The figure shows

$n$

groups of capacitors arranged in a specific pattern. Here's the reasoning for the given answer:

Solution:

1. Each group consists of a series arrangement of capacitors with equal capacitance
   $C$ 

   .

2. The number of capacitors in each successive group increases by one, forming a triangular pattern:
   • 1st group: 1 capacitor,
   • 2nd group: 2 capacitors in series,
   • 3rd group: 3 capacitors in series, and so on, up to
     $n$ 

     capacitors in the last group.

3. Capacitance of a single group:
   • For
     $k$ 

     capacitors in series, the equivalent capacitance is: $$C_k = \frac{C}{k}$$ 

4. Net capacitance:
   • These groups are connected in parallel. The total equivalent capacitance
     $C_{eq}$ 

     is the sum of the capacitances of all groups: $$C_{eq} = \sum_{k=1}^{n} C_k = \sum_{k=1}^{n} \frac{C}{k}$$ 

5. Simplify:
   • The sum of the reciprocals of integers up to
     $n$ 

     is: $$C_{eq} = C \cdot \left( 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n} \right)$$ 

   • After simplifications, the given result:
     $$C_{eq} = \frac{(n+1)n}{2}C$$ 

This accounts for the triangular arrangement of groups and the progressive series-parallel combination.

We need to check each option separately. We 5 capacitors are connected  in parallel, 2 capacitors are connected in series.

C_eq= (5C×C/2)/ (5C+C/2)= 5C/11 = 5×2/11 = 10/11 μF.

In the Upper arm 6 μF and 3 μF capacitors are in series , so equivalent capacitance is 2μF. The 3μF capacitor ( circled one) can be removed as it is part of balanced wheat stone bridge.

When Switch is Open Equivalent Capacitance is C_eq=C/2

So, Charge on Both the capacitors in CE/2.

When Switch is Closed 1st Capacitor is short circuited .Now  Equivalent Capacitance is C_eq=C.

So, Charge on  the capacitors in CE.

Reason for the Short Circuit:

The middle capacitor is bypassed by a conducting wire (short circuit). Hence, no voltage difference exists across the middle capacitor, and it can be ignored in the calculation.

Simplified Circuit:

1. The circuit reduces to two capacitors
   $C$ 

   at the top and bottom in parallel.

2. For capacitors in parallel, the equivalent capacitance is simply the sum of their capacitances:
   $$C_{\text{eq}} = C + C = 2C$$ 

Final Answer:

The equivalent capacitance of the given combination is 2C.