In the previous section we saw some cases where there is proportionality between two quantities. In this section we will see some solved examples.
Solved example 24.2
For each pair of quantities given below, check whether the first is proportional to the second. For the proportional quantities, calculate the constant of proportionality.
(a) Perimeter and radius of circles
(b) Area and radius of circles
(c) Distance travelled and the number of rotations of a circular ring along a line
(d) Interest got in an year and the amount deposited in a scheme. Interest is compounded annually
(e) Volume of water poured into a hollow prism and the height of the water level
Solution:
(a) The two quantities are:
(i) Perimeter of the circle
(ii) Radius of the circle
1. Let the perimeter be 'p' and radius be 'r'
2. We know that p = 2πr
3. We can put whatever value we like for r on the right side of the above equation. So r is a variable
4. What ever value we give for r, that value will be multiplied by a constant 2π. So p will always be proportional to r
5. The constant of proportionality is p⁄r = 2π
(b) The two quantities are:
(i) Area of the circle
(ii) Radius of the circle
1. Let the area be 'a' and the radius be 'r'
2. We know that a = πr2 = π × r × r
3. We can put whatever value we like for r on the right side of the above equation. So r is a variable.
4. Whatever value we give for r, that value will be multiplied by a 'constant π' and the 'variable r'. That means, r is not multiplied by the same constant every time.
5. So a and r are not proportional
(c) The two quantities are:
(i) Distance travelled
(ii) Number of rotations
1. Let the distance travelled be 'd' and the number of rotations be 'n'
2. We know this: The distance travelled in one rotation = 2πr. (where r is the radius of the ring) Details here. So distance travelled in n rotations = d = 2nπr
3. We can put whatever value we like for n on the right side of the above equation. So n is a variable. But r is a constant because, we are considering the movement of a single ring.
4. Whatever value we give for n, that value will be multiplied by a constant 2πr. So d will always be proportional to n
5. The constant of proportionality is d⁄n = 2πr.
Note that, the constant of proportionality is the perimeter of the ring under consideration.
(d) The two quantities are:
(i) Interest got in an year
(ii) Amount deposited
1. Let the interest got in an year be 'i' and the amount deposited be 'p'
2. We know that i = p×r (where r is the rate of interest)
3. We can put whatever value we like for p on the right side of the above equation. So p is a variable. But r is a constant for a scheme
4. Whatever value we give for p, that value will be multiplied by a constant r. So i will always be proportional to p
5. The constant of proportionality is i⁄p = r
(e) The two quantities are:
(i) Volume of water poured
(ii) Height of water level
1. Let volume be 'v' and the height of the water level from the base be 'h'
2. Let the base area of the prism be 'a'
3. The poured volume = volume of the water column formed in the prism
4. But volume of water column in the prism = base area × height = ah
5. So we get v = ah ⇒ h = v⁄a
6. We can put whatever value we like for v on the right side of the above equation. So v is a variable. But a is a constant for a prism
7. Whatever value we give for v, that value will be multiplied by a constant 1⁄a. So h will always be proportional to p
8. The constant of proportionality is h⁄v = 1⁄a
Solved example 24.3
During rainfall, the volume of water falling in each square metre may be considered equal.
(a) Prove that, the volume of water falling in a region is proportional to the area of that region
(b) Explain why the heights of rain water collected in different sized hollow prisms kept near one another are equal.
Solution:
1. Consider the blue surface in fig.24.7 below. It is a surface on the ground.
2. On this surface, we can mark a random number of squares, each of side 1 m. So each of such squares will have an area of 1 m2. Three such squares are marked.
3. It is stated in the question that, “the volume of water falling in each square metre may be considered equal”.
4. So each of the three squares will receive equal volume of water.
5. If this water is not allowed to run off and if there is no seepage into the ground, the water received will become a water column. There will be 3 water columns as shown in the fig.24.7.
6. These water columns are prisms with square base. And they have the same volume (v), because, according to (3), volume of water are the same.
