Class 10

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 14 Statistics Ex 14.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 14 Statistics Ex 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:

Now, by drawing the points on the graph
i.e. (120, 12); (140, 26); (160, 34); (180, 40); (200, 50).
We get graph of less than type cumulative frequency.

Scale chosen:
On x-axis 10 units = Rs. 10
On y-axis 10 units = 5 workers.

Question 2.
During the medial 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 verWy the result by using the formula.
Solution:

Now, By drawing the points on the graph i.e., (38, 0); (40, 3); (42, 5); (44, 9); (46, 14); (48, 28) ; (50, 32) ; (52, 35) we get graph of less than type cumulative frequency.

Scale Chosen:
On x-axis, 10 units = 2 kg
On y-axis units = 5 students

From above graph, it is clear that
Median = 46.5 kg ; which lies in class interval 46 – 48.
Now, in the given table
\(\Sigma f_{i}\) = n = 35

∴ \(\frac{n}{2}=\frac{35}{2}\) = 17.5 ; which lies in the interval 46 – 48.

∴ Median class = 46 – 48
So, l = 46; n = 35; f = 14; cf = 14 and h = 2

Using formula, Median = l + \(\left\{\frac{\frac{n}{2}-c f}{f}\right\}\) × h

Median = 46 + \(\left\{\frac{\frac{35}{2}-14}{14}\right\}\) × 2

= 46 + \(\left\{\frac{\frac{35-28}{2}}{14}\right\}\)

= 46 + \(\frac{7}{2} \times \frac{1}{14}\) = 46 + \(\frac{1}{2}\)

= 46 + 0.5 = 46.5
From above discussion and graph; it is clear that median is same in both cases. Hence, Median weight of students is 46.5 kg.

Question 3.
The following table gives production yield per hectare of wheat of 1(X) farms of a village.

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

Solution:

Now, by drawing the points on the graph i.e. (50, 100); (55, 98); (60, 90); (65, 78); (70, 54); (75, 16)
we get graph of more than type cumulative frequency.

Scale chosen:
On x-axis 10 units = 5 kg/ha
On y-axis 10 units = 10 forms

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter Statistics Ex 14.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.3

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

Solution:

Here, \(\Sigma f_{i}\) = 68 then \(\frac{n}{2}=\frac{68}{2}\) = 34
Which lies in interval 125 – 145
Median class = 125 – 145
So, l = 125; n = 68; f = 20; çf = 22 and h = 20

Using formula, Median = l + \(\left[\frac{\frac{n}{2}-c f}{f}\right]\) × h

= 125 + \(\left\{\frac{\frac{68}{2}-22}{20}\right\}\) × 20

=125+ \(\frac{34-22}{20}\) × 20

= 125 + 12 = 137

For mean:

From above data, assumed mean (a) = 135
Width of class (h) = 20
∴ \(\bar{u}=\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}=\frac{7}{68}\) = 0.102
Using formula, Mean \((\overline{\mathrm{X}})=a+h \bar{u}\)
\(\overline{\mathrm{X}}\) = 135 + 20 (0.102)
= 135 + 2.04 = 137.04.

For Mode:
In the given data,
Maximum frequency is 20 and it correspond to 125 – 145.
∴ Modal class = 125 – 145
So l = 125; f₁ = 20; f₀ = 13; f₂ = 14and h = 20
Using formula, Mode = l + \(\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right)\) × h

Mode = 125 + \(\left(\frac{20-13}{2(20)-13-14}\right)\) × 20

= 125 + \(\frac{7}{40-27}\) × 20

= 125 + \(\frac{140}{13}\)
= 125 + 10.76923
= 125 + 10.77 = 135.77.
Hence. median, mean and mode of given data is 137 units: 137.04 units and 135.77 units.

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

Solution:

In thegiven data, \(\Sigma f_{i}\) = n = 60
∴ \(\frac{n}{2}=\frac{60}{2}\) = 30
Also, median of the distribution = 28.5 ………….(Given)
which lies in the class interval 20 – 30
Median class = 20 – 30
So, l = 20; f = 20; cf = 5 + x; h = 10
From table, it is clear that 45 + x + y = 60
x + y = 60 – 45 = 15
or x + y = 15 ……………….(1)
Now, using formula, Median = l + {\(\left\{\frac{\frac{n}{2}-c f}{f}\right\}\)

28.5 = 2o + \(\left\{\frac{30-(5+x)}{20}\right\}\)

or 28.5 = 20 + \(\frac{30-5-x}{2}\)

or 28.5 = \(\)

or 2(28.5) = 65 – x
or 57.0 = 65 – x
or x = 65 – 57 = 8
∴ x = 8
Substitute this value of x in (1), we get
8 + y = 15
Hence, values of x and y is 8 and 7.

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

Solution:

Here, \(\Sigma f_{i}\) = n = 100
then, \(\frac{n}{2}=\frac{100}{2}\) = 50, which lies in the interval 35 – 40
∴ Median class = 35 – 40
So, l = 35; n = 100; f = 33; cf = 45 and h = 5
Using formula, Median = l + \(\left\{\frac{\frac{n}{2}-c f}{f}\right\}\) × h

= 35 + \(\left\{\frac{\frac{100}{2}-45}{33}\right\}\) × 5

= 35 + \(\frac{50-45}{33}\) × 5

= 35 + \(\frac{25}{33}\)
= 35 + 0.7575 = 35 + 0.76 (approx.) = 35.76
Hence, median age of given data is 35.76 years.

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.
Solution:
Since the frequency distribution is not continuous, so firstly we shall make it continuous.

Here, \(\Sigma f_{i}\) = n = 40
then, \(\frac{n}{2}=\frac{40}{2}\) = 20, which lies in the interval 144.5 – 153.5
∴ Median class = 144.5 – 153.5
So, l = 144.5; f = 12; cf = 17; h = 9
Using formula, Median = l + \(\left\{\frac{\frac{n}{2}-c f}{f}\right\}\) × h

Median = 144.5 + \(\left\{\frac{20-17}{12}\right\}\) × 9

= 144.5 + \(\frac{3 \times 9}{12}\)
= 144.5 + 225 = 146.75
Hence, median length of the leaves is 146.75 mm.

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

Find the median life time of a lamp.

Solution:

Here, \(\Sigma f_{i}\) = n = 400
∴ \(\frac{n}{2}=\frac{400}{2}\) = 200; which lies in the interval 3000 – 3500.
∴ Median class = 3000 – 3500
So, l = 3000; n = 400; f = 86; cf = 130 and h = 500
Using formula, Median = l + \(\left\{\frac{\frac{n}{2}-c f}{f}\right\}\) × h

Median = 3000 + \(\left\{\frac{\frac{400}{2}-130}{86}\right\}\) × 500

= 3000 + \(\left(\frac{200-130}{86}\right)\) × 500

= 3000 + \(\frac{70 \times 500}{86}\) + 406.9767441

= 3000 + 406.98 (approx.) = 3406.98
Hence, median life time 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 alphabets 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.
For Median:

Here, Here, \(\Sigma f_{i}\) = n = 100
∴ \(\frac{n}{2}=\frac{100}{2}\) = 50, which lies in interval 7 – 10.
∴ Median class = 7 – 10
So, l = 7; n = 100; f = 40; cf = 36 and h = 3
Using formula, Median = l + \(\left\{\frac{\frac{n}{2}-c f}{f}\right\}\) × h

Median = 7 + \(\left\{\frac{\frac{100}{2}-36}{40}\right\}\) × 3

= 7 + \(\left\{\frac{50-36}{40}\right\}\) × 3

= 7 + \(\frac{14 \times 3}{40}\)

= 7 + \(\frac{21}{20}\) = 7 + 1.05 = 8.05
Hence, the median of letters in the surnames is 8.05.

For Mean:

From above data, Assumed Mean (a) = 8.5
Width of class (h) = 3
∴ \(\bar{u}=\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}=\bar{u}=\frac{-6}{100}\) = – 0.06

Using formula, Mean \((\overline{\mathrm{X}})=a+h \bar{u}\)
\(\bar{X}\) = 8.5 + 3 (- 0.06) = 8.5 – 0.18 = 8.32
Hence, mean number of letters in the surnames is 8.32.

For Modal:
In the given data Maximum frequency is 44 and it corresponds to interval 7 — 10
∴ Modal class = 7 – 10
So l = 7; f₁ = 40; f₀ = 30; f₂ = 16 and h = 3
Using formula, Mode = l + \(\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right)\) × h

Mode = 7 + \(\left(\frac{40-30}{2(40)-30-16}\right)\) × 3

= 7 + \(\frac{10}{80-46}\) × 3

= 7 + \(\frac{30}{34}\) = 7 + 0.882352941

= 7 + 0.88 (approx.) = 7.88.
Hence. modal size of the surnames is 7.88 letters.

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

Solution:

Here, \(\Sigma f_{i}\) = n = 30
∴ \(\frac{n}{2}=\frac{30}{2}\) = 15; which lies in the interval 55 – 60.
∴ Median class = 55 – 60
So, l = 55; n = 30; f = 6; cf = 13 and h = 5
Using formula, Median = l + \(\left\{\frac{\frac{n}{2}-c f}{f}\right\}\) × h

Median = 55 + \(\left\{\frac{\frac{30}{2}-13}{6}\right\}\) × 5

= 55 + \(\left\{\frac{15-13}{6}\right\}\) × 5

= 55 + \(\frac{2 \times 5}{6}\)

= 55 + \(\frac{5}{3}\) = 55 + 1.66666
= 55 + 1.67 (approx.) = 56.67
Hence, median weight of the students are 56.67 kg.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter Statistics Ex 14.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.2

Question 1.
The following table shows the ages of the patients admitted in a hospital during a year:

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

Solution:
For mode:
In the given data, Maximum frequency is 23 and it corresponds to the class interval 35 – 45
Modal class = 35 – 45
So, l = 35; f₁ = 23; f₀ = 21; f₂ = 14 and h = 10
Using fonnula, Mode l = l + \(\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right)\) × h
Mode = 35 + \(\left[\frac{23-21}{2(23)-21-14}\right]\) × 10
= 35 + \(\frac{2}{46-35}\) × 10
= 35 + \(\frac{20}{11}\) = 35 + 1.8 = 36.8.

