Physics - Gravitation Question with Solution | TestHub

$T=2\pi \sqrt{\frac{l}{g}}=2\pi \sqrt{\frac{\ell (R+h{)}^2}{GM}}$
$T\propto (R+h)$
or$\frac{T}{T}=\frac{R+h}{R}=1+\frac{h}{R}=1+\frac{1}{100}\therefore \frac{T^{\text{'}}}{T}-1=\frac{1}{100}$
$\therefore \left(\frac{T^{\text{'}}-T}{T}\right)\times 100=1$
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