Please note that the question appeared in NEET 2016 Phase II exam. As per the options of the question, no option will match if we follow the correct approach and concepts. Therefore, we have given both correct as well as incorrect approaches below. The marked option is as per the incorrect approach.

Correct Approach:

Hint: The potential difference across

K_{1}, K_{2} and K_{3}

will be different.

Step 1: Find the capacitance of each part.

$C_{1} = \frac{2 K_{1} ε_{0} A}{3 d}, C_{2} = \frac{2 K_{2} ε_{0} A}{3 d} C_{3} = \frac{2 K_{3} ε_{0} A}{3 d}, C_{4} = \frac{2 K_{4} ε_{0} A}{d}$

Step 2: Find the equivalent capacitance of the combination.

Break $C_{4}$ in three equal parts and each part will be in series with $C_{1}, C_{2} & C_{3}$ respectively.

$C' = \frac{2 ε_{o} A}{3 d} (\frac{K_{1} K_{4}}{K_{1} + K_{4}}), C'' = \frac{2 ε_{o} A}{3 d} (\frac{K_{2} K_{4}}{K_{2} + K_{4}}) & C''' = \frac{2 ε_{o} A}{3 d} (\frac{K_{3} K_{4}}{K_{3} + K_{4}})$
$Now C', C'' and C''' are in parallel .$
$Step 3 : Find the equivalent dielectric constant .$
$C_{net} = \frac{2 ε_{o} A}{3 d} (\frac{K_{1} K_{4}}{K_{1} + K_{4}} + \frac{K_{2} K_{4}}{K_{2} + K_{4}} + \frac{K_{3} K_{4}}{K_{3} + K_{4}})$
$\frac{{Kε}_{o} A}{d} = \frac{2 ε_{o} A}{3 d} (\frac{K_{1} K_{4}}{K_{1} + K_{4}} + \frac{K_{2} K_{4}}{K_{2} + K_{4}} + \frac{K_{3} K_{4}}{K_{3} + K_{4}})$
$K = \frac{2}{3} K_{4} (\frac{K_{1}}{K_{1} + K_{4}} + \frac{K_{2}}{K_{2} + K_{4}} + \frac{K_{3}}{K_{3} + K_{4}})$

Incorrect approach:

The potential difference across K1, K2 and K3K1, K2 and K3 is assumed same.

1.	They are eukaryotic
2.	All fungi possess a purely cellulosic cell wall
3.	They are heterotrophic
4.	They are both unicellular as well as multicellular

NEET 2016 Phase 2 Paper