Question

a piece of gold weighs 10g in air and 9g in water what is volume of cavity​

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

Here is your answer ⤵⤵⤵

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 IT HELPS YOU ☺ !!