Time period of satellite $T=2 \pi \sqrt{\frac{r^{3}}{g R^{2}}}$ where $R=6400 \mathrm{km}=6.4 \times 10^{6} \mathrm{m}$
$\therefore 24 \times 3600=2 \pi \sqrt{\frac{r^{3}}{9.8\left(6.4 \times 10^{6}\right)^{2}}}$
OR $\frac{r^{3}}{401.408 \times 10^{12}}=1.89 \times 10^{8} \quad \Longrightarrow r^{3}=76 \times 10^{21}$
$\Rightarrow r=42400 \mathrm{km}$
Thus height of satellite above earth surface $\quad h=42400-6400=36000 \mathrm{km}$
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$I.$ The speed of the wave is $4n \times ab$
$II.$ The medium at $a$ will be in the same phase as $d$ after $\frac{4}{{3n}}s$
$III.$ The phase difference between $b$ and $e$ is $\frac{{3\pi }}{2}$
Which of these statements are correct
