# A test tube is loaded with lead shots so that it floats in a liquid immersed to a mark on the tube. The total weigh of tube and lead shots is 30 gf. the tube is then placed in water and lead shots are added to sink the tube to the same mark. Now, the tube and lead shots weigh 35 gf. Calculate the relative density of liquid.

According to principle of floatation,
Buoyant force = weight of displaced liquid
In the first case
buoyant force = ρgV = mV
here,
ρ = density of the test tube plus lead balls,V= volume
and weight of displaced liquid = mg= ρ'gV'
ρ'  = density of liquid
V' is the volume of liquid displaced
so, using the condition, we get
mV =  ρ'gV'
now, as m =30g = 0.030 kg and g = 9.81 m/s2
by rearranging and solving further, we get
ρ' = 0.00305 x (V / V').........................(1)
and
In the second case,
ρ'gV = m'V =  ρ''gV''
here,
ρ' is the density of test tube
m' = 35g = 0.035 kg
ρ'' is the density of water
V''=volume of water displaced
so,
ρ'' = 0.00357 x (V / V'')............................(2)
thus, the relative density of liquid would be
Density of liquid / Density of water=ρ'/ρ'' =   [0.00305 x (V / V')]/[0.00357 x (V / V'')]
=0.85 x (V / V')
Hope this helps.

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