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

November 27, 2020January 17, 2019 by Veerendra

NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.1 are part of NCERT Solutions for Class 10 Maths. Here are we have given Chapter 13 Surface Areas and Volumes Class 10 NCERT Solutions Ex 13.1.

Students find NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes quite helpful for their CBSE Board Preparation. Refer to our Textbook Solutions any time, while doing your homework or while preparing for the exam. All questions and answers from the NCERT Book of class 10 Math Chapter 13 are provided here for you for free.

Board	CBSE
Textbook	NCERT
Class	Class 10
Subject	Maths
Chapter	Chapter 13
Chapter Name	Surface Areas and Volumes
Exercise	Ex 13.1
Number of Questions Solved	9
Category	NCERT Solutions

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

NCERT Solutions for Class 10 Maths

Question 1.
2 cubes each of volume 64 cm³ are joined end to end. Find the surface area of the resulting cuboid.

Solution:
NCERT Solutions for Class 10 Maths Chapter 13

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:
NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes ex 13.1 2s

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:
NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes ex 13.1 2s1

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:
NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes ex 13.1 4s1

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

Solution:
NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes ex 13.1 5s

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:
NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes ex 13.1 6s

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:
NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes ex 13.1 7s

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:
NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes ex 13.1 8s

Question 9.
A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in 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.
NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes ex 13.1 9q

Solution:
NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes ex 13.1 9s

We hope the NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.1 help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.1, drop a comment below and we will get back to you at the earliest.

NCERT Solutions For Class 10 Maths – AplusTopperhttps://t.co/q7hSKh3xmf #NCERTSolutionsForClass10Maths #NCERTSolutions #AplusTopper
— ObulReddy cbse (@ObulReddyCBSE) August 3, 2018

NCERT Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.1

November 27, 2020January 17, 2019 by Veerendra

NCERT Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.1 are part of NCERT Solutions for Class 10 Maths. Here are we have given Chapter 14 Statistics Class 10 NCERT Solutions Ex 14.1.

Students find NCERT Solutions for Class 10 Maths Chapter 14 Statistics quite helpful for their CBSE Board Preparation. Refer to our Textbook Solutions any time, while doing your homework or while preparing for the exam. All questions and answers from the NCERT Book of class 10 Math Chapter 14 are provided here for you for free.

Board	CBSE
Textbook	NCERT
Class	Class 10
Subject	Maths
Chapter	Chapter 14
Chapter Name	Statistics
Exercise	Ex 14.1
Number of Questions Solved	9
Category	NCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.1

NCERT Solutions for Class 10 Maths

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.

Number of plants	0 – 2	2 – 4	4 – 6	6 – 8	8 – 10	10 – 12	12 – 14
Number of houses	1	2	1	5	6	2	3

Which method did you use for finding the mean, and why?
Solution:
Let us find class marks (x_i) for each interval by using the relation.

Now we may compute x_i and f_ix_i as following

Number of plants	Number of houses (f_i)	x_i	f_ix_i
0 – 2	1	1	1— 1 = 1
2 – 4	2	3	2 — 3 = 6
4 – 6	1	5	1 — 5 = 5
6 – 8	5	7	5 — 7 = 35
8 – 10	6	9	6 — 9 = 54
10 – 12	2	11	2 —11 = 22
12 – 14	3	13	3 — 13 = 39
Total	20		162

From the table we may observe that
NCERT Solutions for Class 10 Maths Chapter 14 Statistics 1s1
So, mean number of plants per house is 8.1.
We have used here direct method as values of class marks (x_i) and f_i are small.

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

Daily wages (in Rs)	100 – 120	120 – 140	140 -160	160 – 180	180 – 200
Number of workers	12	14	8	6	10

Solution:
Let us find class mark for each interval by using the relation.

Class size (h) of this data = 20
Now taking 150 as assured mean (a) we may calculate di, u_i and f_iu_i as following.

