Bhagya

CBSE Class 9 Maths Lab Manual – Area of Parallelograms on the Same Base and between the Same Parallels

Objective
To show that the parallelograms on the same base and between the same parallel lines are equal in area by paper cutting and pasting.

Prerequisite Knowledge

1. Area of parallelogram = base x height.
2. Shortest distance between two parallel lines is the perpendicular distance between parallel lines and it remains same for that pair of parallel lines.

Materials Required
Glazed paper, a pair of scissors, gluestick, geometry box.

Procedure

1. Draw a parallelogram by paper folding activity.
2. Cut this parallelogram using scissors and name it as ABCD, [fig.(i)].
   
3. Mark any point E on DC.
4. From A fold the parallelogram till the point E (any point on DC). A crease AE is formed, fill another colour in ∆ADE. [fig. (ii)].
   
5. Cut the parallelogram ABCD along AE to get ∆AED.
6. Paste the ∆AED on the other side along BC of parallelogram ABCD as shown in fig. (iii).
   
7. We get new parallelogram AEPB.

Observation

1. We observe that the two parallelograms ABCD and ABPE have same base AB.
2. Two parallelograms lie between same parallel lines, i.e., AB and CD and height between them is same at all points.
   By formula, area of parallelogram = base x height
   ar(||gm ABCD) = AB x EH
   ar(||gm ABPE) = AB x EH

Result
We have verified that two parallelograms lying on same base and between same parallel lines are equal in area.

Learning Outcome
Students follow that the parallelograms on the same base and between the same parallels have same area.

Activity Time

1. This theorem can be proved by using a graph paper. Students will try this and verify the theorem.
2. Prove that areas of a rectangle and a square of same height and on the same base are equal by using paper cutting and pasting method.

Viva Voce

Question 1.
How many kinds of parallelogram are possible ?
Answer:
3 main kinds (rectangle, square, rhombus).

Question 2.
What is the altitude of a parallelogram ?
Answer:
Altitude of a parallellogram is a perpendicular distance between the vertex to its opposite side.

Question 3.
If the two squares are lying on the same base and between the same parallel lines and area of the one square is 64 cm².
What will be the area of second square ?
Answer:
64 cm²

Question 4.
Two squares are lying on the same base and having same height and area of one of square is 121 cm . What will be the side of another square ?
Answer:
11cm.

Question 5.
In a triangle, a median divides it into two triangles. What will be ratio of areas of two small triangles so formed ?
Answer:
Areas of two triangles will be same.

Multiple Choice Questions

Question 1.
A parallelogram ABCD and a rectangle ABPQ are on the same base AB and between the same parallels AB and CQ. If AB = 8 cm and AQ = 6 cm, find the area of parallelogram ABCD:
(i) 48 cm²
(ii) 64 cm²
(iii) 36 cm²
(iv) none of these

Question 2.
ABCD is a parallelogram. If AB = 8 cm, altitude DM = 6 cm and altitude DN = 4.8 cm, find the perimeter of parallelogram ABCD:

(i) 36 cm
(ii) 10.8 cm
(iii) 14 cm
(iv) none of these

Question 3.
ABCD is a parallelogram, in which AB ⊥DE and CF ⊥ AD. If AB = 16 cm, DE = 8 cm and CF = 10 cm, find AD:
(i) 12 cm
(ii) 12.8 cm
(iii) 128 cm
(iv) none of these

Question 4.
In the given figure, if ar(||gm ABCD) = 80 cm², then ar(∆BEF) is

(i) 80 cm²
(ii) 160 cm²
(iii) 40 cm²
(iv) none of these

Question 5.
PQRS and ABRS are parallelograms and X is any point on the side BR. What is the ratio of ar(∆AXS) and ar(PQRS)?
(i) 2:1
(ii) 1:2
(iii) 3:2
(iv) none of these

Question 6.
Find the area of a rhombus, if the lengths of its diagonals are 16cm and 24cm respectively:
(i) 192 cm²
(ii) 384 cm²
(iii) 24 cm²
(iv) none of these

Question 7.
Calculate the area of a trapezium PQRS:

(i) 64 cm²
(ii) 156 cm²
(iii) 180 cm²
(iv) none of these

Question 8.
Areas of two parallelograms ABCD and PQRS are equal 20 cm². Are these two parallelograms congruent?

