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Jharkhand Board JAC Class 10 Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.2 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 5 Arithmetic Progressions Exercise 5.2

Question 1.
Fill in the blanks in the following table, given that a is the first term, d the common difference and a, the nth term of the AP:

Solution:
1. Here, a = 7, d = 3, n = 8 and a_n is to be found.
We have a_n = a + (n – 1)d
a₈ = 7 + (8 – 1) 3 = 7 + 21 = 28

2. Here, a = -18, n = 10, a_n = a₁₀ = 0 and d is to be found.
a_n = a + (n – 1)d
∴ 0 = – 18 + (10 – 1)d
∴ 18 = 9d ∴ d = 2

3. Here, d = -3, n = 18, a_n = a₁₈ = -5 and a is to be found.
an = a + (n – 1)d
∴ -5 = a + (18 – 1)(-3)
∴ -5 = a – 51
∴ a = 51 – 5 ∴ a = 46

4. Here, a = -18.9, d = 2.5, a_n = 3.6 and n is to be found.
a_n = a + (n – 1)d
∴ 3.6 = – 18.9 + (n – 1)(2.5)
∴ 22.5 = 2.5 (n – 1)
∴ (n – 1) = \(\frac{22.5}{2.5}\)
∴ n – 1 = 9 ∴ n = 10

5. Here, a = 3.5, d = 0, n = 105 and a_n is to be found.
a_n = a + (n – 1) d
∴ a₁₀₅ = 3.5 + (105 – 1) (0)
∴ a₁₀₅ = 3.5

Question 2.
Choose the correct choice in the following and justify:
1. 30th term of the AP: 10, 7, 4, ……, is
(A) 97
(B) 77
(C) -77
(D) -87
2. 11th term of the AP: -3, –\(\frac{1}{2}\), 2, ….. is
(A) 28
(B) 22
(C) -38
(D) -48\(\frac{1}{2}\)
Solution:
1. For the given AP 10, 7, 4,… a = 10,
d = 7 – 10 = -3 and n = 30
a_n = a + (n – 1)d
∴ a₃₀ = 10 + (30 – 1) (-3)
∴ a₃₀ = 10 – 87
∴ a₃₀ = -77
Thus, the correct choice is (C) -77.

2. For the given AP -3, –\(\frac{1}{2}\), 2, ….., a = -3.
d = –\(\frac{1}{2}\) – (-3) = 2\(\frac{1}{2}\) = \(\frac{5}{2}\) and n = 11.
a_n = a + (n – 1)d
∴ a₁₁ = -3 + (11 – 1)(\(\frac{5}{2}\))
∴ a₁₁ = – 3 + 25
∴ a₁₁ = 22
Thus, the correct choice is (B) 22.

Question 3.
In the following APs, find the missing terms in the boxes:

Solution:
For the given AP, first term = a = 2 and third term = a + 2d = 26.
a = 2 and a + 2d = 26 gives d = 12.
Then, second term = a + d = 2 + 12 = 14
Thus, the missing term in the box is

2. For the given AP,
second term = a + d = 13 …….(1)
fourth term = a + 3d = 3 …….(2)
Solving equations (1) and (2) we get d = -5 and a = 18.
Now, first term = a = 18 and third term = a + 2d
= 18 + 2(-5) = 8.
Thus, the missing terms in the boxes are

Alternative Method:
Let the terms of the given AP be a₁, a₂, a₃, a₄.
Here, a₂ = 13 and a₄ = 3.
Now, a₄ – a₃ = a₃ – a₂ = d
∴ 3 – a₃ = a₃ – 13
∴ 2a₃ = 16
∴ a₃ = 8
Again, a₂ – a₁ = a₃ – a₂
∴ 13 – a₁ = 8 – 13
∴ 13 – a₁ = -5
∴ a₁ = 18
Thus, the missing terms in the boxes are

3. For the given AP
first term a = 5 ……….(1)
fourth term = a + 3d = 9\(\frac{1}{2}\) ……….(2)
From equations (1) and (2), we get a = 5 and d = 1\(\frac{1}{2}\).
Now, second term = a + d = 5 + 1\(\frac{1}{2}\) = 6\(\frac{1}{2}\)
and third term = a + 2d = 5 + 2 (1\(\frac{1}{2}\)) = 8
Thus, the missing terms in the boxes are

4. For the given AP
first term a = -4 ……….(1)
sixth term = a + 5d = 6 ……….(2)
From equations (1) and (2), we get
a = -4 and d = 2
Now, second term = a + d(-4) + 2 = -2,
third term = a + 2d = (-4) + 2 (2) = 0.
fourth term = a + 3d = (-4) + 3(2) = 2 and
fifth term = a + 4d = (-4) + 4(2) = 4.
Thus, the missing terms in the boxes are

5. For the given AP,
second term = a + d = 38 ……….(1)
sixth term = a + 5d = -22 ……….(2)
Solving equations (1) and (2), we get
d = -15 and a = 53.
Now, first term = a = 53,
third term = a + 2d = 53 + 2(-15) = 23,
fourth term = a + 3d = 53 + 3(-15) = 8 and
fifth term = a + 4d = 53 + 4(-15) = -7.
Thus, the missing terms in the boxes are

Question 4.
Which term of the AP: 3, 8, 13, 18, … is 78 ?
Solution:
Suppose nth term of the AP 3, 8, 13, 18, … is 78.
Here, a = 3, d = 8 – 3 = 5, a_n = 78 and n is to be found.
a_n = a + (n – 1)d
∴ 78 = 3 + (n – 1)5
∴ 75 = 5 (n-1)
∴ 15 = n – 1 ∴ n = 16
Thus, the 16th term of the AP 3, 8, 13, 18, …….., is 78.

Question 5.
Find the number of terms in each of the following APs:
1. 7, 13, 19, …….., 205
2. 18, 15\(\frac{1}{2}\), 13, ……., -47
Solution:
1. For the given finite AP 7, 13, 19, ….. 205, a = 7, d = 13 – 76 and last term l = 205.
Let us consider that the last term is the nth term.
a_n = a + (n – 1)d
∴ 205 = 7 + (n – 1)6
∴ 198 = 6 (n – 1)
∴ n – 1 = 33
∴ n = 34
Thus, there are 34 terms in the given finite AP.

2. For the given finite AP 18, 15\(\frac{1}{2}\), 13….. -47, a = 18, d = 15\(\frac{1}{2}\) – 18 = -2\(\frac{1}{2}\) = –\(\frac{5}{2}\) and last term l = -47.
Let us consider that the last term is the nth term.
a_n = a + (n – 1) d
∴ -47 = 18 + (n-1) (-\(\frac{5}{2}\))
∴ -65 = –\(\frac{5}{2}\)(n – 1)
∴ n – 1 = 26
∴ n = 27
Thus, there are 27 terms in the given finite AP.

Question 6.
Check whether -150 is a term of the AP: 11, 8, 5, 2…
Solution:
If possible, let -150 be the nth term of the AP 11, 8, 5, 2,…
Here, a = 11; d = 8 – 11 = -3 and a_n = -150.
a_n = a + (n – 1)d
∴ -150 = 11 + (n – 1)(-3)
∴ -161 = -3 (n – 1)
∴ n – 1 = \(\frac{161}{3}\)
∴ n = \(\frac{164}{3}\)
∴ But n must be positive integer for an AP.
Hence, -150 cannot be a term of the AP 11, 8, 5, 2…..

Question 7.
Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.
Solution:
For any AP, a_n = a + (n – 1)d.
a₁₁ = a + 10 d
∴ a + 10 d = 38 ………(1)
∴ a₁₆ = a + 15d
∴ a + 15d = 73 ………(2)
Solving equations (1) and (2), we get
d = 7 and a = -32.
Now, 31st term = a₃₁ = a + 30d
= -32 + 30(7)
= -32 + 210
= 178
Thus, the 31st term of the given AP is 178.
Note: d = \(\frac{a_{16}-a_{11}}{16-11}=\frac{73-38}{5}=\frac{35}{5}=7\) can also be used.

Question 8.
An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
Solution:
The given finite AP has 50 terms, hence its last term is a₅₀.
So, a₃ = 12 and a₅₀ = 106.
Now, a_n = a + (n – 1)d.
That gives, a₃ = a + 2d = 12 ………(1)
and a₅₀ = a + 49d = 106 ………(2)
Solving equations (1) and (2), we get
d = 2 and a = 8.
Now, 29th term = a₂₉ = a + 28d
∴ a₂₉ = 8 + 28 (2)
∴ a₂₉ = 64
Thus, the 29th term of the given AP is 64.

Question 9.
If the 3rd and the 9th terms of an AP are 4 and -8 respectively, which term of this AP is zero?
Solution:
For the given AP, a₃ = 4 and a9 = -8
We know, a_n = a + (n – 1)d.
∴ a₃ = a + 2d = 4 ………(1)
and a₉ = a + 8d = -8 ………(2)
Solving equations (1) and (2), we get
d = -2 and a = 8.
Now, let nth term of the AP be 0.
a_n = a + (n – 1)d
∴ 0 = 8 + (n – 1) (-2)
∴ 2(n – 1) = 8
∴ n – 1 = 4
∴ n = 5
Thus, the 5th term of the given AP is zero.

Question 10.
The 17th term of an AP exceeds its 10th term by 7. Find the common difference.
Solution:
For the given AP,
a₁₇ + 16d = a + 9d + 7a [∵ a_n = a + (n – 1) d]
∴ 7d = 7
∴ d = 1.
Thus, the common difference of the given AP is 1.

Question 11.
Which term of the AP:3, 15, 27, 39, …….. will be 132 more than its 54th term?
Solution:
For the given AP 3, 15, 27, 39, …….., a = 3
and d = 15 – 3 = 12.
Suppose nth term of the AP is 132 more than its 54th term.
∴ a_n = a₅₄ + 132
∴ a + (n – 1)d = a + 53d + 132
∴ 3 + (n – 1) (12) = 3 + 53(12) + 132
∴ 12(n – 1) = 12 (53 + 11)
∴ 12(n – 1) = 12 × 64
∴ n – 1 = 64
∴ n = 65
Thus, the 65th term of the given AP is 132 more than its 54th term.

Question 12.
Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?
Solution:
Let the first term of two given APs be a₁ and a₂ respectively and let the same common difference be d.
Also, let a₁ > a₂
Then, 100th term of the first AP = a₁ + 99d
(a_n = a + (n – 1)d)
100th term of the second AP = a₂ + 99d.
The difference between their 100th terms is 100.
∴ (a₁ + 99d) – (a₂ + 99d) = 100 (a₁ > a₂)
∴ a₁ – a₂ = 100 ……(1)
Now, 1000th term of the first AP = a₁ + 999d
1000th term of the second AP = a₂ + 999d.
Then, the difference between their 1000th terms
= (a₁ + 999d) – (a₂ + 999d)
= a₁ – a₂
= 100 (by (1))
Thus, the difference between the 1000th terms of the two APs is 100.

Question 13.
How many three digit numbers are divisible by 7?
Solution:
The list of three digit numbers divisible by 7 is as below:
105, 112, 119, …….., 987, 994.
These numbers form a finite AP with a = 105, d = 112 – 105 = 7 and last term l = 994.
Suppose the last term of the AP is its nth term.
∴ l = a_n
∴ 994 = a + (n – 1)d
∴ 994 = 105 + (n – 1)7
∴ 7(n – 1) = 889
∴ n – 1 = 127
∴ n = 128
Hence, there are 128 terms in the AP.
Hence, 128 three digit numbers are divisible by 7.

Question 14.
How many multiples of 4 lie between 10 and 250?
Solution:
The multiples of 4 lying between 10 and 250 give rise to following finite AP:
12, 16, 20, ….., 244, 248.
Here, a = 12, d = 16 – 12 = 4 and last term l = 248.
If the last term is the nth term of the AP then l = a_n.
∴ l = a + (n – 1) d
∴ 248 = 12 + (n – 1)4
∴ 236 = 4 (n – 1)
∴ n – 1 = 59
∴ n = 60
Thus, there are 60 terms in the AP.
Thus, 60 multiples of 4 lie between 10 and 250.

Question 15.
For what value of n, are the nth terms of two APs 63, 65, 67, …… and 3, 10, 17,… equal?
Solution:
For the first AP 63, 65, 67, ….., a = 63, d = 65 – 63 = 2.
Then, nth term of the first AP a is given by a_n = a + (n – 1)d = 63 + (n – 1)(2).
For the second AP 3, 10, 17, ……, A = 3, D = 10 – 3 = 7.
Then, nth term of the second AP An is given by
A_n = A + (n – 1) D = 3 + (n – 1) (7).
Now, a_n = A_n
∴ 63 + (n – 1) (2) = 3 + (n – 1)(7)
∴ 63 – 3 = (n – 1)(7 – 2)
∴ 60 = 5(n – 1)
∴ n – 1 = 12
∴ n = 13
Thus, for n = 13, the nth term of two given APs are equal.

Question 16.
Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.
Solution:
For the given AP a₃ = 16 and a₇ = a₅ + 12.
For any AP, a_n = a + (n – 1) d.
∴ a + 2d = 16 and a + 6d = a + 4d + 12
a + 6d = a + 4d + 12 gives 2d = 12,
i.e., d = 6.
Substituting d = 6 in a + 2d = 16, we get a = 4.
Then, the required AP is 4, 4 + 6, 4 + 2 (6), 4 + 3(6), …..
Hence, the required AP is 4, 10, 16, 22, ……

Question 17.
Find the 20th term from the last term of the AP: 3, 8, 13, ….., 253.
Solution:
For the given finite AP 3, 8, 13, …., 253,
a = 3, d = 8 – 3 = 5 and last term l = 253.
Let the last term be its nth term.
∴ l = a_n
∴ l = a + (n – 1)d
∴ 253 = 3 + (n – 1)(5)
∴ 250 = 5 (n – 1)
∴ n – 1 = 50
∴ n = 51
Thus, there are in all 51 terms in the AP.
Now, the 20th term from the last term is (51 – 20 + 1)th term = 32nd term from the beginning.
a₃₂ = a + 31d
∴ a₃₂ = 3 + 31(5)
∴ a₃₂ = 158
Thus, the 20th term from the last term is 158.

Question 18.
The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.
Solution:
For any AP, a_n = a + (n – 1) d.
a₄ = a + 3d, a₈ = a + 7d, a₆ = a + 5d and a₁₀ = a + 9d.
Now, a₄ + a₈ = 24 (Given)
∴ (a + 3d) + (a + 7d) = 24
∴ 2a + 10d = 24
∴ a + 5d = 12 ……..(1)
Again, a₆ + a₁₀ = 44 (Given)
∴ (a + 5d) + (a + 9d) = 44
∴ 2a + 14d = 44
∴ a + 7d = 22 …….(2)
Solving equations (1) and (2), we get d = 5 and a = -13.
Then, a₂ = a + d = 13 + (5) = -8 and
a₃ = a + 2d = -13 + 2(5) = -3.
Thus, the first three terms of the AP are -13, 8, 3.

Question 19.
Subba Rao started work in 1995 at an annual salary of ₹ 5,000 and received an increment of ₹ 200 each year. In which year did his income reach ₹ 7,000?
Solution:
Subba Rao’s income in first year = ₹ 5,000
His income in second year = ₹ 5,000 + ₹ 200
= ₹ 5200
His income in third year = ₹ 5200 + ₹ 200
= ₹ 5400
and so on.
These numbers of his income (in rupees) form the AP 5000, 5200, 5400, …….
Here, a = 5000; d = 5200 – 5000 = 200;
a_n = 7000 and n is to be found.
a_n = a + (n – 1)d
∴ 7000 = 5000 + (n – 1)(200)
∴ 2000 = 200(n – 1)
∴ n – 1 = 10
∴ n = 11
Thus, Subba Rao’s income will reach ₹ 7000 in 11th year, i.e., in the year 2005.

Question 20.
Ramkali saved ₹ 5 in the first week of a year and then increased her weekly savings by ₹ 1.75. If in the nth week, her weekly savings become ₹ 20.75, find n.
Solution:
Ramkali’s savings in first week = ₹ 5
her savings in second week = ₹ 5 + ₹ 1.75
= ₹ 6.75,
her savings in third week = ₹ 6.75 + ₹ 1.75
= ₹ 8.50.
and so on.
Thus, the weekly savings (in rupees) of Ramkalt form the AP 5, 6.75, 8.50, …..
Here, a = 5; d = 6.75 – 5 = 1.75; a_n = 20.75 and n is to be found.
a_n = a + (n – 1)d
∴ 20.75 = 5 + (n – 1)(1.75)
∴ 1.75(n – 1) = 15.75
∴ n – 1 = 9
∴ n = 10
Thus, if Ramkali’s weekly savings is ₹ 20.75 in nth week, then n = 10.

Jharkhand Board JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.1 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 14 Statistics Exercise 14.1

Question 1.
A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

Which method did you use for finding the mean, and why?
Solution:

Question 2.
Consider the following distribution of daily wages of 50 workers of a factory:

Find the mean daily wages of the workers of the factory by using an appropriate method.
Solution:

Question 3.
The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs. 18. Find the missing frequency f.

Solution:

752 + 20f = 792 + 18.1
20f – 18f = 792 – 752
2f = 40
f = \(\frac{40}{2}\) = 20.

Question 4.
Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute were recorded and summarised as follows. Find the mean heart beats per minute for these women, choosing a suitable method.

Solution:

Question 5.
In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes:

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?
Solution:

Question 6.
The table below shows the daily expenditure on food of 25 households in a locality.

Find the mean daily expenditure on food by a suitable method.
Solution:

Question 7.
To find out the concentration of SO₂ in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:

Find the mean concentration of SO₂ in the air.
Solution:

Mean concentration of SO₂ in air = 0.099 ppm

Question 8.
A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

Solution:


Mean number of days a student was absent is 12.48.

Question 9.
The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

Solution:

Jharkhand Board JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.4 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 14 Statistics Exercise 14.4

Question 1.
The following distribution gives the daily income of 50 workers of a factory:

Convert the distribution above to a less than type cumulative frequency distribution, and draw its ogive.
Solution:

Question 2.
During the medical check-up of 35 students of a class, their weights were recorded as follows:

Draw a less than type ogive for the given data. Hence obtain the median weight from the graph and verify the result by using the formula.
Solution:

\(\frac{\mathrm{n}}{2}=\frac{35}{2}=17.5\)
Plot the points (38, 0) (40, 3) (42, 5) (44, 9) (46, 14) (48, 28) (50, 32) (52, 35)
Median = l + \(\left[\frac{\frac{\mathrm{n}}{2}-\mathrm{cf}}{\mathrm{f}}\right]\) × h
= 46 + \(\left[\frac{17.5-14}{14}\right]\) × 2
= 46 + \(\frac{3.5 \times 2}{14}\)
= 46 + 0.5= 46.5.

Question 3.
The following table gives production yield per hectare of wheat of 100 farms of a village:

Change the distribution to a more than type distribution, and draw its ogive.
Solution:

∴ Co-ordinate points are (50, 100), (55, 98), (60, 90), (65, 78), (70, 54) and (75, 16).

JAC Board Class 9th Science Important Questions Chapter 15 Improvement in Food Resources

Multiple Choice Questions

Question 1.
Nitrogen, phosphorus, and potassium are examples of …………….
(a) micronutrients
(b) macronutrients
(c) fertilizers
(d) both (a) and (c)
Answer:
(b) macronutrients

Question 2.
Cyprinus and Parthenium are types of …………….
(a) diseases
(b) pesticides
(c) weeds
(d) pathogens
Answer:
(c) weeds

Question 3.
Using fertilisers in farming is an example of …………… .
(a) no cost production
(b) low – cost production
(c) high – cost production
(d) none of these
Answer:
(c) high – cost production

Question 4.
What is the other name for Apis cerana indica?
(a) Indian cow
(b) Indian buffalo
(c) Indian honeybee
(d) None of these
Answer:
(c) Indian honeybee

Question 5.
The management and production of fish is called …………… .
(a) pisciculture
(b) apiculture
(c) sericulture
(d) aquaculture
Answer:
(a) pisciculture

Question 6.
Pasturage is related to …………… .
(a) cattle
(b) fishery
(c) apiculture
(d) sericulture
Answer:
(c) apiculture

Question 7.
What is the process of growing two or more crops in a definite pattern called?
(a) Crop rotation
(b) Intercropping
(c) Mixed cropping
(d) Organic cropping
Answer:
(b) Intercropping

Question 8.
The kharif season extends from …………… .
(a) November to April
(b) June to October
(c) March to November
(d) December to March
Answer:
(b) June to October

Question 9.
For mixed cropping, which of the following combinations of crops is not suitable?
(a) Wheat + maize
(b) Wheat + gram
(c) Wheat + mustard
(d) Groundnut + sunflower
Answer:
(a) Wheat + maize

Question 10.
Catla, Rohu and Mrigals constitute …………… .
(a) marine fishes
(b) brackish water fishes
(c) fresh water fishes
(d) both (a) and (b)
Answer:
(c) fresh water fishes

Analysing & Evaluating Questions

Question 11.
Madhu visited a dairy farm with her friends. There they saw the various kinds of cattle kept in sheds, food given to them and so on. What are the main components of feed provided to the cattle?
(a) Roughage
(b) Concentrates
(c) Water
(d) All of these
Answer:
(d) All of these

Question 12.
Find out the correct sentences.
(i) Hybridisation means crossing between genetically dissimilar plants.
(ii) Cross between two varieties is called interspecific hybridisation.
(iii) Introducing genes of desired character into a plant gives genetically modified crop.
(iv) Cross between plants of two species is called intervarietal hybridisation.
(a) (i) and (iii)
(b) (ii) and (iv)
(c) (ii) and (iii)
(d) (iii) and (iv)
Answer:
(a) (i) and (iii)

Question 13.
The characteristic which is not chosen for selective breeding in dairy animals is
(a) lactation period
(b) resistance to diseases
(c) good shelter
(d) nutritional requirement
Answer:
(c) good shelter

Assertion Reason Questions

Directions: In the following questions, the Assertions and the Reasons have been put forward. Read the statements carefully and choose the correct alternative from the following:
(A) Both the assertion and the reason are correct and the reason is the correct explanation of the assertion.
(B) The assertion and the reason are correct but the reason is not the correct explanation of the assertion.
(C) The assertion is true but the reason is false.
(D) Both the statements are false.
1. Assertion: Organic matter is important for crop production.
Reason: Organic matter provides major essential nutrients to the plant.
Answer:
(C) The assertion is true but the reason is false.

