Question:
Expression for time in terms of G (universal gravitational constant), h (Planck constant) and c (speed of light) is proportional to :
Expression for time in terms of G (universal gravitational constant), h (Planck constant) and c (speed of light) is proportional to :
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Write t = G^a h^b c^d and equate the powers of M, L and T on both sides.
Updated On: Nov 5, 2026
- \(\sqrt{\frac{hc^5}{G}}\)
- \(\sqrt{\frac{c^3}{Gh}}\)
- \(\sqrt{\frac{Gh}{c^5}}\)
- \(\sqrt{\frac{Gh}{c^3}}\)
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The Correct Option is C
Solution and Explanation
\([G] = M^{-1}L^{3}T^{-2}\), \([h] = ML^{2}T^{-1}\), \([c] = LT^{-1}\), \([t] = T\).
Let \(t \propto G^a h^b c^d\). Then:
M: \(-a + b = 0 \Rightarrow b = a\)
L: \(3a + 2b + d = 0 \Rightarrow 5a + d = 0 \Rightarrow d = -5a\)
T: \(-2a - b - d = 1 \Rightarrow -3a + 5a = 1 \Rightarrow a = \frac12\).
So \(a = b = \frac12\), \(d = -\frac52\), giving \(t \propto \sqrt{\frac{Gh}{c^5}}\).
Answer: (3) \(\sqrt{\frac{Gh}{c^5}}\)
Let \(t \propto G^a h^b c^d\). Then:
M: \(-a + b = 0 \Rightarrow b = a\)
L: \(3a + 2b + d = 0 \Rightarrow 5a + d = 0 \Rightarrow d = -5a\)
T: \(-2a - b - d = 1 \Rightarrow -3a + 5a = 1 \Rightarrow a = \frac12\).
So \(a = b = \frac12\), \(d = -\frac52\), giving \(t \propto \sqrt{\frac{Gh}{c^5}}\).
Answer: (3) \(\sqrt{\frac{Gh}{c^5}}\)
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