In the previous section, we saw the 'rules for arithmetic operations with significant figures'. We saw some solved examples also. In this section, we will see a few more solved examples.
Solved example 2.15
Fill in the blanks
(Note: In stating numerical answers, take care of significant figures)
(a) The volume of a cube of side 1 cm is equal to .....m3
(b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to ...(mm)2
(c) A vehicle moving with a speed of 18 km h-1 covers ....m in 1 s
(d) The relative density of lead is 11.3. Its density is ....g cm-3 or ....kg m-3
Solution:
Part (a):
1. We have: Volume = s × s × s
• Where 's' is the length of side
2. The side is given in cm. But we want the volume in m3
• 1 cm = 0.01 m
♦ 1 cm have 1 significant figure
♦ 0.01 m also have 1 significant figure
3. So volume in m3 = (0.01)3 = 10-6 m3
Part (b):
1. We have: Total surface area of a solid cylinder = 2𝞹r2 + 2𝞹rh = 2𝞹r(r + h)
See details here
2. First we evaluate the quantity within the brackets
• r = 2.0 cm
• h = 10.0 cm
(r + h) = 2.0 + 10.0 = 12.0 cm
(Since both values have the same number of digits after the decimal point, we do not need to draw the magenta line that we saw in fig.2.18 above)
3. So total surface area = 2𝞹r × 12.0 = (2 × 𝞹 × 2.0 × 12.0) cm2
• 2 is a factor
It has infinite number of significant figures
• 2.0 has 2 significant figures
• 10.0 has 3 significant figures
• So the final result must have only 2 significant figures
• We will take the value of 𝞹 in such a way that, it has 3 (one more than what is required in the final result) significant figures
• So 𝞹 = 3.14
4. Thus total surface area = (2 × 3.14 × 2.0 × 12.0) cm2 = 150.72 cm2
5. one cm = 10 mm
• So 1 cm2 = 100 mm2
• Here 100 is a factor. It has infinite number of significant figures
6. We can now convert the 'area in cm2' to 'area in m2'
• Total surface area in m2 = 150.72 × 100
• In scientific notation, this value is 1.5072 × 104 m2
7. But in (2) we saw that, the final result must have only 2 significant digits. This is valid even if units change
• So total surface area = 1.5 × 104 m2.
Part (c):
1. Distance traveled in 1 h = 18 km
• '18' has 2 significant figures
• 1 h = (60 × 60) = 3600 s
• So distance traveled in 1 s = (18⁄3600) km
2. 3600 is a factor. It has infinite number of significant figures
• So the result of (18⁄3600) must have only 2 significant figures
3. One km = 1000 m
• So (18⁄3600) km = [(18⁄3600) × 1000] m
4. 1000 is a factor. It has infinite number of significant figures
• So the result of [(18⁄3600) × 1000] must have only 2 significant figures
5. We have: [(18⁄3600) × 1000] = 5
• If '5' is to have two significant figures, we must write it as 5.0
• So we get:
A vehicle moving with a speed of 18 km h-1 covers 5.0 m in 1 s
Part (d):
• Relative density of a material is it's density expressed in relation to the 'density of another material' (usually water)
• Relative density of a material A is given by: Density of A⁄Density of water
■ Numerator and denominator must be in the same units: g cm-3 OR kg m-3
• So relative density does not have a unit. It is just a ratio
• Now we can write the steps:
1. When the unit is g cm-3:
• From data book, density of water is 1.000 g cm-3
• 1.000 has 4 significant figures
2. We can write: Density of lead⁄Density of water = 11.3
⇒ Density of lead⁄1.000 = 11.3
⇒ Density of lead = (11.3 × 1.000) g cm-3
• 11.3 has 3 significant figures
• 1.000 has 4 significant figures
• So the result (11.3 × 1.000) must have only 3 significant figures
3. In scientific notation, we can write (11.3 × 1.000) as 1.13 × 101
• (1.13 × 101) = 11.3
It has only 3 significant figures
• So we get: Density of lead = 11.3 g cm-3
4. When the unit is kg m-3:
• From the data book, density of water is 1000 kg m-3
• 1000 has 4 significant figures
6. We can write: Density of lead⁄Density of water = 11.3
⇒ Density of lead⁄1000 = 11.3
⇒ Density of lead = (11.3 × 1000) kg m-3
• 11.3 has 3 significant figures
• 1000 has 4 significant figures
• So the result (11.3 × 1000) must have only 3 significant figures
7. In scientific notation, we can write (11.3 × 1000) as 1.13 × 104
• It has only 3 significant figures
• So we get: Density of lead = 1.13 × 104 kg m-3
Solved example 2.16
The length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m, and 2.01 cm respectively. Give the area and volume of the sheet to correct significant figures.
