| The viscosity of a fluid u, can be determined by measuring the terminal velocity \[{{V}_{T}}\] of a sphere when it descends in the fluid. |
| The fluid has a density \[{{\rho }_{f}}\] while the sphere has a density \[{{\rho }_{s}}\] and a diameter of d. The viscosity can then be calculated by the formula \[\mu =\frac{5({{\rho }_{s}}-{{\rho }_{f}})}{9{{V}_{T}}}{{d}^{2}}\] |
| The values measured are: |
| \[{{V}_{T}}=(1.60\pm 0.04)m{{s}^{-1}}\] |
| \[{{p}_{s}}=(2700\pm 20){{m}^{-3}}\] |
| \[{{p}_{f}}=(900\pm 10)kg\,{{m}^{-3}}\] |
| \[d=(20.0\pm 0.4)mm\] |
| \[=\left( 2700\pm 20 \right)\text{ }kg\text{ }{{m}^{-3}}\] |
A) \[6.2%\]
B) \[7.1%\]
C) \[8.2%\]
D) \[8.4%\]
Correct Answer: C
Solution :
[c] \[\frac{\Delta u}{u}\times 100=\left( \frac{2\Delta d}{d}+\frac{\Delta v}{v}+\frac{\Delta \ell }{\ell } \right)\times 100\] \[\frac{\Delta \ell }{\ell }=\frac{30}{1800}=\left[ 2\times \frac{0.04}{1-6}+\left( \frac{30}{1800} \right) \right]\times 100\]
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