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Asked in JEE Main 2010 · Equilibrium and small oscillations
In air, tan θ=F/(mg) with F=(kq²)/(x²).
In the liquid two things change: the force becomes F/K, and the weight becomes the apparent weight mg(1-σ/ρ) because of buoyancy.
The angle is unchanged, so F/(mg)=(F/K)/(mg(1-σ/ρ)), which gives K=ρ/(ρ-σ).
K=(1.6)/(1.6-0.8)=(1.6)/(0.8)=2.
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