A rectangular block of wood floats in water with two-thirds of its volume immersed. When placed in another liquid, it floats with half of its volume immerse...

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When an object is immersed in a liquid, it displaces some volume of the liquid. According to Archimedes' principle, the weight of the liquid displaced is equal to the weight of the object. Since the object floats, the weight of the liquid displaced must be equal to the weight of the object. Let V be the volume of the rectangular block of wood and W its weight. Also, let the density of water be ρ1 and the density of the other liquid be ρ2. When the block is immersed in water, two-thirds of its volume is immersed. Therefore, the volume of water displaced is two-thirds of the volume of the block, which is 2V/3. According to Archimedes' principle, the weight of the water displaced is equal to the weight of the block, so: ρ1 x (2V/3) x g = W where g is the acceleration due to gravity. Rearranging, we get: ρ1 = W / [(2V/3) x g] Similarly, when the block is immersed in the other liquid, half of its volume is immersed. Therefore, the volume of the other liquid displaced is half of the volume of the block, which is V/2. According to Archimedes' principle, the weight of the other liquid displaced is equal to the weight of the block, so: ρ2 x (V/2) x g = W where g is the acceleration due to gravity. Rearranging, we get: ρ2 = W / [(V/2) x g] Dividing the second equation by the first equation gives: ρ2/ρ1 = [(V/2) x g] / [(2V/3) x g] Simplifying, we get: ρ2/ρ1 = 3/4 Therefore, the relative density of the liquid is 0.75. So the correct option is (E) 0.75.