In the circuit shown below, the key is pressed at time t = 0. Which...

(i) At t = 0, capacitor acts as short - circuit

The the reading of voltmeter will be -5 V
At $t \to \infty,$ capacitor acts as open circuit and no current flows through voltmeter.

If it is ideal voltmeter then its resistance is infinite & therefore it will read +5V
(ii)

The Charge on the capacitor is given by
$q = {q}_{0} ( 1 - {e}^{- t / RC} )$
as q₀ = RC is the max charge stored

For C1=2C=40 $\mu$F, R1=R=25 k${q}_{x} = 2CV\left(1 - e\dfrac{- t}{2CR}\right)$

The current through resistance of R=25 k is 

$x = \dfrac{V}{R}e\dfrac{- t}{2CR}$

For C₂=C=20 $\mu$F, R₂=2R=50 k
${q}_{y} = CV\left(1 - e\dfrac{- t}{2CR}\right)$

The current through resistance of 2R=50k is 

$y = \dfrac{V}{2R}e\dfrac{- t}{2CR}$

Reading of voltmeter ($∆$V) = Voltage across 2C - Voltage across 2R

$∆ V = \dfrac{{q}_{x}}{2C} - y ( 2R ) = V\left[1 - 2e\dfrac{- t}{2CR}\right] = 0$

(iii)

$τ = 2RC = 2 \times 25 \times {10}^{3} \times 40 \times {10}^{- 6} = 1sec$
Now Current is given by$i = {i}_{0}{e}^{\left(- \dfrac{t}{τ}\right)}$

at t= 1 sec, Current $i = \dfrac{{i}_{o}}{e}$

(iv) After long time no current flows since capacitor act as a open circuit and voltmeter acts as an ideal it's resistance is infinite, therefore current does not allow.