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Question: The following observations were take for determining surface tension of water by capillary tube meth...

The following observations were take for determining surface tension of water by capillary tube method. Diameter of capillary, D = 1.25 x 10−210^{-2}m and rise of water in capillary. h = 1.46 × 10−210^{-2}m

takingg = 9.80ms−2s^{-2} and using the relation T = (rgh2)(\frac{rgh}{2}) x 103Nm−1^{-1}, what is the possible error in surface tension T?

A

2.4%

B

15%

C

1.6%

D

0.15%

Answer

1.6%

Explanation

Solution

The surface tension formula is T∝DghT \propto Dgh. The percentage error in T is the sum of the percentage errors in D, h, and g.

Percentage error in D=ΔDD×100D = \frac{\Delta D}{D} \times 100. Given D=1.25×10−2D=1.25 \times 10^{-2} m, precision is 0.01×10−20.01 \times 10^{-2} m. Taking ΔD=0.01×10−2\Delta D = 0.01 \times 10^{-2} m, percentage error in D=10−41.25×10−2×100=0.8%D = \frac{10^{-4}}{1.25 \times 10^{-2}} \times 100 = 0.8\%.

Percentage error in h=Δhh×100h = \frac{\Delta h}{h} \times 100. Given h=1.46×10−2h=1.46 \times 10^{-2} m, precision is 0.01×10−20.01 \times 10^{-2} m. Taking Δh=0.01×10−2\Delta h = 0.01 \times 10^{-2} m, percentage error in h=10−41.46×10−2×100≈0.685%h = \frac{10^{-4}}{1.46 \times 10^{-2}} \times 100 \approx 0.685\%.

Percentage error in g=Δgg×100g = \frac{\Delta g}{g} \times 100. Given g=9.80g=9.80 m/s², precision is 0.01. Taking Δg=0.01\Delta g = 0.01, percentage error in g=0.019.80×100≈0.102%g = \frac{0.01}{9.80} \times 100 \approx 0.102\%.

Total percentage error in T=0.8%+0.685%+0.102%≈1.587%T = 0.8\% + 0.685\% + 0.102\% \approx 1.587\%.

Rounding to one decimal place gives 1.6%.