WBJEE · Physics · Gravitation
A modified gravitational potential is given by \(V=-\frac{G M}{r}+\frac{A}{r^2}\). If the constant \(A\) is expressed in terms of gravitational constant (G), mass ( \(M\) ) and velocity of light (c), then from dimensional analysis, \(A\) is,
- A \(\frac{\mathrm{G}^2 \mathrm{M}^2}{\mathrm{c}^2}\)
- B \(\frac{\mathrm{GM}}{\mathrm{c}^2}\)
- C \(\frac{1}{\mathrm{c}^2}\)
- D Dimensionless
Answer & Solution
Correct Answer
(A) \(\frac{\mathrm{G}^2 \mathrm{M}^2}{\mathrm{c}^2}\)
Step-by-step Solution
Detailed explanation
Hint : \(V=-\frac{G M}{r}+\frac{A}{r^2}\) \([A]=\frac{[G M]}{[r]}\left[r^2\right]=[G M][r]\) now, we know, \(\frac{G M}{r}\) gives dimension of \(c^2\)…
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