Question

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 1

Answer: 500 kg/m³

Step-by-step 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

Answer 2

500 kg/m³

Step-by-step 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