(i) no
(ii) yes
(iii) can’t say
(iv) none of these

Question 9.
ABCD is a parallelogram and P is any point on the side CD. If ar(∆BAP) = 40 cm², find ar (||gm ABCD):
(i) 60 cm²
(ii) 160 cm²
(iii) 80 cm²
(iv) none of these

Answers

1. (i)
2. (i)
3. (ii)
4. (i)
5. (ii)
6. (i)
7. (iii)
8. (i)
9. (iii)

Math Lab ManualMath Labs with ActivityMath LabsScience LabsScience Practical Skills

CBSE Solved Papers For Class 12 Computer Science (C++) Paper 5

SECTION ‘A’

Question 1:
(a) Differentiate between a Logical Error and Syntax Error. Also , give a suitable example of each in C++. [2]
(b) Observe the following program very carefully and write names of those header files, which are essentially needed to compile and execute the following program successfully: [1]
void main ( )
{
float number;
cout«”number”;
cin>>number,
cout<<sqrt(Number)<<endl;
}
(c) Rewrite the following program after removing the syntactical error(s), if any. Underline each correction. [2]
#include<iostream.h>
const int Max 10;
void main ()
{
int Numbers [Max];
Numbers ={20,50,10,30,40};
for (Loc =Max -1; Loc = > 0; Loc –)
cout >>Number [Loc];
(d) Write the output of the following C++ program code :
Note :
Assume all required header files are already being included in the program : [3]
void Draw (int &N, char mark=’#’)
{
for (int 1=1; I<=N; I++) cout<<mark«endl ;
N++;
}
void main()
{
int count=3; char sign=’*’;
Draw(cout);
Draw(count, sign);
Draw(count, ‘&’);
}
(e) Write the output of the following C++ program code : [2]
Note :
Assume all required header files are already being included in the program,
class Item
{
int Ino;
char Iname [20];
float price; public :
Item( )
{
Ino=100;
strcpy (Iname, “None”); price = 100;
}
void Assign (int I, char Name [],float P)
{
Ino+=I;
strcpy (Iname, Name); price=P;
} ,
void Raise (float P)
{
Price +=P;
}
void Reduce (float P)
{
Price -=P;
}
void DispO
{
cout«Ino«”: “«IName«”: “«Price«endl ;
}
} ;
void main()
{
Item One, Two;
One.Disp() ;
One.Assign (1,”Pen”,95);
Two.Assign (2,”Pencil”,55);
One.Raise(10);
Two.Reduce(5) ;
Two.Disp();
One.Disp();
}
(f) Oberve the following program Score.CPP carefully, if the value of Num entered by the user is 5. Choose the correct possible output(s) from the option from (i) to (iv) and justify your option. [2]
// Program : Score. CPP #include<stdlib.h>
#include<iostream.h>
void main()
{
randomize ();
int Num, Rndnum;
cin>>Num;
Rndnum =random (NUM)+5;
for (int N=l; N<=Rndnum;N++)
cout «N« “”;
}
Output option:
(i) 1 2 3 4
(ii) 1 2
(iii) 123456789
(iv) 1 2 3
Solution:
(a) The differences between logical errors and syntax errors are as follows:
(b)

1. iostream.h
2. math.h

(c) #include<iostream.h>
const int Max=10;
void main ()
{
int Numbers [Max];
Numbers []={20,50,10,30,40};
for (int Loc=Max-1; Loc>=0; Loc – -)
cout « Numbers [Loc];} [2]
(d) Output:
#
#
#
*
*
*
*
&
&
&
(e)
One
Two
One Ass
Two Ass
One Reduce
Two Reduce

Output:
100 : None : 100
102 : Pencil: 50
101: Pen : 105
(f) (iii) 123456789,
The minimum value Rndnum will take is 5.

Question 2:
(a) Differentiate between public and private visibility modes in context of object oriented programming using a suitable example illustrating each. [2]
(b) Explain copy constructor with an example. [2]
(c) Define a class named ADMISSION in C++ with the following descriptions: [4]
Private Members:
AD_No integer (Ranges 10-2000)
NAME Array of characters (String) (store 20 characters)
CLASS Character (store 10 characters)
FEES Float Public Members:

• Function Read_Data () to read an object of ADMISSION type Function Display () to display the details of an object.
• Function Draw_Nos () to choose 2 students randomly, and display the details. Use Random function to generate admission numbers to match with AD_No.