2. Assertion: Manure is better than fertilisers in maintaining soil fertility.
Reason: Manure improves soil structure and increases the water holding capacity of soil.
Answer:
(A) Both the assertion and the reason are correct and the reason is the correct explanation of the assertion.

3. Assertion: It is better to grow soyabean with maize in the same field.
Reason: Root nodules of soyabean plants have nitrogen fixing bacteria which enrich the soil with nitrogen.
Answer:
(A) Both the assertion and the reason are correct and the reason is the correct explanation of the assertion.

4. Assertion: Mixed cropping is a good practice in agriculture.
Reason: By mixed cropping, number of weeds in the field can be reduced.
Answer:
(C) The assertion is true but the reason is false.

5. Assertion: Grains to be stored should have low moisture level.
Reason: Low moisture level in grains inhibits the growth of bacteria and fungi.
Answer:
(A) Both the assertion and the reason are correct and the reason is the correct explanation of the assertion.

Very Short Answer Type Questions

Question 1.
State one demerit with composite fish culture system.
Answer:
Many fishes breed only during monsoon so hormonal stimulation has to be given. Also, good quality fish seeds are not available.

Question 2.
State one importance of photoperiod in agriculture.
Answer:
Photoperiod in agriculture provides adequate light for flowering.

Question 3.
Name two micronutrients and two macronutrients which plants take from the soil.
Answer:
(a) Macronutrients are: calcium (Ca), magnesium (Mg)
(b) Micronutrients are: boron (B), chloride (Cl)

Question 4.
How does catla differ from mrigal?
Answer:
Catla belongs to genus Catla while mrigal belongs to genus Cirrhinus. Catla is a surface feeder and native to the Northern waters of India while mrigal is a bottom – feeder and native to the Ganges and Brahm putra rivers of India.

Question 5.
Name the two vitamins which are added in the poultry feed.
Answer:
Vitamins A and K.

Question 6.
From where do plants acquire the following nutrients?
(a) Nitrogen
(b) Hydrogen
Answer:
(a) Soil (b) Water

Question 7.
Which nutrients are supplied by cereals and pulses?
Answer:
Carbohydrates and proteins are supplied by cereals and pulses, respectively.

Question 8.
Name any two weeds of crop field.
Answer:
Xanthium (chota dhatura), Parthenium (gajar ghas), Cyperinus rotundus (motha).

Question 9.
Define animal husbandry.
Answer:
Animal husbandry is the practice of management and care of farm animals by humans for profit.

Question 10.
Mention two exam pies of crop combinations that are grown in mixed cropping.
Answer:
Some combinations of mixed cropping are:
(a) Wheat and mustard
(b) Maize and urad (pulse)
(c) Groundnut and sunflower

Question 11.
(a) Name an exotic variety of honeybee grown in India.
(b) What is the rearing of fish on a large scale called?
Answer:
(a) Apis cerana indica
(b) Pisciculture

Question 12.
Name two exotic cattle breeds with long lactation periods?
Answer:
The period of milk production after the birth of a calf is called lactation period. Jersey and Brown Swiss are two exotic cattle breeds having long lactation periods.

Question 13.
Between broiler and layer, which one matures earlier?
Answer:
Broilers have fast growth rate.

Question 14.
State the difference between compost and vermicompost.
Answer:

Question 15.
Name two varieties of food required for milch animals.
Answer:
(a) Food to keep animals healthy,
(b) Food to increase lactation.

Question 16.
Define apiculture.
Answer:
Keeping bee for obtaining honey commercially is called apiculture.

Question 17.
Define hybridisation.
Answer:
Hybridisation refers to crossing between genetically dissimilar plants to obtain better variety of crops.

Analysing & Evaluating Questions

Question 18.
A farmer wants to use a crop variety that can give a good yield. What should he do to select the variety to get the desired result?
Answer:
A good crop yield can be obtained by selecting varieties having useful characteristics such as disease resistance, response to fertilizers and high yields.

Question 19.
A bee – keeper tries to collect a good yield of honey from his apiaries. However, he is unable to collect adequate honey. Suggest him a way to produce more honey.
Answer:
The bee – keeper should maintain his apiaries in between the fields of flowering plants or pasturage. This will allow bees to collect plenty of nectar and pollen from the variety of flowers. Also, the taste of honey depends on the variety of flowers available to the bees.

Question 20.
A dairy farmer wants to maintain a good and clean shelter for his dairy animals. How can he do this so that his animals stay healthy and produce clean milk?
Answer:
The shelter for dairy animals should have following features.

1. It should be well – ventilated to allow fresh air to enter.
2. It should have leakage – proof roof to protect them from rain, heat and cold.
3. The floor of cattle shed should be sloping for easy cleaning and keeping it dry.

Short Answer Type Questions

Question 1.
Distinguish between a mullet and a prawn.
Answer:
Mullet is a type of fish while prawn is a crustacean. Both live in water and serve as a food supplements worldwide. Prawn belongs to the phylum arthropoda, whereas mullet belongs to the group of pisces. So one can use their characteristic features to distinguish between the two.

Question 2.
What are genetically modified (GM) crops?
Answer:
GM (Genetically Modified) crops are the crops in which a gene from some other organism, like another plant or a microorganism, is inserted to get desired characteristics such as disease resistance, response to fertilisers, product quality and high yields. For example, varieties of cotton, maize, papaya, soyabean, sugar beet, squash, etc., have been modified genetically.

Question 3.
Give the technical terms for milk -producing females and farm labour animals.
Answer:
Milk – producing females are called milch animals (dairy animals), while the ones used for farm labour are called draught animals.

Question 4.
Mention the preventive and control measures used before the grains are stored.
Answer:
Cleaning of the produce before storage, proper drying of the produce first in sunlight and then in shade, and fumigation using chemicals that can kill pests.

Question 5.
What is the effect of deficiency of nutrients?
Answer:
Deficiency of nutrients affects physiological processes in plants including reproduction, growth, susceptibility to diseases, yield, etc. General health of the plants depends on the nutrients.

Question 6.
In what way does manure help in soil fertility?
Answer:
Manure helps in enriching the soil with mainly organic matter and small quantities of nutrients. The bulk of organic matter in the form of manure helps in increasing water holding capacity in sandy soil. In clayey soil, the large quantities of organic matter help in drainage and avoiding waterlogging.

Question 7.
Give two advantages of using chemical fertilisers over manure.
Answer:
Two advantages of using chemical fertilisers over manure are as follows:

1. Chemical fertilisers are ‘nutrient specific’ and can provide specific elements like nitrogen, phosphorus or potassium to the soil in any desired quantity. Manure is, however, not nutrient specific.
2. Chemical fertilisers, being soluble in water, are readily absorbed by the crops. This is not so in the case of manures.

Question 8.
What is green revolution?
Answer:
Bumper production of cereals (grains) using high-yielding varieties (HYV), higher dose of fertiliser and better modes of irrigation is known as green revolution.

Question 9.
What are pesticides? Give four methods of pest control.
Answer:
Pesticides are the chemicals used to control weeds, insects, rodents, fungi and diseases of plants. They include weedicides, insecticides and fungicides. Some methods of pest control are:

1. Use of resistant varieties
2. Optimum time of sowing the seeds
3. Follow crop rotation and cropping pattern
4. Deep ploughing of the field in summers to destroy undesirable weeds and pathogens.

Question 10.
Define organic farming.
Answer:
It is the farming in which no chemical fertilisers, pesticides or herbicides are used. It uses all organic matter for the growth of plants like manure, neem leaves as pesticides during grain storage, etc.

Question 11:
What desirable traits are focused to develop hybrids by cross-breeding indigenous and exotic breeds of fowl?
Answer:
Desirable traits are focused to develop hybrids by cross-breeding indigenous and exotic breeds of fowl:

1. Number and quality of chicks
2. Dwarf broiler parent for commercial chick production as they require less space and food.
3. Summer adaptation capacity a tolerance to high temperature
4. Low maintenance requirements
5. Reduction in the size of the layer with ability to utilise more fibrous and cheaper diets which are formulated using agricultural by – products.

Question 12.
What decides the quality and quantity of honey production in an apiary?
Answer:
The quality and quantity of honey production in an apiary:

1. For quality of honey: The pasturage, i.e., the kind of flowers available to the bees for nectar and pollen collection will determine the taste of the honey.
2. For quantity of honey: Variety of bee used for the collection of honey. For example, A. mellifera is used to increase yield of honey.

Analysing & Evaluating Questions

Question 13.
What would happen if poultry birds are larger in size and have no summer adaptation capacity? In order to get small-sized poultry birds having summer adaptability, what method will be employed?
Answer:
The maintenance of optimum temperature is required for better egg production in poultry farming. The large size of birds with no adaptability to high temperature may cause decline in egg production. To obtain small-size birds with high – temperature adaptability during summer season, cross – breeding of poultry birds for desired characteristics can be done. Small size is also needed for better housing and less feed.

Question 14.
Figure below shows the two crop fields (plots A and B) that have been treated by manures and chemical fertilisers respectively, keeping other environmental factors same. Observe the graph and answer the following questions:
(a) Why does plot B show sudden increase and then gradual decrease in yield?
(b) Why is the highest peak in plot A graph slightly delayed?
(c) What is the reason for the different pattern of the two graphs?

Answer:
(a) The addition of chemical fertilisers initially leads to rise in crop yield because of the release of the NPK (nitrogen, phosphorus and potassium) and some other nutrients in high quantity. The gradual decline in yield, as shown in plot B, is due to the continuous use of these fertilisers, which cause killing of useful microbes in the soil and alter the chemical composition of soil.

(b) Manures supply nutrients to the soil slowly, as these contain organic matter in high amount. Therefore, manures enrich the soil with nutrients slowly and continuously for a long time. This is the reason that the highest peak in plot A is delayed but maintained for longer period.

(c) In case of plot A, it indicates that the use of manure remains beneficial for longer duration in terms of crop yield and remains high even when the quantity of manure is increased. In case of plot B, chemical fertilisers when used for longer period cause various problems. The loss of soil fertility occurs due to killing of useful microbes in the soil that reduces decomposition of organic matter.

Long Answer Type Questions

Question 1.
Name two fresh initiatives taken to save water and increase the water availability for agriculture.
Answer:
Two new irrigation systems have been developed to save water and increase the availability of water to the crops.
These are:
1. Drip irrigation system: Here, water is supplied to the roots of the plants directly in a drop wise manner. This prevents unnecessaiy wastage of water.

2. Sprinkler system: Here, water is sprinkled over the crops like it happens in rain. So, water is distributed uniformly and absorbed by the soil in a better way.

Question 2.
What are the factors for which variety improvement of crop is done?
Answer:
The factors for which variety improvement of crop is done are as follows:

1. Higher yield: To increase productivity of the crop per acre.
2. Improved quality: The quality of crop products varies from crop to crop, e.g., protein quality is important in pulses, oil quality in oilseeds, longer shelf life in fruits and vegetables.
3. Biotic and abiotic resistance: Biotic factors are the diseases, insects and nematodes while abiotic factors are drought, salinity, water logging, heat, cold and frost which affect the crop productivity. Varieties resistant to these factors can increase the crop production.
4. Change in maturity duration: Shorter maturity period of crop reduces the cost of crop production and makes the variety economical. Uniform maturity makes the harvesting process easy and reduces losses during harvesting.
5. Wider adaptability: It allows the crops to be grown under different climatic conditions in different areas.
6. Desirable agronomic characteristics: It increases productivity, (e) g., tallness and profuse branching are desirable characters for fodder crops; while dwarfness is desired in cereals, so that less nutrients are consumed by these crops.

Question 3.
Define manure. What are its three different types?
Answer:
Manure contains large quantities of organic matter and also supplies small quantities of nutrients to the soil. It is prepared by the decomposition of animal excreta and plant waste. It helps in enriching the soil with nutrients and organic matter and increasing soil fertility. On the basis of the kind of biological waste used to make manure, it can be classified into three types:

1. Compost
2. Vermicompost
3. Green manure

1. Compost: It can be farm waste material such as livestock excreta (cow dung, etc.), vegetable waste, animal refuse, domestic waste, sewage, straw, eradicated weeds, etc. These materials are decomposed in pits and this process of decomposition is called composting.

2. Vermicompost: The compost which is made by the decomposition of plant and animal refuse with the help of redworms is called vermicompost.

3. Green manure: Prior to the sowing of the crop seeds, some plants like sun hemp or guar are grown and then mulched by ploughing them into the soil. These green plants thus turn into green manure which helps in enriching the soil in nitrogen and phosphorus.

Question 4.
What are fertilisers? Excessive use of fertilisers is not advisable. Explain.
Answer:
Fertilisers are commercially produced plant nutrients. Fertilisers supply nitrogen, phosphorus and potassium to the crops. They are used to ensure good vegetative growth (leaves, branches and flowers), giving rise to healthy plants. Fertilisers are an important factor in the higher yields of high – cost farming.
Excessive use of fertilisers is not advisable as:

1. It leads to soil and water pollution.
2. It can destroy the fertility of soil.

As the soil is not replenished, microorganisms in the soil are harmed by fertilisers.

Question 5.
How does intercropping differ from mixed cropping?
Or
What are the different cropping systems?
Answer:
It includes different ways of growing crops so as to get the maximum benefit. These different ways include the following:
1. Mixed cropping: Mixed cropping is growing two or more crops simultaneously on the same piece of land, e.g., wheat + gram, or wheat + mustard, or groundnut + sunflower. This reduces disease risk and gives some insurance against failure of one of the crops.

2. Intercropping: It involves growing two or more crops simultaneously on the same field in a definite proportion or pattern. A few rows of one crop alternate with a few rows of the other crop, e.g., soyabean + maize, or finger millet (bajra) + cowpea (lobia). The crops are selected such that their nutrient requirements are different. This ensures maximum utilisation of the nutrients supplied and also prevents pests and diseases from spreading to all the plants belonging to one crop in a field. This way, both the crops can give better yield.

3. Crop rotation: The growing of different crops on a piece of land in a pre-planned succession is known as crop rotation. Depending upon the duration, crop rotation is done for different crop combinations. The availability of moisture and irrigation facilities decide the choice of the crop to be cultivated after one harvest. If crop rotation is done properly, two or three crops can be grown in a year with good harvest.

Question 6.
Explain the various methods of irrigation in India.
Answer:
Proper irrigation is very important for the success of crops. Different kinds of irrigation systems include wells, canals, rivers and tanks.

1. Wells: These are of two types, viz., dug wells and tube wells. In a dug well, water is collected from water bearing strata Tube wells can tap water from the deeper strat(a) From these wells, water is lifted by pumps for irrigation.
2. Canal system: Water from the main river or reservoir is carried by canal into the field which is divided into branch canals having further distributaries to irrigate the field.
3. River lift system: In areas where canal flow is insufficient or irregular due to inadequate reservoir release,
   the lift system is more rational. Water is directly drawn from the rivers for supplementing irrigation in areas close to rivers,
4. Tanks: These are small storage reservoirs which intercept and store the run-off of smaller catchment areas.

Question 7.
Describe the different types of fisheries.
Answer:
The different types of fisheries are marine fisheries, mariculture, inland fisheries, aquaculture and capture fishing.

1. Marine fisheries: They are caught using fishing nets. Large schools of fishes are located by satellites. Some are farmed in sea water.
2. Mariculture: They are cultured in seawater. This culture of fisheries is called mariculture.
3. Inland fisheries: The fisheries in fresh water resources like canals, ponds, reservoirs and rivers are called inland fisheries.
4. Aquaculture: Culture of fish done in different water bodies is called aquaculture.
5. Capture fishing: It is the method of obtaining fishes from natural resources, both marine and fresh water.

Question 8.
List six facilities that must be provided to cattle to ensure their good health and production of clean milk.
Answer:
Following facilities must be provided to cattle:

1. Regular brushing to remove dirt and loosen hair.
2. Well – ventilated roofed sheds for shelter that can protect them from rain, heat and cold.
3. The floor of the cattle shed needs to be sloping so as to keep them dry and to facilitate cleaning and spraying of disinfectants at regular intervals.
4. A balanced diet should be given which contains:
   • roughage which provides high amount of fibre, and
   • concentrate that provides high levels of proteins and other nutrients.
5. Certain food additives containing micronutrients that promote the health and milk output of dairy animals.
6. Vaccinations of farm animals, at proper time, against major viral and bacterial diseases.

Analysing & Evaluating Questions

Question 9.
Meena belongs to an agricultural family. She attended a seminar of agricultural practices organised by her school. By listening to the research work of scientists, she learned that spraying pesticides on crops is very harmful for the environment. Next day, she saw the stored tanks of pesticides at her home and told her parents not to use these in excessive quantity.
1. Why are pesticides used in crop fields?
2. What are the various types of pesticides used by the farmers?
3. How can Meena convince her parents to stop using pesticides in large quantities?
4. What alternatives could Meena suggest to her parents instead of using pesticides?
Answer:

1. The pesticides are used in fields to protect the plants from disease – causing organisms, i.e., bacteria, fungi, viruses, nematodes and mycoplasmas.
2. Depending on the type of organisms, they destroy, pesticides can be of the following types.
   • Herbicides (for weeds)
   • Insecticides (for insects)
   • Fungicides (for fungi)
   • Bactericides (for bacteria)
3. Meena can tell her parents that regular and excessive use of pesticides contaminates water and soil, causing pollution in the environment. The pesticides affect the quality of food and leave residues on food items which may affect the health of consumers.
4. She could suggest the use of biological control methods or use of disease-resistant varieties of crops.

Activity 1
Visit a weed – infested field in the month of July or August and make a list of the weeds and insect pests in the field.

Observations:

1. Do it yourself.
2. Weeds are unwanted plants in the cultivated field, e.g., Xanthium (chota dhatura), Parthenium (gajar ghas) and Cyperinus rotundus (motha). They compete for food, space and light.
3. Some insect pests of crop fields include aphids, blister beetles, common stalk borer, com borer, flour beetle, etc.

Activity 2
Visit a local poultry farm. Observe the types of breeds and note the type of ration, housing and lighting facilities given to them. Identify the layers and broilers.

Observations:

1. Do it yourself and note down:
   • types of breeds of poultry: Aseel, white Leghorn, Rhode Island Red.
   • types of ration, housing and lighting facilities given to them.
2. Identify the layers, for example, White leghorn, Rhode Island Red, and broilers, for example, Plymouth Rock or Aseel or any other.

Value Based Questions

Question 1.
A group of eco – club students made a compost pit in the school, they collected all the biodegradable waste from the school canteen and used it to prepare the compost.
1. Name two wastes that can be used for the compost and two wastes obtained from canteen which cannot be used for the compost making?
2. What is the other important component required for making the compost?
3. What values of eco – club students are reflected in this act?
Answer:
The compost:

1. Two wastes used for compost are vegetable peels and fruit peels. Two waste materials that cannot be used as compost are polythene bags and plastic items.
2. Bacteria and fungi present in soil are the other important component for making compost.
3. Eco – club students reflect the value of group work and responsible citizens.

Question 2.
Large number of Bhetki fish died and got crushed in the turbines of hydroelectric power stations while
they migrated from river to sea The environmentalist gave power plant the solution of this problem. Now all Bhetki fish is removed with the help of a special technique and hence do not enter the turbines to crush and die.
1. Suggest two different varieties of fish.
2. What value of environmentalist is reflected in the above case?
Answer:
The turbines to crush and die:

1. Two varieties of fish are bony and cartilaginous.
2. Environmentalist showed the value of concern and caring individuals.

