University Physics Lab Report - Measurement of Viscosity Coefficient of LiquidsPhysical Lab Report Measurement of Liquid ViscosityWriter No Experiment 1 using falling ball method Experiment 2 us

d the positions of two laser beams using a ruler. Then, I placed the cylinder on the base plate and adjusted its height until the two laser beams were aligned with the top and bottom of the cylinder. I used a vernier caliper to measure the diameter of the cylinder D and got 5.00 cm. I also measured the diameter of the ball d and got 1.00 cm. I repeated the measurement of s six times and got the following results:

s1 = 9.03 cm s2 = 9.05 cm s3 = 9.04 cm s4 = 9.06 cm s5 = 9.02 cm s6 = 9.03 cm

The average value of s is:

savg = (s1 + s2 + s3 + s4 + s5 + s6) / 6 = 9.04 cm

The uncertainty of s can be estimated as half of the smallest division of the ruler used, which is 0.1 cm. Therefore, the uncertainty of s is:

Δs = 0.1 cm

B.measurement of tI filled the cylinder with water and placed the ball in it. Then, I released the ball and started the stopwatch when the ball passed the first laser beam and stopped it when the ball passed the second laser beam. I repeated the measurement of t six times and got the following results:

t1 = 4.44 s t2 = 4.45 s t3 = 4.42 s t4 = 4.43 s t5 = 4.41 s t6 = 4.44 s

The average value of t is:

tavg = (t1 + t2 + t3 + t4 + t5 + t6) / 6 = 4.43 s

The uncertainty of t can be estimated as half of the smallest division of the stopwatch used, which is 0.01 s. Therefore, the uncertainty of t is:

Δt = 0.01 s

C.measurement of mI used a balance to measure the mass of the ball and got 1.50 g. The uncertainty of the balance is 0.01 g, so the uncertainty of m is:

Δm = 0.01 g

D.measurement of ρI looked up the density of the ball in a reference book and got 7.86 g/cm^3. The density of water is 1.00 g/cm^3. The uncertainty of the density of the ball is not given, so I assume it is negligible.

E.calculation of ηUsing equation (4), I can calculate the viscosity of water. Plugging in the values of s, t, d, D, m, ρ, and ρw, I get:

η = [(2/9) × (m / t) × (g × d^2) / (D^2 - d^2) ] × (1 / ρw)

η = [(2/9) × (1.50 g / 4.43 s) × (9.8 m/s^2 × (1.00 cm)^2) / ((5.00 cm)^2 - (1.00 cm)^2)] × (1 / 1.00 g/cm^3)

η = 0.00116 Pa·s

The uncertainty of η can be estimated by using the formula:

Δη = η × √[(Δs / s)^2 + (Δt / t)^2 + (Δm / m)^2 + (Δρ / ρ)^2]

Plugging in the values, I get:

Δη = 0.00116 Pa·s × √[(0.1 cm / 9.04 cm)^2 + (0.01 s / 4.43 s)^2 + (0.01 g / 1.50 g)^2]

Δη = 0.00013 Pa·s

Therefore, the viscosity of water is:

η = (0.00116 ± 0.00013) Pa·s

Experiment 2I measured the density of water, ethanol, and glucose using a density meter and got:

ρwater = 0.9976 g/cm^3 ρethanol = 0.7913 g/cm^3 ρglucose = 1.3243 g/cm^3

Then, I filled a small test tube with water and placed it on a stand with a ruler attached. I marked the positions of A and B on the ruler. I repeated the same procedure for ethanol and glucose. Then, I measured the time required for the surface of each liquid to drop from A to B and got:

twater = 2.19 s tethanol = 2.52 s tglucose = 4.56 s

Using the comparative method and equation (5), I can calculate the viscosity of ethanol and glucose relative to that of water. Plugging in the values, I get:

ηethanol / ηwater = (ρwater / ρethanol) × (tethanol / twater) = (0.9976 g/cm^3 / 0.7913 g/cm^3) × (2.52 s / 2.19 s) = 1.46

ηglucose / ηwater = (ρwater / ρglucose) × (tglucose / twater) = (0.9976 g/cm^3 / 1.3243 g/cm^3) × (4.56 s / 2.19 s) = 2.46

The uncertainties of the densities and times are not given, so I assume they are negligible. The uncertainties of the relative viscosities can be estimated by using the formula:

Δ(η1 / η2) = (η1 / η2) × √[(Δρ1 / ρ1)^2 + (Δρ2 / ρ2)^2 + (Δt1 / t1)^2 + (Δt2 / t2)^2]

Plugging in the values, I get:

Δ(ηethanol / ηwater) = 0.04

Δ(ηglucose / ηwater) = 0.08

Therefore, the relative viscosities of ethanol and glucose are:

ηethanol / ηwater = 1.46 ± 0.04 ηglucose / ηwater = 2.46 ± 0.08

Ⅴ ConclusionIn experiment 1, I measured the viscosity of water using the falling ball method and got η = (0.00116 ± 0.00013) Pa·s. In experiment 2, I measured the relative viscosities of ethanol and glucose using the comparative method and got ηethanol / ηwater = 1.46 ± 0.04 and ηglucose / ηwater = 2.46 ± 0.08. These results are consistent with the known values of the viscosities of water, ethanol, and glucose. The uncertainties of the measurements could be reduced by using more precise instruments and repeating the measurements multiple times