For Mean:

From above data,
Assumed mean (a) = 30
Width of class (h) = 10
∴ \(\bar{u}=\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}=\frac{43}{80}\) = 0.5375
Using formula, Mean \((\overline{\mathrm{X}})=a+h \bar{u}\)
\(\bar{X}\) = 30 + 10 (0.5375)
= 30 + 5.375 = 35.375 = 35.37
Hence, mode of given data is 36.8 years and mean of the given data is 35.37 years. Also, it is clear from above discussion that average age of a patient admitted in the hospital is 35.37 years and maximum number of patients admitted in the hospital are of age 36.8 years.

Question 2.
The following data gives the information on the observed lifetimes (in hours) of 225 electrical components:

Determine the ¡nodal lifetimes of the components.
Solution:
In the given data.
Maximum frequency is 61 and it corresponds to the class interval 60 – 80.
∴ Model class = 60 – 80
So, l = 60; f₁ = 61 ; f₀ = 52; f₂ = 38 and h = 20
Using formula, Mode = l + \(\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right)\) × h

Mode = 60 + \(\left(\frac{61-52}{2(61)-52-38}\right)\) × 20

= 60 + \(\frac{9}{122-52-38}\) × 20

= 60 + \(\frac{9}{32}\) × 20

= 60 + \(\frac{180}{32}\)

= 60 + 5.625 = 65.625
Hence, modal Lifetimes of the components is 65.625 hours.

Question 3.
The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure:

Solution:
For Mode: In the given data.
Maximum frequency is 40, and it corresponds to the class interval 1500 – 2000.
∴ Model class = 1500 – 2000
So, l = 1500; f₁ = 40; f₀ = 24; f₂ = 33 and h = 500
Using formula, Mode = l + \(\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right)\) × h

= 1500 + \(\left\{\frac{40-24}{2(40)-24-33}\right\}\) × 500

= 1500 + \(\left\{\frac{16}{80-24-33}\right\}\) × 500

= 1500 + \(\frac{16 \times 500}{23}\)

= 1500 + \(\frac{8000}{23}\) = 1500 + 347.83 = 1847.83

For Mean:

From above data,

Assumed Mean (a) = 2750
Length of width (h) = 500
\(\bar{u}=\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}=-\frac{35}{200}\) = – 0.175
Using formula, Mean \((\overline{\mathrm{X}})=a+h \bar{u}\)
\(\overline{\mathbf{X}}\) = 2750 + 500 (- 0.175)
= 2750 – 87.50 = 2662.50
Hence, the modal monthly expenditure of family is 1847.83 and the mean monthly expenditure is 2662.50.

Question 4.
The following distribution gives the slate-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.

Solution:
For Mode:
In the given data,
Maximum frequency is 10 and it corresponds to the class interval is 30 – 35.
∴ Modal class = 30 – 35.
So, l = 30; f₁ = 10; f₀ = 9; f₂ = 3 and h = 5
using formula, Mode = l + \(\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right)\) × h
Mode = 30 + \(\left(\frac{10-9}{2(10)-9-3}\right)\) × 5
= 30 + \(\frac{1}{20-12}\) × 5
= 30 + \(\frac{5}{8}\) = 30 + 0.625 = 30.625 = 30.63 (approx.)

For mean:

From above data, Assumed Mean (a) = 32.5
Width of class (h) = 5
∴ \(\bar{u}=\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}=-\frac{23}{35}\) = – 0.65

Using formula, Mean \((\overline{\mathrm{X}})=a+h \bar{u}\)
\(\overline{\mathbf{X}}\) = 32.5 + 5 (- 0.65)
= 32..5 – 3.25 = 29.25 (approx.)
Hence, mode and mean of given data is 30.63 and 29.25. Also, from above discussion, it clear that states/U.T. have student per teacher is 30.63 and on average, this ratio is 29.25.

Question 5.
The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches

Find the mode of the data.
Solution:
In the given data,
Maximum frequency is 18 and it corresponds to the class interval 4000 – 5000.
∴ Modal class = 4000 – 5000
So, l = 4000; f₁ = 18; f₀ = 4; f₂ = 9 and h = 1000
Using formula, Mode = l + \(\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right)\) × h

Mode = 4000 + \(\left(\frac{18-4}{2(18)-4-9}\right)\) × 1000

= 4000 + \(\frac{14}{36-13}\) × 1000

= 4000 + \(\frac{14000}{23}\) = 4000 + 608.6956

= 4000 + 608.7 = 4608.7 (approx.)
Hence, mode of the given data is 4608.7.

Question 6.
A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised It in the table given below. Find the mode of the data:

Solution:
In the given data,
Maximum frequency is 20 and it corresponds to the class interval 40 – 50
∴ Modal Class = 40 – 50
So, l = 40; f₁ = 20; f₀ = 12; f₂= 11 and h = 10
Using formula, Mode = l + \(\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right)\) × h

Mode = 40 + \(\left(\frac{20-12}{2(20)-12-11}\right)\) × 10

= 40 + \(\frac{8}{40-23}\) × 10

= 40 + \(\frac{80}{17}\) = 40 + 4.70588

= 40 + 4.7 = 44.7 (approx.)
Hence, mode of the given data is 44.7 cars.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter Statistics Ex 14.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.1

Question 1.
A survey was conducted by a group of students as a part of their enviroment 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:
Since, number of plants and houses are small in their values; so we must use direct method

Mean X = \(\overline{\mathrm{X}}=\frac{\Sigma f_{i} x_{i}}{\Sigma f_{i}}\)

\(\frac{162}{20}\) = 8.1

Hence, mean number of plants per house is 8.1.

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:

From given data,
Assumed Mean (a) = 150
and width of the class (h) = 20
∴ \(\bar{u}=\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}\)
= \(\frac{-12}{50}\) = – 0.24

Using formula, Mean \((\overline{\mathrm{X}})=a+h \bar{u}\)
= 150 + (20) (- 0.24)
= 150 – 4.8 = 145.2
Hence,mean daily wages of the workers of factory is ₹ 145.20.

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

Solution;

From the above data,
Assumed mean (a) = 18
Using formula, Mean \((\overline{\mathrm{X}})=a+\frac{\sum f_{i} d_{i}}{\Sigma f_{i}}\)

\(\bar{X}=18+\frac{2 f-40}{44+f}\)
But. Mean of data \((\bar{x})\) = 18 …(Given)
∴ 18 = 18 + \(\frac{2 f-40}{44+f}\)
or \(\frac{2 f-40}{44+f}\) = 18 – 18 = 0
or 2f – 40 = 0
or 2f = 40
or f = \(\frac{40}{2}\) = 20
Hence, missing frequency f is 20.

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

Solution:

From above data,
Assumed Mean (a) = 75.5
Width of Class (h) = 3
∴ \(\bar{u}=\frac{\sum f_{i} u_{i}}{\sum f_{i}}\)
= \(\frac{4}{30}\) (approx.)
Using formula,
Mean \((\overline{\mathrm{X}})=a+h \bar{u}\)
= 75.5 + 3 (0.13) = 75.5 + 0.39
\(\overline{\mathrm{X}}\) = 75.89
Hence, mean heart beats per minute for women is 75.89.

Question 5.
In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number 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:
Since, values of number of mamgoes and number of boxes are large numerically. So, we must use step-deviation method.

From above data,
Assumed Mean (a) = 57
Width of class (h) = 3
∴ \(\bar{u}=\frac{\sum f_{i} u_{i}}{\sum f_{i}}\)

\(\bar{u}=\frac{25}{400}\) = 0.0625

Using formula, Mean \((\overline{\mathrm{X}})=a+h \bar{u}\)
\(\bar{X}\) = 57 + 3(0.0625)
= 57 + 0.1875 = 57.1875 = 57.19.
Hence, mean number of mangoes kept in a packing box is 57.19.

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:

From above data,
Assumed mean (a) = 225
Width of class (h) = 50
∴ \(\bar{u}=\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}\)

\(\bar{u}=-\frac{7}{25}\) = 0.28
Using formula,
Mean \((\overline{\mathrm{X}})=a+h \bar{u}\)
\(\bar{X}\) = 225 + 50 (- 0.28)
\(\bar{X}\) = 225 – 14
\(\bar{X}\) = 211
Hence, mean daily expenditure on food is ₹ 211.

Question 7.
To find out the concentration of SO₂ in the air (in parts per million, 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:

From above data,
Assumed mean (a) = 0.10
Width of class (h) = 0.04
∴ \(\bar{u}=\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}\)

\(\bar{u}=-\frac{1}{30}\) = – 0.33(approx.)
Using formula, Mean \((\overline{\mathrm{X}})=a+h \bar{u}\)
\(\bar{X}\) = 0.10 + 0.04 (- 0.33)
= 0.10 – 0.0013 = 0.0987 (approx.)
Hence, mean concentration of SO₂ in the air is 0.0987 ppm.

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

Solution:

From above data,
Assumed Mean (a) = 17
Using formula, Mean(\(\bar{X}\)) = a + \(\frac{\Sigma f_{i} d_{i}}{\Sigma f_{i}}\)
\(\bar{X}\) = 17 + \(-\frac{181}{40}\)
= 17 – 4.52 = 12.48.
Hence, mean 12.48 number of days a student was absent.

