Chapter 02 Polynomials 


Problem No 01 :- Find the Zeros of the Polynomials and verify the relationship between the zeros and coefficients :-


x2 + 7x + 12


To find the zeros of polynomial, It should be equated to zero first

Hence,


x2 + 7x + 12 = 0


Or, x2 + 4x + 3x + 12 = 0


Or, x (x + 4)+ 3 (x + 4 ) = 0


Or, (x + 3) * (x + 4 ) = 0


It implies that x = - 3 & -4 Ans ............................................... (1)


Comparing the Given Eqn with Stnd Eqn i.e. ax2 + bx + c = 0


We Get, a=1 , b = 7 & c = 12


α + β = -3 + ( - 4) = - 7................................... (2)


-b /a = -7 / 1 = -7 .............................................(3)


Also α * β = -3 * -4 = 12 .................................(4)


c / a = 12 / 1 ......................................................(5)


From (2), (3) (4) & (5) relation between the zeros and coefficients can be verified.

 

CLASS 10 MATHEMATICS

Welcome All, 

In this Section,  a collection of problems have been presented from various Books of CBSE Mathematics Class 10. 

Problems can be seen by clicking on the list of chapters given below :- 

  1. Real Numbers
  2. Polynomials
  3. Linear Equation in Two Variables
  4. Triangles
  5. Trigonometric Ratios. 
  6. T-Ratios of some Particular Angles. 
  7. Trigonometric Ratios of Complementary Angles. 
  8. Trigonometric Identities. 
  9. Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive. 
  10. Quadratic Equations
  11. Arithmetic Progressions.
  12. Constructions
  13. Circles
  14. Heights and Distances
  15. Probability
  16. Coordinate Geometry 
  17. Perimeter and Area of Plane Figures. 
  18. Area of Circle, Sector and Segment
  19. Volume and Surface Area of Solids
  20. Misc Problems


Feel Free to give your suggestions and feedback in the comment section below. 

Thank you. 





Class 10 Area


Answer:- 

As per the Diagram
EC = BC (Radius in Sector BGEC) 
Hence EC = 7 Cm. 
DE = 4 cm Given 
DC = DC + EC 
= 4 CM + 7 CM = 11 cm 
Area of trapezium ABCD 
= (1/2) *( Sum Of ||el sides) * (Height)
= (1/2) * (AB+DC) *(BF) 
=(1/2) * (7 cm + 11 cm ) * (3.5 cm ) 
= 0.5 * 18 cm * 3.5 cm 
= 9 cm * 3.5 cm 
= 31.50 cm2 
Area of Sector BGEC :- 
(O/360) *( pi *r*r) 
= (30/360) *(22/7*7*7)
(1/12) * (22*7) 
= 12.83 cm2 
Area Of shaded portion = Area of trapezium - Area of Sector 
= 31.50 cm2 - 12.83 cm2 
= 18.67 cm2 
Hence the Right answer is (d). 
Please like share and comment. 
Do write us for any other numerical related to Class 10 Mathematics. 
Thank you. 






Hydrostatic Forces on Surfaces

Numerical No. 1


Numerical No 02 

  Plumbing 

 

Plumbing notes based on SP-35 1987 are as follows :- 

 1. General 

  •  Plumbing System includes 
    • water supply and distributing pipes 
    • plumbing fixtures for the use in water supply
    • sanitary drainage system to carry the wastes 
    • Anti-siphonage system which carry only the air for the purpose of ventilation  
    • Storm water drainage system to collect and carry rain water 
  • Planning for Plumbing
    • The layout of the building should be such as to allow for good and economical plumbing to be carried out.   
    • Cross connection between water supply and waste water pipes should be avoided. 
    • Noise associated with Plumbing system (Water Pumps) should be kept in mind. 
    • When a water pipe is to be concealed, it must be wrapped with Hassian cloth dipped in bitumen. Hassian Cloth can be seen in the picture given below :- 
    •   
       
      • Single Pipe system is now generally preferred over two pipe systems. 
      • Indian Standards should be adopted while selecting materials, equipment, construction or testing the fixtures so that uniformity can be maintained. 
      •  

       

      Principles of Plumbing. 