Solution:
1. Area of top and bottom faces = length × breadth = 4.234 × 1.005 = 4.25517 m2.
• 4.234 has 4 significant figures
• 1.005 has 4 significant figures
• So the result 4.25517 should have only 4 significant figures
• Rounding off, we get: 4.255
• So area of the top and bottom faces = 4.255 m2.
2. Area of front and rear faces = length × thickness
• Thickness is given to us in cm. We have to convert it into m
• 1 cm = 0.01 m
• So we have to multiply each cm by 10-2
• So 2.01 cm = 2.01 × 10-2 m
3. Now we can find the area:
• length × thickness = 4.234 × (2.01×10-2) = 8.51034 ×10-2 m3.
• 4.234 has 4 significant figures
• (2.01×10-2) has 3 significant figures
• So the result (8.51034 ×10-2) should have only 3 significant figures
• Rounding off, we get: (8.51 ×10-2)
• So area of the front and rear faces = (8.51 ×10-2) m2.
4. Area of left and right faces = breadth × thickness
= 1.005 × (2.01×10-2) = 2.02005 ×10-2 m3.
• 1.005 has 4 significant figures
• (2.01×10-2) has 3 significant figures
• So the result (2.02005 ×10-2) should have only 3 significant figures
• Rounding off, we get: (2.02 ×10-2)
• So area of the left and right faces = (2.02 ×10-2) m2
5. So total area = 2 [4.255 + (8.51 ×10-2) + (2.02 ×10-2)]
= 2 [4.255 + 0.0851 + 0.0202]
6. The addition inside the square brackets is shown in fig.2.19 below:
• The digits on the right side of the magenta line should be rounded off
• After rounding off, we get: 4.360
7. Thus the total area becomes: 2 × [4.360] = 8.72
• 2 is a factor. It has infinite number of significant figures
• 4.360 has 4 significant figures
• So the result 8.72 must also have 4 significant figures
• Thus the total area becomes: 8.720 m2.
• We can write:
Total surface area of the metal sheet = 8.720 m2.
8. Calculation of volume:
• We have: Volume = area of top face × thickness
= 4.255 × (2.01×10-2) = 8.55255 ×10-2
• 4.255 has 4 significant figures
• (2.01×10-2) has 3 significant figures
• So the result (8.55255 ×10-2) should have only 3 significant figures
• Rounding off, we get: (8.55 ×10-2)
• So volume = (8.55 ×10-2) m3.
Solved example 2.17
The mass of a box measured by a grocer’s balance is 2.3 kg. Two gold pieces of masses 20.15 g and 20.17 g are added to the box. What is (a) the total mass of the box, (b) the difference in the masses of the pieces to correct significant figures?
Solution:
Part (a):
1. Mass of the box = 2.3 kg
• Mass of gold piece 1 = 20.15 g
• Mass of gold piece 2 = 20.17 g
2. We will convert g into kg
• In this way, we will get measurements with decimal places
• If we convert kg into g, the decimal point will vanish
(2.3 kg = 2300 g)
• For problems involving addition/subtraction, we need to have numbers with decimal places. This will enable us to apply the rule more easily
3. 20.15 g = (20.15 ×10-3) kg = 0.02015 kg
♦ There is no change in the number of significant figures
20.17 g = (20.17 ×10-3) kg = 0.02017 kg
♦ There is no change in the number of significant figures
4. Sum of all the weights is shown in fig.2.20 (a) below:
• The digits on the right side of the magenta line should be rounded off
• After rounding off, we get: 2.3 kg
■ This result does not show the effects of adding the gold pieces to the box. This is because, the box is weighed with a device of low precision
Part (b):
• The difference in weights is shown in fig.2.20(b) above
• There are no digits on the right side of the magenta line. So there is no rounding off to be done
• We get: The difference in the masses of the pieces = 0.02 g
Solved example 2.18
The radius of a circle is 2.12 m. What is it's area according to rules of significant rules?
Solution:
1. We have: Area of circle = 𝞹r2
2. So in our present case, area = (𝞹 × 2.12 × 2.12) m2
• 2.12 has 3 significant figures
• So the final result must have 3 significant figures
• We will take the value of 𝞹 in such a way that, it has 4 (one more than what is required in the final result) significant figures
• So 𝞹 = 3.141
3. Substituting the value, we get:
• Area = (3.141 × 2.12 × 2.12) = 14.1169104 m2
• Rounding off to 3 significant figures, we get:
Area = 14.1 m2.