(d) Answer the questions (i) to (iv) based on the following code:
class Shape
{
int length;
protected: int width, height;
public: void getDimension (int, int, int);
void dispDimension();
} ;
class sideShape: protected Shape
{
int sideLength, sideWidth;
protected: void getSide (int, int);
public: void dispSide ();
} ;
class SubShape : public sideShape
{
int SubLength;
void display (void) ;
public: void enter ();
} ;

1. Name the base class and derived class of the class SideShape. [1]
2. Name the data member(s) that can be accessed from function dispSide(). [1]
3. Name the private member function(s) of class SubShape. [1]
4. Is the member function dispSide () accessible by the object of SubShape? [1]

Solution:
(a) Public and Private visibility modes in Object Oriented Programming:
(i) A key feature of object oriented programming is data hiding, i.e., the data is concealed within a class. So that it carmot be accessed mistakenly by functions outside the class.
Private visibility modes in Object Oriented Programming:
(i) The primary mechanism for hiding data is to put it in a class and make it private.
Private data or functions can only be accessed from within the class.
Public Visibility Modes in Object Oriented Programming:
(i) Public data or functions, are accessible from outside the class, by using object of the respective class.

Example:
class small_obj
{
private: int somedata; public:
void setdata (intd);
{ somedata = d;
}
void showdata ()
{
cout« “Data is” «somedata;
}
} ;
Data: Data member some data cannot be acessed outside the class because it is declared as private but functions setdata () and showdata () can be accessed outside the class because they are declared as public.
(b) A copy constructor is used to copy the values of one object into another.
Syntax : classname (classname &obj)
Example: class Student {
int a; public:
Student ()
{
a=10;
}
Student (student &x)//copy constructor
{
a=x.a;
}
} ;
void main()
{
Student S;
Student Sl=S;//call to copy contructor }
(c) class ADMISSION {
int AD_No;
char NAME [20];
char CLASS [10];
float FEES; public:
void Read_Data ();
void Display ();
void Draw_Nos () ;
} ;
void ADMISSION :: Read_Data ()
{ cout<<“\n Enter admission No”;
cin»AD_No;
cout«”\n Enter Name”;
gets(NAME);
cout«”\n Enter class”;
gets(CLASS);
cout«”\n Enter Fees”;
cin»FEES;
}
void ADMISSION : Display ()
{
cout «”\n Admission No :” « AD_No;
cout <<“\n Name :”;
puts(NAME);
cout <<“\n Class ;
puts(CLASS);
cout «”\n Fees :” « FEES;
}
void ADMISSION :: Draw_Nos ()
{ int adno;
adno=random (AD_No); if (AD No= = adno)
Display ();
}
(d)

1. Base class is Shape and derived class is Subshape. [1]
2. width, height, sidelength, sidewidth. [1]
3. display (), [1]
4. Yes, because it is publically derived from sideshape. [1]

SECTION ‘B’

Question 3:
(a) An Array A [10] [20] is stored in the memory with each element requiring 2 bytes of storage. If the base address of array in the memory is 800, determine the location of A[9] [11] when the array is stored as .

1. Row major
2. Column major [3]

(b) Write a C++ function to sort an array of integer using insertion sort. The function should have two parameters; name of the array and number of elements in the array. [3]
(c) Give the necessary declarations for a queue containing float type numbers. Write a user- defined function in C++ to insert a float type number in the queue and you should use linked representation of queue. [4]
(d) Write a function in C++ to print the product of each column of a two-dimesional integer array passed as the argument of the function. [2]
(e) Evaluate the following postfix notation of expression: 15 3 2 + / 7 + 2 * [2]
Solution:
(a) Array: A [10] [20]
Base Address, B=800.
Size of element, W : 2 bytes A [I] [J]=A [9] [11] & 1=9, J=ll Total Rows, R=10 Total Column, C=20 Lowest Row in C++, lr=0 Lowest Column in C + +, Ic=0
Row Major: [1]
A [I] [] = B+W [c(I-Ic) + (J-Ic)]
= 800+2[20(9-0) + (11-0)]
= 800+2[20×9+ll]
= 800+2(180+11)
= 800+2(191)
= 800+382 = 1182
Column Major: [1]
= B+W [(I—Ir)+R (J-lc)]
= 800+2[(9-0)+10 (11-0)]
= 800+2(9+10×11)
= 800+2(9+110)
= 800+2×119 = 800+238
= 1038 [3]
(b) void insertsort (int a[],int n)
{
int p, ptr;
//Assuming a[0]=int_min i.e. smallest integer for (p=j; p < n; P++)
{
temp=a[p] ptr=p-l;
while (temp < a[Ptr])
{
a[ptr+l] = a[ptr];//move element forward .
}
a[ptr+1]=temp://inserting element at proper place
}
int x;
for (x = 0; x <n; x++)
{
cout « a [x] ;
}
cout «”\n”;
}
(c) struct node
{
float data; node * link;
};
node* Insert Q (node*rear, float val)
{
node*temp=new node;
temp—>data=val;
temp—»1ink=NULL;
if (rear==NULL)
rear=temp;
else
{
rear—>link=temp;
rear=temp;
}
return (rear);
}
(d) void product (int arr [] [], int m, int n)
{
int product;
for (int i = 0; i<n; i++)
{
product = 1;
for (j = 0; j<m; j++)
{
product = product * arr [j][i];
}
cout«”product of column” «i + l «”=” « product;
}
}
(e) Postfix expression:

1532 + /7 + 2*
Ans: 20.

Question 4:
(a) Find the output of the following C++ code considering that the binary file GAME.DAT exists on the hard disk with information of around 200 games. [1]
class GAME
{
int Gno;
char GName [20];
public : void GetlnO;
void showName () ;
} ;
void main ()
{
fstream GF;
GF.open(“GAME.DAT”, ios::binary:ios::in);
GAME G;
GF.seekg(sizeof(G)*5) ;
GF.read((char*)&G,sizeof (G);
GF.read((char*)&G,sizeof(G));
int RECORDNO=GF.tellg()/sizeof(G); cout«”RECORDNO: “«RECORDNO«endl ;
GF.close() ;
}
(b) Write a function in C++ to count the number of lines starting with a digit in a text file “Diary.txt” [2]
(c) Given is a binary file PHONE.DAT, containing records of the following structure type: [3]
class Phonelist
{
char Name [20];
char Address [30];
char AreaCode [5] ;
char PhoneNo. [15];
public:
void Register ();
void Show ();
int CheckCode (char AC [])
{
return strcmp (AreaCode, AC) ;
}
};
Write a function TRANSFER () in C++, that would copy all those records which are having Area Code as “DEL” from PHONE. DAT to PHONE BACK. DAT
Solution:
(a) RECORDNO : 7
(b) int countNum() .
{
ifstream fin (“Diary. txt”) ;
char ch[80];
int count=0;
while (!fin. eof () )
{
fin. get line (ch, 80) ;
if(isdigit(ch[0]))
count++;
}
fin. close () ; return count;
}
(c) void Transfer()
{
Phonelist P; ifstream fin; ofstream fout;
fin.open(“PHONE.DAT”,IOS::binary);//IOS & ios both are same
fout.open(“PHCNEBACK.DAT”, ios::binary);
while (fin. read ( (char*) &P, sizeof (P) )
{
if(P.Checkcode(“DEL”)==0)
fout.write((char*)&P, sizeof(P));
}
fin . close () ;
font.close (c);
}

SECTION ‘C’

Question 5:
(a) Observe the following table carefully and find the degree and cordinality of the table : [2]

(b) Give the following table for a database LIBRARY:
Table: Books

Table: Issued

Write SQL queries for (a) to (d):

1. To show Book name, Author name and Price of Books of First Publ. [1]
2. To list the names from books of Text type. [1]

(c) To display the names and price from books in ascending order of their price. [1]
(d) To increase the price of all books of EPB publishers by 50. [1]
(e) Give the output of the following queries based on the above data:

1. SELECT COUNT (*) FROM BOOKS; VA]
2. SELECT MAX (Price) FROM BOOKS WHERE Quantity >=15;
3. SELECT Book Name, Author_Name FROM BOOKS WHERE Publishers = “EPB”;
4. SELECT COUNT (DISTINCT Publishers) FROM BOOKS WHERE Price>400;

Solution:
(a) Degree = no. of columns = 3
Cardinality = no. of rows = 5
(b)

1. SELECT Book Name, Author_name, Price FROM BOOKS
   WHERE Publishers = “First Publ.”;
2. SELECT * FROM BOOKS
   WHERE Type = “Text”;

(c) SELECT Book Name, Price
FROM Books
ORDER BY price ASC;
(d) UPDATE BOOKS SET Price – Price+50.
WHERE Publishers = “EPB”;
(e)

1. 5.
2. 750
3. Fast cook Lata Kapoor
   My first C++ Brain & Brooke
4. 2

Question 6:
(a) State and verify Associative law. [2]
(b) Write the equivalent expression for the following logical circuit: [2]

(c) Write the SOP form of a Boolean function F, which is represented by the following truth table:

(d) Reduce the following Boolean Expression using K-map:
F (P,Q,R,S) = Σ(0,3,5,6,7,11,12,15) [3]
Solution:
(a) Associative : This law states that:
(A+B)+C= A+(B+C) Or (A.B).C=A.(B.C)
Proof:

From the above table, (A+B) +C = A+ (B+C) Similarly, we can prove A.(B.C) = (A.B).C

(b) Exp is= A.C + B.A + B.C
(c) (A’B’C) + (A’BC) + (ABC’) + (ABC)
(d) F(P, Q, R, S) = Σ(0, 3, 5, 6, 7,11,12,15)

Minimized Expression:
= RS+P.Q.S.+ P.Q.R. + P.Q.R.S.+ P.Q.R. S.

