What is the greatest length x such that $3\frac{1}{2}\text{m}$ and $8\frac{3}{4}\text{m}$ are integral multiples of x?

The correct option is $1\frac{3}{4}\text{m}$.

To find the greatest length $x$ such that both given lengths are integral multiples of $x$, we need to determine the Highest Common Factor (HCF) of the two lengths.

First, let's convert the given mixed fractions into improper fractions:

$3\frac{1}{2}=\frac{3\times 2+1}{2}=\frac{7}{2}\text{m}$

$8\frac{3}{4}=\frac{8\times 4+3}{4}=\frac{35}{4}\text{m}$

The formula to find the HCF of fractions is given by:
$\text{HCF of fractions}=\frac{\text{HCF of numerators}}{\text{LCM of denominators}}$

Step 1: Find the HCF of the numerators (7 and 35):
The factors of 7 are 1, 7.
The factors of 35 are 1, 5, 7, 35.
The Highest Common Factor of 7 and 35 is 7.

Step 2: Find the LCM of the denominators (2 and 4):
The multiples of 2 are 2, 4, 6, 8...
The multiples of 4 are 4, 8, 12...
The Least Common Multiple of 2 and 4 is 4.

Step 3: Combine these results to calculate $x$:

$x=\frac{\text{HCF(7, 35)}}{\text{LCM(2, 4)}}=\frac{7}{4}\text{m}$

Converting $\frac{7}{4}$ back to a mixed fraction:

$\frac{7}{4}=1\frac{3}{4}\text{m}$

Thus, the greatest length $x$ is $1\frac{3}{4}\text{m}$.