If the size of an atom (1Å) were enlarged to the tip of a sharp pin (10⁻⁵ m), how large would the height of Mount Everest be ?

The correct option is 10⁹ m.

To find the enlarged height of Mount Everest, we first determine the magnification factor by which the atom is enlarged. Let us denote the original size of the atom as $d_{\text{atom}}$ and its enlarged size (the tip of a pin) as $D_{\text{atom}}$.

Given values:
Original size of the atom, $d_{\text{atom}}=1\text{Å}={10}^{-10}\text{m}$
Enlarged size of the atom, $D_{\text{atom}}={10}^{-5}\text{m}$

The magnification factor, $M$, is the ratio of the enlarged size to the original size:
$M=\frac{D_{\text{atom}}}{d_{\text{atom}}}=\frac{{10}^{-5}}{{10}^{-10}}={10}^5$

Now, let us consider the actual height of Mount Everest. The height of Mount Everest is approximately 8,848 meters, which is on the order of 10⁴ m (specifically, $\approx {10}^4\text{m}$ when rounded to the nearest power of ten). Let this actual height be $h_{\text{Everest}}\approx {10}^4\text{m}$.

If Mount Everest is enlarged by the same magnification factor $M$, its new height $H_{\text{Everest}}$ would be:
$H_{\text{Everest}}=h_{\text{Everest}}\times M$
$H_{\text{Everest}}\approx {10}^4\text{m}\times {10}^5={10}^9\text{m}$

Therefore, the enlarged height of Mount Everest would be on the order of 10⁹ m.