Question 7:
(a) What is spam mail? [1]
(b) What is the significance of Cyber Law? [1]
(c) Differentiate FTP and HTTP. [1]
(d) What is the URL in networking? [1]
(e) “Hindustan Connecting World Association” is planning to start their offices in four major cities in India to provide regional IT Infrastructure support in the field of Educations Culture. The company has planned to set up their head office in New Delhi in three locations and have named their New Delhi offices as “Sales office”, “Head office” and “Teach office”. The company’s regional offices are located at: “Coimbatore”, “Kolkata” and “Ahmedabad”.
A rough layout of the same is as follows:

Approximate distances between these offices as per network survey team is as following:

In continuation of the above, the company experts have planned to install the following number of computers in each of their offices:

1. Suggest network type (out of LAN, MAN, WAN) for connecting each of the following set of their offices: [1]
   • Head office and Teach office
   • Head office and Coimbatore office
2. Which device will you suggest to be procured by the company for connecting all the
   computers within each of their offices out of the following device? [1]
   • Modem
   • Telephone
   • Switch/ Hub
3. Which of the following Communication Media, will you suggest to be procured by the
   company for connecting their local offices in New Delhi for very effective and fast communication? [1]
   • Ethernet Cable
   • Optical Fibre
   • Telephone Cable
4. Suggest a cable/wiring layout for connecting the company’s local offices located in New Delhi. Also, suggest an effective method/technology for connecting the company’s regional offices at “Kolkata”, “Combatore” and “Ahmedabad”.

(f) What is RJ-45 connector?
(g) What is video conferencing?
Solution:
(a) Spam mail is an unwanted mail. It is mostly commercial advertising. Sometimes it contains virus. [1]
(b) Significance of Cyber Law: [1]
In Internet, the communication technology uses the means of transferring textual messages, pic¬tures and many more.
Each time, there may be threats on either from the senders or receivers side which creates a bridge between networking communication.
(c) To shortcut these problems, the Internet Security Council made number of precautions i.e rules. These predefined rules are called Cyber Law or Law of Internet.

(d) Importance of URL in Networking :

1. URL stands for UNIFORM RESOURCE LOCATOR.
2. Each page that is created for web browsing is assigned a URL that effectively serves as the pages worldwide name or address.
3. URL has three parts: the protocol, the DNS name of the machine on which the page is located and a local name uniquely indicating the specific page (generally the filename), for e.g. http://www.cluestch. con [1]

(e)

1. LAN is used for connecting Head office and Teach office. WAN is used for connecting Head office and Coimbatore office.
2. Switch/ Hub to be procured by the company for connecting all the computers within each of their offices. [1]
3. Optical cable to be procured by the company for connecting their local offices in New Delhi for very effective and fast communication. [1]
4. A cable/wiring layout for connecting the company’s local offices located in New Delhi is Optical Fibre Cable.
   An effctive method/technology for connecting the company’s regional offices at “Kolkata”, “Coimbatore” and “Ahmedabad” is as follows: [1]

(f) RJ-45 connector: RJ-45 is abbreviation for Registered Jack-45. It is an eight wire connector used to connect computers on LAN’s, especially Ethernets. [1]
(g) Video Conferencing: When two or more participants are able to converse in real time through video-phones, it is called Video Conferencing.

CBSE Previous Year Solved  Papers  Class 12 Chemistry Outside Delhi 2014

Time allowed: 3 hours                                                                                      Maximum Marks: 70

General Instructions:

1. All questions are compulsory.
2. Questions number 1 to 5 are very short-answer questions and carry 1 mark each.
3. Questions number 6 to 10 are short-answer questions and carry 2 marks each.
4. Questions number 11 to 22 are also short-answer questions and carry 3 marks each.
5. Questions number 23 is a value based question and carry 4 marks.
6. Questions number 24 to 26 are long-answer questions and carry 5 marks each.
7. Use log tables, if necessary. Use of calculators is not allowed.

