Problem Dated 15 Nov 2021. 

Class 11 Physics

Chapter No 03 
Question No 3.8 

3.8 On a two-lane road, car A is travelling with a speed of 36 km h–1. Two cars B and C approach car A in opposite directions with a speed of 54 km h–1 each. At a certain instant, when the distance AB is equal to AC, both being 1 km, B decides to overtake A before C does. What minimum acceleration of car B is required to avoid an accident ?
3.8 दो-लेन वाली किसी सड़क पर कार A 36 km h–1 की चाल से चल रही है । एक दूसरे की विपरीत दिशाओं में चलती दो कारें B व C जिनमें से प्रत्येक की चाल 54 km h–1 है, कार A तक पहुँचना चाहती हैं । किसी क्षण जब दूरी AB दूरी AC के बराबर है तथा दोनों 1km है, कार B का चालक यह निर्णय करता है कि कार C के कार A तक पहुँचने के पहले ही वह कार A से आगे निकल जाए । किसी दुर्घटना से बचने के लिए कार B का कितना न्यूनतम त्वरण जरूरी है ?


Problems Dated Nov 05 2021. 




Problem No 09 

First Number = x 

Second Number =y 

Case 1 
(x+2)/(y+2) = 1/2

Case 2 
(x-4)/(y-4) = 5/11

From 1 

2x - y = -2 ...... 3 

from 2 

11x - 5y = 24 .... 4 

Multiplying Eqn 1 by 5 

10 x - 5y = -10   .......5

Eqn 4 -Eqn 5 

11x - 5y -10 x + 5y = 24 + 10 

x = 34 

From Eqn 1 

2 (34) - y = -2 
or y = 68 + 2 = 70  Answer. 


 

Exercise 3E R S Agarwal 

Problem No 01 

Chair = x 

Tables = y 


5 x + 4 y = 5600 ..... 1 

4 x + 3 y = 4340 ....... 2 

1 * 4  &  2 * 5 

20x + 16 y = 22400.......... a 

20x + 15y = 21700............b 

a-b 

y = 700

5x + 2800 = 5600 

5x = 2800 

x = 560 Ans 


Problem No 02

 

MISC

Maths Problems dated 03 Nov 2021


Problem No 01

In the given figure APB and AQB are two semi circle. (Given) 

Perimeter of the given shape will be equal to 

= Circumference of half semi circle (APB) + Circumference of quarter semicircle AQO + QO RADIUS + OB RADIUS 

= ( PI * R ) + (0.5 * PI * R) + R + R 

But this perimeter is given equal to the 47 cm.

Hence, 

( PI * R ) + (0.5 * PI * R) + R + R  = 47 cm 

R * ( 22/7 + 11/7 + 1 + 1 ) = 47 CM 

R * ( 47/7) = 47 CM 

It implies that R = 7 cm 

Area of Shaded Region 


Area of total Circle - Area of Quarter CIrcle 

(PI * R * R ) - (0.25 * PI * R* R) = (0.75 * PI * R* R) 

= 0.75 * 22/7 * 7 * 7 = 115.50 CM2 ANS 
It seems that answer given is incorrect. 

CHAPTER O2 PART B POLYNOMIALS

In the Figure given below, find the number of zeroes in each case. 

How to Calculate Zeroes, when the Graph of a Polynomial has been given ? 

The Zeroes of a Polynomial indicates the values of 'x' at which the value of polynomial p(x) becomes zero. Hence In a Grapgh plotted for the Polynomial, the points at which the value of Polynomials becomes zero, or the points at which graph either interset or touch the 'x' axis gives the zeroes of that Polynomial. 

If a Polynomial does not intersect or touch the 'x' axis at any point, it indicates that the Polynomial does not have any zeroes. 

Figure 1. As the given graph of the Polynomial does not intersect 'x' axis at any point, it indicates that the Polynomial does not have any zeroes. 

Figure 2. 
The Polynomials Grapgh intersect the 'x' axis at exactly one point, so Polynomial has one zeroes.

Figure 3 
The Polynomial interset the 'x' axis at three points. It indicates that the Polynomial have three zeroes. 

Figure 4 
The Polynomial Graph intersect the 'x' axis at two exact points. So the Polynomial have two zeroes. 

Figure 5 
The Polynomial Graph intersect the 'x' axis at 04 exact points, so the Polynomial have four zeroes. 

Figure 06
The Polynomial Grapgh touches the 'x' axis at two points and intersect the 'x' axis at one point. It implies that Polynomial have zeroes out of which one is unique and other two are similar. 

Thank you Very Much 

Please like share and comment. 

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





 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.