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3 years ago

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Two cylinders with the same mass density rhoC = 713 kg / m3 are floating in a container of water (with mass density rhoW = 1025

kg / m3). Cylinder #1 has a length of L1 = 20 cm and radius r1 = 5 cm. Cylinder #2 has a length of L2 = 10 cm and radius r2 = 10 cm. If h1 and h2 are the heights that these cylinders stick out above the water, what is the ratio of the height of Cylinder #2 above the water to the height of Cylinder #1 above the water (h2 / h1)? h2 / h1 =

1 answer:

Tanzania [10]3 years ago

5 0

Answer:

Explanation:

Given

density of cylinder is $\rho _c=713 kg/m^3$

Length of first cylinder is $L_1=20 cm$

radius $r_1=5 cm$

For cylinder 2 $L_2=10 cm$

$r_2=10 cm$

$h_1$ and $h_2$ are the height above water

E

as object is floating so its weight must be balanced with buoyant force

$\rho _c\frac{\pi }{4}d_1^2L_1g=\rho _w\frac{\pi }{4}d_1^2(L_1-h_1)g----1$

For 2nd cylinder

$\rho _c\frac{\pi }{4}d_2^2L_2g=\rho _w\frac{\pi }{4}d_2^2(L_2-h_2)g----2$

Dividing 1 and 2 we get

$\frac{L_1}{L_2}=\frac{L_1-h_1}{L_2-h_2}$

$\frac{20}{10}=\frac{20-h_1}{10-h_2}$

$2h_2=h_1$

$\\\Rightarrow\frac{h_2}{h_1}=\frac{1}{2}$

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$V_{k}$ - Volume of kerosene, measured in cubic meters.

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$T_{k,o}$ , $T_{t,o}$ - Initial temperatures of kerosene and tin, measured in degrees Celsius.

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