A hole of radius 1.0 mm forms in the bottom of a very large water storage tank that holds water to a depth of 15 m. At what volume rate will water flow out of the hole?
A) $5.4 \times 10 ^ { - 5 } \mathrm {~m} ^ { 3 } / \mathrm { s }$
B) $5.4 \times 10 ^ { - 6 } \mathrm {~m} ^ { 3 } / \mathrm { s }$
C) $5.4 \times 10 ^ { - 4 } \mathrm {~m} ^ { 3 } / \mathrm { s }$
D) $5.4 \times 10 ^ { - 7 } \mathrm {~m} ^ { 3 } / \mathrm { s }$

Question

A hole of radius  1.0 mm forms in the bottom of a very large water storage tank that holds water to a depth of  15 m. At what volume rate will water flow out of the hole?
A) $5.4 \times 10 ^ { - 5 } \mathrm {~m} ^ { 3 } / \mathrm { s }$
B) $5.4 \times 10 ^ { - 6 } \mathrm {~m} ^ { 3 } / \mathrm { s }$
C) $5.4 \times 10 ^ { - 4 } \mathrm {~m} ^ { 3 } / \mathrm { s }$
D) $5.4 \times 10 ^ { - 7 } \mathrm {~m} ^ { 3 } / \mathrm { s }$

Quizplus · Accepted Answer

The volume rate of water flowing out of the hole can be calculated using Torricelli's Law, which states that the speed of efflux, $v$, is given by $v = \sqrt{2gh}$, where $g$ is the acceleration due to gravity (approximately $9.8 \, \text{m/s}^2$) and $h$ is the depth of the water (15 m in this case). The volume flow rate, $Q$, is then $Q = Av$, where $A$ is the cross-sectional area of the hole. Given the radius of the hole is 1.0 mm (or 0.001 m), the area $A = \pi r^2 = 3.14 \times (0.001)^2 \, \text{m}^2$. Plugging in the values and solving gives a volume flow rate of approximately $5.4 \times 10^{-5} \, \text{m}^3/\text{s}$.

A Hole of Radius 1 5.4×10−5 m3/s5.4 \times 10 ^ { - 5 } \mathrm {~m} ^ { 3 } / \mathrm { s }5.4×10−5 m3/s

A Hole of Radius 1 $5.4 \times 10 ^ { - 5 } \mathrm {~m} ^ { 3 } / \mathrm { s }$