At what height over the earth's pole does the freefall acceleration decreases by 1% ?

Let the height be ' $h$ '

Given, acceleration is one percent less, which is $0.99 \mathrm{~g}$

$\begin{array}{l} \Rightarrow \quad \frac{99}{100} \mathrm{~g}=\frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})^{2}} \\ \text { As, } \mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}} \\ \frac{99}{100}=\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)^{-2} \end{array}$

Using binomial rule, as $\mathrm{h} \ll\mathrm{R}$

$\begin{array}{l} \frac{99}{100}=\left(1-\frac{2 h}{R}\right) \\ \Rightarrow \quad \mathrm{h}=\frac{\mathrm{R}}{200} \end{array}$

Radius of earth is $6400 \mathrm{~km}$

Therefore, $\mathrm{h}=2400 / 200=32 \mathrm{~km}$