      1. Plumbing Fixtures should be smooth and non absorbent, well ventilated, away from fouling places, easily accessible, leak free. 
      2. Plumbing system should help in using the water economically, should avoid cross connection of water and sewerage line, well ventilated and well designed.

       

      Thank you. 

       

      Further Read in this Series :- 

       

       

       

       

       

 

 

Principle of Effective Stress, Capillarity and Permeability

Numerical 

Total Stress = Sum of all the stresses above a point. 

Neutral Stress = Hydrostatic Pressure above a point. 

Effective Stress = Total Stress - Neutral Stress

Few Numerical are given below based on these concepts. 

Numerical 1 :- 


 


Numerical 2 :- 



In respect of these two numerical, it can be understood that, 

1. A rise in water table results in the decrease of effective stress. 

2.  Lowering of water table results in increase of effective stress. 




NUMERICAL ON SOIL CLASSIFICATION 


1. Numerical of Gopal Ranjan and AS RAO Book.

Given Details as shown below in the numerical. We need to classify the soil as per indian Standard.

Sieve Size (mm)Percent Finer
Sample 1Sample 2Sample 3
4.75100.001001100
2100.0099.25100
199.8098.7599.9
0.4599.659899.7
0.21298.9554.1590
0.1597.557.681.5
0.07596.85663
wL23.00'----'---
wp16.77'----'---
IP6.23Non PlasticNon Plastic

Sample 1. 

A. Look at the % finer in 0.075 mm sieve (75 micron sieve) 

      Which in this case is 96.85 (marked Red) As this fraction is greater than  50 % so soil is fine grained soil and we need to look in to Liquid Limit and Plastic Limit of the soil sample. 


B. Liquid limit and Plastic limits are given. 
          wl=23.00
          wp = 16.77 
          Plasticity Index = 6.23 (Given) 

In the plasticity chart the Liquid Limit and Plasticity Index will be plotted somewhere marked hatched in the image. 



So for the Sample Dual Symbol will be used. 

Sample will therefore will be classified as CL-ML. 


Sample 2 

A. % Finer of 75 micron sieve Size is 6.00 which is less than 50 %. Hence sample will be coarse grained soil. 

B. % finer than 4.75 mm is 100.00 (Given) which is greater than 50 %. Hence sample will be sand. 

C. Now we need to look in to the Grain Size Curve.

Grain Size Curve is as Follows :-

It can be easily observed from the Grain Size Distribution Curve that 

for Sample 2 :- 
 D60 = 0.22 mm. D30 = 0.19 mm, D10 = 0.16 mm

Cu = D60/D10 = 0.22/0.16 = 1.38. (Which is less than 06. 

Cc = (D30*D30)/(D60*D10) = (0.19*0.19)/(0.22*0.16)= 1.03 

For a soil to be well graded Cu > 6 & 1<Cc<3. Both these two conditions should satisfy simultaneously. 

so First Part is SP. 

Also % Finer passing 75 micron is 6 % which lies between 5% and 12 %. However it is also given that fines are Non Plastic. SM will be the classification. 

Hence Sample 2 will be SP-SM. 


Sample 3 


Step 1. More than 50 5 of soil is passing through 75 micron sieve. (63 % Given). Soil is either clay or silt. 

Step 2. Liquid Limit and Plastic Limit is not Given and soil has been described as Non Plastic. So Classification will be termed as ML



Click here For Another Numerical based on Soil Classification. 















Classification of Soil

















Step A . Pass Soil sample through 75 micron sieve









More than 50 % soil sample retained : Coarse Grained Soil.








More than 50 % soil sample passing : Fine Grained Soil.










Step B. For Coarse Grained Soil.









i ) Pass the soil (retained on 75 micron sieve) through 4.75 mm sieve







More than 50 % soil sample retained : Gravel (G)







More than 50 % soil sample passing : Sand (S)











ii) Check the percent fines in the coarse fraction.










If Fines < 5 % -> Check Uniformity Coefficient and Coefficient of Curvature










Soil will termed as Poorly Graded and Well Graded.












