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, \[=2\cdot
2\times {{10}^{9}}N{{m}^{-2}}.\]
\[V=1\]
\[{{10}^{-3}}{{m}^{3}};\]
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