One pipe can fill an empty cistern in 3 hrs less than the another pipe. When both the pipes are opened together, the empty cistern is filled in 2 hrs. The second pipe will fill the empty cistern in :

1. 3 hrs

2. 6 hrs

3. 1 hr

4. 5 hrs

Let time taken by 2nd pipe be x min.

So, in 1 min it will fill $\dfrac{1}{x}$ of cistern.

So, 1st pipe will take 3 hrs i.e. 180 min less = (x - 180) min

So, in 1 min it will fill $\dfrac{1}{x - 180}$ of cistern.

Both pipes will fill the cistern together in 2 hrs i.e. 120 mins.

So, in 1 min they will fill = $\dfrac{1}{120}$ of cistern.

$\therefore \dfrac{1}{x} + \dfrac{1}{x - 180} = \dfrac{1}{120} \\[1em] \Rightarrow \dfrac{x - 180 + x}{x(x - 180)} = \dfrac{1}{120} \\[1em] \Rightarrow \dfrac{2x - 180}{x^2 - 180x} = \dfrac{1}{120} \\[1em] \Rightarrow 120(2x - 180) = x^2 - 180x \\[1em] \Rightarrow 240x - 21600 = x^2 - 180x \\[1em] \Rightarrow x^2 - 180x - 240x + 21600 = 0 \\[1em] \Rightarrow x^2 - 420x + 21600 = 0 \\[1em] \Rightarrow x^2 - 420x + 21600 = 0 \\[1em] \Rightarrow x^2 - 360x - 60x + 21600 = 0 \\[1em] \Rightarrow x(x - 360) - 60(x - 360) = 0 \\[1em] \Rightarrow (x - 60)(x - 360) = 0 \\[1em] \Rightarrow x - 60 = 0 \text{ or } x - 360 = 0 \\[1em] \Rightarrow x = 60 \text{ or } x = 360.$

x cannot be equal to 60 as then x - 180 will be negative, which is not possible as time cannot be negative.

∴ x = 360 mins = 6 hours.

Hence, Option 2 is the correct option.