Solution;
Given;
ρ1=1500kg/m3
ρ2=500kg/m3
ρ12=800kg/m3
g=9.375m/s2
v12=100litres=0.1m3
Let volume of the liquids poured be v1 and v2
So the masses of the individual masses is;
m1=ρ1v1=1500v1
m2=ρ2v2=500v2
The density of the mixture is;
ρ12=vm1+m2
By substitution;
800=0.11500v1+500v2
Rewrite v2 in terms of v1
800=0.11500v1+500(0.1−v1)
80=1500v1+50−500v1
30=1000v1
v1=0.03m3
v2=0.1−0.03=0.07m3
The masses of individual liquids will be;
m1=1500×0.03=45kg
m2=500×0.07=35kg
Total mass of the mixture is;
m12=m1+m2=80kg
The weight of the mixture;
W=m12g=80×9.375=750N
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