JAC Class 9 Science Important Questions

JAC Board Class 9th Science Important Questions Chapter 14 Natural Resources

Multiple Choice Questions

Question 1.
Most of the water on the earth’s surface is found in
(a) lakes
(b) rivers
(c) oceans and seas
(d) underground
Answer:
(c) oceans and seas

Question 2.
Ozone hole was first observed over
(a) Antarctica
(b) Australia
(c) Arctic ocean
(d) America
Answer:
(a) Antarctica

Question 3.
Nitrogen fixing bacteria cannot fix N2 in the presence of
(a) CO₂
(b) N₂
(c) O₂
(d) light
Answer:
(c) O₂

Question 4.
Which of the following compounds is not degraded by any biological process?
(a) CFC_s
(b) CH₄
(c) Glucose
(d) Nitrites
Answer:
(a) CFC_s

Question 5.
Nitrogen – fixing bacteria are found in the roots of
(a) wheat
(b) maize
(c) pulses
(d) sugarcane
Answer:
(c) pulses

Question 6.
Venus and Mars have no life because
(a) they have no atmosphere
(b) their atmosphere has only oxygen
(c) their atmosphere has 95% – 97% carbon dioxide
(d) their atmosphere has 95% – 97% oxygen
Answer:
(c) their atmosphere has 95%-97% carbon dioxide

Question 7.
Which one of the following organisms are very sensitive to the levels of contaminants like sulphur dioxide in air?
(a) Bacteria
(b) Fungi
(c) Algae
(d) Lichens
Answer:
(d) Lichens

Question 8.
Burning of fossil fuels adds
(a) CO₂, SO₂, NO₂ gases in air
(b) C, SO₂, N₂ gases in air
(c) O₂, SO₃, NO₃ gases in air
(d) H₂O, CO₂ NO₂, gases in air
Answer:
(a) CO₂, SO₂, NO₂ gases in air

Question 9.
Nitrogen fixation can be done by
(a) Industries
(b) Rhizobium
(c) Lightening
(d) All of the above
Answer:
(d) All of the above

Question 10.
On moon the temperature ranges from – 190°C to 110°C. This is due to
(a) absence of water bodies
(b) presence of water bodies
(c) absence of biogeochemical cycles
(d) absence of atmosphere
Answer:
(d) absence of atmosphere

Question 11.
Depletion of ozone molecules in the stratosphere is due to
(a) chlorine compounds
(b) fluorine compounds
(c) halogen compounds
(d) none of these
Answer:
(c) halogen compounds

Question 12.
The life supporting zone of the earth is
(a) lithosphere
(b) hydrosphere
(c) atmosphere
(d) biosphere
Answer:
(d) biosphere

Question 13.
The organism which helps in the formation of soil is
(a) bacterium
(b) moss
(c) lichen
(d) both (b) and (c)
Answer:
(d) both (b) and (c)

Question 14.
The outermost crust of the earth is called
(a) atmosphere
(b) exosphere
(c) lithosphere
(d) hydrosphere
Answer:
(c) lithosphere

Analysing & Evaluating Questions

Question 15.
The atmosphere of the earth is heated by radiations which are mainly
(a) radiated by the sun
(b) re – radiated by land
(c) re – radiated by water
(d) re – radiated by land and water
Answer:
(c) re – radiated by water

Question 16.
The term “water pollution” can be defined in several ways. Which of the following statements does not give the correct definition?
(a) The addition of undersirable substances to water bodies
(b) The removal of desirable substances from water bodies
(c) A change in pressure of the water bodies
(d) A change in temperature of the water bodies
Answer:
(c) A change in pressure of the water bodies

Question 17.
When we breathe in air, nitrogen also goes inside along with oxygen. What is the fate of this nitrogen?
(a) It moves along with oxygen into the cells.
(b) It comes out with the CO₂ during exhalation.
(c) It is absorbed only by the nasal cells.
(d) Nitrogen concentration is already more in the cells so it is not at all absorbed.
Answer:
(b) It comes out with the CO₂ during exhalation.

Assertion Reason Questions

Directions: In the following questions, the Assertions and the Reasons have been put forward. Read the statements carefully and choose the correct alternative from the following:
(A) Both the assertion and the reason are correct and the reason is the correct explanation of the assertion.
(B) The assertion and the reason are correct but the reason is not the correct explanation of the assertion.
(C) The assertion is true but the reason is false.
(D) Both the statements are false.
1. Assertion: Plants cannot utilise
nitrogen directly from the atmosphere. Reason: Plants can only use nitrates and nitrites.
Answer:
(B) The assertion and the reason are correct but the reason is not the correct explanation of the assertion.

2. Assertion: Water vapour is not a greenhouse gas.
Reason: Water vapour does not contribute in global warming.
Answer:
(D) Both the statements are false.

3. Assertion: The moon has very cold and very hot temperature variations.
Reason: Moon does not possess the atmosphere.
Answer:
(A) Both the assertion and the reason are correct and the reason is the correct explanation of the assertion.

4. Assertion: It is easier to fly a kite near a sea shore.
Reason: There is a regular unidirectional wind from sea to land.
Answer:
(A) Both the assertion and the reason are correct and the reason is the correct explanation of the assertion.

5. Assertion: Legumes revive the soil fertility.
Reason: Microbes in the root nodules of legumes fix atmospheric nitrogen.
Answer:
(A) Both the assertion and the reason are correct and the reason is the correct explanation of the assertion.

Very Short Answer Type Questions

Question 1.
What is biosphere?
Answer:
The life – supporting zone of earth where atmosphere, hydrosphere and lithosphere interact and make life is known as biosphere.

Question 2.
What are the biotic and abiotic components of biosphere?
Answer:
Abiotic or physical components of biosphere consist of geographical conditions such as the temperature, rainfall, soil, seasons and the climate, while biotic components include animals, plants, fungi and bacteria.

Question 3.
What percentage of oxygen and nitrogen is present in the air?
Answer:
About 21% oxygen and 78% nitrogen is present in the air.

Question 4.
Name the air pollutants released by the industries.
Answer:
Industrial air pollutants are sulphur dioxide, carbon dioxide, oxides of nitrogen, hydrogen sulphide, fumes of acids, dust, particles of unbumt carbon, lead, asbestos and even cement.

Question 5.
What are the two factors that cause changes in our atmosphere?
Answer:
(a) Heating of air, and
(b) formation of water vapour.

Question 6.
State any two harmful effects of air pollution.
Answer:
Two harmful effects of air pollution are:
(a) Respiratory problems
(b) Global warming

Question 7.
What is soil?
Answer:
Soil is a mixture of small particles of rocks of different sizes, humus and microscopic life.

Question 8.
State the major source of minerals in the soil.
Answer:
The mineral nutrients present in a particular soil depend on the rocks it was formed from. This means that the major source of minerals in a soil is their parent rocks.

Question 9.
What is top soil?
Answer:
The topmost layer of the soil that contains humus and living organisms in addition to the soil particles is called top soil.

Question 10.
Name two chemicals that are depleting ozone layer.
Answer:
Chlorofluorocarbons (CFC_s) and halogenated ozone depleting substance (ODS).

Question 11.
Name two greenhouse gases.
Answer:
Methane and carbon dioxide.

Question 12.
Define nitrification.
Answer:
The biological conversion of ammonia into nitrites and then oxidation of nitrites to nitrates is called nitrification.

Question 13.
Name a nitrogen fixing bacterium.
Answer:
Rhizobium

Question 14.
Define humus.
Answer:
The fertile dark substance present in the topmost layer of the soil which contains dead remains of plants and animal wastes, like excreta, that adds nutrients to the soil is called humus.

Question 15.
What is denitrification?
Answer:
Conversion of nitrates into free nitrogen is called denitrification.

Question 16.
How does carbon exist in all life forms?
Answer:
Carbon is present in all life forms in the form of proteins, carbohydrates, fats, nucleic acids and vitamins.

Question 17.
Name two biologically important compounds that contain both oxygen and nitrogen.
Answer:
Proteins and nucleic acids (DNA and RNA).

Question 18.
Name the two gases given out by the burning of fossil fuels, which dissolve in rain to form acid rain.
Answer:
Sulphur dioxide (SO₂) and oxides of nitrogen.

Analysing & Evaluating Questions

Question 19.
A person is burning a huge amount of waste in an open are(a) Which gas is being utilised for burning process and which gas is released into the atmosphere?
Answer:
The process of burning or combustion utilises oxygen and produces carbon dioxide. Thus, oxygen gas is being utilised in the process, and carbon dioxide is released into the atmosphere.

Question 20.
A huge amount of plant and animal waste is being dumped in a lake. What will be the condition of lake after some time?
Answer:
When organic waste (plant and animal residue) is dumped into a water body, the biological oxygen demand of the water increases. This is because the decomposition of organic matter by microorganisms needs more oxygen in water. As a result, there arises oxygen deficiency in water, which leads to the death of other aquatic organisms.

Question 21.
Some industries release hot water or very cold water into water sources directly. Why should this be stopped?
Answer:
When excessive hot water or cold water is released into water bodies, it may affect some aquatic organisms. This is because these organisms live in a certain range of temperature and any change in this range (due to the release of excessive hot water or cold water) may cause threat to their survival.

Short Answer Type Questions

Question 1.
The atmosphere acts as a blanket. How?
Answer:
The blanket of atmosphere, which is covering the earth, keeps the average temperature of the earth fairly steady during the day and even during the course of the whole year. The atmosphere prevents the sudden increase in temperature during the daylight hours. And during the night, it slows down the escape of heat into outer space.

Question 2.
What are the consequences of global warming?
Answer:
(a) An increase in temperature of earth even by 1°C may lead to the melting of ice on the poles.
(b) The melting of ice will result in rise of sea level.
(c) Due to rise in sea level, many coastal cities will be flooded or submerged.
(d) Increase in temperature of earth results in change in weather and may cause excessive raining or drought or extreme hot or cold weather conditions.

Question 3.
Name the various organisms involved in nitrogen cycle.
Answer:
(a) Nitrogen fixing bacteria, e.g., Rhizobium, Azotobacter.
(b) Bacteria which convert complex nitrogenous organic compounds (proteins) into ammonia, e.g., Clostridium, Proteus.
(c) Nitrifying bacteria which convert ammonia into nitrates, e.g., Nitrosomonas and Nitrobacter.
(d) Denitrifying bacteria, e.g., Pseudomonas.

Question 4.
What does the presence of smog in an area indicate?
Answer:
The presence of smog in an area indicates the high percentage of smoke released in the air by combustion of fossil fuels in industries, thermal power plants or automobiles. It is an indicator of air pollution.

Question 5.
Write three ways to prevent soil pollution.
Answer:
(a) By judicious use of fertilizers and pesticides.
(b) By proper management of disposal of household waste.
(c) By practising intensive cropping and terrace farming.

Question 6.
How is greenhouse effect related to global warming?
Answer:
Higher concentration of greenhouse gases, such as carbon dioxide, methane, nitrous oxide, etc., in the atmosphere causes absorption of reflected heat and avoids their escape into the space. This phenomenon is called greenhouse effect. This leads to rise in the temperature of earth’s atmosphere throughout the world causing global warming. This global warming caused by greenhouse gases like CO₂, CH₄, etc., leads to the melting of glacier and polar ice. This would cause rise in the level of sea and other climate changes. Hence, we can say that global warming is a consequence of greenhouse effect.

Question 7.
What is air pollution? How is it caused? Write its two harmful effects.
Answer:
The contamination of air with unwanted gases, particles like dust, etc, which makes it unfit for inhalation is called air pollution.

• Causes:
  1. (a) Burning of fossil fuels releases SO₂, CO₂ and NO₂ gases.
  2. (b) Burning of fuels releases unbumt carbon particles and smoke.
  3. (c) Smoke from industries and vehicles.
• Harmful effects:
  1. (a) It causes respiratory problems.
  2. (b) It causes allergies, asthma, cancer and heart diseases.

Question 8.
What is acid rain? Write its harmful effects.
Answer:
The gases released due to combustion of fossil fuels are SO₂, NO₂ and CO₂. These gases remain suspended in the air. When it rains, the rainwater mixes with these gases to form sulphuric acid, nitrous acid, carbonic acid and comes down on the surface of the earth in the form of acid rain.

(a) It corrodes statues, monuments of marble, buildings, etc.
(b) It makes the soil acidic.
(c) It damages crops and plantations.

Question 9.
What is water pollution? Give its causes and harmful effects.
Answer:
When water is contaminated with unwanted substances and chemicals which make it unfit for use and cause diseases, it is called water pollution.

• Causes:
  1. (a) Sewage from towns, cities is dumped in the water.
  2. (b) Fertilizers, pesticides, insecticides get washed away into the waterbodies from farmlands.
  3. (c) Effluent from industries.
• Harmful effects:
  1. (a) Polluted water, when consumed, causes many diseases which are water – borne, like cholera, typhoid, etc.
  2. (b) Mercury in salts dumped by industries causes a brain disorder called Minamata disease.
  3. (c) Many life – forms which are susceptible to temperature changes die.

Question 10.
What is the difference between fog and smog? Give two harmful effects of smog.
Answer:
The water vapour present in air when condenses due to very low temperature is called fog. The smoke released in the air, due to burning of fuels, mixes with the fog and forms smog. Harmful effects of smog: It decreases the visibility and causes adverse effect on aeroplane landing, railways and road transport.

Question 11.
State in brief the role of photosynthesis and respiration in carbon-cycle in nature.
Answer:
Photosynthesis is performed by green plants in the presence of sunlight and it converts carbon dioxide into carbohydrates that are utilised by other living organisms through food chain. Oxygen is replenished in nature only through the process of photosynthesis. Oxygen enters the living world through the process of respiration, i.e., it oxidises the food material (glucose molecules) and produces energy and carbon dioxide.

Question 12.
Explain the importance of ozone to mankind.
Answer:
Ozone covers the earth’s atmosphere. It is present in stratosphere. It does not allow the harmful ultraviolet radiations coming from the sun to enter our earth. These ultraviolet radiations cause ionizing effect and can cause cancer and genetic disorder in any life form. The ozone is getting depleted at the South Pole near Antarctica The ozone depletion is due to the halogens like CFC_s (Chlorofluorocarbons) released in the air. Chlorine and fluorine react with the ozone and split it, thereby leading to the formation of a big hole called ozone hole.

Question 13.
How are winds caused and what decides the breeze to be gentle, strong wind or a terrible storm?
Answer:
Movement of air, terrible storm and rains, all these phenomena are the result of changes that take place in our atmosphere due to the heating of air and the formation of water vapour. Water vapour is formed due to the heating of water bodies and the activities of living organisms. The atmosphere can be heated from below by the radiation that is reflected back or re-radiated by the land or water bodies. On being heated, convection currents are set up in the air. This causes wind. The pressure gradient or the pressure difference determines the speed and intensity of wind. Larger the gradient, more is the wind speed.

Question 14.
Why is step farming common in hills?
Answer:
At hills, the rainwater flows with a very high speed which provides very less time to absorb rainwater into the soil. So, the fields contain wide steps which slow down the speed of fast flowing water. Therefore, farming fields get more chances to absorb rainwater, i.e, more water can seep into the soil for better farming. Besides this, step farming also reduces soil erosion.

Question 15.
Write the harmful effects of ozone layer depletion.
Answer:
Harmful effects of ozone layer depletion are as follows:

1. Due to depletion of ozone layer, more ultraviolet (UV) radiation will reach the earth. UV radiation causes skin cancer, damage to eyes and immune system.
2. UV radiation kills microorganisms, such as bacteria, even useful ones.
3. Ozone layer depletion may lead to variation in rainfall, ecological disturbances and dwindling of good food supply.

Analysing & Evaluating Questions

Question 16.
Rashmi visited her village during summer vacations and found that there is severe shortage of water for villagers and animals.
1. Give reason of water shortage in villages.
2. Why some areas of earth suffer from the problem of water scarcity?
3. What ways can Rashmi suggest to villagers to solve the problem of water shortage?
Answer:

1. The lack of public water supply in villages is the most prominent reason of water scarcity in villages. Villagers depend on natural sources of water which often dry up due to excessive heat during summer.
2. The water scarcity occurs due to uneven distribution of freshwater on the earth.
3. Rainwater harvesting is an effective way to solve the problem of water scarcity in villages as well as in cities. The rainwater can be stored in underground tanks, check dams and recharges the groundwater. This groundwater can be drawn for use at the time of shortage of water.

Question 17.
A farmer followed the practice of sowing cereal crops regularly in his field for several seasons. After sometime, he found decline in cereal production.
(a) What may be the reason for less cereal production in farmer’s field?
(b) How the farmer can increase production in his field?
Answer:
(a) Sowing a same crop regularly in the field for several seasons makes the soil deficient of certain essential nutrients.
Thus, the fertility of soil declines after sometime and the crop production becomes less.

(b) The farmer can grow leguminous crops after cereal crop. The nitrogen fixing bacteria that live in root nodules of legume crops help in replenishment of lost nitrogen in the soil by fixing atmospheric nitrogen.

Long Answer Type Questions

Question 1.
Describe the carbon – cycle in nature.

Answer:
All living things are made of carbon. Carbon is also a part of the oceans, air, and even rocks. Because the earth is a dynamic place, carbon does not stay still. Carbon – cycle is the series of processes by which carbon compounds are interconverted in the environment, involving the incorporation of carbon dioxide into living tissues by photosynthesis and its return to the atmosphere through respiration, the decay of dead organisms and the burning of fossil fuels.

If we see from the beginning, CO₂ in the atmosphere is taken by the plants which use it to make glucose, water and oxygen. Also, direct CO₂ from air forms carbonates in water which later turns into limestone. Then, these plants are eaten us and other animals, i.e., we indirectly use COus and other animals, i.e., we indirectly use CO₂ present in the plants. Also, we respire CO₂ back in atmosphere.

Then, organic compounds (plants / animals) form coal and petroleum, or we can say fossil fuel. And also, animal body present in the plants. Also, we respire CO₂ back in atmosphere. Then, organic compounds (plants / animals) form coal and petroleum, or we can say fossil fuel. And also, animal body forms inorganic carbonates or shells. But nowadays, we bum fossil fuels too much (and also cut the trees) which is harmful. As a result, we are not getting rid of excess CO₂ from our atmosphere.

Question 2.
Why is replenishment of forests necessary?
Answer:
Forests need to be replenished because of the following reasons:

1. Rainfall: During transpiration, trees give out enormous amount of water vapour. This water vapour helps in the formation of clouds. So, if trees are cut and not replenished, the rainfall in the area will reduce.
2. Natural rate of tree growth: Forests cannot be re – grown in a few days or months as trees take many years to grow fully. Thus, it becomes necessary to replenish the forests periodically.
3. Soil erosion: If a large number of trees are cut, the soil becomes naked. The topsoil, which is rich in organic matter will be washed away by water or carried away by wind. Trees help in binding the soil.
4. Carbon dioxide – oxygen balance: Forests have a very large number of trees which give out 02 and take in C02 by photosynthesis. In this way, they help in maintaining the carbon dioxide-oxygen balance in the atmosphere.
5. Timber and fuel: Forests are the best suppliers of timber for furniture and fuel. So, for their constant supply, forests need to be replenished.

Question 3.
Draw a labelled diagram to show nitrogen cyclele in nature.
Answer:
Nitrogen exists as free nitrogen in the atmosphere. In air, N₂ is about 78%. This free nitrogen is fixed into compounds of ammonia and nitrates. Most of the organisms cannot utilise molecular nitrogen. Fixation of Nitrogen:
Fixation of free nitrogen into compounds takes place by the following means:
1. Certain blue green algae and bacteria can fix atmospheric nitrogen.

2. Nitrogen – fixing bacteria found in the root nodules of legumes such as grams, beans, pulses, etc., fix atmospheric nitrogen into nitrogen containing compounds.

3. Lightning also helps in the formation of nitrogen containing compounds. Nitrogen containing fertilisers produced artificially in factories are the fixed form of nitrogen. Plants take up compounds containing nitrogen from the soil. From plants nitrogen passes into food web Decay of dead plants and animals and excreta like urine, faeces, cause return of nitrogen compounds to the soil. Denitrifying bacteria and fire cause liberation of free nitrogen in the atmosphere.

Importance of Nitrogen – cycle: Nitrogen is an important constituent of tissues, proteins, enzymes, nucleic acids and amino acids. Atmosphere contains about 78 per cent nitrogen but plants and animals cannot use nitrogen in this form. Plants take nitrogen in the form of nitrates, the usable form. From plants, nitrogen travels to animals through food. If nitrogen in the form of proteins, amino acids, enzymes, etc., remains locked up in the bodies of organisms, there will be shortage of usable form of nitrogen. Therefore, circulation of nitrogen in nature is very essential.

Question 4.
Explain the oxygen – cycle in nature.
Answer:
Oxygen is an important component of everyday life. We cannot survive without oxygen. It comprises about 21% of atmospheric air. It is a component of several biological molecules such as carbohydrates, proteins, nucleic acids and fats. Like carbon dioxide, oxygen too is cycled through the process of photosynthesis and respiration. Oxygen is also utilised during combustion or burning.

Oxygen – cycle: Oxygen from the atmosphere is used up in three processes, viz., combustion, respiration and in the formation of oxides of nitrogen.

