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A 5 $\mu$F capacitor is charged fully by a 220 V supply. It is then disconnected from the supply and is connected in series to another uncharged 2.5 $\mu$F capacitor. If the energy change during the charge redistribution is $X\over100$ J then value of X to the nearest integer is : 

Numerical Value Type

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Electrostatics, electrostatic potential and capacitance

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Ans. 4

NTA answer is 36 but JeeNeetQnA answer is 4

Sol. C1 = 5 $\mu$F    V1 = 220 Volt

C2 = 2.5 $\mu$F    V2 = 0

Heat loss: $\Delta H = U_i – U_f = {1\over2}{C_1C_2\over C_1+C_2}(V_1-V_2)^2$

$={1\over2}\times{5\times2.5\over(5+2.5)}(220-0)^2\mu J$

$={5\over2\times3}×22×22×100×10^{-6}J$

$={5×11×22\over3}×10^{-4}J={55×22\over3}×10^{-4}J$

$={1210\over3}×10^{-4}J={121\over3}×10^{-3}J$

$=4×10^{-2}J$

According to questions

${x\over100}=4×10^{-2}$

So, x = 4

Note : But given answer by JEE Main x = 36

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Ans: 4

Explanation:

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