Solved example 2.15
Fill in the blanks
(Note: In stating numerical answers, take care of significant figures)
(a) The volume of a cube of side 1 cm is equal to .....m3
(b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to ...(mm)2
(c) A vehicle moving with a speed of 18 km h-1 covers ....m in 1 s
(d) The relative density of lead is 11.3. Its density is ....g cm-3 or ....kg m-3
Solution:
Part (a):
1. We have: Volume = s × s × s
• Where 's' is the length of side
2. The side is given in cm. But we want the volume in m3
• 1 cm = 0.01 m
♦ 1 cm have 1 significant figure
♦ 0.01 m also have 1 significant figure
3. So volume in m3 = (0.01)3 = 10-6 m3
Part (b):
1. We have: Total surface area of a solid cylinder = 2𝞹r2 + 2𝞹rh = 2𝞹r(r + h)
See details here
2. First we evaluate the quantity within the brackets
• r = 2.0 cm
• h = 10.0 cm
(r + h) = 2.0 + 10.0 = 12.0 cm
(Since both values have the same number of digits after the decimal point, we do not need to draw the magenta line that we saw in fig.2.18 above)
3. So total surface area = 2𝞹r × 12.0 = (2 × 𝞹 × 2.0 × 12.0) cm2
• 2 is a factor
It has infinite number of significant figures
• 2.0 has 2 significant figures
• 10.0 has 3 significant figures
• So the final result must have only 2 significant figures
• We will take the value of 𝞹 in such a way that, it has 3 (one more than what is required in the final result) significant figures
• So 𝞹 = 3.14
4. Thus total surface area = (2 × 3.14 × 2.0 × 12.0) cm2 = 150.72 cm2
5. one cm = 10 mm
• So 1 cm2 = 100 mm2
• Here 100 is a factor. It has infinite number of significant figures
6. We can now convert the 'area in cm2' to 'area in m2'
• Total surface area in m2 = 150.72 × 100
• In scientific notation, this value is 1.5072 × 104 m2
7. But in (2) we saw that, the final result must have only 2 significant digits. This is valid even if units change
• So total surface area = 1.5 × 104 m2.
Part (c):
1. Distance traveled in 1 h = 18 km
• '18' has 2 significant figures
• 1 h = (60 × 60) = 3600 s
• So distance traveled in 1 s = (18⁄3600) km
2. 3600 is a factor. It has infinite number of significant figures
• So the result of (18⁄3600) must have only 2 significant figures
3. One km = 1000 m
• So (18⁄3600) km = [(18⁄3600) × 1000] m
4. 1000 is a factor. It has infinite number of significant figures
• So the result of [(18⁄3600) × 1000] must have only 2 significant figures
5. We have: [(18⁄3600) × 1000] = 5
• If '5' is to have two significant figures, we must write it as 5.0
• So we get:
A vehicle moving with a speed of 18 km h-1 covers 5.0 m in 1 s
Part (d):
• Relative density of a material is it's density expressed in relation to the 'density of another material' (usually water)
• Relative density of a material A is given by: Density of A⁄Density of water
■ Numerator and denominator must be in the same units: g cm-3 OR kg m-3
• So relative density does not have a unit. It is just a ratio
• Now we can write the steps:
1. When the unit is g cm-3:
• From data book, density of water is 1.000 g cm-3
• 1.000 has 4 significant figures
2. We can write: Density of lead⁄Density of water = 11.3
⇒ Density of lead⁄1.000 = 11.3
⇒ Density of lead = (11.3 × 1.000) g cm-3
• 11.3 has 3 significant figures
• 1.000 has 4 significant figures
• So the result (11.3 × 1.000) must have only 3 significant figures
3. In scientific notation, we can write (11.3 × 1.000) as 1.13 × 101
• (1.13 × 101) = 11.3
It has only 3 significant figures
• So we get: Density of lead = 11.3 g cm-3
4. When the unit is kg m-3:
• From the data book, density of water is 1000 kg m-3
• 1000 has 4 significant figures
6. We can write: Density of lead⁄Density of water = 11.3
⇒ Density of lead⁄1000 = 11.3
⇒ Density of lead = (11.3 × 1000) kg m-3
• 11.3 has 3 significant figures
• 1000 has 4 significant figures
• So the result (11.3 × 1000) must have only 3 significant figures
7. In scientific notation, we can write (11.3 × 1000) as 1.13 × 104
• It has only 3 significant figures
• So we get: Density of lead = 1.13 × 104 kg m-3
Solved example 2.16
The length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m, and 2.01 cm respectively. Give the area and volume of the sheet to correct significant figures.
