The correct option is (A) : 100%
P1 = \(\sqrt{2}\)mE1 ; P2 = \(\sqrt{2}\)mE2
=\(\sqrt{2}\)m(\(\frac{E_1+300}{100E_1}\)) =√2m(4E1)=2P1
∴ % change = \(\frac{P_2-P_1}{P_1}\times100\) = \(\frac{2P_2-P_1}{P_1}\times100\)=100%
The output (Y) of the given logic implementation is similar to the output of an/a …………. gate.
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