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

Solution:

From the above data,
Assumed Mean (a) = 70
Width of class (h) = 10
∴ \(\bar{u}=\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}=\frac{-2}{35}\) = – 0.057

Using formula, Mean \((\overline{\mathrm{X}})=a+h \bar{u}\)
\(\bar{X}\) = 70 + 10 (- 0.057)
= 70 – 0.57 = 69.43
Hence, mean literacy rate is 69.43%.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.5 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.5

Question 1.
A copper wire 3 mm in diameter is wound about a cylinder whose length is 12 cm, and diameter 10 cm, so as to cover the curved surface of the cylinder. Find the length and mass of the wire, assuming the density of copper to 8.88 g per cm³.
Solution:
Diametcr of wire (d) = 3 mm
∴ Radius of wire (r) = \(\frac{3}{2}\) mm = \(\frac{3}{20}\) cm

Diameter of cylinder = 10 cm
Radius of cylinder (R) = 5 cm
Height of cylinder (H) = 12 cm
Circumference of cylinder = length of wire used in one turn
2πR = length of wire used in one turn
2 × \(\frac{22}{7}\) × 5 = length of wire used in one turn
\(\frac{220}{7}\) = length of wire used in one turn
Number of turn used = \(\frac{\text { Height of cylinder }}{\text { Diameter of wire }}\)
= \(\frac{12 \mathrm{~cm}}{3 \mathrm{~mm}}=\frac{12}{3} \times 10\) [1 mm = \(\frac{1}{10}\) cm]
= \(\frac{120}{3}\) = 40

∴ length of wire used = Number of turns × length of wire used in one turn
H = 40 × \(\frac{220}{7}\) = 1257.14 cm
Volume of wire used = πr²H
= \(\frac{22}{7} \times \frac{3}{20} \times \frac{3}{20} \times 1257.14\) = 88.89 cm³
Mass of 1 cm³ = 8.88 gm
Mass of 88.89 cm³ = 8.88 × 88.89 = 789.41 gm
Hence, length of wire is 1257.14 cm
and mass of wire is 789.41 gm.

Question 2.
A right triangle, whose sides are 3 cm and 4 cm (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area the double cone so formed. (Choose value of t as found
appropriate.)
Solution:
Let ∆ABC be the right triangle right angled at A whose sides AB and AC measure 3 cm and 4 cm respectively.
The length of the side BC (hypotenuse) = \(\sqrt{3^{2}+4^{2}}=\sqrt{9+16}\) = 5 cm
Here, AO (or A’O) is the radius of the common base of the double cone formed revolving the right triangle about BC.
Height of the cone BAA’ is BO and slant height is 3 cm.
Height of the cone CAA’ is CO and slant height is 4 cm.
Now, ∆AOB ~ ∆CAB (AA similarity)

∴ \(\frac{\mathrm{AO}}{4}=\frac{3}{5}\)

⇒ AO = \(\frac{4 \times 3}{5}=\frac{12}{5}\) cm

Also \(\frac{\mathrm{BO}}{3}=\frac{3}{5}\)

⇒ BO = \(\frac{\mathrm{BO}}{3}=\frac{3}{5}\) cm
Thus CO = BC – OB
= 5 – \(\frac{9}{5}\) = \(\frac{16}{5}\) cm
Now, Volume of double cone = [Volume of Cone ABA’] + [Volume of cone ACA’]

∴ Volume of double cone = 30 cm³.
Also, surface area of double cone = [Surface area of Cone ABA’] + [surface area of Cone ACA’]
= π .AO.AB + π. AO.A’C
= \(\left(\frac{22}{7} \times \frac{12}{5} \times 3\right)+\left(\frac{22}{7} \times \frac{12}{5} \times 4\right)\)

= \(\left(\frac{792}{35} \times \frac{1056}{35}\right)\)
= \(\frac{1848}{35}\) = 52.75 cm²

Question 3.
A cistern, Internally measuring 150 cm × 120 cm × 110 cm, has 129600 cm3 of water in it. Porous bricks are placed in the water until the cistern is full to the brim. Each brick absorbs one-seventeenth of its own volume of water. How many bricks can be put in without the water overflowing, each brick being 22.5 cm × 7.5 cm × 6.5 cm?
Solution:
Volume of one brick = 22.5 × 7.5 × 6.5 cm³ = 1096.87 cm³
Volume of cistern = 150 × 120 × 110 cm³ = 1980000 cm³
Let number of bricks used = n
Total volume of n bricks = n (volume of one brick) = n [1096.871] cm³
Volume of water = 129600 cm³
Volume of water available for bricks = 1980000 – 129600 = 1850400 cm³

Each bricks absorb \(\frac{1}{17}\)th of its volume in water
Volume of water available for bricks = \(\frac{17}{16}\) × volume of water available for bricks
= \(\frac{17}{10}\) × 1850400
Volume of water available for bricks = 1966050 cm³
Total volume of n bricks = Volume of water available for bricks
n[1096.87] cm³ = 1966050 cm³

n = \(\frac{1966050}{1096.87}\)
n = 1792.42
Hence, Number of bricks used = 1792.

Question 4.
In one fortnight of a given month, there was a rainfall of 10 cm in a river valley. If the area of the valley is 97280 km², show that the total rainfall was approximately equivalent to the addition to the normal water of three rivers each 1072 km long, 75 m wide and 3 m deep.
Solution:
Area of valley = 97280 km²
Rainfall in valley = 10 cm
∴ Volume of total rainfall = 97280 × \(\frac{10}{100}\) × \(\frac{1}{1000}\) km³
= 9.728 km³

Question 5.
An oil funnel made of tin sheet consists of a cylindrical portion 10 cm long attached to a frust.un of a cone. if the total height is 22 cm, diameter of the cylindrical portion is 8 cm and the diameter of the top of the funnel is 18 cm, find the area of the tin sheet required to make the funnel.

Solution:
Diameter of top of funnel = 18 cm
∴ Radius of top of funnel (R) = \(\frac{18}{2}\) cm = 9 cm
Diameter of bottom of funnel = 8 cm
Radius of bottom of funnel (r) = 4 cm
Height of cylindrical portion (h) = 10 cm
Height of frustum (H) = (22 – 10) = 12 cm
Slant height of frustum (l)
= \sqrt{\mathrm{H}^{2}+(\mathrm{R}-r)^{2}}\(\)

= \(\sqrt{(12)^{2}+(9-4)^{2}}\)

= \(\sqrt{144+(5)^{2}}\)

= \(\sqrt{144+25}\) = \(\sqrt{169}\)
Slant height of frustum (l) = 13 cm
Area of metal sheet required curved
surface area of cylindrical base + curved surface of frustum = 2πrh + πL [R + r]
= 2 × \(\frac{22}{7}\) × 4 × 10 + \(\frac{212}{7}\) × 13 [9 + 4] cm²
= 251.42 + 531.14 = 782.56 cm²

Question 6.
Derive the formula for the curved surface area and total surface area of a frustum of a cone, given to you in Section 13.5, using the symbols as explamed.
Solution:
A frustum of a right circular cone has two unequal flat circular bases and a curved surface. Let ACDB be the frustum of the cone obtained by removing the portion VCÐ. The line-segment OP joining the centres of two bases is called the height of the frustum. Each of the line-segments AC and BD of frustum ACDB is called its slant height.

Let R and r be the radii of the circular end faces (R > r) of the frustum ACDB of the cone (V, AB). We complete the conical part VCD. Let h and l be the vertical height and slant height respectively. Then OP = h and AC = BD = l.

The frustum of the right circular cone can be viewed as the difference of the two right circular VAB and VCD.
Let the height of the cone VAB be h₁ and its slant height l₁ i.e. VP = h₁ and VA = VB = l₁.

Now from right ∆DEB, we have
DB² = DE² + BE²

⇒ l² = h² + (R – r)²
⇒ l = \(\sqrt{h^{2}+(\mathrm{R}-r)^{2}}\)
Again ∆VOD ~ ∆VPB

⇒ \(\frac{V D}{V B}=\frac{O D}{P B}\)

⇒ \(\frac{l_{1}-l}{l_{1}}=\frac{r}{R}\)

⇒ 1 – \(\frac{l}{l_{1}}=\frac{r}{\mathrm{R}}\)

⇒ \(\frac{l}{l_{1}}=1-\frac{r}{\mathrm{R}}=\frac{\mathrm{R}-r}{\mathrm{R}}\)

⇒ l₁ = \(\frac{l \mathrm{R}}{\mathrm{R}-r}\)

⇒ Now, l₁ – l₁ = \(\frac{l \mathrm{R}}{\mathrm{R}-r}-l=\frac{l r}{\mathrm{R}-r}\)

Curved surface area of the frustum of cone = πRl₁ – πr(l₁ – l)
[Curved surface area of a cone = π × r × 1]

= πR. \(\frac{l R}{R-r}\) – πr. \(\frac{l r}{\mathrm{R}-r}\)

= πl (\(\frac{\mathrm{R}^{2}-r^{2}}{\mathrm{R}-r}\)) = \(\frac{\pi l(\mathrm{R}-r)(\mathrm{R}+r)}{(\mathrm{R}-r)}\)
= πl (R + r) sq. units.
∴ Curved snafaue area of the frustum of right circular cone = πl (R + r) sq. units,
where l = \(\sqrt{h^{2}+(\mathrm{R}-r)^{2}}\)
and total surface area of frustum of right circular cone = Curved surface area + Area of base + Area of top
= πl (R + r) + πR² + πr²
∴ π [R² + r² + l (R + r)] sq. units.

Question 7.
Derive the formula for the volume of the frustum of a cone, given to you in section 13.5, using the symbols as explained.
Solution:
A frustum of a right circular cone has two unequal flat circular bases and a curved surface. Let ACDB be the frustum of the cone obtained by removing the portion VCD. The line-segment OP joining the centres of two bases is called the height of the frustum. Each of the line-segments AC and BD of frustum ACDB is called its slant height.

