A) [a,b,c]
B) 3 [a b c]
C) 2 [a b c]
D) 0
Correct Answer: C
Solution :
\[{{\omega }^{2}}=\frac{2qE}{mL}\] \[T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{mL}{2qE}}\] \[\frac{T}{4}.\] \[\text{t=}\frac{\text{T}}{\text{4}}\text{=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\sqrt{\frac{\text{mL}}{\text{2qE}}}\] \[\text{250Jk}{{\text{g}}^{\text{-1}}}{{\text{K}}^{\text{-1}}}\] \[\Delta \theta =0.4K\] \[\text{U=}\frac{\text{1}}{\text{2}}\text{C}{{\text{V}}^{\text{2}}}\] \[=\frac{1}{2}C\times {{\left( 200 \right)}^{2}}\] \[\text{=2C }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{4}}}\text{J}\]
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