• :00Days
• :00Hours
• :00Mins
• 00Seconds
A new era for learning is coming soon Suggested languages for you:

Europe

Answers without the blur. Sign up and see all textbooks for free! Q36P

Expert-verified Found in: Page 441 ### Physics For Scientists & Engineers

Book edition 9th Edition
Author(s) Raymond A. Serway, John W. Jewett
Pages 1624 pages
ISBN 9781133947271 # A hydrometer is an instrument used to determine liquid density. A simple one is sketched in Figure P14.36. The bulb of a syringe is squeezed and released to let the atmosphere lift a sample of the liquid of interest into a tube containing a calibrated rod of known density. The rod, of length ${\mathbit{L}}$ and average density ${{\mathbit{\rho }}}_{{\mathbf{0}}}$, floats partially immersed in the liquid of density ${\mathbit{\rho }}$. A length ${\mathbit{h}}$ of the rod protrudes above the surface of the liquid. Show that the density of the liquid is given by role="math" localid="1663764227560" ${\mathbit{\rho }}{\mathbf{=}}\frac{{\mathbf{\rho }}_{\mathbf{0}}\mathbf{L}}{\mathbf{L}\mathbf{-}\mathbf{h}}$. It is proved that the density of the liquid is $\rho =\frac{{\rho }_{0}L}{L-h}$.

See the step by step solution

## Given data:

The length of the rod is $L$.

The length of the rod not immersed in the liquid is $h$.

The average density of the rod is ${\rho }_{0}$.

The density of the liquid is $\rho$.

## Buoyancy and gravitational force:

The buoyant force from a liquid of density ${\mathbit{d}}$ when volume of the liquid displaced ${\mathbit{v}}$ is,

${\mathbit{B}}{\mathbf{=}}{\mathbit{g}}{\mathbit{d}}{\mathbit{v}}$ ….. (1)

Here, ${\mathbit{g}}$ is the acceleration due to gravity.

The gravitational force on a body of mass ${\mathbit{m}}$ is,

${\mathbit{F}}{\mathbf{=}}{\mathbit{m}}{\mathbit{g}}$ ….. (2)

## Show that the density of the liquid is ρ=ρ0LL−h:

Length of the rod immersed in the liquid is $L-h$.

Let $A$ be the cross sectional area of the rod. Volume of the rod immersed in the liquid is $A\left(L-h\right)$.

This is also the volume of the liquid displaced. From equation (1), the buoyancy force on the rod directed upward is,

$B=\rho gA\left(L-h\right)$

The total volume of the rod is $AL$ and thus its mass is ${\rho }_{0}AL$. Thus from equation (2), the gravitational force acting on the rod directed downward is,

$F={\rho }_{0}ALg$

At equilibrium, these two forces cancel each other. Thus

$\rho gA\left(L-h\right)={\rho }_{0}ALg\phantom{\rule{0ex}{0ex}}\rho =\frac{{\rho }_{0}L}{L-h}$

Hence it is proved. ### Want to see more solutions like these? 