1. Animals take in oxygen through the process of respiration. They release CO₂ into the atmosphere.
2. Carbon dioxide, released by animals, is used by plants in the process of photosynthesis.
3. Plants release oxygen into the atmosphere as a by-product of photosynthesis.
4. Fuels need oxygen for combustion, so they take oxygen and release CO₂ into the atmosphere as a by-product along with other gases.
5. CO₂ is released into the air in the process of decaying of dead animals and plants.
6. This CO₂ is taken up by plants for the process of photosynthesis and O₂ is released and this process continues.

Question 5.
Describe water – cycle.
Answer:
Water is one of the most important physical components which is essential for survival of life on the earth. The water from the water bodies on evaporation moves up. As the vapours rise up in the atmosphere they become cooler and condense to form clouds which fall down as rain. Rainwater then passes through rivers and gets collected again in the ocean. The circulation of water in this manner is known as water – cycle. The cycle is also performed by living beings like absorption and transpiration of water by plants and drinking by animals. Animals lose water during respiration and perspiration. They lose water through excretion also.

Analysing & Evaluating Questions

Question 6.
There are several sources like respiration by living organisms, combustion, burning of fossil fuels and forest fires which contribute immensely in the carbon dioxide levels of atmosphere. Despite all these activities, the air contains only a mere fraction of carbon dioxide in it. How does it happen?
Answer:
The level of carbon dioxide does not increase in atmosphere because atmospheric carbon dioxide is fixed continuously in different components by the following ways.

• The plants ‘fix’ or capture the carbon dioxide and convert it into glucose through the process known as photosynthesis.
• Animals get their carbon directly by eating plants or indirectly by eating herbivores. Many marine animals use carbonates dissolved in seawater to make their shells.
• The fossil fuels are also the storehouse of carbon. These are the deposits of organic materials formed from decayed plants and animals deep inside the earth. By exposure to heat and pressure in the earth’s crust over hundreds of millions of years, these dead decaying organisms changed into fossil fuels, such as coal and petroleum.
• The water rich in carbon dioxide accumulates at the bottom of water bodies and forms limestone or carbonated rocks. It involves a series of chemical reactions that remove carbon dioxide from the atmosphere and deposits it in the form of rocks.

Activity 1

• Take the following:
  1. (i) a beaker full of water
  2. (ii) a beaker full of soil/sand and
  3. (iii) a closed bottle containing a thermometer.
• Measure the temperature of all these in shade.
• Now, keep them in bright sunlight for three hours and measure the temperature of all three vessels.

Observations:

• The temperature of air is less in shade than the temperature of sand or water.
• The temperature of water does not rise quickly whereas sand gets heated up easily.
• The temperature of air in the closed bottle is more than the temperature of air in open. This is because heat received from the sun has no outlet to reflect back the heat radiation. The glass does not allow to escape reflected heat radiations to go out.

Activity 2

• Place a candle in a beaker or wide mouthed bottle and light it. Light an incense stick and take it to the mouth of the beaker as shown in the figure.
• Now, keep the incense stick near the edge of the mouth, a little above the candle, and in other regions. Record your observations.

Observations:

• When the incense stick is kept near the edge of the mouth, the smoke flows inwards towards the flame. The hot air around the candle rises up and cold air from the surroundings rushes in to fill the space. This air rushing inside the beaker brings smoke towards the flame.
• The smoke rises up along with hot rising air when incense stick is kept a little above the candle.
• In other places, the smoke from the incense stick rises up and then diffuses in the air.

Activity 3
Take two identical trays and fill them with soil. Plant mustard or green gram or paddy in one of the trays and water both the trays regularly for a few days, till the first tray is covered with plant growth. Now, tilt both the trays and fix them in that position. Make sure that both the trays are tilted at the same angle. Pour equal amount of water gently on both trays such that the water flows out of the trays.

1. Study the amount of soil that is carried out of the trays by the water.
2. Now pour equal amounts of water on both the trays from a height. Pour three or four times the amount that you poured earlier.
3. Study the amount of soil that is carried out of the trays now. Record your observations.

Observations:

• The tray without vegetation loses more soil and holds more water than the tray that has plants.
• On pouring water on both the trays from a height, more soil flows along with water in the tray without vegetation.
• Less amount of soil is washed out earlier than that washed out when water was poured from a height.

Value Based Questions

Question 1.
Sudha saw a child sleeping in a car parked, with closed doors and glasses rolled up, in an open area on a sunny day near the market. She immediately raised an alarm and with the help of police she got the window rolled down.
1. Why was it not safe to keep the doors with window glasses rolled up for a child inside the car?
2. Name two gases that can lead to the above effect.
3. What value of Sudha is reflected in the above act?
Answer:

1. It was not safe for the child in the car with locked doors and windows rolled up because the sunlight would result in the greenhouse effect in the car. This would increase the temperature in the car and also result in the increase in CO₂ level which would lead to suffocation.
2. Carbon dioxide gas and methane gas can lead to greenhouse effect.
3. Sudha reflects the value of an aware citizen and responsible behaviour.

Question 2.
After doing a project on “save water”, Sumit realised the problem of shortage of drinking water on the earth. Sumit started checking the misuse of water in his vicinity.
1. What is the percentage of drinking water available on the earth?
2. Give any two practices that one should follow to save water.
3. What value of Sumit is reflected in this act?
Answer:

1. 1% of drinking water is available on the earth.
2. To save water:
   • Do not use shower to take bath every day, instead use a bucket of water.
   • Mop the floor instead of washing.
   • Sumit showed the value of responsible behaviour and participating citizen.
3. A project on “save water”, Sumit realised the problem of shortage of drinking water on the earth.

JAC Class 9 Science Important Questions

Jharkhand Board JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.1 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Exercise 13.1

Unless stated otherwise, take π = \(\frac{22}{7}\)

Question 1.
Two cubes each of volume 64 cm³ are joined end to end. Find the surface area of the resulting cuboid.
Solution:
Volume of the cube = a³ = 64 cm³
Side of the cube \(\sqrt[3]{64}\) = a = 4 cm.

Surface area of the cuboid = 2(lb + bh + lh)
= 2[(8 × 4) + (4 × 4) + (8 × 4)]
= 2[32 + 16 + 32]
= 2 × 80
= 160 cm².

Question 2.
A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.
Solution:

π = \(\frac{22}{7}\), radius of the hemisphere = 7 cm,
height of the hemisphere = 7 cm,
height of the cylinder, h = 13 – 7 = 6 cm.
Inner area of the vessel = Inner area of the hemisphere vessel + Inner area of the cylinder
= 2πr² + 2πrh
= 2 × \(\frac{22}{7}\) × 7 × 7 + 2 × \(\frac{22}{7}\) × 7 × 6
= 2 × 22 × 7+2 × 22 × 6
= 2 × 22(7 + 6)
= 44 × 13 = 572 cm².

Question 3.
A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.
Solution:

π = \(\frac{22}{7}\), r = 3.5, l = ?
Total surface area of the toy = C.S.A. of the cone + C.S.A. of the hemisphere
= πrl + 2πr²
lv = r² + h²
= (3.5)² + (12)²
= 12.25 + 144
= 156.25
l = \(\sqrt{156.25}\)
= 12.5 cm.

h = Ax – Ox
= 15.5 – 3.5
= 12 cm.

Surface area of the toy = πrl + 2πr²
= \(\frac{22}{7}\) × 3.5 × 12.5 + 2 × \(\frac{22}{7}\) × 3.5 × 3.5
= \(\frac{22}{7}\) × 3.5[12.5 + 2 × 3.5]
= 22 × 0.5[12.5 + 7]
= 11[12.5 + 7]
= 11 × 19.5 = 214.5 cm².

Question 4.
A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.
Solution:

Greatest diameter = 7 cm.
Surface area of the block = T.S.A. of the cube – Base area of the hemisphere + C.S.A. of the hemisphere
= 6 × 72 – πr² + 2πr²
= 6 × 49 + πr²
= 6 × 49 + \(\frac{22}{7}\) × 3.5 × 3.5
= 294 + 11 × 3.5
= 294 + 38.5
= 332.5 cm².

Question 5.
A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.
Solution:

Surface area of the remaining solid = Surface area of the cube + Surface area of the hemisphere
= 6l² + 2πr²
= 6l² + 2π\(\left(\frac{l}{2}\right)^2\)
= 6l² + 2π\(\frac{l^2}{4}=\frac{l^2}{4}\)(24 + 2π) sq. units.

Question 6.
A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.
Solution:

Height of the cylindrical portion = 14 – 2.5 – 2.5 = 9m = h
r = 2.5 m.
Surface area of the capsule = Surface area of the cylindrical portion + Areas of the hemispherical regions
= 2πrh + 2πr² + 2πr²
= 2πr(h + r + r)
= 2 × \(\frac{22}{7}\) × 2.5 (9 + 2.5 + 2.5)
= 2 × \(\frac{22}{7}\) × 2.5 (9 + 5)
=2× \(\frac{22}{7}\) × 2.5 × 14
= 44 × 5
= 220 mm².

Question 7.
A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of Rs. 500 per m². (Note that the base of the tent will not be covered with canvas.)
Solution:
r = \(\frac{4}{2}\) = 2m, h = 2.1, l = 2.8

Area of the canvas used = C.S.A. of the cylindrical portion + C.S.A. of the conical region
= 2πrh + πrl
= πr(2h + l)
= \(\frac{22}{7}\) × 2(2 × 2.1 + 2.8)
= \(\frac{22}{7}\) × 2(4.2 + 2.8)
= \(\frac{22}{7}\) × 2 × 7
= 44 m².
Cost of the canvas = Area × Rate
= 44 × 500
= Rs. 22000.

Question 8.
From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm².
Solution:
Height of conical part = Height of cylindrical part h = 2.4 cm.
Diameter of cylindrical part = 1.4 cm, so, the radius of cylindrical part r = 0.7 cm

Slant height of cylindrical part l = \(\sqrt{r^2+h^2}\)
= \(\sqrt{(0.7)^2+(2.4)^2}\)
= \(\sqrt{0.49+5.76}\)
= \(\sqrt{6.25}\)
= 2.5
The total surface area of the remaining solid = CSA of cylindrical part + CSA of conical part + Area of base of cylinder
= 2πrh + πrl + πr²
= 2 × \(\frac{22}{7}\) × 0.7 × 2.4 + \(\frac{22}{7}\) × 0.7 × 2.5 + \(\frac{22}{7}\) × 0.7 × 0.7
= 4.4 × 2.4 + 2.2 × 2.5 + 2.2 × 0.7
= 10.56 + 5.50 + 1.56
= 17.60 cm².

Question 9.
A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in the figure. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the total surface area of the article.
Solution:

Total surface area of the article = C.S.A. of the cylinder + Surface area of the hemisphere at the top + Surface area of the hemisphere at the bottom
= 2πrh + 2πr² + 2πr²
= 2πr(h + r + r)
= 2 × \(\frac{22}{7}\) × 3.5 (10 + 3.5 + 3.5)
= 2 × 22 × 0.5 × 17
= 22 × l × 17
= 374 cm².

Jharkhand Board JAC Class 10 Maths Solutions Chapter 15 Probability Ex 15.1 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 15 Probability Exercise 15.1

Question 1.
Complete the following statements:
i) Probability of an event E+ Probability of the event ‘not E’ =
ii) The probability of an event that cannot happen is …………….. Such an event is called ……………
iii) The probability of an event that is certain to happen is …………. Such an event is called …………….
iv) The sum of the probabilities of all the elementary events of an experiment is ……………
v) The probability of an event is greater than or equal to ………………. and less than or equal to ………………
Solution:
i) 1,
ii) 0, impossible event,
iii) 1, sure or certain,
iv) 1,
v) 0, 1.

Question 2.
Which of the following experiments have equally likely outcomes? Explain.
i) A driver attempts to start a car. The car starts or does not start.
ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.
iii) A trial made to answer a true-false question. The answer is right or wrong.
iv) A baby is born. It is a boy or a girl.
Solution:
(iii) and (iv) have equally likely outcomes. Only two possibilities are there in each of these cases.

Question 3.
Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game?
Solution:
When a coin is tossed, head or tail are equally likely possible events. So the result of an individual coin toss is unpredictable.

Question 4.
Which of the following cannot be the probability of an event:
A) \(\frac{2}{3}\)
B) -1.5
C) 15%
D) 0.7?
Solution:
(B) Because, probability of an event cannot be negative.

Question 5.
If P(E)= 0.05, what is the probability of ‘not E’?
Solution:
P(E) = 0.05.
[P(\(\overline{\mathrm{E}}\)) = Probability of not an event]
We know that P(E) + P(\(\overline{\mathrm{E}}\)) = 1
P(E) = 1 – 0.05 = 0.95.

Question 6.
A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out
i) an orange flavoured candy?
ii) a lemon flavoured candy?
Solution:
i) P(orange flavoured candy) = 0. Impossible event.
ii) P(Lemon flavoured candy) = 1. Sure event.

Question 7.
It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?
Solution:
Let E = Event of 2 students not having the same birthday
∴ P(E) = 0.992
∴ P(E) + P(\(\overline{\mathrm{E}}\)) = 1
∴ 0.992 + P(\(\overline{\mathrm{E}}\)) = 1
⇒ P(\(\overline{\mathrm{E}}\)) = 1 – 0.992
= 0.008
So, the probability of two students having the same birthday is 0.008.

Question 8.
A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is (i) red? (ii) not red?
Solution:
Total number of balls, n(S) = 3 + 5 = 8.
Let E = Event of drawing 1 red ball
∴ n(E) = 3
(i) Probability of drawing a red ball = \(\frac{n(E)}{n(S)}=\frac{3}{8}\)
(ii) Probability of not drawing a red ball = 1 – P(Drawing a red ball)
= \(1-\frac{3}{8}=\frac{5}{8}\)

Question 9.
A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be
(i) red?
(ii) white?
(iii) not green?
Solution:
Total possible outcomes = 5 + 8 + 4 = 17.
P(R) = \(\frac{5}{17}\), P(W) = \(\frac{8}{17}\)
P(Not green) = P(R + W) = \(\frac{5}{17}+\frac{8}{17}=\frac{13}{17}\)

Question 10.
A piggy bank contains hundred 50 p coins, fifty Re. 1 coins, twenty Rs. 2 coins and ten Rs. 5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin (i) will be a 50 p coin? (ii) will not be a Rs. 5 coin?
Solution:
Total possible outcomes: 100 + 50 + 20 + 10 = 180.
P(50 paise coin) = \(\frac{100}{180}=\frac{5}{9}\)
P(Not Rs. 5 coin) = \(\frac{100}{180}+\frac{50}{180}+\frac{20}{180}\)
= \(\frac{170}{180}=\frac{17}{18}\).

Question 11.
Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish. What is the probability that the fish taken out is a male fish?
Solution:
Total number of fish in an aquarium = 5 male fish + 8 female fish = 13 fish
∴ Probability of taking out a male fish = \(\frac{\text { Number of male fish }}{\text { Total number of fish }}=\frac{5}{13}\)

Question 12.
A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8, and these are equally likely outcomes. What is the probability that it will point at
i) 8?
ii) an odd number?
iii) a number greater than 2?
iv) a number less than 9?

Solution:
Total possible outcomes = 8.
i) P(8) = \(\frac{1}{8}\)
ii) P(odd number) = \(\frac{4}{8}=\frac{1}{2}\)
iii) P(no. > 2) = \(\frac{6}{8}=\frac{3}{4}\)
iv) P(no. < 9) = \(\frac{8}{8}\) = 1

Question 13.
A die is thrown once. Find the probability of getting
(i) a prime number, (ii) a number lying between 2 and 6, (iii) an odd number.
Solution:
Total possible outcomes 1, 2, 3, 4, 5, 6 = 6
P(Prime number) (2, 3, 5) = \(\frac{3}{6}=\frac{1}{2}\)
P(Number between 2 and 6) = \(\frac{3}{6}=\frac{1}{2}\)
P(Odd number) = \(\frac{3}{6}=\frac{1}{2}\)

Question 14.
One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting
i) a king of red colour
ii) a face card
iii) a red face card
iv) the jack of hearts
v) a spade
vi) the queen of diamonds.
Solution:
Total number of cards in one deck, n(S) = 52.
i) Let E₁ = Event of getting a king of red colour
∴ n(E₁) = 2
(∵ In a deck of cards, 26 cards are red and 26 cards are black. There are four kings in a deck in which two are red and two are black)
Probability of getting a king of red colour = \(\frac{n\left(E_1\right)}{n(S)}=\frac{2}{52}=\frac{1}{26}\)

ii) Let E₂ = Event of getting a face card
∴ n(E₂) = 12
(∵ In a deck of cards, there are 12 face cards – 4 king, 4 jack, 4 queen)
Probability of getting a face card = \(\frac{n\left(E_2\right)}{n(S)}=\frac{12}{52}=\frac{3}{13}\)

iii) Let E₃ = Event of getting a red face card
∴ n(E₃) = 6
(∵ In a deck of cards, there are 12 face cards – 6 red, 6 black)
Probability of getting a red face card = \(\frac{n\left(E_3\right)}{n(S)}=\frac{6}{52}=\frac{3}{26}\)

iv) Let E₄ = Event of getting a jack of hearts
∴ n(E₄) = 1
(∵ There are four jacks in a deck- 1 heart, 1 club, 1 spade, 1 diamond)
Probability of getting a jack of hearts = \(\frac{n\left(E_4\right)}{n(S)}=\frac{1}{52}\)

v) Let E₅ = Event of getting a spade
∴ n(E₅) = 13
(∵ In a deck of cards, there are 13 spades, 13 clubs, 13 hearts, 13 diamonds)
Probability of getting a spade = \(\frac{n\left(E_5\right)}{n(S)}=\frac{13}{52}=\frac{1}{4}\)

vi) Let E₆ = Event of getting a queen of diamond
∴ n(E₆) = 1
(∵ In 13 diamond cards, there is only one queen)
Probability of getting a queen of diamond = \(\frac{n\left(E_6\right)}{n(S)}=\frac{1}{52}\).

Question 15.
Five cards – the ten, jack, queen, king and ace of diamonds, are well shuffled with their face downwards. One card is then picked up at random.
i) What is the probability that the card is the queen?
ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen?
Solution:
i) Total possible outcomes = 5
P(Queen card) = \(\frac{1}{5}\)
ii) If the queen card is put aside, total possible outcomes = 4.
iii) P(ace) = \(\frac{1}{4}\)
iv) P(queen) = \(\frac{0}{2}\) = 0.

Question 16.
12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.
Solution:
Total possible outcomes = 132 + 12 = 144
No. of good pens = 132.
P(good pen) = \(\frac{132}{144}=\frac{11}{12}\)

Question 17.
i) A lot of 20 bulbs contains 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?
ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?
Solution:
i) Total possible outcomes = 20
P(Defective bulbs) = \(\frac{4}{20}=\frac{1}{5}\)

ii) Total possible outcomes = 20 – 1 = 19
No. of defective bulbs = 4
No. of good bulbs = 15
P(Not defective bulb) = \(\frac{15}{19}\)

Question 18.
A box contains 90 dises which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears (i) a two-digit number, (ii) a perfect square number, (iii) a number divisible by 5.
Solution:
S = {1, 2, 3, 4, 5, … 90}
∴ Total possible outcomes n(S) = 90
i) Number of 2-digit numbers = 90 – 9 = 81
P(a 2-digit number) = \(\frac{81}{90}=\frac{9}{10}\)

ii) Event A = {A perfect square number}
A = (1, 4, 9, 16, 25, 36, 49, 64, 81) = n(A) = 9
Probability of the event P(A) = \(\frac{n(A)}{n(S)}=\frac{9}{90}\)

iii) A number divisible by 5, i.e., multiples of 5.
5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90 = 18.
P(a no. divisible by 5) = \(\frac{18}{90}=\frac{1}{5}\).

