$1,116.05$ cubic millimeters of material was removed by drilling the blind hole shown in the illustration. Find the depth of the hole.
(The volume of a cylinder is given by $V = \pi r ^ { 2 } h$ )

$$a = 7$$
A) $\quad 31$ millimeters
B) 26 millimeters
C) 22 millimeters
D) $\quad 29$ millimeters
E) 37 millimeters

Question

$1,116.05$ cubic millimeters of material was removed by drilling the blind hole shown in the illustration. Find the depth of the hole.
(The volume of a cylinder is given by $V = \pi r ^ { 2 } h$ )

$$a = 7$$
A) $\quad 31$ millimeters 
B) 26 millimeters 
C) 22 millimeters 
D) $\quad 29$ millimeters 
E) 37 millimeters

Quizplus · Accepted Answer

The volume of the removed material equals the volume of the blind hole, so we have:

$$\pi r^2 h = 1,116.05$$

We know that one end of the blind hole is at a distance of 7 mm from the top of the cylinder, so the height of the blind hole is $h - 7$ mm. Substituting this information into the expression above, we get:

$$\pi r^2 (h-7) = 1,116.05$$

Dividing both sides by $\pi$ and simplifying, we get:

$$r^2 (h-7) = 355. \ldots (1)$$

The bottom of the blind hole is a hemisphere of radius $r$, so its volume is:

$$\frac{2}{3} \pi r^3. \ldots (2)$$

Since the blind hole is formed by drilling a cylindrical hole and then a smaller hemispherical hole, we can write:

$$\frac{2}{3} \pi r^3 = \pi r^2 (h-7) - 1,116.05$$

Substituting Equation (1) into the expression above, we get:

$$\frac{2}{3} \pi r^3 = 355$$

Simplifying, we get:

$$r^3 = (355)(\frac{3}{2 \pi})$$

Taking the cube root of both sides, we get:

$$r \approx 5.2 \ldots (3)$$

Substituting Equation (3) into Equation (1), we get:

$$(5.2^2)(h-7) = 355$$

Simplifying, we get:

$$h-7 = \frac{355}{5.2^2}$$

$$h-7 \approx 29$$

Therefore, the depth of the hole is approximately 29 mm, which is answer choice D.

1,116.051,116.051,116.05 Cubic Millimeters of Material Was Removed by Drilling the Blind

$1,116.05$ Cubic Millimeters of Material Was Removed by Drilling the Blind