Let $S$ be the set of prime numbers greater than or equal to $2$ and less than $100$. Multiply all elements of $S$. With how many consecutive zeros will the product end?

The correct option is A. $1$

Since, $S$ is set of all the prime numbers from $2$ to $100)$, we have,

$S=2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97$

Now, we know that the $2$ is the only even prime number and $5$ is the only prime number of the multiples of $5$.

Also, we know that to have $n$ number of $0$ at the end of the number, the number should have factors in the form of $2^n\times 5^n$.

Since, in the set $S$, only one pair of such exists ie $2^1\times 5^1$,
therefore, number of $0$ at the end of desired product will be $1$.