Question

Suppose a 1.90 N force can rupture an eardrum having an area of 1.14 cm².(a) Calculate the maximum tolerable gauge pressure inside the eardrum (in the middle ear) in N/m². (Pressures in themiddle ear may rise when an infection causes a fluid buildup. Use 13.6 x 10³ kg/m³ as the density of mercury.) submit answer in N/m²(a) part 2: Convert this value to mm Hg.mm Hg(b) At what depth in fresh water would this person's eardrum rupture, assuming the gauge pressure in the middle ear iszero?Submit Answer in m

Answer 1

Given:

The force is

`F=1.90\text{ N}`

The area of the eardrum is

`\begin{gathered} A=1.14\text{ cm}^2 \\ =1.14*10^(-4)\text{ m}^2 \end{gathered}`

To find:

The maximum tolerable gauge pressure inside the eardrum

a) the pressure in mm of Hg

b) At what depth in freshwater would this person's eardrum rupture

Step-by-step explanation:

The pressure at the eardrum is

`\begin{gathered} P=(F)/(A) \\ =(1.90)/(1.14*10^(-4)) \\ =16.67*10^3\text{ N/m}^2 \end{gathered}`

Hence, the pressure is

`16.67*10^3\text{ N/m}^2`

a)

We know,

`1\text{ N/m}^2=0.0075\text{ mm of Hg}`

So,

`\begin{gathered} 16.67*10^3\text{ N/m}^2=0.0075*16.67*10^3\text{ mm of Hg} \\ =125.02\text{ mm of Hg} \end{gathered}`

Hence, the pressure is 125.02 mm of Hg.

b)

The depth of fresh water is,

`\begin{gathered} h=(P)/(dg) \\ Here,\text{ d=1000 kg/m}^3 \\ g=9.8\text{ m/s}^2 \end{gathered}`

So,

`\begin{gathered} h=(16.67*10^3)/(1000*9.8) \\ =1.70\text{ m} \end{gathered}`

Hence, the depth of water is 1.70 m.