Question 19.
A child has a die whose six faces show the letters as given below:

The die is thrown once. What is the probability of getting (i) A? (ii) D?
Solution:
Total possible outcomes = 6
No. of A’s = 2
No. of D’s = 1
P(A) = \(\frac{2}{6}=\frac{1}{3}\)
P(D) = \(\frac{1}{6}\)

Question 20.
A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that
i) she will buy it?
ii) she will not buy it?
Solution:
Total number of possible outcomes = 144
No. of good pens = 144 – 20 = 124.
P(of buying) = \(\frac{124}{144}=\frac{31}{36}\)
P(of not buying) = \(\frac{20}{144}=\frac{5}{36}\)

Question 21.
(i) Two dice, one blue and one grey, are thrown at the same time. Write down all possible outcomes. What is the probability that the sum of the two numbers appearing on the top of the dice is (i) 8, (ii) 13, (iii) less than or equal to 12? Complete the following table:

Solution:


1) Sum of 2 dice = 2 (1 + 1)
P(Sum 2) = \(\frac{1}{36}\)

2) Sum of 2 dice = 3 (1 + 2) (2 + 1)
P(Sum 3) = \(\frac{2}{36}\)

3) Sum 4 (1, 3) (2, 2) (3, 1)
P(Sum 4) = \(\frac{3}{36}\)

4) Sum 5 (1, 4) (2, 3) (3, 2) (4, 1)
P(Sum 5) = \(\frac{4}{36}\)

5) Sum 6 (1, 5) (2, 4) (3, 3) (4, 2) (5, 1)
P(Sum 6) = \(\frac{5}{36}\)

6) Sum 7 (1,6) (2, 5) (3, 4) (4, 3) (5, 2) (6, 1)
P(Sum 7) = \(\frac{6}{36}\)

7) Sum 8 (2, 6) (3, 5) (4, 4) (5, 3) (6, 2)
P(Sum 8) = \(\frac{5}{36}\)

8) Sum 9 (3, 6) (4, 5) (5, 4) (6, 3)
P(Sum 9) = \(\frac{4}{36}\)

9) Sum 10 (4, 6) (5, 5) (6, 4)
P(Sum 10) = \(\frac{3}{36}\)

10) Sum 11 (5, 6) (6,5)
P(Sum 11) = \(\frac{2}{36}\)

11) Sum 12 (6,6)
P(Sum 12) = \(\frac{1}{36}\)

ii) A student argues that ‘there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12.
Therefore, each of them has a probability \(\frac{1}{11}\). Do you agree with this argument? Justify your answer.
Solution:
Total possible outcomes of throwing the two dice, S =
{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6,3), (6, 4), (6, 5), (6, 6)}
∴ n(S) = 36

a) Let E₁ = Sum of two dice is 3 = {(1, 2), (2, 1)}
n(E₁) = 2
∴ P(E₁) = \(\frac{n\left(E_1\right)}{n(S)}=\frac{2}{36}\)

b) Let E₂ = Sum of two dice is 4 = {(1, 3), (2, 2), (3, 1)}
n(E₂) = 3
∴ P(E₂) = \(\frac{n\left(E_2\right)}{n(S)}=\frac{3}{36}\)

c) Let E₃ = Sum of two dice is 5 = {(1, 4), (2, 3), (3,2), (4, 1)}
n(E₃) = 4
∴ P(E₃) = \(\frac{n\left(E_3\right)}{n(S)}=\frac{4}{36}\)

d) Let E₄ = Sum of two dice is 6 = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)}
n(E₄) = 5
∴ P(E₄) = \(\frac{5}{36}\)

e) Let E₅ = Sum of two dice is 7 = {(1, 6), (2, 5), (3, 4), (4,3), (5, 2), (6, 1)}
n(E₅) = 6
∴ P(E₅) = \(\frac{n\left(E_5\right)}{n(S)}=\frac{6}{36}\)

f) Let E₆ = Sum of two dice is 8 = {(2, 6), (3, 5), (4, 4), (5, 3), (6,2)}
n(E₆) = 5
∴ P(E₆) = \(\frac{n\left(E_6\right)}{n(S)}=\frac{5}{36}\)

g) Let E₇ = Sum of two dice is 9 = {(3, 6), (4, 5), (5, 4), (6,3)}
n(E₇) = 4
∴ P(E₇) = \(\frac{n\left(E_7\right)}{n(S)}=\frac{4}{36}\)

h) Let E₈ = Sum of two dice is 10 = {(4, 6), (5, 5), (6, 4)}
n(E₈) = 3
∴ P(E₈) = \(\frac{n\left(E_8\right)}{n(S)}=\frac{3}{36}\)

i) Let E₉ = Sum of two dice is 11 = {(6,5), (5, 6)}
n(E₉) = 2
∴ P(E₉) = \(\frac{n\left(E_9\right)}{n(S)}=\frac{2}{36}\)

j) Let E₁₀ = Sum of two dice is 12 = {(6, 6)}
n(E₁₀) = 1
∴ P(E₁₀) = \(\frac{n\left(E_{10}\right)}{n(S)}=\frac{1}{36}\)
No. The eleven events are not equally likely.

Question 22.
A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result i.e., three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.
Solution:
Total possible outcomes (H + T)³
= H³ + 3H²T + 3HT² + T³
HHH HHT HTH THH HTT THT TTH TTT = 8.
Possible losses HHT HTH THH HTT THT TTH = 6
P(of losses) = \(\frac{6}{8}=\frac{3}{4}\)

Question 23.
A die is thrown twice. What is the probability that
(i) 5 will not come up either time? (ii) 5 will come up at least once?
[Hint: Throwing a die twice and throwing two dice simultaneously are treated as the same experiment].
Solution:
i) Total number of cases, n(S) = 6² = 36
Let \(\overline{\mathrm{E}}\) = Event that 5 will come up either time
= {(1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 6)}
⇒ n(\(\overline{\mathrm{E}}\)) = 11
and E = Event that 5 will not come up either time
n(E) = 36 – 11 = 25
∴ Probability that 5 will not come up either time = \(1-\frac{11}{36}=\frac{36-11}{36}\)
= \(\frac{25}{36}\)
ii) Probability that 5 will come up at least once = 12 – 1 = \(\frac{1}{2}\).

Question 24.
Which of the following arguments are correct and which are not correct? Give reasons for your answer.
i) If two coins are tossed simultaneously there are three possible outcomes – two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is \(\frac{1}{3}\).
ii) If a die is thrown, there are two possible outcomes – an odd number or an even number. Therefore, the probability of getting an odd number is \(\frac{1}{2}\).
Solution:
i) Incorrect: We can classify the outcomes like this but they are not then ‘equally likely’. The reason is that ‘one of each’ can result in two ways – from head on first coin and tail on the second coin or from tail on the first coin and head on the second coin. This makes it twice as likely as 2 heads or 2 tails.

ii) Correct. The two outcomes considered in the question are equally likely. Both have the same probability. i.e., \(\frac{1}{2}\).

Jharkhand Board JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.3 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 14 Statistics Exercise 14.3

Question 1.
The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the media, mean and mode of the data and compare them.

Solution:


Mean is 137 units.
Median is 137 units.
Mode is 135.76 units.
The three measures are approximately the same.

Question 2.
If the median of the distribution given below is 28.5, find the values of x and y.

Solution:

n = 60, 45 + x + y = 60
x + y = 60 – 45
x + y = 15
The median is 28.5. It lies in the class interval 20 – 30.
∴ l = 20, f = 20, cf = 5 + x, h = 10

Question 3.
A life insurance agent found the following data for distribution of ages of 100 policyholders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 years.

Solution:

l = 35, \(\frac{n}{2}\) = 50, cf = 45, f = 33, h = 5

Question 4.
The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table:

Find the median length of the leaves.
(Hint: The data need to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 – 126.5, 126.5 – 135.5, ….., 171.5 – 180.5).
Solution:
The data have to be converted to continuous classes for finding the median since the formula. assumes continuous classes.

n = 40, \(\frac{n}{2}\) = 20
The median lies in the class interval 144.5 – 153.5.
l = 144.5, \(\frac{\mathrm{n}}{2}=\frac{40}{2}\) = 20, cf = 17, f = 12, h = 9.
Median = l + \(\left[\frac{\frac{n}{2}-c f}{f}\right]\) × h
= 144.5 + \(\left[\frac{20-17}{12} \times 9\right]\)
= 144.5 + \(\left[\frac{27}{12}\right]\)
= 144.5 + 2.25
= 146.75 mm.

Question 5.
The following table gives the distribution of the lifetime of 400 neon lamps:

Find the median lifetime of a lamp.
Solution:

The median lies in the class interval 3000 – 3500.
\(\frac{\mathrm{n}}{2}=\frac{400}{2}=200\)
l = 3000, \(\frac{n}{2}\) = 200, cf = 130, f = 86, h = 500.
Median = l + \(\left[\frac{\frac{\mathrm{n}}{2}-\mathrm{cf}}{\mathrm{f}}\right]\) × h
= 3000 + \(\left[\frac{200-130}{86}\right]\) × 500
= 3000 + \(\left[\frac{70}{86}\right]\) × 500
= 3000 + \(\frac{35000}{86}\)
= 3000 + 406.976
Median life of a lamp is 3406.98 hours.

Question 6.
100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabet in the surnames was obtained as follows:

Determine the median number of letters in the surnames. Find the mean number of letters in the surnames. Also, find the modal size of the surnames.
Solution:

Hence the modal size of the surnames is 7.88.

Question 7.
The distribution below gives the weights of 30 students of a class. Find the median weight of the students.

Solution:

\(\frac{n}{2}\) = 15
The median lies in the class 55 – 60.
l = 55, \(\frac{n}{2}\) = 15, c.f. = 13, f = 6, h = 5.
Median = l + \(\frac{1}{2}\) × h
= 55 + \(\left[\frac{15-13}{6}\right]\) × 5
= 55 + \(\frac{2}{6}\) × 5
= 55 + \(\frac{5}{3}\)
= 55 + 1.666 = 56.666
∴ Median = 56.67 kg.
Hence, the median weight of the students is 56.67 kg.

JAC Jharkhand Board Class 10th Sanskrit Solutions शेमुषी भाग 2

• Chapter 1 शुचिपर्यावरणम्
• Chapter 2 बुद्धिर्बलवती सदा
• Chapter 3 व्यायामः सर्वदा पथ्यः
• Chapter 4 शिशुलालनम्
• Chapter 5 जननी तुल्यवत्सला
• Chapter 6 सुभाषितानि
• Chapter 7 सौहार्द प्रकृतेः शोभा
• Chapter 8 विचित्रः साक्षी
• Chapter 9 सूक्तयः
• Chapter 10 भूकम्पविभीषिका
• Chapter 11 प्राणेभ्योऽपि प्रियः सुहृद्
• Chapter 12 अन्योक्तयः

JAC Class 10 Sanskrit अपठित-अवबोधनम्

• अपठित गद्यांशः

JAC Class 10 Sanskrit अनुप्रयुक्त-व्याकरणम्

• सन्धिकार्यम्
• समास:
• कारकम्
• प्रत्ययज्ञानम्
• अव्ययपदानि
• उपसर्ग प्रकरणम्
• शब्दरूप प्रकरणम्
• धातुरूप-प्रकरणम्
• संख्याज्ञानम्

JAC Class 10 Sanskrit रचनात्मक कार्यम्

• पत्र-लेखनम्
• चित्राधारित वर्णनम्
• संकेत आधारित वार्तालाप लेखनम्
• अनुवाद कार्यम्
• घटनाक्रमस्य संयोजनम्
• वाक्य निर्माणम्

Jharkhand Board JAC Class 10 Maths Important Questions Chapter 13 पृष्ठीय क्षेत्रफल एवं आयतन Important Questions and Answers.

JAC Board Class 10th Maths Important Questions Chapter 13 पृष्ठीय क्षेत्रफल एवं आयतन

लघुत्तरात्मक / निबन्धात्मक प्रश्न :

प्रश्न 1.
एक अर्धगोलाकार टैंक पानी से भरा है, जिसका पानी एक पाइप द्वारा \(\frac{25}{7}\) लीटर प्रति सेकण्ड की दर से खाली किया जा रहा है। ज्ञात कीजिए कि इस टैंक को आधा खाली करने में कितना समय लगेगा, यदि टैंक के आधार का व्यास 3 मी. है।
हल:
दिया है,
अर्धगोलाकार टैंक का व्यास = 3 मी
∴ अर्धगोलाकार टैंक की त्रिज्या (r) = \(\frac{3}{2}\) मी
अर्धगोलाकार टैंक का आयतन = \(\frac{2}{3}\) πr³

∵ \(\frac{25}{7}\) लीटर पानी को खाली करने में लगा समय
= 1 सेकण्ड
∴ \(\frac{99000}{28}\) लीटर पानी को खाली करने में लगा समय
= \(\frac{7}{25} \times \frac{99000}{28}\)
= 990 सेकण्ड
= \(\frac{990}{60}\) मिनट
= 16.5 मिनट
अतः आधे टैंक को खाली करने में लगा समय = 16.5 मिनट।

प्रश्न 2.
10 सेमी भुजा वाले एक घनाकार ब्लॉक के ऊपर एक अर्धगोला रखा हुआ है। अर्धगोले का अधिकतम व्यास क्या हो सकता है ? इस प्रकार बने ठोस के संपूर्ण पृष्ठीय क्षेत्र को पेंट करवाने का ₹ 5 प्रति 100 वर्ग सेमी की दर से व्यय ज्ञात कीजिए। [π = 3.14 लीजिए]
हल:
अर्द्धगोले का अधिकतम व्यास:
= घन के एक किनारे की ल. (a) = 10 सेमी

∴ अर्द्धगोले की त्रिज्या, r = \(\frac{10}{2}\) = 5 सेमी
अब ठोस का सम्पूर्ण पृष्ठीय क्षे. = घन का सम्पूर्ण पृष्ठीय क्षेत्रफल + अर्द्धगोलीय का वक्र पृष्ठीय क्षेत्रफल – अर्द्धगोले के आधार का क्षेत्रफल
= 6a² + 2πr² – πr²
= 6 × 10² + 2 × 3.14 × (5)² – 3.14 × (5)² सेमी²
= 600 + 157 – 78.5 cm²
= 678.5 cm²
पेंट करवाने का खर्चा = ₹ 5 प्रति 100 सेमी²
∴ ठोस को पेंट करवाने का खर्चा
= 678.5 × \(\frac{5}{100}\)
= ₹ 33.90 (लगभग)
अतः पेंट करवाने का व्यय = ₹ 33.90 (लगभग)।

प्रश्न 3.
एक ठोस धातु के बेलन के दोनों किनारों से उसी व्यास के अर्द्धगोले के रूप में धातु निकाली गई। बेलन की ऊँचाई 10 सेमी तथा इसके आधार की त्रिज्या 4.2 सेमी है। शेष बेलन को पिघलाकर 1.4 सेमी मोटी बेलनाकार तार बनाई गई तार की लम्बाई ज्ञात कीजिए।
हल:
ठोस धातु के बेलन की ऊँचाई = 10 सेमी
ठोस धातु के बेलन की त्रिज्या = 4.2 सेमी
∴ सभी अर्द्धगोले की त्रिज्या = ठोस धातु के बेलन की त्रिज्या; r = 4.2 सेमी
अब शेष बचे हुए बेलन का आयतन = बेलन का आयतन – 2 × अर्द्धगोले का आयतन
= πr²h – 2 × \(\frac{2}{3}\)πr³
= πr²\(\left(h-\frac{4}{3} r\right)\) = π × (4.2)² \(\left(10-\frac{4}{3} \times 4.2\right)\)
= π(4.2)² × 4.4 सेमी³
बेलनाकार तार की मोटाई = 1.4 सेमी
बेलनाकार तार की त्रिज्या = \(\frac{2}{3}\) = 7 सेमी
माना तार की लम्बाई = H सेमी
प्रश्नानुसार-
बेलनाकार तार का आयतन = शेष बचे हुए बेलन का आयतन
⇒ π × .7 × .7 × H = π × (4.2)².4.4
⇒ H = \(\frac{4.2 \times 4.2 \times 4.4}{.7 \times .7}\)
= 158.4
अतः तार की लम्बाई = 138.4 सेमी

प्रश्न 4.
पानी से भरी हुई अर्द्धगोलाकार टंकी को एक पाइप द्वारा 5 लीटर प्रति सेकण्ड की दर से खाली किया जाता है। यदि टंकी का व्यास 3.5 मीटर है तो कितने समय में आधी खाली हो जायेगी।
हल:
दिया है,
अर्द्धगोलाकार टैंकी का व्यास = 13.5 मीटर
∴ त्रिज्या r = \(\frac{3.5}{2}\) मीटर
अर्द्धगोलाकार टंकी की आयतन

…….5 लीटर पानी को खाली करने में लगा समय = 1 सेकण्ड
∴ \(\frac{67375}{21}\) लीटर पानी को खाली करने में लगा समय
= \(\left(\frac{1}{5} \times \frac{67375}{21}\right)\)
\(\left(\frac{1}{5} \times \frac{67375}{21}\right)\) से. = 641.66 से. = 10.69 मिनट
अतः आधी टंकी को खाली करने में लगभग = 10.69 मिनट।

प्रश्न 5.
आकृति में, PQRS एक वर्गाकार लॉन है जिसकी भुजा PQ = 42 मीटर है। दो वृत्ताकार फूलों की क्यारियाँ भुजा PS तथा QR पर हैं जिनका केन्द्र इस वर्ग के विकणों का प्रतिच्छेन बिन्दु O है। दोनों फूलों की क्यारियों (छायांकित भाग) का कुल क्षेत्रफल ज्ञात कीजिए।

हल:
वर्गाकार लॉन का क्षेत्रफल PQRS = 42 मी × 42 मी
माना OP = OS = x मी
इसलिए x² + x² = (42)²
⇒ 2x² = 42 × 42
x² = 21 × 42 …..(1)
अब भाग POS का क्षेत्रफल = \(\frac{90}{360}\) πx² = \(\frac{1}{4}\) πx²
= \(\frac{1}{4} \times \frac{22}{7}\) × 21 × 42 मी² …..(2)
पुन: ΔPOS का क्षे. = \(\frac{1}{4}\) × वर्गाकार लॉन PQRS का क्षे.
= \(\frac{1}{4}\) × 42 × 42 मी² …..(3)
∠POQ = 90°
∴ फूलों की क्यारियों PSP का क्षे. = खण्ड POS भाग का क्षे. – ΔPOS का क्षे.
= \(\frac{1}{4} \times \frac{22}{7}\) × 21 × 42 – \(\frac{1}{4}\) × 42 × 42
= 33 × 21 – 441 = 693 – 441 = 252
दोनों फूलों की क्यारियों का कुल = 2 × 252 = 504 मी²
अतः दोनों फूलों की क्यारियाँ का कुल क्षे. 504 मी² है।

प्रश्न 6.
3.5 सेमी व्यास तथा 3 सेमी ऊँचे 504 शंकुओं को पिघलाकर एक धात्विक गोला बनाया गया। गोले का व्यास ज्ञात कीजिए। अतः इसका पृष्ठीय क्षेत्रफल ज्ञात कीजिए। [π = \(\frac{22}{7}\) लीजिए]
हल:
शंकु का व्यास = 3.5 सेमी
(r) = \(\frac{3.5}{2}\) सेमी
ऊँचाई (h) = 3 सेमी
अब प्रत्येक शंकु का आयतन = \(\frac{1}{3}\) πr²h
= \(\frac{1}{3} \times \frac{22}{7} \times\left(\frac{3.5}{2}\right)^2 \times 3\) = 9.625 सेमी³
∴ 504 शंकुओं का आयतन = 504 × 9.625 = 4851 सेमी³
माना धात्विक गोले की त्रिज्या = R
प्रश्नानुसार,
गोले का आयतन = 504 × (शंकु का आयतन)
\(\frac{4}{3}\) πR³ = 4851
R³ = \(\frac{4851 \times 3}{4 \times 3.14}\) = 1157.625
R = 10.5 सेमी
गोले का व्यास = 2R = 2 × 10.5 = 21 सेमी
∴ गोले का पृष्ठीय क्षेत्रफल = 4πR² = 4 × \(\frac{22}{7}\) × (10.5)²
= 1386 सेमी²

प्रश्न 7.
दो घनों, जिनमें से प्रत्येक का आयतन 27 सेमी³ है, तो संलग्न फलकों को मिलाकर एक ठोस बनाया जाता है। प्राप्त घनाभ का पृष्ठीय क्षेत्रफल ज्ञात कीजिए।
हल:

माना
धन की भुजा = a
घन का आयतन = 27 सेमी³
घन का आयतन = 27 भुजा³
27 सेमी³ = a³
a³ = 27
a = 3 सेमी
∴ घनाभ का सम्पूर्ण पृष्ठीय क्षे.
= 2 (ल. × चौ. + चौ. x ऊँ. + ऊँ. x ल.)
= 2 ( 3 × 3 + 3 × 6 + 6 × 3)
= 3(9 + 18 + 18)
= 3(45) = 135 सेमी²

प्रश्न 8.
एक ठोस अर्धगोले का सम्पूर्ण पृष्ठीय क्षेत्रफल 462 वर्ग सेमी है। इसकी त्रिज्या ज्ञात कीजिए।
हल:
माना अर्धगोले की त्रिज्या r सेमी है।
प्रश्नानुसार,
अर्धगोले का सम्पूर्ण पृष्ठीय क्षेत्रफल = 462 वर्ग सेमी
⇒ 3πR² = 462
⇒ 3 × \(\frac{22}{7}\) × r² = 462
⇒ r² = \(\frac{462 \times 7}{3 \times 22}\) = 49
⇒ r = \(\sqrt{49}\) = 7 सेमी
अतः अर्धगोले की त्रिज्या = 7 सेमी

प्रश्न 9.
एक चाँदी के घनाभ, जिसकी विमाएँ 8 सेमी × 9 सेमी × 11 सेमी है, को पिघलाकर समान त्रिज्या के सात गोले बनाए गए है। एक चाँदी के गोले की त्रिज्या ज्ञात कीजिए।
हल:
माना गोले की त्रिज्या = r सेमी
घनाभ का आयतन = लम्बाई × चौड़ाई × ऊँचाई
= 8 × 9 × 11 घन सेमी
∵ घनाभ को पिघलाकर सात गोले बनाए गए हैं,
∴ घनाभ का आयतन = 7 गोलों का आयतन
⇒ 8 × 9 × 11 = 7 × \(\frac{4}{3}\) πR³
⇒ 8 × 9 × 11 = \(7 \times \frac{4}{3} \times \frac{22}{7} \times r^3\)
⇒ r³ = \(\frac{8 \times 9 \times 11 \times 3}{4 \times 22}\) = 27
⇒ r³ = (3)³
⇒ r = 3 सेमी
अतः एक गोले की त्रिज्या = 3 सेमी

