Question

While an astronaut on planet X, you find a sample of an unknown liquid. Without your fancy equipment, you decided to measure the mass of your spoon (25 grams, using local gravity) in and out of the liquid. In the liquid, the scale reads 17.3 grams and it reads 18.9 grams in water. What is the density and the identity of the unknown liquid?

Answer 1

By force balance on spoon in water,

W - Fb_water = W_water

here, W = Original weight of spoon in sir = m*g

Fb_water = _water*V*g

W_water = (m_water)*g

given, m = mass of spoon = 25 gm = 0.025 kg

m_water = mass of spoon in water = 18.9 gm = 0.0189 kg

V = volume of spoon = ??

_water = density of water = 1000 kg/m^3

So,

0.025*g - 1000*V*g = 0.0189*g

V = (0.025 - 0.0189)/1000

V = 6.1*10^-6 m^3

Now, by force balance in unknown liquid:

W - Fb_liquid = W_liquid

here,

Fb_liquid = _liquid*V*g

W_liquid = (m_liquid)*g

given,

m_liquidr = mass of spoon in liquid = 17.3 gm = 0.0173 kg

V = volume of spoon = 6.1*10^-6 m^3

_liquid = density of liquid = ?? kg/m^3

So,

0.025*g - _liquid*6.1*10^-6*g = 0.0173*g

_liquid = (0.025 - 0.0173)/(6.1*10^-6) = 1262.29

_liquid = 1262 kg/m^3 = density of the unknown liquid

Unknown liquid is Glycerin.

"Let me know if you have any query."