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Q.55

Expert-verifiedFound in: Page 386

Book edition
4th

Author(s)
Randall D. Knight

Pages
1240 pages

ISBN
9780133942651

A plastic "boat" with a $25{\mathrm{cm}}^{2}$ square cross section floats in a liquid. One by one, you place $50g$ masses inside the boat and measure how far the boat extends below the surface. Your data are as follows:

Draw an appropriate graph of the data and, from the slope and intercept of the best-fit line, determine the mass of the boat and the density of the liquid.

Therefore, the length of bottle above the water is $5.2cm$

The expression for Buoyancy force acting on a submerged object in a liquid is,

${F}_{\mathrm{B}}={\rho}_{f}{V}_{\mathrm{dis}}g$

Here, ${\rho}_{f}$ is density of the fluid, ${V}_{dis}$ is displaced volume of the liquid due to object, and $g$ is acceleration due to gravity.

Convert the radius of the soda bottle from centimeter to meter.

$r=\frac{6.2\mathrm{cm}}{2}$

$=3.1\mathrm{cm}\left(\frac{1.0\mathrm{m}}{100\mathrm{cm}}\right)$

$=3.1\times {10}^{-2}\mathrm{m}$

Convert the volume of soda from mile-liters to liters.

$V=355\mathrm{mL}\left(\frac{{10}^{-6}{\mathrm{m}}^{3}}{1.0\mathrm{mL}}\right)$

$=355\times {10}^{-6}{\mathrm{m}}^{3}$

Using Archimedes' Principle, the floating force equals to the weight of the liquid displaced. A half full soda can is floating because it displaces at least half can volume of water plus a bit more to support the can's own weight.

The volume of the soda can $V$ is,

$V=\pi {r}^{2}l$

Here, / is the length of the soda can.

If we change the volume to $\text{Sl unit,}355\mathrm{mL}\text{is}0.355\mathrm{L}$. One liter is ${10}^{-3}{\mathrm{m}}^{3}$, the volume of soda can is $0.355\times {10}^{-3}{\mathrm{m}}^{3}$, so we have calculated the length of the can from the above equation:

$l=\frac{0.355\times {10}^{-3}{\mathrm{m}}^{3}}{\pi {\left(3.1\times {10}^{-2}\mathrm{m}\right)}^{2}}$

$=0.118\mathrm{m}$

As the soda can is half full with water, it is immersed half-length / to support the half full of water and plus a bit more $\mathrm{\Delta}l$ as a result of the weight of the can itself.

$\mathrm{\Delta}/\pi {\left(3.1\times {10}^{-2}\right)}^{2}{\rho}_{W}g=20\times {10}^{-3}\mathrm{g}$

Here, ${\mathit{\rho}}_{\mathbf{W}}$ the mass density of water and $g$ is the gravitational acceleration. Solve the equation for $\mathrm{\Delta}l$.

$\mathrm{\Delta}l=\frac{20\times {10}^{-3}}{\pi {\left(3.1\times {10}^{-2}\right)}^{2}1000}\mathrm{m}$

$=6.6\times {10}^{-3}\mathrm{m}$

The length of the can above water is

$\frac{l}{2}-\mathrm{\Delta}l=\frac{0.118\mathrm{m}}{2}-6.6\times {10}^{-3}\mathrm{m}$

$=0.052\mathrm{m}$

$=0.052\mathrm{m}\left(\frac{100\mathrm{cm}}{1\mathrm{m}}\right)$

$=5.2cm$

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