Question

A piece of gold weighs 10 g in air and 9 g in water. What is the volume of cavity?(Density of gold = 19.3 g cm–3)(1)0.182 cc(2)0.282 cc(3)0.382 cc(4)0.482 cc

Answer 1

Hey mate,

● Answer-
(4) 0.482 cc.

● Explaination-
# Given-
ma = 10 g
mw = 9 g

# Solution-
Let's consider,
ρw = density of water
ρg = density of gold
Vg = volume of gold
Vw = volume of substituted water

Now,
Buoyant force = Weight of water displaced
ma.g - ρw(Vg+Vw).g = mw.g

Here, Vg = ma / ρg
ma - ρw(ma/ρg+Vw) = mw
10 - 1(10/19.3+Vw) = 9
Vw = 1 - 10/19.3
Vw = 1 - 0.518
Vw = 0.482 cm^3

Volume of cavity is 0.482 cm^3.

Hope this helps...

Answer 2

Question

A piece of gold weighs 10 g in air and 9 g in water. What is the volume of cavity?(Density of gold = 19.3 g cm–3)

(1)0.182 cc

(2)0.282 cc

(3)0.382 cc

(4)0.482 cc

Answer:

• ρw= density of water
• ρg =density of gold
• Vg= volume of gold
• Vw= volume of substituted water

NOW,

buoyant force=weight of water displaced

⇒ma.g= ρw(Vg+Vw).g= mw.g

HERE

⇒Vg=ma/ρg

⇒ma-ρw(ma/ρg+ Vw)=9

⇒Vw=1-10/19.3

⇒Vw=1-0.518

⇒Vw=0.482 cm^3

SO, option d is correct