Ler R and r be the radii of the circular end faces (R > r) of the fnistum ACDB of the cone (V, AB). We complete the conical part VCD. Let h and l be the vertical height and slant height respectively. Then OP = h and AC = BD = l. The frustum of the right circular cone can be viewed as the difference of the two right circular VAB and VCD.
Let theheight of the cone VAB be h₁ and its slant height l₁ i.e.VP = h₁ and VA = VB = l₁.
∴ The height of the cone VCD = VP – OP = h₁ – h
Since right ∆s VOD and VPB are similar
⇒ \(\frac{\mathrm{VO}}{\mathrm{VP}}=\frac{\mathrm{OD}}{\mathrm{PB}}=\frac{h_{1}-h}{h_{1}}=\frac{r}{\mathrm{R}}\)

⇒ 1 – \(\frac{h}{h_{1}}=\frac{r}{\mathrm{R}}\)

⇒ \(\frac{h}{h_{1}}=1-\frac{r}{\mathrm{R}}=\frac{\mathrm{R}-r}{\mathrm{R}}\)

⇒ h₁ = \(\frac{h \mathrm{R}}{\mathrm{R}-r}\)

⇒ Height of the cone VCD = \(\frac{h \mathrm{R}}{\mathrm{R}-r}-h=\frac{h \mathrm{R}-h \mathrm{R}+h r}{\mathrm{R}-r}=\frac{h r}{\mathrm{R}-r}\)

Volume of the frustrum ACDB of the cone (V,AB) = Volume of the cone (V, AB) – Volume of the cone (V, CD)
= \(\)

= \(\frac{\pi}{3}\left(\mathrm{R}^{2} \cdot \frac{h \mathrm{R}}{\mathrm{R}-r}-r^{2} \cdot \frac{h r}{\mathrm{R}-r}\right)\)

= \(\frac{\pi h}{3}\left(\frac{\mathrm{R}^{3}-r^{3}}{\mathrm{R}-r}\right)\)

= \(\frac{1}{3} \pi h \times \frac{(\mathrm{R}-r)\left(\mathrm{R}^{2}+\mathrm{R} r+r^{2}\right)}{(\mathrm{R}-r)}\)

= \(\frac{1}{3}\) πh (R² + Rr + r²)
Hence, volume of the frustum of cone is \(\frac{1}{3}\) πh (R² + Rr + r²).

Again if A₁ and A₂ (A₁ > A₂) are the surface areas of the two circular bases, then A₁ = πR² andA₂ = πr²

Now volume of the frustum of cone,

= \(\frac{1}{3}\) πh (R² + r² + Rr)

= \(\frac{h}{3}\) (πR² + πr² + \(\sqrt{\pi \mathrm{R}^{2}} \sqrt{\pi r^{2}}\))

= \(\frac{h}{3}\) (A₁ + A₂ + \(\sqrt{A_{1} A_{2}}\))

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.4

Question 1.
A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass.
Solution:

Radius of upper end (R) = 2 cm
Radius of lower end (r) = 1 cm
Height of glass (H) = 14 cm
Glass is in the shape of frustum
Volume of frustum = \(\frac{1}{3}\) π [R² + r² +Rr]H

= \(\frac{1}{3}\) × \(\frac{22}{7}\) [(2)² + (1)² + 2 × 1] 14

= \(\frac{1}{3}\) × \(\frac{22}{7}\) [4 + 1 + 2] 14

= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 7 × 14

= \(\frac{22 \times 14}{3}\)
Hence, Volume of glass = 102.67 cm³.

Question 2.
The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum.
Solution:
Slant height of frustum = 4 cm

Let radius of upper end and lower ends are R and r
Circumference of upper end = 18 cm
2πR = 18
R = \(\frac{18}{2 \pi}=\frac{9}{\pi}\) cm
Circumference of lower end = 6 cm
2πr = 6 cm
r = \(\frac{6}{2 \pi}=\frac{3}{\pi}\) cm
Curved surface area of frustum = π [R + r] l
= π \(\left[\frac{9}{\pi}+\frac{3}{\pi}\right]\) 4
= π \(\left[\frac{9+3}{\pi}\right]\) 4
= 12 × 4 = 48 cm²
Hence, Curved surface area of frustum = 48 cm².

Question 3.
A fez, the cap used by the Turks, is shaded like the frustum of a cone. If its radius on the open side is 10 cm, radius at the upper base is 4 cm and its slant height is 15 cm, find the area of material used for making it.

Solution:

Radius of lower end of frustum (R) = 10 cm
Radius of upper end of frustum (r) = 4 cm
Slant height of frustum (l) = 15 cm
Curved surface area of frustum = πL [R+ r]
= \(\frac{22}{7}\) × 15 [10 + 4]
= \(\frac{22}{7}\) × 15 × 14
= 22 × 15 × 2 = 660 cm²
Area of the closed side = πr² = \(\frac{22}{7}\) × (4)²
= \(\frac{22}{7}\) × 4 × 4 = \(\frac{352}{7}\) cm²
Total area of the material used = Curved surface area of frustum + Area of the closed side
= 660 + 50.28 = 710.28 cm²
Hence, Total material used = 710.28 cm².

Question 4.
A container opened from the top is made up of a metal sheet ¡s in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm, respectively. Find the cost of the milk which can completely fill the container, at the rate, of ₹ 20 per litre. Also fmd the cost of metal sheet used to make the container, If it costs ₹ 8 per 100 cm². (Take π = 3:14.)
Solution:

Radius of upper end of container (R) = 20 cm
Radius of lower end of container (r) = 8 cm
Height of container (H) = 16 cm
Slant height (l) = \(\sqrt{\mathrm{H}^{2}+(\mathrm{R}-r)^{2}}\)
= \(\sqrt{(16)^{2}+(20-8)^{2}}\) = \(\sqrt{256+144}\)
Slant height (l) = \(\sqrt{400}=\sqrt{20 \times 20}\) = 20 cm

Capacity of the container = \(\frac{1}{3}\) πH[R² + r² + Rr]
= \(\frac{1}{3}\) × 3.14 × 16 [(20)² + (8)² + 20 × 8]
= \(\frac{3.14 \times 16}{3}\) [400 + 64 + 100]
= 3.14 × 16 × 624 = 10449.92 cm³
Milk in the container = 10449.92 cm³
= \(\frac{10449.92}{1000}\) litres [∵ 1 cm³ = \(\frac{1}{1000}\) litres]
∴ Milk in the container = 10.45 litres
Cost of 1 it. milk = ₹ 20
∴ Cost of 10.45 litre = ₹ 20 × 10.45
Total cost of milk = ₹ 209
Curved surface area of frustum = πL [R + r]
= 3.14 × 20[20 + 8]
= 3.14 × 20 × 28 cm² = 1758.4 cm²
Area of base of container = πr²
= 3.14 × (8)π = 3.14 × 64 = 200.96 cm²

Total metal used to make coniainer = curved surface area of frustum + area of base
= (1758. 4 + 200.96) cm² = 1959.36 cm²
Cost of 100 cm² metal sheet used = ₹ 8
Cost of 1 cm² metal sheet used = ₹ \(\frac{8}{100}\)
Cost of 1959.36 cm² metal sheet used = ₹ \(\frac{8}{100}\) × 1959.36
= ₹ 156.748 = ₹ 156.75
Hence, Total cost of sheet used = ₹ 156.75
and Total cost of milk is ₹ 209.

Question 5.
A metallic right circular cone 20 cm high and whose vertical angle Ls 600 is cut into two parts at the middle of its height by a plane parallel to Its base. If the frustum so obtained be drawn into a wire of diameter \(\frac{4}{4}\) cm, find the length of the wire.
Solution:
Vertical angle of cone = 60°
Altitude of cone divide vertical angle ∠EOF = 30°

In ∆ODB,
\(\frac{\mathrm{BD}}{\mathrm{OD}}\) = tan 30°

\(\frac{r}{10}=\frac{1}{\sqrt{3}}\)

r = \(\frac{10}{\sqrt{3}}\) cm

In ∆OEF,

\(\frac{E F}{O E}\) = tan 30°

\(\frac{\mathrm{R}}{20}=\frac{1}{\sqrt{3}}\)

R = \(\frac{20}{\sqrt{3}}\) cm

Volume of frustum = \(\frac{\pi}{3}\)h[R² + r² + Rr]
= \(\frac{22}{73} \times \frac{10}{3}\left[\left(\frac{20}{\sqrt{3}}\right)^{2}+\left(\frac{10}{\sqrt{3}}\right)^{2}+\frac{20}{\sqrt{3}} \times \frac{10}{\sqrt{3}}\right]\)

= \(\frac{22}{7} \times \frac{10}{3}\left[\frac{400}{3}+\frac{100}{3}+\frac{200}{3}\right]\)

= \(\frac{22}{7} \times \frac{10}{3}\left[\frac{400+100+200}{3}\right]\)

Volume of frustum = \(\frac{22}{7} \times 10 \times \frac{700}{9}\) cm³

= \(\frac{22}{7} \times \frac{7000}{9}\) cm³
Frustum is made into wiie, which is in shape of cylinder having diameter \(\frac{1}{16}\) cm
∴ Radius of cylinderical wire (r₁) = \(\frac{1}{2} \times \frac{1}{16} \mathrm{~cm}=\frac{1}{32} \mathrm{~cm}\)

Let height of cylinder so formed be H cm
On recasting volume remain same
Volume of frustum = Volume of cylindrical wire

\(\frac{22}{7} \times \frac{7000}{9}\) = πr₁²H

\(\frac{22}{7} \times \frac{7000}{9}=\frac{22}{7} \times\left(\frac{1}{32}\right)^{2} \times \mathrm{H}\)

H = \(\frac{\frac{22}{7} \times \frac{7000}{9}}{\frac{22}{7} \times \frac{1}{32} \times \frac{1}{32}}\)

= \(\frac{7000}{9}\) × 32 × 32

H = 796444.44 cm

H = \(\) = 7964.44 m
Hence, Length of cylindrical wire (H) = 7964.44 m

Punjab State Board PSEB 6th Class Punjabi Book Solutions Chapter 1 ਪੰਜਾਬ ਦੀ ਮਿੱਟੀ Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Punjabi Chapter 1 ਪੰਜਾਬ ਦੀ ਮਿੱਟੀ

ਪਾਠ-ਅਭਿਆਸ ਪ੍ਰਸ਼ਨ-ਉੱਤਰ

1. ਛੋਟੇ ਉੱਤਰ ਵਾਲੇ ਪ੍ਰਸ਼ਨ

ਪ੍ਰਸ਼ਨ 1.
‘ਪੰਜਾਬ ਦੀ ਮਿੱਟੀ ਕਵਿਤਾ ਵਿਚ ਕਿਸ ਪੁੱਤ ਦਾ ਜ਼ਿਕਰ ਹੈ ?
ਉੱਤਰ :
ਪੰਜਾਬ ਦਾ ।

ਪ੍ਰਸ਼ਨ 2.
“ਪੰਜਾਬ ਦੀ ਮਿੱਟੀ ਕਵਿਤਾ ਦਾ ਕਵੀ ਕੌਣ ਹੈ ?
ਉੱਤਰ :
ਸਰਦਾਰ ਅੰਜੁਮ ।

2. ਸੰਖੇਪ ਉੱਤਰ ਵਾਲੇ ਪ੍ਰਸ਼ਨ

ਪ੍ਰਸ਼ਨ 1.
ਹੇਠ ਲਿਖੀ ਕਾਵਿ-ਸਤਰ ਦਾ ਭਾਵ-ਅਰਥ ਲਿਖੋ :
‘ਇਹ ਮਿੱਟੀ ’ਤੇ ਘਰ-ਘਰ ਜਾ ਕੇ ਪਿਆਰ ਦਾ ਬੂਟਾ ਲਾਵੇ ।
ਉੱਤਰ :
ਪੰਜਾਬੀ ਲੋਕਾਂ ਦਾ ਮੁਢਲਾ ਸੁਭਾ ਸਾਰੀ ਮਨੁੱਖਤਾ ਵਿਚ ਪਿਆਰ ਦਾ ਪਸਾਰ ਕਰਨ ਵਾਲਾ ਹੈ ।

