The output in the circuit of figure is taken across a capacitor. It is as shown in figure
A
B
C
D
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C
As $R \mathrm{C}$ time constant of the capacitor is quite large $\left(\tau=R C=10 \times 10^3 \times 10 \times 10^{-6}=0.1 \mathrm{sec}\right)$, if will not discharge appreciably.
Hence voltage remains nearly constant.
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