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)
the gold piece will weigh more
the silver piece will weigh more
both silver and gold pieces weigh equal
weighing will depend on their masses
Answer is Right!
Answer is Wrong!
This question was previously asked in
UPSC CDS-1 – 2019
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 < V_silver. To determine which one weighs more in air, consider the relation derived from the apparent weight equation: W_air,gold - W_air,silver = (V_gold - V_silver) * density_water * g. Since V_gold < V_silver, (V_gold - V_silver) is negative. Therefore, W_air,gold - W_air,silver is negative, meaning W_air,gold < W_air,silver. The silver piece weighs more in air.