ਪ੍ਰਸ਼ਨ 2.
‘ਪੰਜਾਬ ਦੀ ਮਿੱਟੀ ਕਵਿਤਾ ਦੀਆਂ ਚਾਰ ਸਤਰਾਂ ਜ਼ਬਾਨੀ ਲਿਖੋ ।
ਉੱਤਰ :
ਇਹਦਾ ਰੰਗ ਸੰਧੂਰੀ ਹੈ, ਇਹ ਗੋਰੀ ਚਿੱਟੀ ਹੈ ।
ਇਹਨੂੰ ਮੈਲੀ ਨਾ ਕਰਨਾ, ਮੇਰੇ ਪੰਜਾਬ ਦੀ ਮਿੱਟੀ ਹੈ ।
ਇਸ ਮਿੱਟੀ ਵਿਚ ਕੋਈ ਕੁੜੱਤਣ, ਕਿੱਦਾਂ ਕੋਈ ਉਗਾਵੇ ।
ਇਹ ਮਿੱਟੀ ‘ਤੇ ਘਰ-ਘਰ ਜਾ ਕੇ ਪਿਆਰ ਦਾ ਬੂਟਾ ਲਾਵੇ ।

ਪ੍ਰਸ਼ਨ 3.
“ਪੰਜਾਬ ਦੀ ਮਿੱਟੀ ਕਵਿਤਾ ਨੂੰ ਟੋਲੀ ਬਣਾ ਕੇ ਆਪਣੀ ਜਮਾਤ ਵਿਚ ਗਾਓ ।
ਉੱਤਰ:
(ਨੋਟ :-ਵਿਦਿਆਰਥੀ ਆਪ ਕਰਨ ।)

3. ਅਧਿਆਪਕ ਲਈ

ਪ੍ਰਸ਼ਨ 1.
ਪੰਜਾਬ ਦੀ ਮਹਾਨਤਾ ਬਾਰੇ ਬੱਚਿਆਂ ਨੂੰ ਹੋਰ ਜਾਣਕਾਰੀ ਦਿੱਤੀ ਜਾਵੇ ।
ਉੱਤਰ :
ਪੰਜਾਬ ਪੰਜਾਂ ਦਰਿਆਵਾਂ ਦੀ ਧਰਤੀ ਹੈ । 1947 ਵਿਚ ਇਹ ਭਾਰਤੀ ਪੰਜਾਬ ਤੇ ਪੰਜਾਬ ਦੀ ਮਿੱਟੀ ਪਾਕਿਸਤਾਨੀ ਪੰਜਾਬ ਵਿਚ ਵੰਡੀ ਗਈ ਹੈ , ਜਿਸ ਕਰਕੇ ਹੁਣ ਢਾਈ ਦਰਿਆ, ਸਤਲੁਜ, ਬਿਆਸ ਤੇ ਅੱਧਾ ਰਾਵੀ ਇਧਰ ਰਹਿ ਗਏ ਹਨ ਤੇ ਢਾਈ ਦਰਿਆ ਚਨਾਬ, ਜਿਹਲਮ ਤੇ ਅੱਧਾ ਰਾਵੀ ਉਧਰ । ਇਹ ਦਸ ਗੁਰੂ ਸਾਹਿਬਾਨ ਤੇ ਸੂਫ਼ੀ ਫ਼ਕੀਰਾਂ ਦੀ ਧਰਤੀ ਹੈ । ਇਸੇ ਧਰਤੀ ਉੱਤੇ ਸਿੱਖ ਧਰਮ ਦਾ ਜਨਮ ਹੋਇਆ ਤੇ ਸ੍ਰੀ ਗੁਰੂ ਗ੍ਰੰਥ ਸਾਹਿਬ ਦੀ ਰਚਨਾ ਹੋਈ । ਸੰਸਾਰ ਦੇ ਸਭ ਤੋਂ ਪੁਰਾਤਨ ਧਰਮ ਗੰਥ ਰਿਗਵੇਦ ਦੀ ਰਚਨਾ ਇੱਥੇ ਹੀ ਹੋਈ, । ਮਹਾਂਭਾਰਤ ਦਾ ਯੁੱਧ ਤੇ ਸੀ । ਮਦ ਭਗਵਦ ਗੀਤਾ ਦੀ ਰਚਨਾ ਵੀ ਪੁਰਾਤਨ ਪੰਜਾਬ ਦੀ ਧਰਤੀ ‘ਤੇ ਹੀ ਹੋਈ ।ਇਸਦੀ ਬੋਲੀ ਪੰਜਾਬੀ ਹੈ । ਅਜੋਕੇ ਪੰਜਾਬ ਦੇ ਪੱਛਮੀ ਪਾਸੇ ਪਾਕਿਸਤਾਨ ਲਗਦਾ ਹੈ । ਉੱਤਰੀ ਪਾਸੇ ਕਸ਼ਮੀਰ, ਪੂਰਬੀ ਪਾਸੇ ਹਿਮਾਚਲ, ਦੱਖਣ-ਪੂਰਬੀ ਪਾਸੇ ਹਰਿਆਣਾ ਤੇ ਦੱਖਣ-ਪੱਛਮੀ ਪਾਸੇ ਰਾਜਸਥਾਨ ਲਗਦੇ ਹਨ । ਇਸਦਾ ਖੇਤਰਫਲ 50,362 ਵਰਗ ਕਿਲੋਮੀਟਰ ਹੈ । ਕਣਕ ਤੇ ਝੋਨਾ ਇਸਦੀਆਂ ਮੁੱਖ ਫ਼ਸਲਾਂ ਹਨ । ਇਸਦੇ 22 ਜ਼ਿਲ੍ਹੇ ਹਨ । ਇਸਦੀ ਆਬਾਦੀ 2 ਕਰੋੜ 80 ਲੱਖ ਹੈ । ਪੰਜਾਬੀ ਲੋਕ ਆਪਣੇ ਖੁੱਲ੍ਹੇ-ਡੁੱਲੇ, ਮਿਹਨਤੀ ਤੇ ਅਣਖੀਲੇ ਸੁਭਾ ਕਰਕੇ ਸੰਸਾਰ ਭਰ ਵਿਚ ਪ੍ਰਸਿੱਧ ਹਨ । ਉਹ ਕਿਰਤ ਕਰਨ, ਨਾਮ ਜਪਣ ਤੇ ਵੰਡ ਛਕਣ ਵਿਚ ਵਿਸ਼ਵਾਸ ਰੱਖਦੇ ਹਨ । ਅਫ਼ਸੋਸ ਕਿ ਅੱਜ ਦੀ ਨੌਜਵਾਨ ਪੀੜੀ ਖੇਡਾਂ ਵਿਚ ਨਾਮਣਾ ਕਮਾਉਣ ਤੇ ਸਰੀਰ ਪਾਲਣ ਦੇ ਸ਼ੌਕ ਛੱਡ ਕੇ ਨਸ਼ਿਆਂ ਵਿਚ ਗ਼ਰਕ ਹੋ ਚੁੱਕੀ ਹੈ ।

 

ਪਸ਼ਨ 1.
ਹੇਠ ਲਿਖੇ ਕਾਵਿ-ਟੋਟੇ ਦੇ ਸਰਲ ਅਰਥ ਕਰੋ :
(ਉ) ਇਹਦਾ ਰੰਗ ਸੰਧੂਰੀ ਹੈ ਇਹ ਗੋਰੀ ਚਿੱਟੀ ਹੈ ।
ਇਹਨੂੰ ਮੈਲੀ ਨਾ ਕਰਨਾ, ਮੇਰੇ ਪੰਜਾਬ ਦੀ ਮਿੱਟੀ ਹੈ ।
ਉੱਤਰ :
ਕਵੀ ਅੰਜੁਮ ਕਹਿੰਦਾ ਹੈ ਕਿ ਪੰਜਾਬ ਦੀ ਮਿੱਟੀ ਦਾ ਰੰਗ ਦਿਲ-ਖਿਚਵਾਂ ਸੰਧੂਰੀ ਹੈ । ਇਹ ਮਿੱਟੀ ਗੋਰੀ-ਚਿੱਟੀ ਅਰਥਾਤ ਸਾਫ਼-ਸੁਥਰੀ ਹੈ । ਸਾਡਾ ਸਭ ਦਾ ਫ਼ਰਜ਼ ਹੈ ਕਿ ਪੰਜਾਬ ਦੀ ਇਸ ਸਾਫ਼-ਸੁਥਰੀ ਮਿੱਟੀ ਨੂੰ ਕਿਸੇ ਤਰ੍ਹਾਂ ਵੀ ਮੈਲੀ ਨਾ ਕਰੀਏ ।