SET I

Question.1. What is the effect of temperature on chemisorption ?
Answer : Chemisorption initially increases then’ decreases with rise in temperature. The initial increase is due to the high energy of activation and the decrease afterwards is due to the exothermic nature of adsorption equilibrium.

Question.2. What is the role of zinc metal in the extraction of silver ?
Answer : Zinc is used as a reducing agent to recover silver from its cyanide complex. It reduces Ag⁺ to Ag and itself get oxidised to Zn²⁺.

Question.3. What is the basicity of H₃PO₃ ?
Answer : The H₃PO₃ ionises to give two H⁺ ions. Its basicity is two.

Question.4. Identify the chiral molecule in the following pair :

Question.5. Which of the following is a natural polymer? Buna-S, Proteins, PVC
Answer : Protein, is a natural polymer having amino acid as a monomer.

Question.6. The conversion of primary aromatic amines into diazonium salts is known as ——.
Answer : Diazotisation reaction.

Question.7. What are the products of hydrolysis of sucrose?
Answer : The products of Hydrolysis of sucrose are : Glucose and Fructose

Question.8. Write the structure of p-methylbenzaldehyde.

Question.9. An element with density 2.8 g cm^-3 forms a F.C.C. unit cell with edge length 4 x 10^-8cm. Calculate the molar mass of the element. (Given NA = 6.022 x 10²³ mol^-1)
Answer:

Question.10. (i) What type of non-stoichiometric point defect is responsible for the pink colour of LiCl?
(ii) What type of stoichiometric defect is shown bv NaCl?
OR
How will you distinguish between the following pairs of terms?
(i) Tetrahedral and octahedral voids
(ii) Crystal lattice and unit cell
Answer : (i) Metal excess defect due to anionic vacancies is responsible for pink colour of LiCl
(ii) Schottky defect is shown by NaCl.
OR

Question.11. State the Kohlrausch law of independent migration of ions. Why does the conductivity of a solution decrease with dilution?
Answer : It states that the limiting molar conductirily of an eledrolyte can be expressed as sum of individual contribution of anion and cation of the electrolyte.
With dilution, the number of ions per unit volume of electrolyte decrease and hence, conductivity decreases.

Question.12. For a chemical reaction R—>P, of solution the variation in the concentration (R) Vs. time (t) plot is given as

(i) Predict the order of the reaction.
(ii) What is the slope of the curve?
Answer : (i) zero order reaction
(ii) Slope of the curve is (-K). i.e., negative slope.

Question.13. Explain the principle of the method of electrolytic refining of metals. Give one example.
Answer : In electrolytic method, the impure metal is made to act as anode. A strip of the same; metal in pure form is used as cathode. They are put in a suitable electrolytic bath containing soluble salt of the same metal. The more basic metal remains in the solution and the less basic ones go to the anode mud.
For e.g. : Copper is refined using an electrolytic method, the net result of electrolysis is the transfer of copper in pure form from the anode to the cathode.
Anode : Cu ——>Cu²⁺ + 2e^–
Cathode : Cu²⁺ + 2e^– ——> Cu

Question.14. Complete the following equations:
(i) P₄ + H₂O —–>
(ii) XeF₄ + 0₂F₂ —–>
Answer : (i) P₄+ H₂0 —> No reaction.
(ii) XeF₄ + O₂F₂ —> XeF₆ + O₂

Question.15. Draw the structures of the following: (i) XeF₂ (ii) BrF₃
Answer:

Question.16. Write the equations involved in the following reactions:
(i) Reimer-Tiemann reaction
(ii) Williamson synthesis
Answer : (i) Reimer Tiemann reaction :

Question.17. Write the mechanism of the following reaction :

Question.18. Write the name of monomers used for getting the following polymers:
(i) Bakelite (ii) Neoprene
Answer: (i) Phenol and formaldehyde are the monomer used for getting Bakelite.
(ii) Chloroprene is the monomer used to get Neoprene.

Question.19.

Question.20. The following data were obtained during the first order thermal decomposition of S0₂Cl₂ at a constant volume :

Calculate the rate constant. (Given : log 4 = 0.6021, log 2 =log 2 = 0.3010)

Question.21. What are emulsions? What are their different types? Give one example of each type.
Answer : Emulsions are colloids in which both the dispersed phase and dispersing medium are liquid. It is a mixture of two or more liquids that are normally immiscible. They are of two types :
(i) Oil in water type emulsions (O/W) : In such emulsions, oil is the dispersed phase and water is the dispersed medium, e.g. Milk, Vanishing cream.
(ii) Water in oil type emulsions (W/O) : In such emulsions, water is the dispersed phase and oil is the dispersed medium, e.g. Butter, Cod liver oil.

