StudySmarter AI is coming soon!

- :00Days
- :00Hours
- :00Mins
- 00Seconds

A new era for learning is coming soonSign up for free

Suggested languages for you:

Americas

Europe

Q36P

Expert-verifiedFound in: Page 441

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 ${\mathit{L}}$ and average density ${{\mathit{\rho}}}_{{\mathbf{0}}}$, floats partially immersed in the liquid of density ${\mathit{\rho}}$. A length ${\mathit{h}}$ of the rod protrudes above the surface of the liquid. Show that the density of the liquid is given by **

It is proved that the density of the liquid is **$\rho =\frac{{\rho}_{0}L}{L-h}$**.** **

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 $.

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

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

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

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

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

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,

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

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

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

Hence it is proved.

94% of StudySmarter users get better grades.

Sign up for free