1. We need the smallest equivalence relation on $S=\{1,2,3,4\}$ containing $(1,2), (1,3).$

2. Recall: an equivalence relation must be

   • reflexive: $(a,a)\in R$ for all $a\in S$
   • symmetric: if $(a,b)\in R$, then $(b,a)\in R$
   • transitive: if $(a,b),(b,c)\in R$, then $(a,c)\in R$

3. Start with the given pairs: $(1,2), (1,3).$

4. Since $R$ must be symmetric, we must also include: $(2,1), (3,1).$

5. Since $R$ must be reflexive on $\{1,2,3,4\}$, we must include: $(1,1), (2,2), (3,3), (4,4).$

6. Now apply transitivity:

   • From $(2,1)$ and $(1,3)$, we get $(2,3)$
   • From $(3,1)$ and $(1,2)$, we get $(3,2)$

   So we must also include: $(2,3), (3,2).$

7. Thus the elements $1,2,3$ all become equivalent to each other, forming one equivalence class: $\{1,2,3\}$ and $4$ remains alone in the class $\{4\}.$

8. For an equivalence class of size $3$, all ordered pairs within that class belong to $R$, giving $3^2 = 9$ pairs.

   For the singleton class $\{4\}$, we get $1^2 = 1$ pair.

9. Therefore total number of elements in $R$ is $9+1=10.$

10. So the correct option is $\boxed{\text{B}}.$

Verification with stored answer

Stored correct answer: $\text{B}$

Our derived answer is also $\text{B}$, so they agree.