प्रश्न 10.
कोई बर्तन एक खोखले अर्धगोले के आकार का है। अर्धगोले का व्यास 14 सेमी है। इस बर्तन का आन्तरिक पृष्ठीय क्षेत्रफल ज्ञात कीजिए।
हल:
दिया है, अर्धगोले का व्यास = 14 सेमी
∴ त्रिज्या = 7 सेमी
∵ गोला खोखला है,
∴ अर्धगोले का आन्तरिक पृष्ठीय क्षेत्रफल = 2πr²
= 2 × \(\frac{22}{7}\) × (7)² वर्ग सेमी
= 308 वर्ग सेमी

प्रश्न 11.
7 मी व्यास वाला एक कुआँ खोदा जाता है और खोदने से निकली हुई मिट्टी को समान रूप से फैलाकर 22 मी × 14 मी × 2.5 मी वाला एक चबूतरा बनाया गया है। कुएं की गहराई ज्ञात कीजिए।
हल:
कुआँ बेलनाकार होता है।
माना कुएँ की गहराई h मी है।
∴ कुएँ से निकली मिट्टी का आयतन = बेलनाकार कुएँ का आयतन
= πr²h
= \(\frac{22}{7}\) × (7)² × घन मी
घनाभाकार चबूतरे का आयतन = 22 × 14 × 2.5 घन मी
प्रश्नानुसार,
\(\frac{22}{7}\) × (7)² × h = 22 × 14 × 2.5
⇒ h = \(\frac{22 \times 14 \times 2.5}{22 \times 7}\)
⇒ h = 17.5 मी
अतः कुएँ की गहराई = 17.5 मी

प्रश्न 12.
एक ठोस बेलन के आकार का है जिसके दोनों सिरे अर्धगोलाकार है। ठोस की कुल लम्बाई 20 सेमी है तथा बेलन का व्यास 7 सेमी है। ठोस का कुल आयतन ज्ञात कीजिए। (π = \(\frac{22}{7}\) प्रयोग कीजिए)
हल:
माना ABCD एक ठोस बेलन है जिसके दोनों सिरों पर दो एक-समान अर्धगोले हैं।

दिया है, पूरे ठोस की लम्बाई = 20 सेमी हैं, 2r = 7
∴ AB की लम्बाई = 20 – 2r
= 20 – 2 × \(\frac{7}{2}\)
h = 13 सेमी
अब, बेलन ABCD का आयतन = πr²h
= \(\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 13\)
= 500.5 सेमी³
दोनों अर्धगोलों का आयतन = 2 × एक अर्धगोले का आयतन
= 2 × \(\frac{2}{3}\)πr³
= \(2 \times \frac{2}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times \frac{7}{2}\)
= 179.67 सेमी³
ठोस का कुल आयतन = बेलन ABCD का आयतन + दोनों अर्धगोलों का आयतन
= 179.67 + 500.5
= 680.17 सेमी³

प्रश्न 13.
एक ही धातु के दो गोलों का भार 1 किलोग्राम तथा 7 किलोग्राम है। छोटे गोले की त्रिज्या 3 सेमी है। दोनों गोलों को पिघलाकर एक बड़ा गोला बनाया गया। नए गोले का व्यास ज्ञात कीजिए।
हल:
माना छोटे गोले की त्रिज्या r₁ तथा बड़े गोले की त्रिज्या r₂ है और माना इनसे मिलकर बनने वाले नए गोले की त्रिज्या R है।
छोटे गोले का आयतन = \(\frac{4}{3}\)πr₁³ [∵ r₁ = 3 ( दिया है)]
= \(\frac{4}{3}\)π(3)³ = 36π घन सेमी
अब,
छोटे गोले के पदार्थ का घनत्व = गोले का द्रव्यमान / गोले का आयतन
छोटे गोले के पदार्थ का आयतन = \(\frac{1}{36 \pi}\)
दोनों गोले एक ही पदार्थ से बने हैं, अतः इनके पदार्थ का घनत्व भी समान होगा।

अतः छोटे गोले का आयतन + बड़े गोले का आयतन = नए गोले का आयतन
⇒ \(\frac{4}{3} \pi r_1^3+\frac{4}{3} \pi r_2^3=\frac{4}{3} \pi R^3\)
⇒ r₁³ + r₂³ = R³
⇒ 27 + 189 = R³
⇒ R³ = 216
R = 6 ⇒ D = 12 सेमी
अतः नए गोले का व्यास 12 सेमी है।

प्रश्न 14.
त्रिज्या 6 सेमी और ऊँचाई 15 सेमी वाले एक लंबवृत्तीय बेलन के आकार का बर्तन आइसक्रीम से पूरा भरा हुआ है। इस आइसक्रीम को 10 बच्चों में बाँटने के लिए बराबर-बराबर शंकुओं में भरा जाना है, जिनका ऊपरी सिरा अर्धगोले के आकार का है। यदि शंक्वाकार भाग की ऊँचाई इसके आधार की त्रिज्या का 4 गुना है, तो आइसक्रीम शंकु की त्रिज्या ज्ञात कीजिए।
हल:
माना R तथा H क्रमशः लंबवृत्तीय बेलन की त्रिज्या तथा ऊँचाई है।
दिया है, R = 6 सेमी तथा H = 15 सेमी
लंब वृत्तीय बेलन में आइसक्रीम का आयतन
= πR²H
= π × 36 × 15
= 540π सेमी³
अब माना कि शंकु के आधार की त्रिज्या r सेमी हैं. तब इसकी ऊँचाई (h) = 4r

शंकु में आइसक्रीम का आयतन = शंकु का आयतन + अर्द्धगोले का आयतन
= \(\frac{1}{3}\)πr²h + \(\frac{2}{3}\)πr³
= \(\frac{1}{3}\)πr²(h + 2r)
= \(\frac{1}{3}\)πr²(4r + 2r) (∵ h = 4r)
= \(\frac{1}{3}\)πr² × 6r
= 2πr³
अब, 10 × शंकु में आइसक्रीम का आयतन = बेलन में आइसक्रीम का आयतन
⇒ 10 × 2πr³ = 540π
⇒ r³ = 27
⇒ r = 3 सेमी
अतः आइसक्रीम शंकु की त्रिज्या 3 सेमी है।

प्रश्न 15.
6 मी चौड़ी और 1.5 मी गहरी एक नहर में पानी 10 किमी/घंटा की चाल से बह रहा है। 30 मिनट में, यह नहर कितने क्षेत्रफल की सिंचाई कर पाएगी, जबकि सिंचाई के लिए 8 सेमी गहरे पानी की आवश्यकता होती है।
हल:
दिया है, नहर की चौड़ाई = 6 मी तथा गहराई 1.5 मी है व नहर में बहने वाले पानी की चाल 10 किमी/घंटा है, तो नहर में बहने वाला पानी घण्टे में 10 किमी या 10000 मी दूरी तय करेगा।
तथा माना सिंचाई हेतु क्षेत्रफल x मी है।
तब 30 मिनट में पानी का आयतन = 6 × 1.5 × 10000 × \(\frac{30}{60}\)
= 45000 मी³
अब, सिंचाई क्षेत्र का आयतन = नहर के पानी का आयतन
x × \(\frac{8}{100}\) = 45000
x = \(\frac{45000 \times 100}{8}\)
x = 562500 मी² या 56.25 हेक्टेयर

प्रश्न 16.
शंकु के छिन्नक के आकार की ऊपर से खुली एक बाल्डी का आयतन 12308.8 घन सेमी है। इसके ऊपरी तथा निचले वृत्तीय सिरों की त्रिज्याएँ क्रमशः 20 सेमी तथा 12 सेमी हैं। बाल्टी की ऊंचाई तथा इसके बनाने में लगी धातु की चादर का क्षेत्रफल ज्ञात कीजिए। [π = 1.732 लीजिए]
हल:
दिया है, r₁ = 20 सेमी, r₂ = 12 सेमी तथा छिन्नक का आयतन 12308.8 सेमी³ है।

अब,
छिन्नक का आयतन = \(\frac{1}{3}\) × π × h (r₁² + r₂² + r₁r₂)
12308.8 = \(\frac{1}{3}\) × 3.14 × (20² + 12² + 20 × 12)
= \(\frac{1}{3}\) × 3.14 × h(400 + 144 + 240)
12308.8 = \(\frac{1}{3}\) × 3.14(784) × h
h = \(\frac{12308.8 \times 3}{3.14 \times 784}\)
h = 15 सेमी
धातु का क्षेत्रफल = πl(r₁ + r₃) + πr₂²
यहाँ l = \(\sqrt{\left(r_1-r_2\right)^2+h^2}\)
= \(\sqrt{(20-12)^2+15^2}\)
= \(\sqrt{289}\)
= 17 सेमी
धातु का क्षेत्रफल = 3.14 × 17 (20 + 12) + 3.14 × 12²
= 1708.16 + 452.16
= 2160.32 सेमी²

प्रश्न 17.
लकड़ी के एक ठोस बेलन के प्रत्येक सिरे पर एक अर्थ गोला खोद कर निकालते हुए, एक वस्तु बनाई गई, जैसा कि दी गई आकृति में दर्शाया गया है। यदि बेलन की ऊँचाई 10 सेमी है और आधार की त्रिज्या 3.5 सेमी है, तो इस वस्तु का सम्पूर्ण पृष्ठीय क्षेत्रफल ज्ञात कीजिए।

हल:
वस्तु का सम्पूर्ण पृष्ठीय क्षेत्रफल = बेलन का वक्रपृष्ठीय क्षेत्रफल + 2 × अर्द्धगोले का वक्रपृष्ठीय क्षेत्रफल

= 2πrh + 2 × 2πr²
= 2πr[h + 2r]
= 2 × \(\frac{22}{7} \times \frac{35}{10}\)[10 + 2 × 3.5]
= 22[17]
= 374 सेमी²

प्रश्न 18.
चावल की एक ढेरी शंकु के आकार की है जिसके आधार का व्यास 24 मी तथा ऊँचाई 3.5 मी है। चावलों का आयतन ज्ञात कीजिए इस ढेरी को पूरा-पूरा ढकने के लिए कितने कैनवस की आवश्यकता हैं?
हल:
शंक्वाकार ढेरी का व्यास = 24 मी
त्रिज्या, r = 12 मी
ऊँचाई, h = 35 मी
तिर्यक ऊँचाई, l = \(\sqrt{h^2+r^2}\)
= \(\sqrt{(3.5)^2+(12)^2}\)
= \(\sqrt{12.25+144}\)
= \(\sqrt{156.25}\)
= 12.5 मी
चावलों का आयतन = \(\frac{1}{3}\) πr²h
= \(\frac{1}{3} \times \frac{22}{7} \times(12)^2 \times(3.5)\)
= \(\frac{11}{3}\) × 144
= \(\frac{1584}{3}\)
= 528 मी³
ढेरी को ढकने के लिए कैनवास की आवश्यकता = शंकु का वक्रपृष्ठीय क्षेत्रफल
= πrl
= \(\frac{22}{7}\) × 12 × 12.5
= \(\frac{3300}{7}\)
= 471 \(\frac{3}{7}\) मी²

प्रश्न 19.
शंकु के छिन्नक के आकार की एक बाल्टी के निचले तथा ऊपरी किनारों के व्यास क्रमशः 10 सेमी तथा 30 सेमी हैं। यदि बाल्टी की ऊँचाई 24 सेमी है, तो ज्ञात कीजिए :
(i) बाल्टी को बनाने में लगने वाली धातु की शीट का क्षेत्रफल।
(ii) बाल्टी बनाने में सामान्य प्लास्टिक को क्यों नहीं लगाना चाहिए? [π = 3.14 लीजिए]
हल:
बाल्टी के ऊपरी सिरे का व्यास = 30 सेमी
बाल्टी के ऊपरी सिरे की त्रिज्या R = 15 सेमी
बाल्टी के निचले सिरे का व्यास = 10 सेमी
बाल्टी के निचले सिरे की त्रिज्या, r = 5 सेमी
बाल्टी की ऊँचाई, h = 24 सेमी
तिर्यक ऊँचाई, l = \(\sqrt{h^2+(R-r)^2}\)
= \(\sqrt{(24)^2+(15-5)^2}\)
= \(\sqrt{576+100}\)
= \(\sqrt{676}\)
= 26 सेमी

(i) बाल्टी बनाने में लगी धातु की शीट का क्षेत्रफल = छिन्नक का वक्रपृष्ठीय क्षेत्रफल + आधार का क्षेत्रफल
= πl(R + r) + πr²
= π[l(R + r) + r²]
= 3.14[26(15 + 5) + 5²]
= 3.14[520 + 25]
= 3.44(545)
= 1711.3 सेमी²
(ii) बाल्टी बनाने में सामान्य प्लास्टिक इसलिए नहीं लगानीं चाहिये, क्योंकि उसकी शक्ति तथा गलनांक बहुत कम होता है और प्लास्टिक वातावरण के लिए हानिकारक है।

प्रश्न 20.
5.4 मी चौड़ी और 1.8 मी गहरी एक नहर में पानी 25 किमी / घण्टा की गति से बह रहा है। इससे 40 मिनट में कितने प्रतिशत क्षेत्रफल की सिंचाई हो सकती है, यदि सिंचाई के लिए 10 सेमी गहरे पानी की आवश्यकता है।
हल:
नहर की चौड़ाई = 5.4 मी
नहर की गहराई = 1.8 मी
नहर में 1 घण्टे में बहे पानी की लम्बाई = 25 किमी = 25000 मी
नहर में 1 घण्टे में बहे पानी का आयतन
= l × b × h
= 5.4 × 1.8 × 25000
= 243000 मी³
40 मिनट में पानी का आयतन = 243000 × \(\frac{40}{60}\)
= 162000 मी³
10 सेमी ऊँचाई के सींचे जाने वाले खेत का का क्षेत्रफल

= 162000 मी²
= 162 हेक्टेयर

प्रश्न 21.
एक शंकु के छिन्नक की तिर्यक ऊँचाई 4 सेमी है तथा इसके वृत्तीय सिरों की परिमाप 18 सेमी और 6 सेमी है। इस छिन्नक का व्रक पृष्ठीय क्षेत्रफल ज्ञात कीजिए।
हल:
छिन्नक को तिर्यक ऊँचाई = 4 सेमी
ऊपरी हिस्से का परिमाप = 18 सेमी
⇒ 2πR = 18 सेमी
⇒ R = \(\frac{9}{4}\) सेमी
निचले हिस्से का परिमाप = 6 सेमी
2πr = 6 ⇒ r = \(\frac{3}{8}\) सेमी
छिन्नक का वक्रपृष्ठ = πl[R + r]
= π × 4 × \(\left[\frac{9}{\pi}+\frac{3}{\pi}\right]\)
= π × 4 × \(\frac{12}{\pi}\) = 48 सेमी²

प्रश्न 22.
एक ठोस लोहे के घनाभ की विमाएँ 4.4 मी × 2.6 मी × 2.0 मी है। इसे पिघलाकर 30 सेमी आन्तरिक त्रिज्या और 5 सेमी मोटाई का एक खोखला बेलनाकार पाइप बनाया गया है। पाइप की लंबाई ज्ञात कीजिए।
हल:
पाइप की आन्तरिक त्रिज्या, r = 30 सेमी
पाइप की मोटाई = 5 सेमी
पाइप की बाहरी त्रिज्या = 30 + 5
R = 35 सेमी
माना, पाइप की लम्बाई = h सेमी
खोखले पाइप की आयतन = घनाभ की आयतन
πrh[R² – r²] = l × b × h
\(\frac{22}{7}\) × h[35² – 30²] = 4.4 × 2.6 × 1 × 100 × 100 × 100
\(\frac{22}{7}\) × h × 65 × 5 = 44 × 26 × 1 × 100 × 100
h = \(\frac{44 \times 26 \times 100 \times 100 \times 7}{22 \times 65 \times 5}\)
= 11200 सेमी
⇒ h = 112 मीटर

प्रश्न 23.
किसी वर्षा जल संग्रहण तन्त्र में, 22 मी × 20 मी की छत से वर्षा जल बहकर 2 मी आधार के व्यास तथा 3.5 मी ऊँचाई के एक बेलनाकार टैंक में आता है। यदि टैंक भर गया हो, तो ज्ञात कीजिए कि सेमी में कितनी वर्षा हुई। जल संरक्षण पर अपने विचार व्यक्त कीजिए।
हल:
एकत्रित पानी का आयतन = बेलनाकार टैंक का आयतन
L × B × H = πr²h
20 × 20× H = \(\frac{22}{7}\) × 1 × 1 × 3.5
22 × 20 × H = 11
H = \(\frac{11}{22 \times 20}=\frac{1}{40}\) मीटर
= \(\frac{1}{40} \times 100\)
= \(\frac{5}{2}\) = 2.5 सेमी
अतः 2.5 सेमी वर्षा हुई।
जल संरक्षण अति आवश्यक है, वह सूखे कि स्थिति के समय में कारगर सिद्ध होता है।

प्रश्न 24.
5 सेमी आंतरिक त्रिज्या तथा 24 सेमी ऊँचाई के एक शंक्वाकार बर्तन का \(\frac{3}{4}\) भाग पानी से भरा है। इस पानी को 10 सेमी आंतरिक त्रिज्या के बेलनाकार बर्तन में खाली किया जाता है। बेलनाकार बर्तन में पानी की ऊँचाई ज्ञात कीजिए।
हल:
प्रश्नानुसार,
शंक्वाकार बर्तन में \(\frac{3}{4}\) भाग का आयतन = बेलनाकार बर्तन का आयतन

अतः बेलनाकार बर्तन में पानी की ऊँचाई 1.5 सेमी है।

वस्तुनिष्ठ प्रश्न :

रिक्त स्थानों की पूर्ति कीजिए :

प्रश्न (क).
1. ऐसी वस्तुओं को जो स्थान घेरती हैं ………….. आकृतियाँ लाती हैं।
2. ठोस आकृति की सीमा बनाने वाली समतल आकृतियों का क्षेत्रफल ……………. क्षेत्रफल कहलाता है।
3. कोई वस्तु जितना स्थान घेरती है, उसे उस वस्तु का ……………… कहते है।
4. एक …………… के आयताकार समतल फलक होते हैं।
5. किसी आयत के उसकी एक स्थिर भुजा के परितः घुमाने से प्राप्त आकृति लम्बवृत्तीय ………….. कहलाती हैं।
उत्तर:
1. ठोस,
2. पृष्ठीय,
3. आयतन,
4. घनाभ,
5. बेलन।

निम्न में सत्य / असत्य बताइए :

प्रश्न (ख).
1. बेलन का आयतन 2πrh वर्ग इकाई होता है।
2. खोखले बेलन में लगे पदार्थ का आयतन दोनों बेलनों के आयतनों के योग के बराबर होता है।
3. गोले का आयतन 4πr² घन इकाई होता है।
4. अर्द्धगोले का सम्पूर्ण पृष्ठ = 3πr² वर्ग इकाई ।
5. शंकु का आयतन = πrl घन इकाई
उत्तर:
1. असत्य,
2. असत्य,
3. असत्य,
4. सत्य,
5. असत्य।

(ग) बहुविकल्पीय प्रश्न :

प्रश्न 1.
12π घन सेमी आयतन वाले गोले की त्रिज्या (सेमी में) है :
(A) 3
(B) 3\(\sqrt{3}\)
(C) 3^2/3
(D) 3^1/2
हल:
गोले का आयतन = \(\frac{4}{3}\)πr³
12π = \(\frac{4}{3}\)πr³
r³ = \(\frac{12 \pi \times 3}{4 \pi}\) = 9
r³ = 3²
⇒ r = 3^2/3
अत: सही विकल्प (C) है।

प्रश्न 2.
एक ठोस अर्धगोले का कुल पृष्ठीय क्षेत्रफल हैं:
(A) 3πr²
(B) 2πr²
(C) 4πr²
(D) \(\frac{2}{3}\)πr³
हल:
कुल पृष्ठीय क्षेत्रफल = 2πr² + πr² = 3πr²
अत: सही विकल्प (A) है।