Daily wages (in Rs)	Number of workers (f_i)	x_i	d_i = x_i– 150		f_iu_i
100 -120	12	110	– 40	-2	– 24
120 – 140	14	130	– 20	-1	– 14
140 – 160	8	150	0	0	0
160 -180	6	170	20	1	6
180 – 200	10	190	40	2	20
Total	50				-12

From the table we may observe that
NCERT Solutions for Class 10 Maths Chapter 14 Statistics 2s1
So mean daily wages of the workers of the factory is Rs.145.20

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.

Daily pocket allowance (in Rs)	11 – 13	13 – 15	15 -17	17 – 19	19 – 21	21 – 23	23 – 25
Number of workers	7	6	9	13	f	5	4

Solution:
We may find class mark (x_i) for each interval by using the relation.

Given that mean pocket allowance \(\overline { x } \) = Rs.18
Now taking 18 as assured mean (a) we may calculate d_i and f_id_i as following.

Daily pocket allowance (in Rs.)	Number of children f_i	Class mark x_i	d_i = x_i – 18	f_id_i
11 – 13	7	12	– 6	– 42
13 – 15	6	14	– 4	– 24
15 – 17	9	16	– 2	– 18
17 – 19	13	18	0	0
19 – 21	f	20	2	2 f
21 – 23	5	22	4	20
23 – 25	4	24	6	24
Total				2f – 40

From the table we may obtain
NCERT Solutions for Class 10 Maths Chapter 14 Statistics 3s.2
Hence the 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 summarized as follows. Fine the mean heart beats per minute for these women, choosing a suitable method.

Number of heart beats per minute	65 – 68	68 – 71	71-74	74 – 77	77 – 80	80 – 83	83 – 86
Number of women	2	4	3	8	7	4	2

Solution:
We may find class mark of each interval (x_i) by using the relation.

Class size h of this data = 3
Now taking 75.5 as assumed mean (a) we may calculate d_i, u_i, f_iu_i as following.

Number of heart beats per minute	Number of women f_i	x_i	d_i = x_i -75.5		f_iu_i
65 – 68	2	66.5	– 9	– 3	– 6
68 – 71	4	69.5	– 6	– 2	– 8
71 – 74	3	72.5	– 3	– 1	– 3
74 – 77	8	75.5	0	0	0
77 – 80	7	78.5	3	1	7
80 – 83	4	81.5	6	2	8
83 – 86	2	84.5	9	3	6
Total	30				4

Now we may observe from table that
NCERT Solutions for Class 10 Maths Chapter 14 Statistics 4s.2
So mean heart beats per minute for these women are 75.9 beats per minute.

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.

Number of mangoes	50 – 52	53 – 55	56 – 58	59 – 61	62 – 64
Number of boxes	15	110	135	115	25

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

Number of mangoes	Number of boxes f_i
50 – 52	15
53 – 55	110
56 – 58	135
59 – 61	115
62 – 64	25

We may observe that class intervals are not continuous. There is a gap of 1 between two class intervals. So we have to add 1/2 to upper class limit and subtract 1/2 from lower class limit of each interval.
And class mark (x_i) may be obtained by using the relation

Class size (h) of this data = 3

Now taking 57 as assumed mean (a) we may calculate d_i, u_i, f_iu_i as following –

Class interval	f_i	x_i	d_i = x_i – 57		f_iu_i
49.5 – 52.5	15	51	-6	-2	-30
52.5 – 55.5	110	54	-3	-1	-110
55.5 – 58.5	135	57	0	0	0
58.5 – 61.5	115	60	3	1	115
61.5 – 64.5	25	63	6	2	50
Total	400				25

Now we may observe that

NCERT Solutions for Class 10 Maths Chapter 14 Statistics 5s.2

Clearly, mean number of mangoes kept in a packing box is 57.19.
We have chosen step deviation method here as values of f_i, d_i are big and also there is a common multiple between all d_i.

