Question

When a hydrometer (see Fig. 2) having a stem diameter of 0.30 in. is placed in water, the stem protrudes 3.15 in. above the water surface. If the water is replaced with a liquid having a specific gravity of 1.10, how much of the stem would protrude above the liquid surface

Answer 1

Answer:

5.79 in

Explanation:

We are given that

Diameter,d=0.30 in

Radius,r=$\frac{d}{2}=\frac{0.30}{2}=0.15 in$

Weight of hydrometer,W=0.042 lb

Specific gravity(SG)=1.10

Height of stem from the water surface=3.15 in

Density of water=$62.4lb/ft^3$

In water

Volume  of water displaced $V=\frac{mass}{density}=\frac{0.042}{62.4}=6.73\times 10^{-4} ft^3$

Volume of another liquid displaced=$V'=\frac{V}{SG}=\frac{6.73\times 10^{-4}}{1.19}=5.66\times 10^{-4}ft^3$

Change in volume=V-V'

$V-V'=\pi r^2 l$

Substitute the values

$6.73\times 10^{-4}-5.66\times 10^{-4}=3.14\times (\frac{0.15}{12})^2l$

By using

1 ft=12 in

$\pi=3.14$

$l=\frac{6.73\times 10^{-4}-5.66\times 10^{-4}}{3.14\times (\frac{0.15}{12})^2}$

l=2.64 in

Total height=h+l=3.15+2.64= 5.79 in

Hence, the height of the stem protrude above the liquid surface=5.79 in