Solve this fluid dynamics question

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Hi
For A) 1.57* 10^-3N
B) It will float. Vs = 0.88V
C) 6.17 * 10^2 kg/m3

Weight of wax = mg or rho Vg
if u use w = mg then,

m= w/g
m = 0.1572/9.8
m= 0.016kg

rho = m/V
rho = 0.016/26 * 10^-6
rho = 6.17 * 10^2 kg/m3

if u use w = rho V g then,

rho = w/ Vg
rho = 0.1572/26 * 10^-6 * 9.8
rho = 6.17 * 10^2

w gasoline = rho gasoline V g
rho gasoline = w gasoline/Vg
= 6.87 * 10^-3/ 1 * 10^-6* 9.8
= 7.01 * 10^2 kg/m3

weight of wax = weight of liquid displaced
rho V g = rho gasoline Vs(submerged volume) g

Cancel g on both sides,

rho V = rho gas Vs
rho/ rho gas = Vs/ V
Vs/V = 6.17 * 10^2/7.01 * 10^2
Vs/V = 0.88
Vs = 0.88 V
Vs = 0.88 (26 * 10^-6)
Vs = 2.288 * 10^-5 m3

weight of liquid displaced
w = rho gasoline Vs g
w = 7.01 * 10^2 * 2.288 * 10^-5 * 9.8
w = 1.57 * 10^-3N

since rho gas > rho wax, the wax will float.