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

Daily expenditure (in Rs)	100 – 150	150 – 200	200 – 250	250 – 300	300 – 350
Number of households	4	5	12	2	2

Find the mean daily expenditure on food by a suitable method.
Solution:
We may calculate cla

mark (x_i) for each interval by using the relation

Class size = 50

Now taking 225 as assumed mean (a) we may calculate d_i, u_i, f_iu_i as following

Daily expenditure (in Rs)	f_i	x_i	d_i = x_i – 225		f_iu_i
100 – 150	4	125	-100	-2	-8
150 – 200	5	175	-50	-1	-5
200 – 250	12	225	0	0	0
250 – 300	2	275	50	1	2
300 – 350	2	325	100	2	4
Total	25				-7

Now we may observe that –

NCERT Solutions for Class 10 Maths Chapter 14 Statistics 6s.2

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:

concentration of SO₂ (in pmm)	Frequency
0.00 – 0.04	4
0.04 – 0.08	9
0.08 – 0.12	9
0.12 – 0.16	2
0.16 – 0.20	4
0.20 – 0.24	2

Find the mean concentration of SO₂ in the air.

Solution:

NCERT Solutions for Class 10 Maths Chapter 14 Statistics 7s

Concentration of SO₂ (in ppm)	Frequency	Class mark x_i	d_i = x_i– 0.14		f_iu_i
0.00 – 0.04	4	0.02	-0.12	-3	-12
0.04 – 0.08	9	0.06	-0.08	-2	-18
0.08 – 0.12	9	0.10	-0.04	-1	-9
0.12 – 0.16	2	0.14	0	0	0
0.16 – 0.20	4	0.18	0.04	1	4
0.20 – 0.24	2	0.22	0.08	2	4
Total	30				-31

NCERT Solutions for Class 10 Maths Chapter 14 Statistics 7s.2

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.

Number of days	0 – 6	6 – 10	10 – 14	14 – 20	20 – 28	28 – 38	38 – 40
Number of students	11	10	7	4	4	3	1

Solution:
We may find class mark of each interval by using the relation

NCERT Solutions for Class 10 Maths Chapter 14 Statistics 8s
Now taking 16 as assumed mean (a) we may calculate d_i and f_id_i as following

Number of days	Number of students *f_i*	*x_i*	*d_i= x_i* – 16	*f_id_i*
0 – 6	11	3	-13	-143
6 -10	10	8	-8	-80
10 – 14	7	12	-4	-28
14 – 20	4	16	0	0
20 – 28	4	24	8	32
28 – 38	3	33	17	51
38 – 40	1	39	23	23
Total	40			-145

Now we may observe that

NCERT Solutions for Class 10 Maths Chapter 14 Statistics 8s.1
So, mean number of days is 12.38 days, for which a student was absent.

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

Literacy rate (in %)	45 – 55	55 – 65	65 – 75	75 – 85	85 – 95
Number of cities	3	10	11	8	3

Solution:
We may find class marks by using the relation

Class size (h) for this data = 10

Now taking 70 as assumed mean (a) we may calculate d_i, u_i, and f_iu_i as following

Literacy rate (in %)	Number of cities *f_i*	*x_i*	*d_i= x_i* – 70	*u_i*	*f_iu_i*
45 – 55	3	50	-20	-2	-6
55 – 65	10	60	-10	-1	-10
65 – 75	11	70	0	0	0
75 – 85	8	80	10	1	8
85 – 95	3	90	20	2	6
Total	35				-2

Now we may observe that
NCERT Solutions for Class 10 Maths Chapter 14 Statistics 9s.1
So, mean literacy rate is 69.43%.

We hope the NCERT Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.1 help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.1, drop a comment below and we will get back to you at the earliest.

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— ObulReddy cbse (@ObulReddyCBSE) August 3, 2018

What are Concave and Convex Mirrors?

December 18, 2020December 18, 2018 by Veerendra

What are Concave and Convex Mirrors?