ਔਖੇ ਸ਼ਬਦਾਂ ਦੇ ਅਰਥ-ਸੰਧੂਰੀ = ਸੰਧੂਰ ਦੇ ਰੰਗ ਵਰਗੀ ।

ਪ੍ਰਸ਼ਨ 2.
ਹੇਠ ਲਿਖੇ ਕਾਵਿ-ਟੋਟੇ ਦੇ ਸਰਲ ਅਰਥ ਕਰੋ :
(ਅ) ਇਸ ਮਿੱਟੀ ਵਿਚ ਕੋਈ ਕੁੜੱਤਣ, ਕਿੱਦਾਂ ਕੋਈ ਉਗਾਵੇ ।
ਇਹ ਮਿੱਟੀ ’ਤੇ ਘਰ-ਘਰ ਜਾ ਕੇ, ਪਿਆਰ ਦਾ ਬੂਟਾ ਲਾਵੇ ।
ਉੱਤਰ :
ਕਵੀ ਅੰਜੁਮ ਕਹਿੰਦਾ ਹੈ ਕਿ ਇਹ ਕਿਵੇਂ ਹੋ ਸਕਦਾ ਹੈ ਕਿ ਕੋਈ ਮੇਰੇ ਪੰਜਾਬ ਦੀ ਮਿੱਟੀ ਵਿਚ ਫ਼ਿਰਕੂਪੁਣੇ ਦੀ ਕੁੜੱਤਣ ਪੈਦਾ ਕਰ ਦੇਵੇ, ਜਦਕਿ ਇਹ ਮਿੱਟੀ ਤਾਂ ਘਰ-ਘਰ ਜਾ ਕੇ ਆਪਸੀ ਪਿਆਰ ਦਾ ਬੂਟਾ ਲਾਉਣ ਦਾ ਕੰਮ ਕਰਦੀ ਹੈ, ਅਰਥਾਤ ਪੰਜਾਬੀ ਲੋਕਾਂ ਦਾ ਮੁਢਲਾ ਸੁਭਾ ਸਾਰੀ ਮਨੁੱਖਤਾ ਵਿਚ ਆਪਸੀ ਪਿਆਰ ਦਾ ਪਸਾਰਾ ਕਰਨ ਵਾਲਾ ਹੈ ।

ਪ੍ਰਸ਼ਨ 3.
ਹੇਠ ਲਿਖੇ ਕਾਵਿ-ਟੋਟੇ ਦੇ ਸਰਲ ਅਰਥ ਕਰੋ :
(ਈ) ਪੁੱਛ ਲਓ ਫ਼ਕੀਰਾਂ ਤੋਂ, ਇਹਦੀ ਬਾਣੀ ਮਿੱਠੀ ਹੈ ।
ਇਹਨੂੰ ਮੈਲੀ ਨਾ ਕਰਨਾ, ਮੇਰੇ ਪੰਜਾਬ ਦੀ ਮਿੱਟੀ ਹੈ ।
ਉੱਤਰ :
ਕਵੀ ਅੰਜੁਮ ਕਹਿੰਦਾ ਹੈ ਕਿ ਇਹ ਗੱਲ ਸੂਫ਼ੀ ਫ਼ਕੀਰਾਂ ਤੇ ਗੁਰੂ ਸਾਹਿਬਾਂ ਤੋਂ ਪੁੱਛ ਕੇ ਸਮਝੀ ਜਾ ਸਕਦੀ ਹੈ ਕਿ ਪੰਜਾਬ ਦੀ ਬੋਲੀ ਬਹੁਤ ਮਿੱਠੀ ਹੈ, ਇਸੇ ਕਰਕੇ ਹੀ ਉਨ੍ਹਾਂ ਨੇ ਆਪਣੇ ਭਾਵਾਂ ਤੇ ਵਿਚਾਰਾਂ ਦਾ ਪ੍ਰਗਟਾਵਾ ਇਸ ਬੋਲੀ ਵਿਚ ਕੀਤਾ ਹੈ । ਜਿਸ ਪੰਜਾਬ ਦੀ ਧਰਤੀ ਉੱਤੇ ਇਹ ਬੋਲੀ ਬੋਲੀ ਜਾਂਦੀ ਹੈ, ਉਸਦੀ ਮਿੱਟੀ ਨੂੰ ਕਿਸੇ ਤਰ੍ਹਾਂ ਵੀ ਮੈਲੀ ਨਹੀਂ ਕਰਨਾ ਚਾਹੀਦਾ

ਔਖੇ ਸ਼ਬਦਾਂ ਦੇ ਅਰਥ-ਬਾਣੀ = ਬੋਲੀ ।

ਪ੍ਰਸ਼ਨ 4.
ਹੇਠ ਲਿਖੇ ਕਾਵਿ-ਟੋਟੇ ਦੇ ਸਰਲ ਅਰਥ ਕਰੋ :
(ਸ) ਕਰੋ ਦੁਆ ਕਦੇ ਇਸ ਮਿੱਟੀ ’ਤੇ ਉਹ ਮੌਸਮ ਨਾ ਆਵੇ।
ਇਸ ਤੇ ਉੱਗਿਆ ਹਰ ਕੋਈ ਸੂਰਜ, ਕਾਲਖ਼ ਹੀ ਬਣ ਜਾਵੇ ।
ਉੱਤਰ :
ਕਵੀ ਅੰਜੁਮ ਸਾਨੂੰ ਸਭ ਨੂੰ ਰੱਬ ਅੱਗੇ ਇਹ ਅਰਦਾਸ ਕਰਨ ਲਈ ਕਹਿੰਦਾ ਹੈ ਕਿ ਪੰਜਾਬ ਦੀ ਧਰਤੀ ਦੀ ਮਿੱਟੀ ਉੱਤੇ ਕਦੇ ਵੀ ਫ਼ਿਰਕੂ ਜ਼ਹਿਰ ਨਾਲ ਭਰਿਆ ਉਹ ਮੌਸਮ ਨਾ ਆਵੇ ਕਿ ਇਸ ਉੱਤੇ ਉੱਗਿਆ ਹਰ ਇਕ ਸੁਰਜ ਭਾਵ ਇਸ ਉੱਤੇ ਚੜ੍ਹਨ ਵਾਲਾ ਹਰ ਦਿਨ ਚਾਨਣ ਦੀ ਥਾਂ ਨਫ਼ਰਤ ਦੀ ਕਾਲਖ਼ ਦਾ ਪਸਾਰ ਕਰ ਦੇਵੇ ।

ਔਖੇ ਸ਼ਬਦਾਂ ਦੇ ਅਰਥ-ਦੁਆ = ਬੇਨਤੀ, ਅਰਦਾਸ ।

ਪ੍ਰਸ਼ਨ 5.
ਹੇਠ ਲਿਖੇ ਕਾਵਿ-ਟੋਟੇ ਦੇ ਸਰਲ ਅਰਥ ਕਰੋ :
(ਹ) ਰੱਬ ਦੇ ਘਰ ਤੋਂ ਅੰਜੁਮ, ਆਈ ਇਹ ਚਿੱਠੀ ਹੈ ।
ਇਹਨੂੰ ਮੈਲੀ ਨਾ ਕਰਨਾ, ਮੇਰੇ ਪੰਜਾਬ ਦੀ ਮਿੱਟੀ ਹੈ ।
ਉੱਤਰ :
ਕਵੀ ਅੰਜੁਮ ਕਹਿੰਦਾ ਹੈ ਕਿ ਰੱਬ ਦੇ ਘਰ ਤੋਂ ਵੀ ਸਾਨੂੰ ਚਿੱਠੀ ਰਾਹੀਂ ਇਹ ਸੰਦੇਸ਼ ਪੁੱਜਾ ਹੈ ਕਿ ਅਸੀਂ ਆਪਣੇ ਪੰਜਾਬ ਦੀ ਸਾਫ਼-ਸੁਥਰੀ ਮਿੱਟੀ ਨੂੰ ਨਫ਼ਰਤ ਦੀ ਜ਼ਹਿਰ ਭਰ ਕੇ ਇਸਨੂੰ ਮੈਲੀ ਨਾ ਕਰੀਏ, ਸਗੋਂ ਸਾਫ਼-ਸੁਥਰੀ ਹੀ ਰਹਿਣ ਦੇਈਏ ।. ਔਖੇ ਸ਼ਬਦਾਂ ਦੇ ਅਰਥ-ਅੰਜੁਮ = ਕਵੀ ਦਾ ਨਾਂ, ਸਰਦਾਰ ਅੰਜੁਮ ॥

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.3

Question 1.
A metallic sphere of radius 4.2 cm is melted and recast into the shape of cylinder of radius 6 cm. Find the height of the cylinder.
Solution:

Radius of sphere (r) = 4.2 cm
Radius of cylinder (R) = 6 cm
Let height of cylinder be H cm
On recasting volume remain same
Volume of sphere = Volume of cylinder
\(\frac{4}{3}\) πR³ = πR²H

\(\frac{4}{3}\) × \(\frac{22}{7}\) × 4.2 × 4.2 × 4.2 = \(\frac{22}{7}\) × 6 × 6 × H
∴ H = \(\frac{\frac{4}{3} \times \frac{22}{7} \times 4.2 \times 4.2 \times 4.2}{\frac{22}{7} \times 6 \times 6}\)

= \(\frac{4}{3} \times \frac{42}{10} \times \frac{42}{10} \times \frac{42}{10} \times \frac{1}{6 \times 6}\)

= \(\frac{2744}{1000}\) = 2.744 cm

Hence, Height of cylinder (H) = 2.744 cm.

Question 2.
Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.
Solution:

Radius of first sphere (r₁) = 6 cm
Radius of second sphere (r₂) = 8 cm
Radius of third sphere (r₃) = 10 cm
Let radius of new sphere formed be R cm
On recasting volume remain same
Volume of all three spheres = Volume of big sphere

\(\frac{4}{3}\) πr₁³ + \(\frac{4}{3}\) πr₂³ + \(\frac{4}{3}\) πr₃³ = \(\frac{4}{3}\) πR³

\(\frac{4}{3}\) π[(6)³ + (8)³ + (10)³] = \(\frac{4}{3}\) πR³

R³ = \(\frac{\frac{4}{3} \pi[216+512+1000]}{\frac{4}{3} \pi}\)

R³ = 1728
R = \(\sqrt[3]{1728}\) = \(\sqrt[3]{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3}\)

= 2 × 2 × 3
R= 12 cm
Hence, Radius of sphere = 12 cm.

Question 3.
A 20 m deep well with-diameter 7 m is dug and the earth from digging is evenly spread out to form of platform 22 m by 14 m. Find the height of the platform.
Solution:
Diameter of well = 7 m
Radius of well (cylinder) R = \(\frac{7}{2}\) m
Height of well (H) = 20 m
Length of Platform (L) =22 m
Width of Platform (B) = 14 m
Let height of Platform be H m

Volume of earth dug out from well = Volume of platform formed

πR²H = L × B × H
× × × 20 = 22 × 14 × H
∴ H = \(\frac{\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 20}{22 \times 14}\)
H = 2.5 cm
Hence, Height of Platform H = 2.5 cm.

