A) \[=\frac{force}{mass}=\frac{eE}{m}\]
B) \[{{v}^{2}}={{u}^{2}}+2ax\]
C) \[u=0\]
D) Nope of these
Correct Answer: B
Solution :
Key Idea: Soap film has two surfaces, hence total increase in surface area is twice of increase. Surface tension \[(\tau )\] of a liquid is equal to the work (W) required to increases the surface area \[F=ma\] of the liquid film by unity at constant temperature. \[\Rightarrow \] \[a=\frac{F}{m}=\frac{0.2}{5}=0.04m/s\] Initial area of film \[v=u+at\] Final area of film \[v=2+\text{ }0.04\times 10=2.4m/s\] Increase in surface area is \[\Delta K=\frac{1}{2}m{{v}^{2}}-\frac{1}{2}m{{u}^{2}}\] Since soap film has two surfaces hence total increase in surface area. \[=\frac{1}{2}\times 5\times {{(2.4)}^{2}}-\frac{1}{2}\times 5\times {{(2)}^{2}}\] \[=\frac{1}{2}\times 5\times 1.76=4.4J\] \[\therefore \] \[{{v}^{2}}={{u}^{2}}+2gh\]
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