23 of 149

Question 23 of 149

Dimensions of universal gravitational constant (G)(G) in terms of Planck's constant (h)(h), distance (L)(L), mass (M)(M) and time (T)(T) are _______.

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Answer: (B) [hT−1LM−2][hT^{-1}LM^{-2}]

The dimensional formula of the universal gravitational constant GG is obtained from Newton's law of gravitation F=Gm1m2r2F = \dfrac{G m_1 m_2}{r^2}.

[G]=[F][r2][m1m2]=[MLT−2][L2][M2]=[M−1L3T−2][G] = \dfrac{[F][r^2]}{[m_1 m_2]} = \dfrac{[MLT^{-2}][L^2]}{[M^2]} = [M^{-1}L^3T^{-2}]

The dimensional formula of Planck's constant hh is obtained from the energy of a photon E=hνE = h\nu.

[h]=[E][ν]=[ML2T−2][T−1]=[ML2T−1][h] = \dfrac{[E]}{[\nu]} = \dfrac{[ML^2T^{-2}]}{[T^{-1}]} = [ML^2T^{-1}]

Let the dimensions of GG be expressed as [G]=[h]a[L]b[M]c[T]d[G] = [h]^a [L]^b [M]^c [T]^d.

Substituting the dimensions, we get:

[M−1L3T−2]=[ML2T−1]a[L]b[M]c[T]d[M^{-1}L^3T^{-2}] = [ML^2T^{-1}]^a [L]^b [M]^c [T]^d

[M−1L3T−2]=[Ma+cL2a+bT−a+d][M^{-1}L^3T^{-2}] = [M^{a+c} L^{2a+b} T^{-a+d}]

Equating the powers of MM, LL, and TT on both sides:

a+c=−1a + c = -1

2a+b=32a + b = 3

−a+d=−2-a + d = -2

Assuming a=1a = 1 (as per the given options), we get:

c=−1−1=−2c = -1 - 1 = -2

b=3−2(1)=1b = 3 - 2(1) = 1

d=−2+1=−1d = -2 + 1 = -1

Therefore, the dimensions of GG are [h1L1M−2T−1][h^1 L^1 M^{-2} T^{-1}], which can be written as [hT−1LM−2][hT^{-1}LM^{-2}].

Answer: [hT−1LM−2][hT^{-1}LM^{-2}]

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