Question 4.
A well of diameter 3 m is dug 14 m deep. The earth taken out of ¡t has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.
Solution:
Depth of well (h) = 14 m
Radius of well (r) = \(\frac{3}{2}\) m

Embankment is in the shape of hollow cylinder whose inner radius is same as radius of well and width of embañkment 4 m
Timer radius of embankment = Radius of well(r) = \(\frac{3}{2}\) m
Outer radius of embankment (R) = (\(\frac{3}{2}\) + 4) m
R = \(\frac{11}{2}\) = 5.5 m
Volume of earth dug out = Volume of embankment (so formed)
πR²h = volume of outer cylinder – volume of inner cylinder
πr²h = πR²H – πr²H
= πH[R² – r²]

\(\frac{22}{7} \times \frac{3}{2} \times \frac{3}{2} \times 14\) = \(\frac{22}{7}\) × H[(5.5)² – (1.5)²]

H = \(\frac{\frac{22}{7} \times \frac{3}{2} \times \frac{3}{2} \times 14}{\frac{22}{7} \times(5.5-1.5)(5.5+1.5)}\)

= \(\frac{1.5 \times 1.5 \times 14}{4 \times 7}\) = 1.125 m.

Hence, Height of embankment H = 1.125 m.

Question 5.
A container shaped like a right circular cylinder having 4iameter 12 cm and height 15 cm Is full of ice-cream. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones which can be tilled with ice-cream.

Dinieter of cvlrnder (D) = 12 cm
.. Radius of cylinder (R) = 6 cm
Height of cylinder (H) = 15 cm
Diameter of cone = 6 cm
Radius of cone (r) = 3 cm
Radius of hemisphere (r) = 3 cm
Height of cone (h) = 12 cm
Let us suppose number of cones used to fill the ice-cream = n
Volume of ice cream in container = n [Volume of ice cream in one cone]

n = 10
Hence, Number of cones formed = 10.

Question 6.
How many silver coins 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions
5.5 cm × 10 cm × 3.5 cm.
Solution:
Silver coin is in the form of cylinder
Diameter of silver coin = 1.75 cm
∴ Radius of silver coin (r) = \(\frac{1.75}{2}\) cm
Thickness of silver coin = Height of cylinder (H) = 2 mm

i.e., h = \(\frac{2}{10}\) cm.
Length of cuboid (L) = 5.5 cm
Width of cuboid (B) = 10 cm
Height of cuboid (H) = 3.5 cm
Let number of coins melted to form cuboid = n
∴ Volume of cuboid = n [volume of one silver coin]
= n[πr²h]
5.5 × 10 × 3.5 = n × \(\frac{22}{7} \times \frac{1.75}{2} \times \frac{1.75}{2} \times \frac{2}{10}\)

\(\frac{\frac{55}{10} \times 10 \times \frac{35}{10}}{\frac{22}{7} \times \frac{175}{200} \times \frac{175}{200} \times \frac{2}{10}}\) = n
n = 400
Hence, Number of corns so formed = 400.

Question 7.
A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.
Solution:

Radius of cylindrical bucket (R) = 18 cm
Height of cylindrical bucket (H) = 32 cm
Height of conical heap (h) = 24 cm
Let ‘r’ cm and ‘l’ cm be the radius and slant height of cone
Volume of sand in bucket = Volume of sand in cone
πR²H = \(\frac{1}{3}\) πr²h

\(\frac{22}{7}\) × 18 × 18 × 32 = \(\frac{1}{3}\) × \(\frac{22}{7}\) × r² × 24
\(\frac{\frac{22}{7} \times 18 \times 18 \times 32}{\frac{1}{3} \times \frac{22}{7} \times 24}\) = r²
r² = \(\frac{18 \times 18 \times 32}{8}\)
r² = 1296
r = \(\sqrt{1296}\)
r = 36
∴ Radius of cone (r) = 36 cm
As we know,
(Slant height)² = (Radius)² + (Height)²
l = r² + h²
l = \(\sqrt{(36)^{2}+(24)^{2}}\)

= \(\sqrt{1296+576}\) = \(\sqrt{1872}\)

= \(\sqrt{12 \times 12 \times 13}\)

l = 12\(\sqrt{13}\) cm

∴ Slant height of cone (1) = 12\(\sqrt{13}\) cm.

Question 8.
Water in a canal 6m wide and 1.5m deep is flowing with a speed of 10 km/k How much area will it irrigate in 30 minutes, If 8 cm of standing water is needed?
Solution:
Width of canal = 6 m
Depth of water in canal = 1.5 m
Velocity at which water is flowing = 10km/hr
Volume of water discharge in one hour = (Area of cross section) velocity
= (6 × 1.5m2) × 10 km
= 6 × 1.5 × 10 × 6 × 1.5 × 10 × 1000 × 1000 m³
∴ Volume of water discharge in \(\frac{1}{2}\) hour = \(\frac{1}{2}\) × \(\frac{6 \times 15}{10}\) × 1000 = 45000 m³
Let us suppose area to be irrigate = (x) m²

According to question, 8 cm standing water is required in field
∴ Volume of water discharge by canal in \(\frac{1}{2}\) hours = Volume of water in field 45000 m³ = (Area of field) × Height of water
45000 m³ = x × (\(\frac{8}{100}\) m)
\(\frac{4500}{8}\) × 100 = x
x = 562500 m²
x = \(\frac{562500}{10000}\) hectares
[1 m² = \(\frac{1}{10000}\) hectares]
x = 56.25 hectares
Hence, Area of field = 56.25 hectares.

Question 9.
A farmer connects a pipe of Internal diameter 20 cm from a canal into a cylindrical tank in her field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate 0f 3km/h, in how much time will the tank be filled?
Solution:
Velocity of water = 3 km/hr
Diameter of pipe = 20 cm
Radius of pipe (r) = 10 cm = \(\frac{10}{100}\) m = \(\frac{1}{10}\) m
Diameter of tank = 10 m
Radius of tank (R) = 5 m
Depth of tank (H) = 2 m
Let us suppose pipe filled a tank in n minutes
Volume of water tank = Volume of water through the pipe in n minutes
πR²H = n[Area of cross section × Velocity of water]

πR²H = n[(πr²) × 3 km/h]

\(\frac{22}{7}\) × (5)² × 2 = n[latex]\frac{22}{7} \times \frac{1}{10} \times \frac{1}{10} \times \frac{3 \times 1000}{60}[/latex]

25 × 2 × \(\frac{22}{7}\) = \(\frac{11}{7}\) n

n = \(\frac{25 \times 2 \times 22 \times 7}{11 \times 7}\)
n = 100 minutes
Hence, Time taken to fill the tank = 100 minutes.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.2

Question1.
A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of it.
Solution:
Radius of cone = Radius of hemisphere = 1 cm

R = 1 cm
Height of cone (H) = 1 cm
Volume of solid = volume of cone + volume of hemisphere
= \(\frac{1}{3}\) πR²H + \(\frac{2}{3}\) πR³
= \(\frac{1}{3}\) πR²2 [H + 2R]
= \(\frac{1}{3}\) π × 1 × 1 [1 + 2 × 1]
= \(\frac{1}{3}\) π × 3 = π cm³
Hence, Volume of solid = π cm³.

Question 2.
Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model the Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)
Solution:

Radius of cone = Radius of cylinder (R) = \(\frac{3}{2}\) cm
∴ R = 1.5 cm
Height of eah cone (h) = 2 cm
∴ Height of cylinder = 12 – 2 – 2 = 8 cm
Volume of air in cylinder = volume of cylinder + 2 (volume of cone)

Volume of air in cylinder = \(\frac{22}{7} \times \frac{3}{2} \times \frac{3}{2} \times \frac{28}{3}\)
= 22 × 3 = 66 cm³

Hence, Volume of air in cylinder =66 cm³.

Question 3.
A gulab jamun, contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm.

Solution:
Gulab Jamun is in the shape of cylinder
Diameter of cylinder = Diameter of hemisphere = 2.8 cm

Radius of cylinder = Radius of hemisphere (R)
= \(\frac{28.2}{2}\) = 1.4 cm
R = 1.4 cm
Height of cylindrical part = 5 – 1.4 – 1.4
= (5 – 2.8) cm = 2.2 cm.
Volume of one gulab Jamun = Volume of cylinder + 2 [Volume of hemisphere]
= πR²H + 2 \(\frac{2}{3}\) πR³

= πR² H + \(\frac{4}{3}\) R

= \(\frac{4}{4}\) × 1.4 × 1.4 2.2 + \(\frac{4}{3}\) × 1.4

= \(\frac{22}{7}\) × \(\frac{14}{10}\) × \(\frac{14}{10}\) 2.2 + \(\frac{4}{3}\) × 1.4

= \(\frac{22}{7}\) × \(\frac{196}{100}\) [2.2 + 1.86]

= \(\frac{22 \times 28}{100}\) [4.06]

Volume of one gulab Jamun = 25.05 cm³
Now volume of 45 gulab Jamuns = 45 × 25.05 cm³ = 1127.25 cm³
Volume of sugar syrup = 30% volume of 45 gulab Jamuns
= \(\frac{30 \times 1127.25}{100}\) = 338.175 cm3
Hence, Approximately sugar syrup = 338 cm³.

Question 4.
A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand.

Solution. Length of cuboid (L) = 15 cm
Width of cuboid (B) = 10 cm
Height of cuboid (H) = 3.5 cm
Radius of conical cavity (r) = 0.5 cm
Height of conical cavity (h) = 1.4 cm
Volume of wood in Pen stand = volume of cuboid – 4 [volume of cone]
= LBH – 4 \(\frac{1}{3}\) πr²h

= 15 × 10 × 3.5 – \(\frac{4}{3}\) × \(\frac{22}{7}\) × 0.5 × 0.5 × 1.4

= \(\frac{15 \times 10 \times 35}{10}-\frac{4}{3} \times \frac{22}{7} \times \frac{5}{10} \times \frac{5}{10} \times \frac{14}{10}\)

= \(15 \times 35-\frac{22}{3 \times 5}\)
= 525 – 1.466 = 523.534 cm³
Hence, Volume of wood in Pen stand = 523.53 cm³.

