Skip to main content

for Sample Paper Summative Assessment- II, March 2014 Class IX - X

Image
By



Modified syllabus -2013 , Design of Assessment and sample questions including questions based on values in the subjects Hindi (Course A and B), English (Communicative), English(Language and Literature), Mathematics, Science, Social Sciences and Sanskrit to be used for Summative Assessment- II, March 2013 Class IX.
Hindi Course A    |    Hindi Course B    |    English (Communicative)    | English(Language and Literature)    |   Mathematics    |    Science    |    Social Sciences    |    Sanskrit
CBSE : Modified syllabus for Class X March 2014
Click here to download CBSE Circular
English (Communicative)
Hindi (Course B)
Mathematics
Sanskrit
Science
Social Sciences





support-material-on-psa-and-value-based

Class X  Model   Answer  CBSE Exam.2013
Class X Model  Answer CBSE Exam.2012
MARKING SCHEME FOR [2013] SUMMATIVE  ASSESSMENT – II
1.Science
1.Science
Hindi Part: A        
 Hindi Part: B
2.Social Science
2.Social Science
 English Communicative   
3.Mathematics
3.Mathematics
 English Language and Literature   
4.Hindi-A
4.Hindi-A
Mathematics   
5.Hindi-B
5.Hindi-B
Science   
6.Eng.Lang & Literature
6.Eng.Lang & Litrature
Social Science   
7.Eng.communicative
7.Eng.communicative
Sanskrit 
Labels:

Comments

  1. UnknownFebruary 23, 2013

    good

    ReplyDelete
  2. UnknownFebruary 28, 2013

    nice but it is of old syllabus

    ReplyDelete
  3. J.Sunil SirDecember 22, 2013

    CBSE is a board of education that provides high quality syllabus to aide students’ development of their skills. In this endeavour, they constantly change the syllabus and exam methodology. Since the pattern is constantly changed, students are often faced with the dilemma and confusion of the pattern of the question paper. To reduce the anxiety, we prepared sample question papers that help students in preparing for the current year’s exams by way of reviewing and revising the questions available therein.

    ReplyDelete

Post a Comment

CBSE ADDA :By Jsunil Sir : Your Ultimate Destination for CBSE Exam Preparation and Academic Insights

Class 10 Chapter 02 Acid Bases and Salts NCERT Activity Explanation

By
NCERT Activity Chapter 02 Acid Bases and Salt Class 10 Chemistry Activity 2.1 Indicator Acid Base Red litmus No Change Blue Blue Litmus Red No change Phenolphthalein Colourless Pink Methyl Orange Pink   Yellow Indictors are substance which change colour in acidic or basic media. Activity 2.2 There are some substances whose odour changes in in acidic or basic media. These are called olfactory indicators. Like onion vanilla, onion and clove. These changes smell in basic solution. Activity 2.3 Take about 5 mL of dilute sulphuric acid in a test tube and add few pieces of zinc granules to it. => You will observe bubbles of hydrogen gas on the surface of zinc granules. Zn + H2SO4 --> ZnSO4 + H2 => Pass the Hydrogen gas through the soap solution. Bubbles formed in the soap solution as Hydrogen gas it does not get dissolved in it
Post a Comment
Read more

CBSE I NCERT 10th Numerical Problem solved Reflection and reflection of light

By
Q. 1. A concave mirror of focal length 20cm is placed 50 cm from a wall. How far from the wall an object be placed to form its real image on the wall?  Solution: V= -50 cm F= -20cm From mirror formula 1/u = 1/f – 1/v = -1/20+ 1/50 = - 3/100  U = - 33.3 cm Therefore, the distance of the object from the wall x =  50 – u X = 50 – 33.3 = 16.7 cm. Q.2. An object is placed at a distance of 40cm from a concave mirror of focal length 15cm. If the object is displaced through a distance of 20 cm towards the mirror, By how much distance is the image displaced? Answer: Here f = - 15 cm, u = - 40 cm Now 1/f = 1/u + 1/v Then 1/v = 1/f – 1/u Or V= uf/u-f =( - 40 x -15)/25 = -24 cm Then object is displaced towards the mirror let u1 be the distance object from the Mirror in its new position. Then u1 = -(40-20) = -20cm If the image is formed at a distance u1 from the mirror then v1 = u1f/u1-f = -20X-15/-20+15 = -60 cm. = - 20 x-15/-20+15 = -60 cm. Therefor
2 comments
Read more

Class 10 Metal and Non MetalsChapter 03 NCERT Activity Solutions

By
X Class 10 NCERT Activity Explanation Class 10 Metals and Non Metals Activity 3.1 Page No. 37 Take samples of iron, copper, aluminium and magnesium. Note the appearance of each sample. They have a shining surface. Clean the surface of each sample by rubbing them with sand paper and note their appearance again. They become more shiny. => Freshly cut Metal have shiny surface Activity 3.2 Page No. 37 Take small pieces of iron, copper, aluminium, and magnesium. Try to cut these metals with a sharp knife and note your observations. They are very hard to cut. Hold a piece of sodium metal with a pair of tongs and try to cut it with a knife. Sodium can be cut easily with knife. Hence K and Na are soft metal cut with knife Activity 3.3 Page No. 38 Take pieces of iron, zinc, lead and copper try to strike it four or five times with a hammer. These metals are beaten into thin sheet on hammering. This property of metal is called malleability and metals are called malleable. Activity 3.4 Page
Post a Comment
Read more

Living science ratna sagar class 6 solutions

By
Ratna sagar living science 6 answers by jsunil. Class6 Living science solution Term-1 Living Science Solution chapter-1 Source of food Download File Living Science Solution chapter-2 Component of food Download File Living Science Solution chapter-3 Fibre to fabric Download File Living Science Sol ch-4 Sorting of material into group Download File Living Science Soln ch-5 Separation of substance Download File Living Science Solution chapter-6 Change around Us Download File Living Science Solution ch-7 Living and Non Living Download File Living Science Solution ch-8 Getting to Know Plants Download File Living Science Sol ch-9 The Body and Its movements Download File Visit given link for full answer Class6 Living science solution Term-II
Post a Comment
Read more

Electricity numerical for class 10 CBSE Trend Setter 50 Problems

By
1. The current passing through a room heater has been halved. What will happen to the heat produced by it? 2. An electric iron of resistance 20 ohm draws a current of 5 amperes. Calculate the heat produced in 30 seconds. 3. An electric heater of resistance 8 ohm takes a current of 15 A from the mains supply line. Calculate the rate at which heat is developed in the heater. 4. A resistance of 40 ohms and one of 60 ohms are arranged in series across 220 volt supply. Find the heat in joules produced by this combination in half a minute. 5. A resistance of 25 ohm is connected to a 12 V battery. Calculate the heat energy in joules generated per minute. 6. 100 joules of heat is produced per second in a 4 ohm resistor. What is the potential difference across the resistor? 7. An electric iron is connected to the mains power supply of 220 V. When the electric iron is adjusted at minimum heating’ it consumes a power of 360 W but at ‘maximum heating’ it takes a power of 840 W. Ca
Post a Comment
Read more