Question.22. Given reasons for the following:
(i) (CH₃)₃ P = O exists but (CH₃)₃N = O does not.
(ii) Oxygen has less electron gain enthalpy with negative sign than sulphur.
(iii) H₃P0₂ is a stronger reducing agent than H₃P0₃.
Answer : (i) Due to the absence of d-orbitals, nitrogen cannot form pπ-dπ multiple bonds. Thus, N cannot expand its covalency beyond four. So, the compound (CH₃)₃N= O does not exist. On the other hand, P due to the presence of (7-orbitals forms, pπ-dπ multiple bonds and hence forms (CH₃)₃P = O in which covalency of P is 5.
(ii) Due to small size and high electro negativity of oxygen compared to sulphur, oxygen has less electron gain enthalpy.
(iii) Acids which contain P-H bonds have reducing character, Since H₃PO₂ contains two P-H bonds while H₃ P0₃ contains only one P-H bond therefore, H₃P0₂ is a stronger reducing agent than H₃PO₃.

Question.23. (i) Write the IUPAC name of the complex [Cr(NH₃)₄Cl₂] Cl.
(ii) What type of isomerism is exhibited by the complex [Co(en)₃]³⁺?
(en = ethane-1,2-diamine)
(iii) Why is [N1Cl₄]^2- paramagnetic but [Ni(CO)₄] is diamagnetic
(At. nos.: Cr = 24, Co = 27, Ni = 28)
Answer: (i) Tetraammine dichlorido chromium(III) chloride
(ii) Optical isomerism is shown by the complex [CO(en)₃]³⁺

(iii) [Ni(Cl)₄]^2-, Ni is in +2 oxidation state with the electronic configuration 3d⁸4s⁰. As Cl^– is a weak ligand it cannot pair up the electrons in 3d orbitals therefore, [NiCl₄]^2- is paramagnetic. In [Ni(CO)₄], Ni is in O oxidation state with the electronic configuration 3d⁸4s². Co is a strong ligand it causes As electrons to shift to id and pair up id electrons. Therefore, Ni(CO)₄ is diamagnetic.

Question.24. (a) Draw the structure of major monohalo products ii each of the following reactions :

(b) Which halogen compound in each of the following pairs will react faster in S_N2 reaction:
(i) CH₃Br or CH₃I
(ii) (CH₃)₃C-Cl or CH₃-Cl

(b) (i) CH₃-I will react faster because I-is better leaving group.
(ii) CH₃-CI will react faster because it is a primary halide and they undergo S_N2 reaction faster.

Question.25. Account for the following:
(i) Primary amines (R-NH₂) have higher boiling point than tertiary amines (R₃N).
(ii) Aniline does not undergo Friedel-Crafts reaction.
(iii) (CH₃)₂NH is more basic than (CH₃)₃N in an aqueous solution.
OR
Give the structures of A, B and C in the following reactions :

Answer : (i) Due to intermolecular hydrogen bonding in primary amines (presence of more number of H-atoms). They have high boiling point in comparison to tertiary amines.
(ii) Aniline does not undergo Friedel-Crafts reaction because Aniline is basic in nature reacts with AlCl₃ to form a salt. The positive charge on the N-atom, electrophilic substitution in the benzene ring is deactivated.

Question.26. Define the following terms related to proteins :
(i) Peptide linkage
(ii) Primary structure
(iii) Denaturation
Answer : (i) Peptide linkage is the amide linkage -COOH group of one α-amino acid and -NH₂ group of other a-amino acid by lose of a water molecule.
(ii) The sequence in which various amino acids are arranged ” in a linear structure with no intermediate bonding is called primary structure of a protein.
(iii) When a protein in its native form is subjected to a change, such as temperature, pit, etc. Due to this protein coss its biological activity is called denaturation of a protein e.g. curdling of.

Question.27. On the occasion ofWorld Health Day, Dr. Satpal organized a ‘health camp’ for the poor farmers living in a nearby village. After check-up, he was shocked to see that most of the farmers suffered from cancer due to regular exposure to pesticides and many were diabetic. They distributed free medicines to them. Dr. Satpal immediately reported the matter to the National Human Rights Commission (NHRC). On the suggestions of NHRC, the government decided to provide medical care, financial assistance, setting up of super-speciality hospitals for treatment and prevention of the deadly disease in the affected villages all over India.
(i) Write the values shown by (a) Dr. Satpal (b) NHRC?
(ii) What type of analgesics are chiefly used for the relief of pains of terminal cancer?
(iii) Give an example of artificial sweetener that could have been recommended to diabetic patients.
Answer : (i)(a) Dr. Satpaf is caring and compassionate for the farmers of the village. He distributed free medicines to them, (b) NHRC is also compassionate towards the farmers and appreciated Dr. Satpal’s work.
(ii) Narcotic analgesics like morphine and heroin.
(iii) Aspartame.