Solution:
1. Area of top and bottom faces = length × breadth = 4.234 × 1.005 = 4.25517 m2.
• 4.234 has 4 significant figures
• 1.005 has 4 significant figures
• So the result 4.25517 should have only 4 significant figures
• Rounding off, we get: 4.255
• So area of the top and bottom faces = 4.255 m2.
2. Area of front and rear faces = length × thickness
• Thickness is given to us in cm. We have to convert it into m
• 1 cm = 0.01 m
• So we have to multiply each cm by 10-2
• So 2.01 cm = 2.01 × 10-2 m
3. Now we can find the area:
• length × thickness = 4.234 × (2.01×10-2) = 8.51034 ×10-2 m3.
• 4.234 has 4 significant figures
• (2.01×10-2) has 3 significant figures
• So the result (8.51034 ×10-2) should have only 3 significant figures
• Rounding off, we get: (8.51 ×10-2)
• So area of the front and rear faces = (8.51 ×10-2) m2.
4. Area of left and right faces = breadth × thickness
= 1.005 × (2.01×10-2) = 2.02005 ×10-2 m3.
• 1.005 has 4 significant figures
• (2.01×10-2) has 3 significant figures
• So the result (2.02005 ×10-2) should have only 3 significant figures
• Rounding off, we get: (2.02 ×10-2)
• So area of the left and right faces = (2.02 ×10-2) m2
5. So total area = 2 [4.255 + (8.51 ×10-2) + (2.02 ×10-2)]
= 2 [4.255 + 0.0851 + 0.0202]
6. The addition inside the square brackets is shown in fig.2.19 below:
 |
| Fig.2.19 |
• After rounding off, we get: 4.360
7. Thus the total area becomes: 2 × [4.360] = 8.72
• 2 is a factor. It has infinite number of significant figures
• 4.360 has 4 significant figures
• So the result 8.72 must also have 4 significant figures
• Thus the total area becomes: 8.720 m2.
• We can write:
Total surface area of the metal sheet = 8.720 m2.
8. Calculation of volume:
• We have: Volume = area of top face × thickness
= 4.255 × (2.01×10-2) = 8.55255 ×10-2
• 4.255 has 4 significant figures
• (2.01×10-2) has 3 significant figures
• So the result (8.55255 ×10-2) should have only 3 significant figures
• Rounding off, we get: (8.55 ×10-2)
• So volume = (8.55 ×10-2) m3.
Solved example 2.17
The mass of a box measured by a grocer’s balance is 2.3 kg. Two gold pieces of masses 20.15 g and 20.17 g are added to the box. What is (a) the total mass of the box, (b) the difference in the masses of the pieces to correct significant figures?
Solution:
Part (a):
1. Mass of the box = 2.3 kg
• Mass of gold piece 1 = 20.15 g
• Mass of gold piece 2 = 20.17 g
2. We will convert g into kg
• In this way, we will get measurements with decimal places
• If we convert kg into g, the decimal point will vanish
(2.3 kg = 2300 g)
• For problems involving addition/subtraction, we need to have numbers with decimal places. This will enable us to apply the rule more easily
3. 20.15 g = (20.15 ×10-3) kg = 0.02015 kg
♦ There is no change in the number of significant figures
20.17 g = (20.17 ×10-3) kg = 0.02017 kg
♦ There is no change in the number of significant figures
4. Sum of all the weights is shown in fig.2.20 (a) below:
 |
| Fig.2.20 |
• After rounding off, we get: 2.3 kg
■ This result does not show the effects of adding the gold pieces to the box. This is because, the box is weighed with a device of low precision
Part (b):
• The difference in weights is shown in fig.2.20(b) above
• There are no digits on the right side of the magenta line. So there is no rounding off to be done
• We get: The difference in the masses of the pieces = 0.02 g
Solved example 2.18
The radius of a circle is 2.12 m. What is it's area according to rules of significant rules?
Solution:
1. We have: Area of circle = 𝞹r2
2. So in our present case, area = (𝞹 × 2.12 × 2.12) m2
• 2.12 has 3 significant figures
• So the final result must have 3 significant figures
• We will take the value of 𝞹 in such a way that, it has 4 (one more than what is required in the final result) significant figures
• So 𝞹 = 3.141
3. Substituting the value, we get:
• Area = (3.141 × 2.12 × 2.12) = 14.1169104 m2
• Rounding off to 3 significant figures, we get:
Area = 14.1 m2.
In the next section, we will learn about errors
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