All numbers in this solution will be in hundredths of a millimeter.
The thinnest coin is the dime, with thickness $135$. A stack of $n$ dimes has height $135n$.
The other three coin types have thicknesses $135+20$, $135+40$, and $135+60$. By replacing some of the dimes in our stack by other, thicker coins, we can clearly create exactly all heights in the set $\{135n, 135n+20, 135n+40, \dots, 195n\}$.
If we take an odd $n$, then all the possible heights will be odd, and thus none of them will be $1400$. Hence $n$ is even.
If $n10$ it will be too high. Thus we are left with cases $n=8$ and $n=10$.
If $n=10$ the possible stack heights are $1350,1370,1390,\dots$, with the remaining ones exceeding $1400$.
Therefore there are $\boxed{\mathrm{(B)\ }8}$ coins in the stack.
Using the above observation we can easily construct such a stack. A stack of $8$ dimes would have height $8\cdot 135=1080$, thus we need to add $320$.
This can be done for example by replacing five dimes by nickels (for $+60\cdot 5 = +300$), and one dime by a penny (for $+20$).
In the United States, coins have the following thicknesses: penny, $1.55$ mm; nickel, $1.95$ mm; dime, $1.35$ mm; quarter, $1.75$ mm. If a stack of these coins is exactly $14$ mm high, how many coins are in the stack?
$\mathrm{(A) \ } 7 \qquad \mathrm{(B) \ } 8 \qquad \mathrm{(C) \ } 9 \qquad \mathrm{(D) \ } 10 \qquad \mathrm{(E) \ } 11$