When immersed in water, the apparent weight of an object is its true weight in air minus the buoyant force. The buoyant force is equal to the weight of the water displaced, which is (volume of the object) (density of water) g. The two pieces have equal apparent weight in water: W_air,gold - B_gold = W_air,silver - B_silver. Since both are solid and immersed, B_gold = V_gold density_water g and B_silver = V_silver density_water g. Given that the density of gold is greater than the density of silver, and they displace the same weight of water to have equal apparent weight difference from their air weight, the object with lower density (silver) must have a larger volume to displace the amount of water needed to satisfy the equality in apparent weight. W_air = density Volume g. W_air,gold - V_gold density_water g = W_air,silver - V_silver density_water g V_gold (density_gold - density_water) = V_silver (density_silver - density_water) Since density_gold > density_silver, it follows that (density_gold - density_water) > (density_silver - density_water). For the equation to hold, V_gold must be less than V_silver. Now comparing weight in air: W_air,gold = density_gold V_gold g and W_air,silver = density_silver V_silver g. We know V_gold
Two identical solid pieces, one of gold and other of silver, when immersed completely in water exhibit equal weights. When weighed in air (given that density of gold is greater than that of silver)
Asked in UPSC CDS 1 2019 · Topic: Mechanics (Fluid Statics and Dynamics (Continuity Equation, Bernoulli's Theorem))
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