We are used to plane mirrors. But, when they are curved, they produce images that are different from those formed by plane mirrors. Common examples of curved mirrors stainless steel spoons, ladles, vessels with round bases, car mirrors, shaving and hub caps of car wheels. These curved objects have two types of reflecting surfaces one that bulges out to form a convex surface, and one that curves inwards to concave surface. The first one is called a convex mirror and the latter is a concave mirror.

A curved mirror is actually a mirror with a curved reflective surface. A curved mirror has the same shape as a section of a large sphere as shown in Figure.
If the outside surface of the section reflects, the mirror is a convex mirror. If the inside surface reflects, the mirror is a concave mirror.
The principal axis of a curved mirror is the line passing through its vertex, P and the centre of curvature, C.
The centre of curvature, C of a curved mirror is the centre of the sphere that forms the curved mirror.
The radius of curvature, R of a curved mirror is the radius of the sphere.
A concave mirror curves inwards. Parallel light rays that hit the concave surface are reflected inwards as shown in Figure.
The focal point or the principal focus, F of a concave mirror is the point on the principal axis where the reflected rays converge, or in other words, meet and intersect. This is a real focus.
A concave mirror is also known as a converging mirror.
A convex mirror curves outwards. Parallel light rays that hit the convex surface are reflected outwards as shown in Figure.
The focal point, F of a convex mirror is the point on the principal axis where the reflected rays diverge, or in other words, appear to spread out from behind the mirror. This is a virtual focus.
A convex mirror is also known as a diverging mirror.
The focal length, f of a curved mirror is the distance from its vertex, P to its principal focus, F.
The focal length of a curved mirror is equal to half of the radius of curvature, that is = R/2

People also ask

Uses of Curved Mirrors

Convex mirrors make things look smaller but you can see a lot more of the surroundings.
In other words, they give you a wider range of view. Some uses of convex mirrors are given below.
• As car wing mirrors, to see through a wide angle behind the car.
• In supermarkets and big shops to look out for shoplifters.
• In buses, so that the driver can view the whole bus.

When objects are kept very close to a concave mirror, objects look larger. This makes concave mirrors useful for the following applications:
• As shaving mirrors or while putting on make-up.
• As a dentist’s mirror for examination.

Concave mirrors can also focus a parallel beam of light to a point. This property is used in solar-powered projects where they reflect the heat rays of the sun so that the rays are concentrated onto a smaller area.

How do you Draw a Ray Diagram for Concave and Convex Mirrors?

1. The characteristics of the images formed by curved mirrors can be determined by drawing ray diagrams using three key rays as described:
  (a) A parallel ray (ray P) that is incident along a path parallel to the principal axis and is reflected through (or appears to go through) the focal point.
  (b) A chief ray (ray C) or radial ray is the ray that incidents through the centre of curvature, C of the curved mirror. Since ray C is incident at normal to the surface of the mirror, it is reflected back along its incident path through point C.
  (c) A focal ray (ray F) is the ray that passes through (or appears to go through) the focal point and is reflected parallel to the axis.
2. Figure shows the three key rays used in the construction of ray diagrams for a concave mirror.

Figure shows the three key rays used in the construction of ray diagrams for a convex mirror.
For both concave and convex, mirrors, ray P and ray F can be reversed to travel back along the same path. This property is known as the reversibility of light.
The point where any of the two construction lines cross or intersect is the position of the image formed by the curved mirrors.

