A) 300 \[\mu F\]
B) 200 \[\mu F\]
C) 400 \[\mu F\]
D) 500 \[\mu F\]
Correct Answer: D
Solution :
Given that V = 200 V m = 100g = 0.1 kg s = \[\text{ }\!\!\alpha\!\!\text{ =}\frac{\text{ }\!\!\tau\!\!\text{ }}{\text{I}}\text{=}\frac{\text{qEL }\!\!\theta\!\!\text{ }}{\frac{\text{m}{{\text{L}}^{\text{2}}}}{\text{2}}}\] \[\Rightarrow \] C = ? The energy stored in the capacitor is given by \[{{\text{ }\!\!\omega\!\!\text{ }}^{\text{2}}}\text{ }\!\!\theta\!\!\text{ =}\frac{\text{2qEL }\!\!\theta\!\!\text{ }}{\text{m}{{\text{L}}^{\text{2}}}}\] \[\because \] \[\alpha ={{\omega }^{2}}\theta \] This energy is used to heat up the block. Let \[\Rightarrow \] be the rise in temperature, then heat energy is given by \[{{\omega }^{2}}=\frac{2qE}{mL}\] = 10 J Now, \[T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{mL}{2qE}}\] \[\frac{T}{4}.\] \[\text{t=}\frac{\text{T}}{\text{4}}\text{=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\sqrt{\frac{\text{mL}}{\text{2qE}}}\] \[\text{250Jk}{{\text{g}}^{\text{-1}}}{{\text{K}}^{\text{-1}}}\] \[\Delta \theta =0.4K\]
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