A body floats in water with one-third of its volume above the surface of water. It is placed in oil it floats with half of: Its volume above the surface of the oil. The specific gravity of the oil is. (A) \((5 / 3)\) (B) \((4 / 3)\) (C) \((3 / 2)\) (D) 1

Let the volume of the body be \(V\). When the body is in water, \(\frac{2}{3}\) of its volume is submerged which means the body has displaced \(\frac{2}{3}V\) of water. Similarly, when the body is in oil it has displaced \(\frac{1}{2}V\) of oil. The weight of the body is the same in both cases(because weight is a property of the body). According to Archimedes' principle, this weight is equal to the weight of the fluid displaced.

We have two situations thus we can define two equations. 1) For the instance when the body is in water: The weight of the body = Weight of water displaced. Weight of water displaced is given by: volume displaced * density of water * g (acceleration due to gravity) So, we have: \(Volume_{body} * Density_{body} * g = \frac{2}{3}V * Density_{water} * g\), where \(V\) is the volume of the body. The g on both sides will cancel out. We call this Equation 1. 2) For the instance when the body is in oil: Again, from Archimedes' principle, the weight of the body = weight of oil displaced So, we get: \(Volume_{body} * Density_{body} * g = \frac{1}{2}V * Density_{oil} * g\). The g on both sides will cancel out. We will call this Equation 2.

From Equation 1, we can infer that, \(Density_{body} = \frac{2}{3} * Density_{water}\) Similarly, from Equation 2, we can infer that, \(Density_{body} = \frac{1}{2} * Density_{oil}\) Using these two equations, we can derive, \(\frac{2}{3} * Density_{water} = \frac{1}{2} * Density_{oil}\) Therefore, \(Density_{oil} = \frac{4}{3} * Density_{water}\) As specific gravity of oil is the ratio of density of oil to the density of water, we get: Specific gravity of oil = \(\frac{Density_{oil}}{Density_{water}} = \frac{4}{3}\) So, the answer to the problem is (B) \(\frac{4}{3}\).