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Solution

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Correct option is D)

$Voltage=V=100V$

$C_{1}=2μF$ (Initial capacitance)

$C_{2}=10μF$ (Final capacitance)

As $E=21 CV_{2}$

Now

$E_{1}=21 C_{1}V_{2}=21 ×2×10_{−6}×V_{2}$

$E_{2}=21 C_{2}V_{2}=21 ×10×10_{−6}×V_{2}$

$ΔE=E_{2}−E_{1}=21 ×10×10_{−6}×V_{2}−21 ×2×10_{−6}×V_{2}$

$ΔE=21 ×8×10_{−6}×V_{2}$

$ΔE=21 ×8×10_{−6}×100_{2}$

$ΔE=4×10_{−2}J$

Hence the correct option is (D).

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