Correct Answer - C
At equilibrium
`tan thet a//2 = F_a /(mg) = 1/([ 4pi varepsilon_0) q^2/(l sin 9 theta //2)]^2 1 /(mg)`

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When supened in liwuid
`tan (theta)/2 = 1/(4 pi in_0 K) q^2/([ l sin 9 theta //2)]^2 1/((mg - F_B))`
`= 1/(4 pi in_0 k) q^2 /((K l sin (theta //2)]^2)`.
`1/(( mg - (m)/( 1.6) xx 0. 8 g))`
on comparing the two equation we get
` K (1 - 0.8)/( 1.6) =1 rArr K = 2` .