MHT CET · Physics · Laws of Motion
A particle moves in a circular orbit of radius ' \(r\) ' under a central attractive force, \(\mathrm{F}=-\frac{\mathrm{k}}{\mathrm{r}}\), where \(\mathrm{k}\) is a constant. The periodic time of tis motion is
- A \(\mathrm{r}^{\frac{1}{2}}\)
- B \(\mathrm{r}^{\frac{2}{3}}\)
- C \(\mathrm{r}\)
- D \(\mathrm{r}^{\frac{3}{2}}\)
Answer & Solution
Correct Answer
(C) \(\mathrm{r}\)
Step-by-step Solution
Detailed explanation
\(\operatorname{mr} \omega^2=\frac{\mathrm{k}}{\mathrm{r}}\)
\(\omega^2=\frac{\mathrm{k}}{\mathrm{mr}}\)
\(\omega^2 \propto \frac{1}{r^2}\)
\(\omega \propto \frac{1}{\mathrm{r}}\)
\(\therefore \mathrm{T} \propto \mathrm{r}\)
\(\omega^2=\frac{\mathrm{k}}{\mathrm{mr}}\)
\(\omega^2 \propto \frac{1}{r^2}\)
\(\omega \propto \frac{1}{\mathrm{r}}\)
\(\therefore \mathrm{T} \propto \mathrm{r}\)
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