Q150 / 155 · WBJEE 2023 · Physics

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,

  • Correct answer
    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

Solution

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 \(c^2\) \(\frac{[\mathrm{GM}]}{[\mathrm{r}]}=\left[\mathrm{c}^2\right] \Rightarrow[\mathrm{r}]=\frac{[\mathrm{GM}]}{\left[\mathrm{c}^2\right]}\) \(\Rightarrow[\mathrm{A}]=\frac{[\mathrm{GM}][\mathrm{GM}]}{\left[\mathrm{c}^2\right]}\) \([\mathrm{A}]=\frac{\mathrm{G}^2 \mathrm{M}^2}{\mathrm{c}^2}\)

Hence, the correct answer is option: 'A'.

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