7. We know that volume = base area × height. But base area of all prisms = 1 m2.
So we can write: v = 1 × h ⇒ v = h
8. But v is same for all the three prisms.
■ So h will also be same for all the three prisms
Now we will first consider part (ii) of the question:
9. Fig.24.8 below shows two hollow prisms kept near to one another. They are shown in yellow colour. They have different base areas.
10. We can think of random number of water columns inside these prisms. Let the base of these columns be squares of area 1 cm2.
11. Then each water column will be a square prism of base area 1 cm2. Two such water prisms are shown inside each of the yellow prisms.
12. We have seen in (8) that the heights of all the water prisms will be equal. Regardless of whether they are in the first yellow prism or the second yellow prism. Let this height be ‘h’.
13. Let the base area of first yellow prism be a1
Let the base area of second yellow prism be a2
14. Then total number of water prisms (each with base area 1 cm2) in first yellow prism = a1/1 = a1
Total number of water prisms (each with base area 1 cm 2) in second yellow prism = a2/1 = a2
15. All the a1 number of water prisms in the first yellow prism will have the same height ‘h’
All the a2 number of water prisms in the second yellow prism will also have the same height ‘h’
That means, the water level in both the yellow prisms will be ‘h’
16. So, whatever number of yellow hollow prisms (of different base sizes) we place near one another, after the rainfall, the height of water in all will be the same.
Now we take up the part (i):
17. Consider two paddy fields in a locality. Let their areas be a1 and a2. If the water is not allowed to run off and if there is no seepage into the ground, there will be two water prisms. Each will cover the entire area of the respective field. From (8), both water prisms will have the same height ‘h’.
18. The volume of water in the first field = v1 = a1 h
The volume of water in the second field = v2 = a2 h
19. ‘h’ is a constant. So, if area increases, volume increases
if area decreases, volume decreases
20. That means volume is proportional to the area.
Solved example 24.4
When a weight is suspended by a spring, the extension is proportional to that weight. Explain how this property can be used to make markings on a spring balance.
Solution:
1. Fig.24.9 below shows the spring used inside a spring balance.
• Position 0 shows the situation when no load is applied on the spring.
• Position 1 shows the situation when a load of w1 is applied on the spring.
♦ We can see that there is an extension of x1 cm from the initial position.
• Position 2 shows the situation when a load of w2 is applied on the spring.
♦ We can see that there is an extension of x2 cm from the initial position.
2. It is given in the question that, the extension is proportional to the applied load.
3. So any extension x will be proportional to the weight w that produces that extension.
We can write: x = a constant 'k' × w ⇒ x = kw ⇒ k = x⁄w
4. The constant 'k' is called the spring constant. Every spring will have a particular value of spring constant. We want to find this constant for our spring.
5. For that, we adopt the following procedure:
(i) Put a known weight w1. Measure the extension x1. Then k = x1 ⁄w1
(ii) Put another known weight w2. Measure the extension x2. Then k = x2 ⁄w2
(iii) Since k is a constant, we will get the same value for k in both (i) and (ii)
(iv) Repeat the trial with several known weights. In all cases we will get the same k
6. Once k is obtained, we can make the markings on the spring balance.
(i) The spring is fixed inside an outer casing. Markings are made on this casing.
(ii) When the spring is at zero load position, mark that position on the casing as '0 kg'
(iii) Now we want to mark the 1 kg position.
• Let the extension for a load of 1 kg be 'x1 kg'
• x1 kg = k × 1 kg = k. We have already calculated k. So we will get 'x1 kg'
• Measure this 'x1 kg' from the '0 kg' mark on the casing. And mark it as 1 kg
(iv) Now we want to mark the 2 kg position.
• Let the extension for a load of 1 kg be 'x2 kg'
• x2 kg = k × 2 kg = 2k. We have already calculated k. So we will get 'x2 kg'
• Measure this 'x2 kg' from the '0 kg' mark on the casing. And mark it as 2 kg
(v) In this way mark 3 kg, 4 kg, 5 kg etc., Fractions between them can also be marked in this way.
