Measurement Of Mass
Units And Dimensions of Class 11
Measurement of Inertial Mass
Inertial mass of a body is measured using a device which is known as inertial balance. It consists of a long metal strip. One end of the strip is clamped to a table such that its flat face is vertical, and it can easily vibrate horizontally. The other end of strip supports a pan in which the object whose inertial mass is to be found can be kept. It is found that the square of time period of vibration is directly proportional to total mass of the pan and the body placed in it.
Measurement of Time: The following methods are used
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Quartz Crystal Clock
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Atomic Clock
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Radioactive dating
Significant figures:
Each measurement involves errors. The measure results has a number that includes all reliably known digits and first unknown digit. The combination of reliable digits and first uncertain digit are significant figures.
Example: If a length is measured as 2.43 cm then 2 and 4 are reliable while 3 is uncertain. Thus the measured value has three significant figures.
Common rules for counting significant figures
(1) All non zero digits are significant.
For example: 1745 has four significant digits.
(2) All zeros present between 2 non zero digits are significant, irrespective of the position of the decimal point.
Example: 208005 has 6 significant figures.
(3) If there is no decimal point, all zeros to the right of the right−most non zero digit are considered to be significant only if they come from a measurement.
Example: 41000 has only 2 significant digits while 41000 m has 5 significant digits.
(4) All zeros to the right of a decimal point but to the left of non−zero digits are considered to be non significant, provided there should be no non zero digit to the left of the decimal point.
Example: 0.00305 has 3 significant figures.
(5) All zeros are significant if they are placed to the right of a decimal point and to the right of a non zero digit.
Example: 0.04080 has 4 significant figures
50.000 has 5 significant figures
(6)The number of significant figures does not alter in different units.
If we want to write 450 m in different units, we can write it 4.5 x 104 cm or 4.5 x 105 mm etc. in which all of them are having 3 significant figures.
Illustration 9. State the number of significant figures in the following –
(a) 6.500310
(b) 754400
(c) 15000 kg
(d) 8.314 x 10+2 J
(e) 1.6 x 10-19 C
(f) 0.0065050
Solution: (a) 7 (b) 4 (c) 5 (d) 4 (e) 2 (f) 5
Exercise 3: The number of significant figures in 0.0160 is
(a) 2 (b) 3 (c) 4 (d) 5
Rounding off
(1) If all the digits to be discarded are such that the first discarded digit is less than 5, the remaining digits are left unchanged.
Example: 7.499498 can be written in 4 significant figures as 7.499
(2) If the digit to be discarded is 5 followed by digits other than zero, then the preceding digit should be raised by 1.
Example: 7.45001, on being rounded off to first decimal, became 7.5
(3) If the digits to be discarded is 5 or 5 followed by zero the preceding digit remains unchanged if it is even and the preceding digit is raised by 1 if it is odd.
Example: 3.6500 will become 3.6 and 4.7500 will become 4.8 in 2 significant figures.
Arithmetic operations with significant figures:
(1) Addition and subtraction In addition and subtraction, the number of decimal places in the result is the smallest number of decimal places of terms in the operation.
Let us consider the sum of following measurements.
3.45 kg., 7.6 kg. and 10.055 kg.
3.45
7.6
10.055
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
21.105
−−−−−−−−−−−−−−−−−−−−−−−−−−−−
So the sum will be 21.1 kg as 7.6 kg has only 1 digit after the decimal point while the others are having more than one digit.
Multiplication and Division:
In the result of multiplication or division, the number of significant figures is same as the smallest number of significant figures among the numbers.
Illustration 9: Multiply 1.21 and 1.1.
Solution: 1.21 x 1.1= 1.331
So the result is 1.3 as there are only 2 significant digits in 1.1
The same procedure is followed for division.
Illustration10. Multiply 107.88 by 0.610 and express the result upto the correct number of significant figure.
(A) 65.8068 (B 65.807
(C) 65.81 (D) 65.8
Solution: Number of significant figures in multiplication is three corresponding to the minimum number 107.88 by 0.610
= 65.8068 = 65.8
∴ (D)
Exercise 4: Value of 1.2 + 1.34 + 2.342 is
(a) 4.88 (b) 4.8 (c) 4.90 (d) 5
Accuracy and Precision of measuring instrument: It is impossible to measure any physical quantity perfectly. It is due to imperfection in manufacturing and working of measuring instruments.
Accuracy: It is the degree of correctness of the measured quantity, i.e. how much close the result is to the true value of the physical quantity.
Precision: It is the degree of repeatability & refinement of a measurement.
