How many prime numbers less than $100$ can be represented as sum of squares of consequtive positive integers?

$ \textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 7$
To determine how many prime numbers less than $100$ can be represented as the sum of squares of consecutive positive integers, we need to check each prime number below $100$ and see if it can be expressed in the form of sums of squares of consecutive integers.

1. **List of prime numbers less than $100$:**
   \[
   2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97
   \]

2. **Identify perfect squares less than $100$:**
   \[
   1, 4, 9, 16, 25, 36, 49, 64, 81
   \]

3. **Check each prime number to see if it can be expressed as the sum of squares of consecutive integers:**

   - **Prime number $5$:**
     \[
     5 = 4 + 1 = 2^2 + 1^2
     \]
     Thus, $5$ can be represented as the sum of squares of consecutive integers.

   - **Prime number $13$:**
     \[
     13 = 9 + 4 = 3^2 + 2^2
     \]
     Thus, $13$ can be represented as the sum of squares of consecutive integers.

   - **Prime number $29$:**
     \[
     29 = 16 + 9 + 4 = 4^2 + 3^2 + 2^2
     \]
     Thus, $29$ can be represented as the sum of squares of consecutive integers.

   - **Prime number $41$:**
     \[
     41 = 25 + 16 = 5^2 + 4^2
     \]
     Thus, $41$ can be represented as the sum of squares of consecutive integers.

   - **Prime number $61$:**
     \[
     61 = 36 + 25 = 6^2 + 5^2
     \]
     Thus, $61$ can be represented as the sum of squares of consecutive integers.

4. **Count the number of such prime numbers:**
   The prime numbers less than $100$ that can be represented as the sum of squares of consecutive positive integers are $5, 13, 29, 41, 61$. There are $5$ such prime numbers.

Conclusion:
\[
\boxed{5}
\]