Steps for Drawing Ray Diagrams

The following steps are used to draw a ray diagram for a concave mirror.
- Step 1: Draw a straight line as the principal axis, curved line as the curved mirror and dotted vertical line at vertex P as the construction line for the curved mirror.
- Step 2: Mark the positions of F and C on the axis with CF = FP.
- Step 3: Draw an arrow as the object standing upright on the axis at a given distance.
- Step 4: Draw ray P coming from the top of the object.
- Step 5: Draw ray F coming from the top of the object. Note that you can also use ray C coming from the top of the object. The top of the image is where any two of the reflected rays meet again after the reflection.
- Step 6: Draw the image with its tail on the axis. The head is the point where the two reflected rays meet.
The characteristics of the image formed can be:
(a) real or virtual,
(b) inverted or upright,
(c) magnified, diminished or of the same size.
For a convex mirror, the ray diagram is as shown in Figure.
Dotted lines are used for extending the rays backward to locate the virtual image behind the surface of the mirror.
A virtual image is also drawn using a dotted line with an arrow.
Ray diagrams showing the images formed by a concave mirror and a convex mirror when the object is placed at different distances are shown in the next section. The characteristics of the image formed are stated in each case.

ISC Specimen Papers for Class 12 Chemsitry 2020, 2019, 2018, 2017

September 11, 2019December 1, 2018 by Veerendra

Download ISC Specimen Papers 2020 Solved for Class 12 Chemsitry and Marking Scheme PDF. Here we have given ISC Chemsitry Question Papers 2020 Solved. Students can view or download the Specimen Papers for ISC 2020 with Answers Class 12 Chemsitry for their upcoming examination.

These ISC Board Sample Papers are useful to understand the pattern of questions asked in the board exam. Know about the important concepts to be prepared for ISC Class 12 Chemsitry board exam and Score More marks.

Board – Indian School Certificate Examinations (CISCE), www.cisce.org
Class – Class 12
Subject – Chemsitry
Year of Examination – 2020, 2019, 2018, 2017.

ISC Class 12 Chemsitry Specimen Papers Solved

www.cisce.org ISC Sample Papers for Class 12 Chemsitry are part of ISC Specimen Papers Solved for Class 12 Here we have given ISC Class 12 Chemsitry Sample Question Papers for Class 12 Chemsitry.

Year of Examination	ISC Chemistry Question Paper
2018	Download PDF (Theory) Download PDF (Practical)
2017	Download PDF (Theory) Download PDF (Practical)

Last 10 Years ISC Question Papers Class 12 Chemistry Solved

The above ISC Model Paper for Class 12 Chemsitry is the official sample paper released by ISC Board as per latest syllabus of Class 12 Indian Certificate of Secondary Education, India.

We hope the ISC Specimen Papers for Class 12 Chemsitry, help you. If you have any query regarding ISC Class 12 Chemsitry Question Papers Solved, drop a comment below and we will get back to you at the earliest.

ISC Specimen Papers for Class 12 Accounts 2020, 2019, 2018, 2017

September 11, 2019December 1, 2018 by Veerendra

Download ISC Specimen Papers 2020 Solved for Class 12 Accounts and Marking Scheme PDF. Here we have given ISC Accounts Question Papers 2020 Solved. Students can view or download the Specimen Papers for ISC 2020 with Answers Class 12 Accounts for their upcoming examination.

These ISC Board Sample Papers are useful to understand the pattern of questions asked in the board exam. Know about the important concepts to be prepared for ISC Class 12 Accounts board exam and Score More marks.

Board – Indian School Certificate Examinations (CISCE), www.cisce.org
Class – Class 12
Subject – Accounts
Year of Examination – 2020, 2019, 2018, 2017.

ISC Class 12 Accounts Specimen Papers Solved

www.cisce.org ISC Sample Papers for Class 12 Accounts are part of ISC Specimen Papers Solved for Class 12 Here we have given ISC Class 12 Accounts Sample Question Papers for Class 12 Accounts.

Year of Examination	ISC Accounts Question Paper
2020	Download PDF
2018	Download PDF
2017	Download PDF

The above ISC Model Paper for Class 12 Accounts is the official sample paper released by ISC Board as per latest syllabus of Class 12 Indian Certificate of Secondary Education, India.

We hope the ISC Specimen Papers for Class 12 Accounts, help you. If you have any query regarding ISC Class 12 Accounts Question Papers Solved, drop a comment below and we will get back to you at the earliest.