We will learn more details about the procedure in higher classes.
In the next section we will see a few more solved examples.
Solved example 24.2
For each pair of quantities given below, check whether the first is proportional to the second. For the proportional quantities, calculate the constant of proportionality.
(a) Perimeter and radius of circles
(b) Area and radius of circles
(c) Distance travelled and the number of rotations of a circular ring along a line
(d) Interest got in an year and the amount deposited in a scheme. Interest is compounded annually
(e) Volume of water poured into a hollow prism and the height of the water level
Solution:
(a) The two quantities are:
(i) Perimeter of the circle
(ii) Radius of the circle
1. Let the perimeter be 'p' and radius be 'r'
2. We know that p = 2πr
3. We can put whatever value we like for r on the right side of the above equation. So r is a variable
4. What ever value we give for r, that value will be multiplied by a constant 2π. So p will always be proportional to r
5. The constant of proportionality is p⁄r = 2π
(b) The two quantities are:
(i) Area of the circle
(ii) Radius of the circle
1. Let the area be 'a' and the radius be 'r'
2. We know that a = πr2 = π × r × r
3. We can put whatever value we like for r on the right side of the above equation. So r is a variable.
4. Whatever value we give for r, that value will be multiplied by a 'constant π' and the 'variable r'. That means, r is not multiplied by the same constant every time.
5. So a and r are not proportional
(c) The two quantities are:
(i) Distance travelled
(ii) Number of rotations
1. Let the distance travelled be 'd' and the number of rotations be 'n'
2. We know this: The distance travelled in one rotation = 2πr. (where r is the radius of the ring) Details here. So distance travelled in n rotations = d = 2nπr
3. We can put whatever value we like for n on the right side of the above equation. So n is a variable. But r is a constant because, we are considering the movement of a single ring.
4. Whatever value we give for n, that value will be multiplied by a constant 2πr. So d will always be proportional to n
5. The constant of proportionality is d⁄n = 2πr.
Note that, the constant of proportionality is the perimeter of the ring under consideration.
(d) The two quantities are:
(i) Interest got in an year
(ii) Amount deposited
1. Let the interest got in an year be 'i' and the amount deposited be 'p'
2. We know that i = p×r (where r is the rate of interest)
3. We can put whatever value we like for p on the right side of the above equation. So p is a variable. But r is a constant for a scheme
4. Whatever value we give for p, that value will be multiplied by a constant r. So i will always be proportional to p
5. The constant of proportionality is i⁄p = r
(e) The two quantities are:
(i) Volume of water poured
(ii) Height of water level
1. Let volume be 'v' and the height of the water level from the base be 'h'
2. Let the base area of the prism be 'a'
3. The poured volume = volume of the water column formed in the prism
4. But volume of water column in the prism = base area × height = ah
5. So we get v = ah ⇒ h = v⁄a
6. We can put whatever value we like for v on the right side of the above equation. So v is a variable. But a is a constant for a prism
7. Whatever value we give for v, that value will be multiplied by a constant 1⁄a. So h will always be proportional to p
8. The constant of proportionality is h⁄v = 1⁄a
Solved example 24.3
During rainfall, the volume of water falling in each square metre may be considered equal.
(a) Prove that, the volume of water falling in a region is proportional to the area of that region
(b) Explain why the heights of rain water collected in different sized hollow prisms kept near one another are equal.
Solution:
1. Consider the blue surface in fig.24.7 below. It is a surface on the ground.
2. On this surface, we can mark a random number of squares, each of side 1 m. So each of such squares will have an area of 1 m2. Three such squares are marked.
 |
| Fig.24.7 |
4. So each of the three squares will receive equal volume of water.
5. If this water is not allowed to run off and if there is no seepage into the ground, the water received will become a water column. There will be 3 water columns as shown in the fig.24.7.
6. These water columns are prisms with square base. And they have the same volume (v), because, according to (3), volume of water are the same.
