First slide

Fluid statics

Question

Two solid pieces, one of gold and the other of silver when immersed completely in water have equal weighs. When weighed in air

Moderate

A

the gold piece will weigh more
B

the silver piece will weigh more
C

both will have the same weight
D

both will weigh less than they weigh in water

Solution

Suppose,
Weight of gold in air = xg,
Weight of silver in air = yg
Density of gold = $d_{1}$
Density of gold = $d_{2} (where d_{1} > d_{2})$
Volume of gold = $\frac{x}{d_{1}}, volume of silver = \frac{y}{d_{2}}$
Weight of gold in water =Weigh of silver in water
or $(x - \frac{x}{d_{1}} \times 1) g = (y - \frac{y}{d_{2}} \times 1) g$
or $x (1 - \frac{1}{d_{1}}) = y (1 - \frac{1}{d_{2}})$
or $\frac{x}{y} = \frac{(1 - \frac{1}{d 2})}{(1 - \frac{1}{d_{1}})} < 1 a s d_{1} > d 2 x < y or y > x$
i.e., weight of silver in air is more than that of gold in air.

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