Question 5.
A vessel is in the form of an inverted cone. Its height ¡s 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.
Solution:

Radius of cone (R) = 5 cm
Height of cone (H) = 8 cm
Radius of each spherical lead shot (r) = 0.5 cm
Let number of shot put into the cone = N
According to Question,
N [Volume of one lead shot] = \(\frac{1}{4}\) Volume of water in cone

= 10 × 10 = 100
Hence, Number of lead shots = 100.

Question 6.
A solid iron pole consists of a cylinder of height 220 cfi and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm3 of Iron has approximately 8 g mass. (Use n = 3.14)
Solution:
Diameter of lower cylinder = 24 cm
Radius of lower cylinder (R) = 12 cm
Height of lower cylinder (H) = 220 cm
Radius of upper cylinder (r) = 8 cm
Height of upper cylinder (h) = 60 cm
Volume of pole = Volume of Lower cylinder + volume of upper cylinder
= πR²H + πr²h
= 3.14 × 12 × 12 × 220 + 3.14 × 8 × 8 × 60
= 99475.2 + 12057.6

Volume of pole = 111532.8 cm³
Mass of 1 cm³ = 8 gm
Mass of 111532.8 cm³ = 8 × 111532.8 = 892262.4 gm
= \(\frac{892262.4}{1000}\) kg = 892.2624 kg
Hence, Mass of Pole = 892.2624 kg.

Question 7.
A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.
Solution:
Radius of cone = Radius of hemisphere = Radius of cylinder

Height of cone (h) = 120 cm
Height of cylinder (H) = 180 cm
Volume of cylindrical vessel = πR²H
= \(\frac{22}{7}\) × 60 × 60 × 180 = 2036571.4 cm³
Volume of solid inserted in cylinder = Volume of hemisphere + Volume of cone
= \(\frac{2}{3}\) πR³ + \(\frac{1}{3}\) πR²h

= \(\frac{1}{3}\) πR² [2R + h]

= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 60 × 60 [2 × 60 + 120]

= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 3600 [120 + 120]

= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 3600 × 240 = 905142.86 cm³
Volume of water flows out = 90514186 cm³
∴ Volume of water left in cylinder = Volume of cylinder – Volume of solid inserted in th’e vessel
= (2036571.4 – 905142.86) cm³ = 1131428.5 cm³
= \(\frac{1131428.5}{100 \times 100 \times 100}\) m³ = 1.131 m³
Hence, Volume of water left in cylinder = 1.131 m³.

Question 8.
A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm³. Check whether she is correct, taking the above as the inside measurements, and π = 3.14.
Solution:
Diameter of neck (cylindrical Portion) = 2 cm

Radius of neck (r) = 1 cm
Height of cylindrical portion (H) = 8 cm
Diameter of spherical portion = 8.5 cm
Radius of spherical portion (R) = \(\frac{8.5}{2}\) cm = 4.25 cm
Volume of water in vessel = Volume of sphere + Volume of cylinder
= \(\frac{4}{3}\) πR³ + πR²h
= \(\frac{4}{3}\) × 3.14 × 4.25 × 4.25 × 4.25 × 3.14 × 1 × 1 × 8
= 321.39 + 25.12 = 346.51 cm³
Hence, Volume of water in vessel = 346.51 cm³ and She is wrong.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.1

Question 1.
2 cubes each of volume 64 cm³ are joined end to end. Find the surface area of the resulting cuboid.
Solution:
Let side of cube = x cm
Volume of cube = 64 cm³
[volume of cube = (side)³]
x³ = 64
x = \(\sqrt[3]{4 \times 4 \times 4}\)
x = 4 cm
∴ side of cube = 4 cm.

When cubes are joined end to end and cuboid is formed
whose Length = 2x cm = 2(4) = 8 cm
Width = x cm = 4 cm
Height = x cm = 4 cm

Surface area of cuboid = 2[LB + Bh + hL]
= 2 [8 × 4 + 4 × 4 + 4 × 8]
= 2 [32 + 16 + 32]
= 2 [80]
∴ Surface area of cuboid = 160 cm².

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

Diameter of hemisphere = Diameter of cylinder
= 14 cm
2R = 14 cm
Radius of hemisphere (R) = 7 cm
Total height of vessel = 13 cm
∴ Height of cylinder = (13 – 7) = 6 cm
Inner surface area of vessel = inner surface area of cylinder + Inner surface area of hemisphere
= 2πRH + 2πR²
= 2πR [H + R]
= 2 × \(\frac{22}{7}\) × 7(16 + 7)
= 44 × 13 = 572 cm²
Hence, Inner surface area of vessel = 572 cm²

Question 3.
A toy is ¡n 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:
Radius of cone = Radius of hemisphere (R) = 3.5 cm
Total height of toy = 15.5 cm
∴ Height of cone (H) = (15.5 – 3.5) = 12 cm

Slant height of cone = \(\sqrt{\mathrm{R}^{2}+\mathrm{H}^{2}}\)

= \(\sqrt{(3.5)^{2}+(12)^{2}}\)

= \(\sqrt{12.25+144}\) = \(\sqrt{156.25}\)
Slain height conk (l) = 12.5 cm
Total surface area of toy = Surface area of cone + Surface area of hemisphere
= πRL + 2πR²
= πR[L + 2R]
= \(\frac{22}{7}\) × 3.5 [12.5 + 2 (3.5)1 cm²
= \(\frac{22}{7}\) × 3.5 [19.5] cm²
= \(\frac{1501.5}{7}\) = 214.5 cm²
∴ Total surface area of toy = 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:
Side of cubical box = 7 cm
Diameter of hemisphere = Side of cubical box = 7 cm
2R = 7
R = \(\frac{7}{2}\) cm

Surface area of solid = (surface area of the cube) – (area of base of hemisphere) + curved surface area of hemisphere)
= 6l² – πR² + 2πR²
= 6l² + πR²
= 6(7)² + \(\frac{22}{7}\) \(\frac{7}{2}\)²
= [6 × 49 + 11 × \(\frac{7}{2}\)]cm²
= 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 (he hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.
Solution:
Let each side of cube = a
∴ Diameter of hemisphere = Side of cube
2R = a
R = \(\frac{a}{2}\)

Surface area of remaining solid = Total surface area of euboid – Area of the top of cube + Inner curved Surface area of hemisphere
= 6 (side)² – πR² + 2πR²
= 6(a)² + πR²
= 6(a)² + π \(\frac{a}{2}\)²
= 6a² + π \(\frac{a^{2}}{4}\)
= a² 6 + \(\frac{\pi}{4}\) cm²

Question 6.
A medicine capsule is ¡n 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:
Diameter of capsule = Diameter of hemisphere = Diameter of cylinder = 5 mm
∴ 2R = 5 mm
R = \(\frac{5}{2}\) mm

Length of entire capsule = 14 mm
Height of cylinderical part = (14 – \(\frac{5}{2}\) – \(\frac{5}{2}\)) mm
= (14 – 5) mm
H = 9 mm
Surface area of capsule = Surface area of cylinder + 2 Surface area of hemisphere
= 2πRH + 2 (2πR²)
= 2πRH + 4πR²
= 2πR [H + 2R]
= 2 × \(\frac{22}{7} \times \frac{5}{2}\left[9+2\left(\frac{5}{2}\right)\right]\)
= 2 × \(\frac{22}{7}\) × \(\frac{5}{2}\) [9 + 5]
= \(\frac{22}{7}\) × 5 × 14
= 220 mm²
Hence, Surface area of capsule = 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 500 per (Note that the base of the tent will not be
covered with canvas.)
Solution:
Diameter of cone = Diameter of cylinder
2R = 4
R = 2 m

Radius of cone = Radius of cylinder
Height of cylinder (H) = 2.1 in
Slant height of cone (L) = 2.8 m
Curved surface area of tent = Curved surface of cylinder + Curved surface of conical part
= 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
∴ Curved surface area of tent = 44 m²
Cost of 1m² canvas = ₹ 500
Cost of 44 m² canvas = 44 × 500 = ₹ 22000
Hence, Total cost of canvas = ₹ 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:
Diameter of cylinder (D) = 1.4 cm = Diameter of cone
∴ Radius of cylinder = Radius of cone (R) = 0.7 cm
Height of cylinder (H) = 2.4 cm

As we know, L² = R² + H² + (2.4)²
L = \(\sqrt{(0.7)^{2}+(2.4)^{2}}\)
= \(\sqrt{0.49+5.76}\) = \(\sqrt{6.25}\)
L = 2.5 cm
Total surface area of remaining solid = curved surface area of cylinder + Area of base of cylinder + Surface area of cone
= 2πRH + πR² + πRL
= πR [2R +R + L]
= \(\frac{22}{7}\) × 0.7 [2(2.4) + 0.7 + 2.5]

=\(\frac{22}{7}\) × \(\frac{7}{10}\) [4.8 + 3.2]

= \(\frac{22}{10}\) [8]

= \(\frac{176}{10}\) = 17.6 cm²

Hence, Total surface area remaining solid to nearest cm² = 18 cm².

Question 9.
A wooden article was made by scooping out a hemisphere form each end of a solid cylinder, as shown in Fig.

If the height of the cylinder is 10 cm, and its base ¡s of radius 3.5 cm, find the total surface area of the article.
Solution:
Height of cylinder (H) = 10 cm
Radius of cylinder = Radius of hemisphere (R) = 3.5 cm

Surface area of article = curved surface area of cylinder + 2 curved surface area of hemisphere
= 2πRH + 2 (2πR²)
= 2πR [H + 2R]
= 2 × \(\frac{22}{7}\) × 3.5 [10 + 2(3.5)]
= \(\frac{44}{7}\) × \(\frac{35}{10}\) [10 + 7]
= 44 × \(\frac{5}{10}\) × 17
= 44 × \(\frac{1}{2}\) × 17
= 22 × 17 = 374 cm²
Hence, total surface area of article = 374 cm².