question_answer 28)

In Millikan's oil drop experiment, what is the terminal speed of a drop of radius \[=2\cdot 74\times {{10}^{-5}}\] m and density \[7\times {{10}^{6}}Pa.\]? Take the viscosity of air at the temperature of the experiment to be \[\text{L=10 cm}=0\cdot \text{10 m; p}=7\times {{10}^{6}}\text{Pa;}\]. How much is the viscous force on the drop at that speed? Neglect buoyancy of the drop due to air.

Answer:

Here, \[\text{ B=140 GPa}=140\times {{10}^{9}}Pa\]\[B=\frac{pV}{\vartriangle V}=\frac{p{{L}^{3}}}{\vartriangle V}\text{ or }\vartriangle V=\frac{p{{L}^{3}}}{B}\] \[=\frac{\left( 7\times {{10}^{6}} \right)\times {{\left( 0\cdot 10 \right)}^{3}}}{140\times {{10}^{9}}}\] Terminal velocity \[=5\times {{10}^{-8}}{{m}^{3}}\] \[=5\times {{10}^{-2}}m{{m}^{3}}\] \[0\cdot 10%\] Viscous force on the drop, \[=2\cdot 2\times {{10}^{9}}N{{m}^{-2}}.\] \[V=1\] \[{{10}^{-3}}{{m}^{3}};\]