GW, SW, GP, SP.











If Fines > 12 % -> Check the nature of Fines from A -Line as Given below.









Step C. For Fine Grained Soil







.
i) Check the Clayey Fraction and Silt Fraction based on the Grain Size. Assign the 









Prefix C and M Accordingly.









ii) Using Atterberg Limits, Arrive at the Suffix, L, I, H and accordingly assign the 








suffiix.









This more or less completes the classification part of the soil. Now We will move towards 








What is the difference between Marble and Granite ?
All the site Engineers use to face the same question during the construction phase. Here we will try to give few points of difference between marble and granite. 


Point No. 01 :- 


Granite is a silicious rock which implies that silica predominates in this stone. 
Marble is a Calcareous Rock which implies that Calcium Carbonate predominates in this stone.

Point No. 02 :- 


Granite due to its chemical composition is not affected by the process of weathering.  
Marble however is not that durable. Due to presence of Calcium Carbonate its durability depends on the surrounding environment. For example Taj Mahal in Agra is made up of White Stones which are believed to be deteriorating due to pollution. 

Point No. 03 :- 


Granite is more strong. Compressive strength of Granite varies from 75 N/mm2 to        127 N/mm2. 
Marble is comparatively less stronger. Its compressive strength is around 71 N/mm2. 

Point No. 04 :- 


Carving (नक्कासी ) is difficult in case of granite and the reason can be better understand in relation with the Point No. 03 Above. 
Carving is easy in case of Marble.  Probably this is the reason of choosing Marble in Historic Structures. 

Point No. 05 :- 


in Terms of Origin, Granite is an Igneous Rock. 

Marble is a Metamorphic Rock.

Point No 06 :- 


In General, Granite is more costly compared to Marble. 



These are the basic differences between Granite and Marble.

If you have any query related to this Topic feel Free to ask in the comment section below.

Thank you.


Click Here to read about other topics related to Building Material and Construction. 


SOIL MECHANICS AND FOUNDATION ENGINEERING 

IES SYLLABUS 



(a) Geo-technical Engineering : Soil exploration - planning & methods, Properties of soil, classification, various tests and inter-relationships; Permeability & Seepage, Compressibility, consolidation and Shearing resistance, Earth pressure theories and stress distribution in soil; Properties and uses of geo-synthetics.
(b) Foundation Engineering: Types of foundations & selection criteria, bearing capacity, settlement analysis, design and testing of shallow & deep foundations; Slope stability analysis, Earthen embankments, Dams and Earth retaining structures: types, analysis and design, Principles of ground modifications.

Click on the link below for further information about the topics : 
  1. Soil Exploration 
  2. Properties of Soil 
  3. Soil Classification 
  4. Permeability and Seepage
  5. Compressibility
  6. Consolidation and Shearing Resistance. 
  7. Earth Pressure
  8. Geo Synthetics. 
  9. Foundation Engineering
  10. Earth Retaining Structures. 
  11. Ground Modification


Happy Reading

Please share your feedback and queries by commenting in the comment box below. 

BUILDING MATERIAL AND CONSTRUCTION


IES Syllabus 



Stone, Lime, Glass, Plastics, Steel, FRP, Ceramics, Aluminum, Fly Ash, Basic Admixtures, Timber, Bricks and Aggregates: Classification, properties and selection criteria;Cement: Types, Composition, Properties, Uses, Specifications and various Tests; Lime & Cement Mortars and Concrete: Properties and various Tests; Design of Concrete Mixes: Proportioning of aggregates and methods of mix design.



Click on the Topics Given below for further information :- 



  1. Lime
  2. Stone
  3. Glass
  4. Plastics
  5. Steel
  6. FRP
  7. Ceramics
  8. Aluminium
  9. Fly Ash
  10. Basic Admixtures
  11. Timber
  12. Bricks
  13. Aggregates
  14. Cement
  15. Mortar
  16. Concrete
  17. What is the difference between Marble and Granite ?

Happy Reading. 

Please Share your feedback and queries in comment section below.