JEE Main 9 Jan 2019 Shift 2 Solved Paper

C12 =

C1C2

C1+C2

=

k1∊0 L 2 ×L d 2 k2[∊0 L 2 ×L] d 2

(k1+k2)[ ∊0 L 2 ×L d 2 ]

C12 =

k1K2

k1+k2

∊0L2

d

in the same way we get, C34 =

k3k4

k3+k4

∊0L2

d

... (i)
Now if keq = k , Ceq =

k∊0L2

d

... (ii)
on comparing equation (i) to equation (ii), weget
keq =

k1k2(k3+k4)+k3k4(k1+k2)

(k1+k2)(k3+k4)

This does not match with any of the optionsso probably they have assumed the wrongcombination
C13 =

k1∊0L L 2

d 2

+ k3∊0

L. L 2

d 2

= (k1+k3)

∊0L2

d

C24 = (k2+k4)

∊0L2

d

Ceq =

C13C24

C13C24

=

(k1+k3)(k2+k4)

(k1+k2+k3+k4)

∊0L2

d

=

k∊0L2

d

k = K =

(K1+K3)(K2+K4)

(K1+K2+K3+K4)

However this is one of the four options.It must be a "Bonus" logically but of the givenoptions probably they might go with (4)