प्रश्न 3.
एक 22 सेमी. आंतरिक किनारे वाले खोखले घन को 0.5 सेमी व्यास वाले गोलाकार कंचो से भरा जाता है तथा यह कल्पना की जाती है कि घन का भाग भरा नहीं जा सकता है। तब घन में समावेशित होने वाले कंचों की संख्या है :
(A) 142296
(B) 142396
(C) 142496
(D) 142596
हल:
दिया है,
घन की आन्तरिक भुजा = 22 सेमी.
तथा गोलाकार कंचे का व्यास = 0.5 सेमी.
∴ गोलाकार कंचे की त्रिज्या (r) = \(\frac{0.5}{2}\) सेमी.
= \(\left(\frac{5}{20}\right)\) सेमी.
∴ 1 गोलाकार कंचे का आयतन = \(\frac{4}{3}\)πr³
= \(\frac{4}{3} \times \frac{22}{7} \times\left(\frac{5}{20}\right)^3\)
= \(\frac{4}{3} \times \frac{22}{7} \times \frac{5 \times 5 \times 5}{20 \times 20 \times 20}\)
माना घन में समावेशित होने वाले कंचों की संख्या x है।
तब x कंचों का आयतन = \(x \times \frac{4}{3} \times \frac{22}{7} \times \frac{5 \times 5 \times 5}{20 \times 20 \times 20}\)
= \(\frac{x \times 55}{21 \times 40}\) घन सेमी. ……….(1)
घन का आन्तरिक आयतन = 22 × 22 × 22 घन सेमी.
……. खोखले घन को कंचों से भरा जाता है, तो पन का \(\frac{1}{8}\) भाग भरा नहीं जा सकता है।
∴ खोखले घन का कंचों से भरा गया भाग = 1 – \(\frac{1}{8}\) = \(\frac{7}{8}\)
खोखले घन का कंचों से भरे भाग का आयतन = \(\frac{7}{8}\) × 22 × 22 × 22 घन सेमी. …..(2)
समीकरण (1) तथा (2) से
\(\frac{x \times 55}{21 \times 40}\) = \(\frac{7}{8}\) × 22 × 22 × 22
⇒ x = \(\frac{7 \times 22 \times 22 \times 22 \times 21 \times 40}{55 \times 8}\)
= 14 × 22 × 22 × 21 = 142296
अत: सही विकल्प (A) है।

प्रश्न 4.
यदि दो शंकुओं की त्रिज्याओं में अनुपात 3 : 1 और ऊँचाइयों में अनुपात 1 : 3 है, तो उनके आयतनों में अनुपात होगा :
(A) 2 : 1
(B) 3 : 1
(C) 1 : 3
(D) 1 : 2
हल:
माना दो शंकुओं की त्रिज्याएँ क्रमशः r₁ और r₂ ऊँचाइयाँ h₁ और h₂ हैं।
प्रश्नानुसार, \(\frac{r_1}{r_2}=\frac{3}{1}\) और \(\frac{h_1}{h_2}=\frac{1}{3}\)
दोनों शंकुओं के आयतनों का अनुपात

दोनों शंकुओं के आयतनों का अनुपात = 3 : 1
अत: सही विकल्प (B) है।

प्रश्न 5.
14 सेमी भुजा के एक घन से एक बड़े से बड़ा शंकु काटा जाता है। शंकु का आयतन है:
(A) 766.18 घन सेमी
(B) 817.54 घन सेमी
(C) 1232 घन सेमी
(D) 718.66 घन सेमी।

हल:
14 सेमी भुजा के घन से बड़े से बड़ा शंकु काटा जाता है।
अतः शंकु की ऊँचाई (h) = 14 सेमी
शंकु के आधार का व्यास = 14 सेमी
अत: शंकु के आधार की त्रिज्या = व्यास / 2 = \(\frac{14}{2}\)
= 7 सेमी
शंकु का आयतन = \(\frac{1}{3}\)πr²

अत: सही विकल्प (D) है।

प्रश्न 6.
दो गोलों के आयतनों का अनुपात 64 : 27 है। उनके पृष्ठीय क्षेत्रफलों का अनुपात है :
(A) 3 : 4
(B) 4 : 3
(C) 9 : 16
(D) 16 : 9
हल दिया है,
माना कि दो गोलों की त्रिज्याएँ क्रमशः r₁ तथा r₂ है। पहले गोले का आयतन : दूसरे गोले का आयतन = 64 : 27

अत: सही विकल्प (B) है।

प्रश्न 7.
क्रमशः आन्तरिक और बाहरी व्यास 4 सेमी और 8 सेमी वाले धातु के गोलाकार खोल को पिघलाकर आधार व्यास 8 सेमी. के एक शंकु के आकार में ढाला जाता है। इस शंकु की ऊँचाई है:
(A) 12 सेमी
(B) 14 सेमी
(C) 15 सेमी
(D) 18 सेमी
हल:
दिया है,
गोलाकार खोल की आन्तरिक व्यास = 4 सेमी
∴ गोलाकार खोल की आन्तरिक त्रिज्या
(r₁) = \(\frac{4}{2}\) = 2 सेमी
गोलाकार खोल की बाहरी व्यास = 8 सेमी
∴ गोलाकार खोल की बाहरी त्रिज्या
(r₂) = \(\frac{8}{2}\) = 4 सेमी
∴ शंकु का व्यास (r) = \(\frac{8}{2}\) = 4 सेमी
माना शंकु की ऊँचाई h सेमी है।
……. गोलाकार खोल को पिघलाकार शंकु के आकार में ढाला जाता है।
∴ गोलाकार खोल का आयतन = शंकु का आयतन
⇒ \(\frac{4}{3}\)π[r₂³ – r₁³] = \(\frac{1}{3}\)πr²h
⇒ 4[4³ – 2³] = 4² × h
⇒ 4 × [64 – 8] = 16 × h
⇒ 4 × 56 = 16 × h
⇒ h = \(\frac{4 \times 56}{16}\) = 14 सेमी
अत: सही विकल्प (B) है।

प्रश्न 8.
आधार व्यास 2 सेमी. और ऊँचाई 16 सेमी. वाले धातु के एक ठोस बेलन को पिघलाकर समान माप के बारह ठोस गोले बनाए जाते हैं। प्रत्येक गोले का व्यास है:
(a) 4 सेमी
(b) 3 सेमी
(c) 2 सेमी
(d) 6 सेमी
हल:
दिया है,
धातु के ठोस बेलन की ऊंचाई (h) = 16 सेमी
और धातु के ठोस बेलन का व्यास = 2 सेमी
∴ धातु के ठोस बेलन की त्रिज्या (r₁) = \(\frac{2}{2}\) = 1 सेमी
माना कि पिघलाकर समान माप के प्रत्येक ठोस गोले की त्रिज्या r₂ सेमी है।
….. ठोस बेलन को पिघलाकर समान माप के 12 ठोस गोले बनाए जाते हैं।
∴ ठोस बेलन का आयतन = 12 ठोस गोलों का आयतन
⇒ πr₁²h = 12 × \(\frac{4}{3}\)πr₂³
⇒ r₁²h = 16r₂³
⇒ 1² × 16 = 16r₂³
⇒ \(\frac{16}{16}\) = r₂³
⇒ r₂³ = 1 सेमी.
⇒ r₂ = \(\sqrt[3]{1}\) = 1 सेमी
∴ ठोस गोले का व्यास = 2 × 1 = 2 सेमी.
अंत: सही विकल्प (C) है।

प्रश्न 9.
यदि 11 सेमी × 3.5 सेमी × 2.4 सेमी मोम के एक घनाभ से 2.8 सेमी व्यास की एक मोमबत्ती बनाई जाती है। मोमबत्ती की लम्बाई होगी:
(A) 14 सेमी
(B) 15 सेमी
(C) 25 सेमी
(D) 12 सेमी
हल:
दिया है,
घनाभ की विमाएँ = 11 सेमी. × 3.5 सेमी × 2.4 सेमी.
∴ घनाभ का आयतन = 11 × 3.5 × 2.4
= 92.4 घन सेमी
माना मोमबत्ती की ऊँचाई h सेमी है।
मोमबत्ती का व्यास = 2.8 सेमी
मोमबत्ती की त्रिज्या (r) = \(\frac{2.8}{2}\) = 1.4 सेमी
प्रश्नानुसार,
मोमबत्ती का आयतन = घनाभ का आयतन
⇒ πr²h = 92.4
⇒ \(\frac{22}{7}\) × (1.4)² × h = 92.4
⇒ \(\frac{22}{7}\) × 1.4 × 14 × h = 92.4
⇒ h = \(\frac{92.4 \times 7}{22 \times 1.4 \times 1.4}\)
⇒ h = \(\frac{4.2}{0.2 \times 1.4}\) = 15 सेमी
मोमबत्ती की ऊँचाई = 15 सेमी
अत: सही विकल्प (B) है।

प्रश्न 10.
एक जलाशय लम्बवृत्तीय शंकु के छिन्नक के आकार में है। इसका ऊपरी सिरा 8 मीटर तथा पेंदी वाला सिरा 4 मीटर चौड़ा है। यदि यह 6 मीटर गहरा हो, तो इसकी क्षमता है:
(A) 176 मीटर³
(B) 196 मीटर³
(C) 200 मीटर³
(D) 110 मीटर³
हल:
दिया है,
छिन्नक के ऊपरी सिरे का व्यास = 8 मीटर
छिन्नक के ऊपरी सिरे की त्रिज्या (r₁) = \(\frac{8}{2}\) = 4 मीटर
छिन्नक के निचले सिरे का व्यास = 4 मीटर
छिन्नक के निचले सिरे की त्रिज्या (r₂) = \(\frac{4}{2}\) = 2 मीटर
छिन्नक की ऊँचाई (h) = 6 मीटर
छिन्नक का आयतन = \(\frac{1}{3}\)π × (r₁² + r₂² + r₁r₂)h
= \(\frac{1}{3} \times \frac{22}{7}\)(4² + 2² + 4 × 2) × 6
= \(\frac{22}{7}\) × (16 + 4 + 8) × 2
= \(\frac{22 \times 56}{7}\) = 22 × 8 = 176 घन मीटर
अत: सही विकल्प (A) है।

प्रश्न 11.
एक तेल की कुप्पी शंकु के छिन्नक को बेलन से जोड़ने पर बनी है। कुप्पी की ऊँचाई 12 सेमी है। बेलनाकार भाग की त्रिज्या 4 सेमी है और कुप्पी के. ऊपरी सिरे की त्रिज्या 9 सेमी है, तो शंकु के छिन्नक की तिर्यक ऊंचाई होगी:
(A) 13 सेमी
(B) 23 सेमी
(C) 26 सेमी
(D) 13.5 सेमी
हल:
कुप्पी के ऊपरी सिरे की त्रिज्या (r₁) = 9 सेमी
कुप्पी के निचले सिरे की त्रिज्या (r₂) = 4 सेमी
कुप्पी की ऊँचाई (h) = 12 सेमी
शंकु के छिनक की तिर्यक ऊँचाई (l) = \(\sqrt{h^2+\left(r_1-r_2\right)^2}\)
= \(\sqrt{(12)^2+(9-4)^2}\)
= \(\sqrt{(12)^2+(5)^2}\)
= \(\sqrt{144+25}\)
= \(\sqrt{169}\)
= 13 सेमी
अतः सही विकल्प (A) है।

प्रश्न 12.
9 सेमी त्रिज्या के धातु के गोले को पिघलाकर 3 सेमी त्रिज्या और 6 सेमी ऊँचाई के शंकु बनाये जा सकने वाले शंकुओं की संख्या है:
(A) 54
(B) 45
(C) 55
(D) 44
हल:
दिया है,
धातु के गोले की त्रिज्या (R) = 9 सेमी
शंकु की त्रिज्या (r) = 3 सेमी
तथा शंकु की ऊँचाई (h) = 6 सेमी
9 सेमी त्रिज्या वाले गोले का आयतन
\(\frac{4}{3}\)πR² = \(\frac{4}{3}\) × π × (9)²
= \(\frac{4}{3}\) × π × 9 × 9 × 9
= 4 × π × 3 × 81
= 972π घन सेमी
अब 3 सेमी त्रिज्या और 6 सेमी ऊँचाई वाले शंकु का आयतन
\(\frac{1}{3}\)πr²h = \(\frac{1}{3}\) × π × (3)² × 6
= \(\frac{1}{3}\) × π × 9 × 6
= π × 3 × 6 = 18π घन सेमी
गोले का आयतन शंकुओं की संख्या

अत: सही विकल्प (A) है।

प्रश्न 13.
एक ठोस लम्ब वृत्तीय शंकु को उसकी ऊँचाई के बीचों बीच से होकर जाते, शंकु के आधार के समान्तर एक तल द्वारा दो भागों में काटा गया है। इस प्रकार छोटे शंकु के आयतन का पूरे शंकु के आयतन से अनुपात है:
(A) 1 : 2
(B) 1 : 4
(C) 1 : 6
(D) 1 : 8
हल:
माना OAB एक शंकु है जिसकी ऊँचाई h तथा त्रिज्या R है। शंकु को बीचों बीच से आधार के समान्तर काटा गया है। अतः छोठे शंकु की ऊँचाई (h’) = \(\frac{h}{3}\)
माना कि छोटे शंकु की त्रिज्या r है।

LD || MB
⇒ ∠OLD = ∠OMB (संगत कोण)
ΔOLD तथा ΔOMB में,
∠OLD = ∠OMB
∠LOD = ∠MOB (उभयनिष्ठ कोण)
ΔOLD ~ ΔOMB (AA समरूपता गुणधर्म से)
⇒ \(\frac{O L}{O M}=\frac{L D}{M B}\)
[समरूप त्रिभुजों की भुजाएँ समानुपाती होती हैं]
⇒ \(\frac{\frac{h}{2}}{h}=\frac{r}{R}\)
⇒ \(\frac{h}{2 h}=\frac{r}{R}\)
⇒ \(\frac{1}{2}=\frac{r}{R}\)
⇒ R = 2r
छोटे शंकु का आयतन : पूरे शंकु का आयतन
= πr²h’ : πR²h
= \(\frac{\pi \times r^2 \times \frac{h}{2}}{\pi \times(2 r)^2 \times h}\)
= \(\frac{r^2 \times h}{2 \times 4 r^2 \times h}=\frac{1}{8}\) = 1 : 8
अतः विकल्प (D) सही है।

प्रश्न 14.
तीनों घनों की कोर क्रमशः 3 सेमी, 4 सेमी और सेमी हैं। इनसे बनने वाले एक घन की भुजा है:
(A) 6 सेमी
(B) 5 सेमी
(C) 7 सेमी
(D) 4 सेमी
हल:
पहले घन का आयतन = (3)³ = 27 घन सेमी
दूसरे घन का आयतन = (4)³ = 64 घन सेमी
तीसरे घन का आयतन = (5)³ = 125 घन सेमी
इन तीनों घनों का कुल आयतन = 27 + 64 + 125
= 216 घन सेमी
∵ तीनों घनों से एक नया घन बनता है।
नये घन का आयतन = तीनों घनों का कुल आयतन
(भुजा)³ = 216
भुजा = \(\sqrt[3]{216}\)
= \(\sqrt[3]{6 \times 6 \times 6}\) = 6 सेमी
अतः सही विकल्प (A) है।

प्रश्न 15.
एक घनाभ की माप 18 सेमी × 12 सेमी × 9 सेमी है। इस घनाभ को पिघलाकार 3 सेमी भुजा वाले कितने घन बनाये जा सकते हैं ?
(A) 60
(B) 55
(C) 69
(D) 72
हल:
दिया है,
घनाभ की माप = 18 सेमी × 12 सेमी × 9 सेमी
दिये गये घनाभ का आयतन = 18 × 12 × 9
= 18 × 108
= 1944 घन सेमी
3 सेमी भुजा वाले घन का आयतन = (भुजा)³
= (3)³ = 27 घन सेमी
घनाभ को पिघलाकर बनाये गये घनों की संख्या = घनाभ का आयतन / 1 घन का आयतन
= \(\frac{1944}{27}\) = 72
अतः सही विकल्प (D) है।

Jharkhand Board JAC Class 10 Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 5 Arithmetic Progressions Exercise 5.3

Question 1.
Find the sum of following APs:
1. 2, 7, 12, ……, to 10 terms.
2. -37, -33, -29, ……, to 12 terms.
3. 0.6, 1.7, 2.8, ……, to 100 terms.
4. \(\frac{1}{15}, \frac{1}{12}, \frac{1}{10}, \ldots\) to 11 terms.
Solution:
1. For the given AP 2, 7, 12, ……., a = 2.
d = 7 – 2 = 5, n = 10 and S_n is to be found.
S_n = \(\frac{n}{2}\)[2a + (n – 1)d]
S₁₀ = \(\frac{10}{2}\)[4 + (10 – 1) 5]
= 5(49) = 245
Thus, the sum of first 10 terms of the given AP is 245.

2. For the given AP -37, -33, -29, a = -37, d = (-33) – (-37) = 4, n = 12 and S_n is to be found.
S_n = \(\frac{n}{2}\)[2a + (n – 1)d]
S₁₂ = \(\frac{12}{2}\)[-74 + (12 – 1)4]
= 6(-30) = – 180
Thus, the sum of first 12 terms of the given AP is -180.

3. For the given AP 0.6, 1.7, 2.8,…… a = 0.6, d = 1.7 – 0.6 = 1.1, n = 100 and S_n is to be found.
S_n = \(\frac{n}{2}\)[2a + (n – 1) d]
∴ S₁₀₀ = \(\frac{100}{2}\)[1.2 + (100 – 1) (1.1)]
= 50[1.2 + 108.9]
= 50 × 110.1 = 5505
Thus, the sum of first 100 terms of the given AP is 5505.

4.

Thus, the sum of first 11 terms of the given AP is \(\frac{33}{20}\).

Question 2.
Find the sums given below:
1. 7 + 10\(\frac{1}{2}\) + 14 + … + 84
2. 34 + 32 + 30 + … + 10
3. (-5) + (-8) + (-11) + … + (-230)
Solution:
1. 7 + 10\(\frac{1}{2}\) + 14 + …… + 84
Here, a = 7; d = 10\(\frac{1}{2}\) – 7 = 3\(\frac{1}{2}\); last term l = 84.
Let the last term be nth term.
a_n = a + (n – 1)d
∴ 84 = 7 + (n – 1) (3\(\frac{1}{2}\))
∴ 77 = \(\frac{7}{2}\)(n – 1)
∴ (n – 1) = 22
∴ n = 23
Now,
S_n = \(\frac{n}{2}\)(a + l), where l is the last term.
= \(\frac{23}{2}\)(7 + 84)
= \(\frac{23}{2}\) × 91 = 1046\(\frac{1}{2}\)
Thus, the required sum is 1046\(\frac{1}{2}\).

2. 34 + 32 + 30 + … + 10
Here, a = 34; d = 32 – 34 = (-2); last term l = 10.
Let the last term be nth term.
a_n = a + (n – 1)d
∴ 10 = 34 + (n – 1)(-2)
∴ -24 = -2(n – 1)
∴ (n – 1) = 12
∴ n = 13
Now,
S_n = \(\frac{n}{2}\)(a + l), where l is the last term.
= \(\frac{13}{2}\)(34 + 10)
= 13 × 22 = 286
Thus, the required sum is 286.

3. (-5) + (-8) + (-11) + … + (-230)
Here, a = (-5); d = (-8) – (-5) = (-3):
last term l = (-230).
Let the last term be nth term.
a_n = a + (n – 1)d
∴ -230 = -5 + (n – 1)(-3)
∴ -225 = -3 (n – 1)
∴ n – 1 = 75
∴ n = 76
Now,
S_n = \(\frac{n}{2}\)(a + l), where l is the last term.
= \(\frac{76}{2}\)[(-5) + (-230)]
= 38(-235) = -8930
Thus, the required sum is -8930.

