This is a classic identification problem where one out of 10 boxes contains heavier coins. To determine which box it is using only one weighing, the trick is to take a unique number of coins from each box.
Take 1 coin from Box 1, 2 coins from Box 2, ..., up to 10 coins from Box 10. Total coins taken:
\[
1 + 2 + 3 + \ldots + 10 = \frac{10 \times 11}{2} = 55
\]
If all coins were 10 gm, total weight = \( 55 \times 10 = 550 \) gm.
But one of the boxes contains 20 gm coins. Suppose the extra weight is \( x \), then
\[
x = (20 - 10) \times i = 10 \times i \Rightarrow i = \frac{\text{Extra Weight}}{10}
\]
Thus, just one weighing is enough to identify the heavy box.