[amp_mcq option1=”0.68 m” option2=”0.88 m” option3=”1.36 m” option4=”1.76 m” correct=”option3″]
The correct answer is $\boxed{\text{B)} 0.88 \text{ m}}$.
The width of a rectangular sewer is twice its depth while discharging 1.5 m/sec. This means that the area of the sewer is $1.5 \times 1.5 = 2.25 \text{ m}^2$. The width of the sewer is twice its depth, so the width is $2x$ and the depth is $x$. The area of the sewer is $wx$, so $wx = 2.25$. Substituting in $x = w/2$, we get $w^2/2 = 2.25$. Solving for $w$, we get $w = \sqrt{2.25 \times 2} = \sqrt{4.5} = 0.88 \text{ m}$.
Option A is incorrect because $0.68 \text{ m}^2 < 2.25 \text{ m}^2$. Option C is incorrect because $1.36 \text{ m}^2 < 2.25 \text{ m}^2$. Option D is incorrect because $1.76 \text{ m}^2 > 2.25 \text{ m}^2$.