Question 3.
In an AP:
1. Given a = 5, d = 3, a_n = 50, find n and S_n.
2. Given a = 7, a₁₃ = 35, find d and S₁₃.
3. Given a₁₂ = 37, d = 3, find a and S₁₂.
4. Given a₃ = 15, S₁₀ = 125, find d and a₁₀.
5. Given d = 5, S₉ = 75, find a and a₉.
6. Given a = 2, d = 8, S_n = 90, find n and a_n.
7. Given a = 8, a_n = 62, S_n = 210, find n and d.
8. Given a_n = 4, d = 2, S_n = -14, find n and a.
9. Given a = 3, n = 8, S_n = 192, find d.
10. Given l = 28, S_n = 144, and there are total 9 terms. Find a.
Solution:
1. a = 5, d = 3, a_n = 50, n = ? S_n = ?
a_n = a + (n – 1)d
∴ 50 = 5 + (n – 1)3
∴ 45 = 3(n – 1)
∴ 15 = n – 1
∴ n = 16
S_n = \(\frac{n}{2}\)(a + l)
∴ S₁₆ = \(\frac{16}{2}\)(5 + 50)
∴ S₁₆ = 8 × 55
∴ S₁₆ = 440

2. a = 7, a₁₃ = 35, d = ?, S₁₃ = ?
a_n = a + (n – 1)d
a₁₃ = a + (13 – 1) d
∴ 35 = 7 + 12d
∴ 28 = 12d
∴ d = \(\frac{28}{12}\)
∴ d = \(\frac{7}{3}\)
S_n = \(\frac{n}{2}\)(a + l)
∴ S₁₃ = \(\frac{13}{2}\)(17 + 35)
∴ S₁₃ = 13 × 21
∴ S₁₃ = 273

3. a₁₂ = 37, d = 3, a = ?, S₁₂ = ?
a_n = a + (n – 1)d
∴ a₁₂ = a + 11d
∴ 37 = a + 11 (3)
∴ a = 4
S_n = \(\frac{n}{2}\)(a + l)
∴ S₁₂ = \(\frac{12}{2}\)(4 + 37)
∴ S₁₂ = 246

4. a₃ = 15, S₁₀ = 125, d = ?, a₁₀ = ?
a_n = a + (n – 1)d
∴ a₃ = a + 2d
∴ a + 2d = 15 …….(1)
S_n = \(\frac{n}{2}\)[2a + (n – 1) d]
∴ S₁₀ = \(\frac{10}{2}\)[2a + 9d]
∴ S₁₂₅ = 5(2a + 9d)
∴ 2a + 9d = 25 …….(2)
Solving equations (1) and (2), we get
d = -1 and a = 17.
a_n = a + (n – 1)d
∴ a₁₀ = a + 9d
∴ a₁₀ = 17 + 9(-1)
∴ a₁₀ = 8

5. d = 5, S₉ = 75, a = ?, a₉ = ?
S_n = \(\frac{n}{2}\)[2a + (n – 1) d]
∴ S₉ = \(\frac{9}{2}\)[2a + (9 – 1) d]
∴ 75 = \(\frac{9}{2}\)[2a + 8(5)]
∴ 75 = 9(a + 20)
∴ \(\frac{25}{3}\) = a + 20
∴ a = \(\frac{25}{3}\) – 20
∴ a = –\(\frac{35}{3}\)
a_n = a + (n – 1) d
∴ a₉ = a + 8d
∴ a₉ = (-\(\frac{35}{3}\)) + 8(5)
∴ a₉ = –\(\frac{35}{3}\) + 40
∴ a₉ = \(\frac{85}{3}\)

6. a = 2, d = 8, S_n = 90, n = ?, a_n = ?
S_n = \(\frac{n}{2}\)[2a + (n – 1) d]
∴ 90 = \(\frac{n}{2}\)[4 + (n – 1)8]
∴ 90 = \(\frac{n}{2}\)[8n – 4]
∴ 90 = n (4n – 2)
∴ 4n² – 2n – 90 = 0
∴ 2n² – n – 45 = 0
∴ 2n² – 10n + 9n – 45 = 0
∴ 2n (n – 5) + 9 (n – 5) = 0
∴ (n – 5) (2n + 9) = 0
∴ n – 5 = 0 or 2n + 9 = 0
∴ n = 5 or n = –\(\frac{9}{2}\)
Since n is a positive integer, n ≠ –\(\frac{9}{2}\)
∴ n = 5
a_n = a + (n – 1)d
∴ a₅ = a + 4d
∴ a₅ = 2 + 4(8)
∴ a₅ = 34

7. a = 8, a_n = 62, S_n = 210, n = ? d = ?
S_n = \(\frac{n}{2}\)(a + l)
∴ S_n = \(\frac{n}{2}\)(a + a_n)
∴ 210 = \(\frac{n}{2}\)(8 + 62)
∴ 420 = n (70)
∴ n = 6
a_n = a + (n – 1)d
∴ a₆ = a + 5d
∴ 62 = 8 + 5d
∴ 54 = 5d
∴ d = \(\frac{54}{5}\)

8. a_n = 4, d = 2, S_n = -14, n = ?, a = ?
a_n = a + (n – 1)d
∴ 4 = a + (n – 1) (2)
∴ 4 = a + 2n – 2
∴ a = 6 – 2n
S_n = \(\frac{n}{2}\)[2a + (n – 1) d]
∴ -14 = \(\frac{n}{2}\)[2 (6 – 2n) + (n – 1) (2)] (by (1))
∴ -14 = \(\frac{n}{2}\)[12 – 4n + 2n – 2]
∴ -14 = \(\frac{n}{2}\)[-2n + 10]
∴ -14 = n(-n + 5)
∴ -14 = -n² + 5n
∴ n² – 5n – 14 = 0
∴ (n – 7)(n + 2) = 0
∴ n – 7 = 0 or n + 2 = 0
∴ n = 7 or n = -2
Since n is a positive integer, n ≠ -2.
∴ n = 7
By (1), a = 6 – 2n
∴ a = 6 – 2(7)
∴ a = -8

9. a = 3, n = 8, S_n = 192, d = ?
S_n = \(\frac{n}{2}\)[2a + (n – 1) d]
∴ 192 = \(\frac{8}{2}\)[6 + (8 – 1) d]
∴ 192 = 4[6 + 7d]
∴ 48 = 6 + 7d
∴ 42 = 7d
∴ d = 6

10. l = 28, S_n = 144, n = 9, a = ?
S_n = \(\frac{n}{2}\)(a + l)
∴ 144 = \(\frac{9}{2}\)(a + 28)
∴ 32 = (a + 28)
∴ a = 4

Question 4.
How many terms of the AP 9, 17, 25, … must be taken to give a sum of 636?
Solution:
Here, a = 9; d = 17 – 9 = 8; S_n = 636, n = ?
S_n = \(\frac{n}{2}\)[2a + (n – 1) d]
∴ 636 = \(\frac{n}{2}\)[18 + (n – 1)8]
∴ 636 = \(\frac{n}{2}\)[10 + 8n]
∴ 636 = n[4n + 5]
∴ 4n² + 5n – 636 = 0
Here, a = 4; b = 5; c = -636
b² – 4ac = (5)² – 4(4)(-636)
= 25 + 10176
= 10201
∴ \(\sqrt{b^2-4 a c}=\sqrt{10201}=101\)
Then, n = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
∴ n = \(\frac{-5 \pm 101}{8}\)
∴ n = \(\frac{96}{2}\) or n = \(\frac{-106}{8}\)
∴ n = 12 or n = \(-\frac{53}{4}\)
As n denotes the numbers of terms, it is a positive integer.
∴ n = \(-\frac{53}{4}\) is not possible.
∴ n = 12
Thus, 12 terms of the AP 9, 17, 25,… must be taken to give a sum of 636.

Question 5.
The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
Solution:
Here, a = 5; l = 45; S_n = 400; n = ?; d = ?
S_n = \(\frac{n}{2}\)(a + l)
∴ 400 = \(\frac{n}{2}\)(5 + 45)
∴ 800 = n (50)
∴ n = 16
l = a_n = a + (n – 1)d
∴ a₁₆ = a + 15d
∴ 45 = 5 + 15d
∴ 40 = 15d
∴ d = \(\frac{40}{15}\)
∴ d = \(\frac{8}{3}\)
Thus, the number of terms is 16 and the common difference is \(\frac{8}{3}\).

Question 6.
The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?
Solution:
Here, a = 17; l = a_n = 350; d = 9; n = ?; S_n = ?
a_n = a + (n – 1)d
∴ 350 = 17 + (n – 1)9
∴ 333 = 9 (n – 1)
∴ n – 1 = 37 ∴ n = 38
Again, S_n = \(\frac{n}{2}\)(a + l)
∴ S₃₈ = \(\frac{38}{2}\)(17 + 350)
∴ S₃₈ = 19 × 367
∴ S₃₈ = 6973
Thus, there are 38 terms and their sum is 6973.

Question 7.
Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.
Solution:
Here, a₂₂ = 149; d = 7; S₂₂ = ?
a_n = a + (n – 1) d
∴ a₂₂ = a + 21d
∴ 149 = a + 21 × 7
∴ a = 149 – 147
∴ a = 2
Now, S_n = \(\frac{n}{2}\)(a + l)
∴ S₂₂ = \(\frac{22}{2}\)(2 + 149)
∴ S₂₂ = 11 × 151
∴ S₂₂ = 1661
Thus, the sum of first 22 terms of the given AP is 1661.

Question 8.
Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Solution:
Here, a₂ = 14; a₃ = 18; S₅₁ = ?
a_n = a + (n – 1)d
∴ a₂ = a + d = 14 ……..(1)
∴ a₃ = a + 2d = 18 ……..(2)
Solving equations (1) and (2), we get
d = 4 and a = 10.
S_n = \(\frac{n}{2}\)[2a + (n – 1)d]
∴ S₅₁ = \(\frac{51}{2}\)[20 + 50 × 4]
∴ S₅₁ = 51 × 110
∴ S₅₁ = 5610
Thus, the sum of first 51 terms of the given AP is 5610.

Question 9.
If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.
Solution:
Here, S₇ = 49; S₁₇ = 289: S_n = ?
S_n = \(\frac{n}{2}\)[2a + (n – 1)d]
∴ S₇ = \(\frac{7}{2}\)[2a + 6d]
∴ 49 = 7(a + 3d)
∴ a + 3d = 7 ……..(1)
Again S₁₇ = \(\frac{17}{2}\)[2a + 16d]
∴ 289 = 17(a + 8d)
∴ a + 8d = 17 ……..(2)
Solving equations (1) and (2), we get d = 2 and a = 1.
∴ S_n = \(\frac{n}{2}\)[2a + (n – 1) d]
∴ S_n = \(\frac{n}{2}\)[2 + (n – 1)2]
∴ S_n = \(\frac{n}{2}\)[2 + 2n – 2]
∴ S_n = \(\frac{n}{2}\)(2n)
∴ S_n = n²
Thus, the sum of first n terms of the given AP is n².

Question 10.
Show that a₁, a₂, ……., a_n, ……. form an AP where a is defined as below:
1. a_n = 3 + 4n
2. a_n = 9 – 5n
Also find the sum of the first 15 terms in each case.
Solution:
1. a_n = 3 + 4n
a₁ = 3 + 4(1) = 7,
a₂ = 3 + 4(2) = 11,
a₃ = 3 + 4(3) = 15,
a₄ = 3 + 4(4) = 19 and so on.
Here, a₄ – a₃ = a₃ – a₂ = a₂ – a₁ = 4.
Hence, a_k+1 – a_k remains the everywhere.
Hence, a₁, a₂, a₃, ….. defined as a_n = 3 + 4n form an AP in which a = 7 and d = 4.
S_n = \(\frac{n}{2}\)[2a + (n – 1)d]
∴ S₁₅ = \(\frac{15}{2}\)[14 + 14 × 4]
∴ S₁₅ = 15 × 35
∴ S₁₅ = 525
The sum of first 15 terms of the given AP is 525.

2. a_n = 9 – 5n
a₁ = 9 – 5(1) = 4,
a₂ = 9 – 5(2) = -1,
a₃ = 9 – 5(3) = -6,
a₄ = 9 – 5(4) = -11 and so on.
Here, a₄ – a₃ = a₃ – a₂ = a₂ – a₁ = -5.
Hence, a_k+1 – a_k remains the same everywhere.
Hence, a₁, a₂, a₃, ….. defined as a_n = 9 – 5n form an AP in which a = 4 and d = -5.
S_n = \(\frac{n}{2}\)[2a + (n – 1) d]
∴ S₁₅ = \(\frac{15}{2}\)[8 + 14 (-5)]
∴ S₁₅ = \(\frac{15}{2}\)(-62)
∴ S₁₅ = 15 × (-31)
∴ S₁₅ = -465
The sum of first 15 terms of the given AP is -465.

Question 11.
If the sum of the first n terms of an AP is 4n – n², what is the first term (that is S₁)? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.
Solution:
For the given AP
S_n = 4n – n²
∴ S₁ = 4(1) – (1)² = 4 – 1 = 3,
S₂ = 4 (2) – (2)² = 8 – 4 = 4,
S₃ = 4(3) – (3)² = 12 – 9 = 3,
S₉ = 4 (9) – (9)² = 36 – 81 = -45,
S₁₀ = 4 (10) – (10)² = 40 – 100 = -60
Now, the first term = a – a₁ = S₁ = 3
The sum of first two terms S₂ = 4
The second term a₂ = S₂ – S₁ = 4 – 3 = 1
The third term a₃ = S₃ – S₂ = 3 – 4 = -1
The tenth term a₁₀ = S₁₀ – S₉
= -60 – (-45) = -15
Now, S_n = 4n – n²
∴ S_n-1 = 4(n – 1) – (n – 1)²
= 4n – 4 – n² + 2n – 1
= -n² + 6n – 5
Now nth term a_n = S_n – S_n-1
∴ a_n = (4n – n²) – (-n² + 6n – 5)
∴ a_n = 4n – n² + n² – 6n + 5
∴ a_n = -2n + 5

Question 12.
Find the sum of the first 40 positive integers divisible by 6.
Solution:
The first 40 positive integers divisible by 6 form the AP 6, 12, 18, ……, 240.
Here, a = 6; d = 12 – 6 = 6; n = 40 and l = 240.
S_n = \(\frac{n}{2}\)(a + l)
∴ S₄₀ = \(\frac{40}{2}\)(6 + 240)
∴ S₄₀ = 20 × 246
∴ S₄₀ = 4920
Thus, the required sum is 4920.

Alternate method:
Required sum
= 6 + 12 + 18 + … + 240
= 6(1 + 2 + 3 + … + 40)
= 6 × \(\frac{40 \times 41}{2}\) (1 + 2 + 3 + …. + n = \(\frac{n(n+1)}{2}\))
= 4920

Question 13.
Find the sum of the first 15 multiples of 8.
Solution:
The first 15 multiples of 8 form the AP 8, 16, 24, ….., 120.
Here, a = 8, d = 16 – 8 = 8, n = 15 and l = 120.
S_n = \(\frac{n}{2}\)(a + l)
∴ S₁₅ = \(\frac{15}{2}\)(8 + 120)
∴ S₁₅ = 15 × 64
∴ S₁₅ = 960
Thus, the required sum is 960.

Alternate method:
Required sum
= 8 + 16 + 24 + … + 120
= 8(1 + 2 + 3 + … + 15)
= 8 × \(\frac{15 \times 16}{2}\) (1 + 2 + 3 + … + n = \(\frac{n(n+1)}{2}\))
= 960.

Question 14.
Find the sum of the odd numbers between 0 and 50.
Solution:
The odd numbers between 0 and 50 form the AP 1, 3, 5, ….., 49.
Here, a = 1, d = 3 – 1 = -2, l = 49.
Let the last term be the nth term.
a_n = a + (n – 1) d
49 = 1 + (n – 1)²
2(n – 1) = 48
n – 1 = 24
n = 25
Now, S_n = \(\frac{n}{2}\)(a + l)
∴ S₂₅ = \(\frac{25}{2}\)(1 + 49)
∴ S₂₅ = 25 × 25
∴ S₂₅ = 625
Thus, the required sum is 625.

Question 15.
A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: ₹ 200 for the first day, ₹ 250 for the second day, ₹ 300 for the third day, etc.. the penalty for each succeeding day being ₹ 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days?
Solution:
The sums (in rupees) of penalty for delay of completion form the AP 200, 250, 300, …..
Here, a = 200; d = 250 – 200 = 50; n = 30 as the contractor has delayed the work by 30 days. The total penalty amount (in rupees) to be paid by the contractor will be given by S₃₀.
S_n = \(\frac{n}{2}\)[2a + (n – 1)d]
∴ S₃₀ = \(\frac{30}{2}\)[400 + (30 – 1)50]
∴ S₃₀ = 15 × 1850
∴ S₃₀ = 27750
Thus, the contractor has to pay a penalty of ₹ 27,750.

Question 16.
A sum of ₹ 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If cach prize is 20 less than its preceding prize, find the value of each of the prizes.
Solution:
₹ 700 is to be distributed as seven prizes such that each prize is ₹ 20 less than its preceding prize. Let the highest prize, i.e., the first prize be ₹ a. Then, the second prize will be of ₹ a – 20, the third prize will be of ₹ a – 40 and so on up to seven prizes. Hence, the amount (in rupees) of these prizes form a finite AP with seven terms as a, a – 20, a – 40, a – 60, a – 80, a – 100 and a – 120.
Here, the first term = a; d = (a – 20) – a = -20;
n = 7 and the sum of all the terms = S₇ = 700.
S_n = \(\frac{n}{2}\)[2a + (n – 1) d]
∴ 700 = \(\frac{7}{2}\)[2a + (7 – 1) (-20)]
∴ 200 = 2a + 6(-20)
∴ 200 = 2a – 120
∴ 2a = 320
∴ a = 160
Then, a – 20 = 140; a – 40 = 120; a – 60 = 100; a – 80 = 80; a – 100 = 60 and a – 120 = 40.
Thus, the values (in rupees) of those seven prizes are 160, 140, 120, 100, 80, 60 and 40.

Question 17.
In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class. will plant, will be the same as the class, in which they are studying, e.g.. a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?
Solution:
The number of trees that the three sections of Class I will plant = 1 + 1 + 1 = 3.
The number of trees that the three sections of Class II will plant = 2 + 2 + 2 = 6.
This system will continue till Class XII.
The number of trees that the three sections of Class XII will plant = 12 + 12 + 12 = 36.
Thus, the number of trees that will be planted will form a finite AP with 12 terms as 3, 6, 9, ….., 36.
Here, a = 3, d = 6 – 3 = 3, n = 12 and S₁₂ will give the total number of trees that will be planted.
S_n = \(\frac{n}{2}\)[2a + (n – 1) d]
∴ S₁₂ = \(\frac{12}{2}\)[6 + (12 – 1)³]
∴ S₁₂ = 6 × 39
∴ S₁₂ = 234
Thus, 234 trees will be planted by the students.

Question 18.
A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, as shown in the given figure. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take π = \(\frac{22}{7}\))

Hint: Length of successive semicircles is l₁, l₂, l₃, l₄, …with centres at A, B, A, B, ….., respectively.]
Solution:
We know that the length of a semicircle = πr, where r is the radius.
Length of 1st semicircle with centre A and radius 0.5 cm = l₁ = π × 0.5 cm.
Length of 2nd semicircle with centre B and radius 1 cm = l₂ = π × 1 cm.
Length of 3rd semicircle with centre A and radius 1.5 cm = l₃ = π × 1.5 cm.
This system continues till 13 semicircles are drawn.
Then, the 13th semicircle will be drawn with centre A and radius 6.5 cm. Length of 13th semicircle with centre A and radius 6.5 cm = l₁₃ = π × 6.5 cm.
Now, the total length of the spiral
= l₁ + l₂ + l₃ + … + l₁₃
= (π × 0.5) + (π × 1) + (π × 1.5) + … + (π × 6.5)
= π(0.5 + 1 + 1.5 + … + 6.5)
The sum inside the brackets is the sum of all the 13 terms of the finite AP 0.5, 1, 1.5, ….., 6.5.
For this AP a = 0.5; d = 1 – 0.5 = 0.5 and n = 13.
S_n = \(\frac{n}{2}\)[2a + (n – 1)d]
∴ S_n = [1 + (13 – 1) (0.5)]
∴ S_n = \(\frac{13}{2}\) × 7
Hence, the total length of the spiral
= π\(\left(\frac{13}{2} \times 7\right)\)
= \(\frac{22}{7} \times \frac{13}{2} \times 7\)
= 143 cm
Thus, the total length of the spiral made up of thirteen consecutive semicircles is 143 cm.

Question 19.
200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on (see the given figure). In how many rows are the 200 logs placed and how many logs are in the top row?

Solution:
The number of logs stacked in the first row from the bottom = 20.
The number of logs stacked in the second row from the bottom = 19.
The number of logs stacked in the third row from the bottom = 18.
This system continues till all the 200 logs are stacked.
Thus, the number of logs stacked in the rows form the finite AP 20, 19, 18,….. up ton-terms and the sum of those n terms is 200. Here, a = 20 and d = 19 – 20 = -1.
S_n = \(\frac{n}{2}\)[2a + (n – 1) d]
∴ 200 = \(\frac{n}{2}\)[40 + (n – 1) (-1)]
∴ 400 = n(40 – n + 1)
∴ 400 = n(41 – n)
∴ 400 = 41n – n²
∴ n² – 41n + 400 = 0
∴ n² – 16n – 25n + 400 = 0
∴ n (n – 16) – 25 (n – 16) = 0
∴ (n – 16) (n – 25) = 0
n – 16 = 0 or n – 25 = 0
n = 16 or n = 25
Here, both the answers are admissible. Hence, we verify by the value of 16th term and 25th term.
a_n = a + (n – 1) d
∴ a₁₆ = 20 + 15(-1) = 5
∴ a₂₅ = 20 + 24(-1) = -4
Thus, for the 25th row, the number of logs in the row becomes negative. This is inadmissible.
Hence, n ≠ 25.
∴ n = 16 and a₁₆ = 5.
Thus, the 200 logs are placed in 16 rows and in the top row there are 5 logs.

Question 20.
In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line (see the given figure).

A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?
[Hint: To pick up the first potato and the second potato, the total distance (in metres) run by a competitor is 2 × 5 + 2 × (5 + 3)].
Solution:
The distance (in metres) to be covered to pick up first potato = 2 × 5 = 10.
The distance (in metres) to be covered to pick up second potato = 2 × (5 + 3) = 16.
The distance (in metres) to be covered to pick up third potato = 2 × (5 + 3 + 3) = 22.
Thus, the distances to be covered to pick up 10 potatoes form the finite AP 10, 16, 22, …. up to 10 terms.
Here, a = 10, d = 16 – 10 = 6, n = 10 and Sn will give the total distance (in metres) that the competitor has to run.
S_n = \(\frac{n}{2}\)[2a + (n – 1) d]
∴ S₁₀ = \(\frac{10}{2}\)[20 + (10 – 1)6]
∴ S₁₀ = 5 × 74
∴ S₁₀ = 370
Thus, the total distance that the competitor has to run is 370 metres.