How many consecutive zeros are there at the end of the integer obtained in the product $1^2\times 2^4\times 3^6\times 4^8\times ...\times {25}^{50}$?

The correct option is 200.

To find the number of consecutive zeros at the end of the integer obtained from the product, we need to find the highest power of 10 that divides the product. Since $10=2\times 5$, the number of trailing zeros is determined by the exponent of 2 or 5 in the prime factorization of the product, whichever is smaller.

In the given product:
$P=1^2\times 2^4\times 3^6\times 4^8\times ...\times {25}^{50}$
the prime factor 2 occurs much more frequently than the prime factor 5. Therefore, the number of trailing zeros is equal to the total exponent of the prime factor 5 in the product.

The terms in the product that contain the prime factor 5 are the multiples of 5, which are 5, 10, 15, 20, and 25. Let us calculate the power of 5 contributed by each of these terms:

1. For the term $5^{10}$:
The number of factors of 5 is 10.

2. For the term ${10}^{20}$:
${10}^{20}={\left(2\times 5\right)}^{20}=2^{20}\times 5^{20}$
The number of factors of 5 is 20.

3. For the term ${15}^{30}$:
${15}^{30}={\left(3\times 5\right)}^{30}=3^{30}\times 5^{30}$
The number of factors of 5 is 30.

4. For the term ${20}^{40}$:
${20}^{40}={\left(4\times 5\right)}^{40}=4^{40}\times 5^{40}$
The number of factors of 5 is 40.

5. For the term ${25}^{50}$:
${25}^{50}={\left(5^2\right)}^{50}=5^{100}$
The number of factors of 5 is 100.

Now, we sum the exponents of 5 from all these terms:
Total power of 5 = $10+20+30+40+100=200$

Since there are at least 200 factors of 2 available in the product, we can form exactly 200 factors of 10. Thus, there are 200 consecutive zeros at the end of the resulting integer.