An object hangs from a spring balance. The balance registers 32 N in air, 22 N when this object is immersed in water, and 27 N when the object is immersed in another liquid of unknown density. What is the density of that other liquid

Answer: 500 kg/m³

Explanation:

Given

Weight of object in air, W = 32 N

Weight of object in water, W (w) = 22 N

Weight of object in unknown liquid, W (u) = 27 N

Using Archimedes principle,

W (w) = W - M (w),

where, M (w) is the mass of water displaced by the object. To get the mass of object now, we do

M (w) = W - W (w)

M (w) = 32 - 22

M (w) = 10 N

Weight of object in unknown liquid is

M (u) = 32 - 27

M (u) = 5 N

if we find the ratio, we have

M (u) / M (w) = 5 / 10 = 0.5

Now, we multiply this by the density of water

We have, 0.5 * 1000 = 500 kg/m³

500 kg/m³

Explanation:

From Archimedes principle,

R. d of the object = Weight of the object in air/upthrust of object in water = Density of the solid/Density of water

W/U = D/D' ... Equation 1

Where W = weight object in air, U = Upthrust of object in water, D = Density of object, D' = Density of water

Make D the subject of the equation

D = (W/U) D' ... Equation 2

Given: W = 32 N, U = 32-22 = 10 N, D' = 1000 kg/m³

Substitute into equation 2

D = (32/10) 1000

D = 3200 kg/m³

Also,

R. d of the object = weight of the object in air/upthrust in liquid = Density of the object / density of the liquid

W/U' = D/D''

Where U' = upthrust of the object in liquid, D'' = Density of the liquid

make D'' the subject of the equation

D'' = (U'/W) D ... Equation 3

Given: D = 3200 kg/m³, U' = 32-27 = 5 N, W = 32 N

substitute into equation 3

D'' = (5/32) 3200

D'' = 500 kg/m³