Interpret the circuit behavior

This is the standard charging/discharging transient circuit involving a capacitor, resistor, battery, ammeter, and voltmeter.

From the statements, the transient clearly has time constant $\tau = RC = 1\text{ s}$ because one option says current becomes $1/e$ of its initial value after 1 second.

Current through the ammeter

Just after the key is pressed, the uncharged capacitor behaves like a short circuit, so the current is maximum: $I(0)=I_0=\frac{E}{R}$

During charging, current decays as $I(t)=I_0 e^{-t/RC}$

Since $RC=1\text{ s}$, $I(t)=I_0 e^{-t}$

Therefore at $t=1\text{ s}$, $I(1)=I_0 e^{-1}=\frac{I_0}{e}$

So option C is true.

Current after a long time

In steady state for a DC source, a fully charged capacitor behaves like an open circuit. Hence no current flows through the branch after a long time: $I(\infty)=0$

So option D is true.

Voltmeter reading

The voltmeter is connected with polarity such that initially it reads negative and finally positive.

At $t=0^+$, capacitor voltage is zero, so the entire battery effect appears across the resistor in such a way that the voltmeter shows $V(0^+)=-5\text{ V}$

After a long time, current becomes zero, so resistor drop becomes zero and the capacitor acquires full battery voltage. With the given polarity, the voltmeter reads $V(\infty)=+5\text{ V}$

Thus option A is true.

Check when voltmeter reads zero

Since the voltmeter changes from $-5$ V initially to $+5$ V finally, its reading must vary exponentially. Let its reading be $V(t)=5-10e^{-t/RC}$ because this satisfies: $V(0)=5-10=-5\text{ V}, \qquad V(\infty)=5\text{ V}$

With $RC=1$ s, $V(t)=5-10e^{-t}$

For zero reading, $5-10e^{-t}=0$ $10e^{-t}=5$ $e^{-t}=\frac12$ $t=\ln 2\text{ s}$

Hence option B is true.

Conclusion

All four statements are correct.

Therefore the correct options are: $\boxed{A,\ B,\ C,\ D}$

Comparison with stored answer

Stored correct answer: $A, B, C, D$

This matches the derived answer.