Question

(C) 0.02 cm (D) 0.001 cm The distance x determined by the difference between two separate position measurements are x; = 10.5 + 0.1 cm and x2 = 26.8 + 0.1 cm. Then x will be recorded as: (A) 16.3 cm (B) 16.3 +0.1 cm (C) 16.3 +0.2 cm (D) 16.3 +0.01cm haluar​

Answer 1

Answer:

Absolute uncertainties are added. For example, the distance “X” found by the difference between two separate position measurements.

X1 = 10.5 ± 0.1 cm and X2 = 26.8 ± 0.1 cm. The difference “X” between them is recorded as

X = X2-X1 = (26.8 ± 0.1) – (10.5 ± 0.1)

So X = 16.3 ± 0.2 cm.

2- For Multiplication and division:

Percentage uncertainties are added.

For Example, The maximum possible uncertainty in the value of resistance R of a conductor determined from the measurements of potential difference “V” and resulting current flow ‘I’ by using R = V/I is found as follows:

V = 5.2 ± 0.1 V

I = 0.84 ± 0.05 A

The %age uncertainty for V =0.1v/5.2v x 100/100 = about 2%

The %age uncertainty for I = 0.05A/0.84A x 100/100 = about 6%

Hence Total uncertainty in the value of resistance ‘R’ when ‘V’ is divided by ‘I’ is 8% The result is thus given as R = 5.2v/0.84A = 6.19A-1 = 6.19 ohms with a %age. uncertainty of 8% because %age uncertainty for V = 2% and for I = 6% so

Total Uncertainty = º2%+6%=8%

Hence R = 6.2 ± 0.5 Ohms (Here result is rounded off to two significant figures)

3- For Power Factor

If absolute uncertainty of a measurement is known and that measurement occurs in power in a formula. Then total percentage uncertainty is calculated by multiplying the power and absolute uncertainty i.e. multiply the %age uncertainty by that power.

For Example:

For the calculation of the volume of a sphere, we use the formula V= 4/3pr3. Percentage uncertainty in volume = 3 x 5age uncertainty in radius ‘r’. When uncertainty is multiplied by power factor, then it increases the precision demand of measurement. If the radius of a small sphere is measured as 2.25 cm by vernier calipers with least count 0.01 cm, then the radius ‘r’ is recorded as

r = 2.25 ± 0.01 cm

Absolute uncertainty = least count = ± 0.01 cm

%age uncertainty in r = 0.01cm/2.25cm x 100/100 = 0.4%

Total percentage uncertainty in V = 3 x 0.4 = 1.2%

Thus volume V= 4/3pr3= 4/3 x 3.14 x (2.25)3 = 47.689 cm3 with 1.2% uncertainty

Hence the result should be recorded as V = 47.7 0.6 cm3.

Explanation:

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