7. We know that volume = base area × height. But base area of all prisms = 1 m2.
So we can write: v = 1 × h ⇒ v = h
8. But v is same for all the three prisms.
■ So h will also be same for all the three prisms
Now we will first consider part (ii) of the question:
9. Fig.24.8 below shows two hollow prisms kept near to one another. They are shown in yellow colour. They have different base areas.
 |
| Fig.24.8 |
11. Then each water column will be a square prism of base area 1 cm2. Two such water prisms are shown inside each of the yellow prisms.
12. We have seen in (8) that the heights of all the water prisms will be equal. Regardless of whether they are in the first yellow prism or the second yellow prism. Let this height be ‘h’.
13. Let the base area of first yellow prism be a1
Let the base area of second yellow prism be a2
14. Then total number of water prisms (each with base area 1 cm2) in first yellow prism = a1/1 = a1
Total number of water prisms (each with base area 1 cm 2) in second yellow prism = a2/1 = a2
15. All the a1 number of water prisms in the first yellow prism will have the same height ‘h’
All the a2 number of water prisms in the second yellow prism will also have the same height ‘h’
That means, the water level in both the yellow prisms will be ‘h’
16. So, whatever number of yellow hollow prisms (of different base sizes) we place near one another, after the rainfall, the height of water in all will be the same.
Now we take up the part (i):
17. Consider two paddy fields in a locality. Let their areas be a1 and a2. If the water is not allowed to run off and if there is no seepage into the ground, there will be two water prisms. Each will cover the entire area of the respective field. From (8), both water prisms will have the same height ‘h’.
18. The volume of water in the first field = v1 = a1 h
The volume of water in the second field = v2 = a2 h
19. ‘h’ is a constant. So, if area increases, volume increases
if area decreases, volume decreases
20. That means volume is proportional to the area.
Solved example 24.4
When a weight is suspended by a spring, the extension is proportional to that weight. Explain how this property can be used to make markings on a spring balance.
Solution:
1. Fig.24.9 below shows the spring used inside a spring balance.
 |
| Fig.24.9 |
• Position 1 shows the situation when a load of w1 is applied on the spring.
♦ We can see that there is an extension of x1 cm from the initial position.
• Position 2 shows the situation when a load of w2 is applied on the spring.
♦ We can see that there is an extension of x2 cm from the initial position.
2. It is given in the question that, the extension is proportional to the applied load.
3. So any extension x will be proportional to the weight w that produces that extension.
We can write: x = a constant 'k' × w ⇒ x = kw ⇒ k = x⁄w
4. The constant 'k' is called the spring constant. Every spring will have a particular value of spring constant. We want to find this constant for our spring.
5. For that, we adopt the following procedure:
(i) Put a known weight w1. Measure the extension x1. Then k = x1 ⁄w1
(ii) Put another known weight w2. Measure the extension x2. Then k = x2 ⁄w2
(iii) Since k is a constant, we will get the same value for k in both (i) and (ii)
(iv) Repeat the trial with several known weights. In all cases we will get the same k
6. Once k is obtained, we can make the markings on the spring balance.
(i) The spring is fixed inside an outer casing. Markings are made on this casing.
(ii) When the spring is at zero load position, mark that position on the casing as '0 kg'
(iii) Now we want to mark the 1 kg position.
• Let the extension for a load of 1 kg be 'x1 kg'
• x1 kg = k × 1 kg = k. We have already calculated k. So we will get 'x1 kg'
• Measure this 'x1 kg' from the '0 kg' mark on the casing. And mark it as 1 kg
(iv) Now we want to mark the 2 kg position.
• Let the extension for a load of 1 kg be 'x2 kg'
• x2 kg = k × 2 kg = 2k. We have already calculated k. So we will get 'x2 kg'
• Measure this 'x2 kg' from the '0 kg' mark on the casing. And mark it as 2 kg
(v) In this way mark 3 kg, 4 kg, 5 kg etc., Fractions between them can also be marked in this way.
We will learn more details about the procedure in higher classes.
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