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.

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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