Question.28. (a) Define the following terms :
(i) Molarity (ii) Molal elevation constant (kb)
(b) A solution containing 15 g urea (molar mass = 60 g mol^-1) per litre of solution in water has the same osmotic pressure (isotonic) as a solution of glucose (molar mass = 190 g mol^-1) in water. Calculate the mass of glucose present in one litre of its solution.
OR
(a) What type of deviation is shown by a mixture of ethanol and acetone? Give reason.
(b) A solution of glucose (molar mass = 180 g mol^-1) in water is labelled as 10% (by mass). What would be the molality and molarity of the solution? (Density of solution = 1.2 g mL^-1)
Answer : (a) (i) The number of moles of a solute present in one litre of solution is known as its molarity.
The elevation in boiling point of a solution when one mole of non-volatile solute is dissolved in one kilogram of a volatite solvent is known as molal elevation constant (Kb).

OR
(a) The mixture of ethanol and acetone shows positive deviation from Raoults law. In pure ethanol hydrogen bond  exist between the molecules. On adding acetone to ethanol, acetone molecules get in between the molecules of ethanol thus breaking some of the hydrogen bonds and weakening molecular interactions leads to increase in vapour pressure , resulting in positive deviation from Raoulr’s law.

Question.29.

(b) (i) Zinc has 3d¹⁰4s² configuration with no unpaired d-orbital electron and hence it is not considered as a transition element.
(ii) Transition metals have vacant orbitals to accommodate lone pairs of electrons for bond formation and have high charge density, therefore, they form complexes.
(iii) Due to high stability of Mn^-2+ (due to its half-filled 3d sub shell) than Mn³⁺ while Cr³⁺ is more stable than Cr²⁺.
OR

Question.30. (a) Write the products formed when CH₃CHO reacts with the following reagents:
(i) HCN (ii) H₂N-OH
(iii) CH₃CHO in the presence of dilute NaOH
(b) Give simple chemical tests to distinguish between the following pairs of compounds.
(i) Benzoic acid and.Phenol
(ii) Propanal and Propanone.
OR
(a) Account for the following:
(i) Cl^–CH₂COOH is a stronger acid than CH₃COOH.
(ii) Carboxylic acids do not give reactions of carbonyl group.
(b) Write the chemical equations to illustrate the following name reactions:
(i) Rosenmund reduction
(ii) Cannizzaro’s reaction
(c) Out of CH₃CH₂-CO-CH₃ and GH₃CH₂-CH₂– CO-CH₃, which give iodoform test?
Answer : (a) (i) On reaction of acetaldehyele with hydrogen cyanide it gives 2-Hydroxypropaneriitrile as product.

SET II

Note: Except for the following questions, all the remaining question have been asked in previous set.

Question.1.Why is adsorption always exothermic?
Answer : Adsorption being a surface phenomenon leads to decrease in surface energy and hence is exothermic.

Question.2. Name the method used for refining of Nickel.
Answer: Mond’s process used for refining of Nickel.

Question.3. Why does NO₂ dimerise?
Answer : NO₂ is an odd-electron molecule and hence dimerises to N₂O₄.

Question.4. Based on molecular forces, What type of polymer is neoprene?
Answer: Elastomer. .

Question.5. What are the products of hydrolysis of maltose?
Answer : Two units a-D-glucose

Question.6. Write the structure of 4-chloropentan-2 one.

Question.9. Write the name of monomers used for getting the following polymers:
(i) Terelyne (ii) Nylon-6, 6
Answer : (i) Ethylene glycol and terephthalic acid.
(ii) Adipic acid and hexamethylene diamine.

Question.10. Describe the role of the following:
(i) SiO₂ in the extraction of copper from copper matte.
(ii) NaCN in froth floatation process.
Answer : (i) SiO₂ (Silica) acts as a flux in the extraction of copper from copper matte to remove ferrous oxide as ferrous silicate slag.
FeO + SiO₂ —> FeSiCO₃ , (Slag)
(ii) NaCN is used as a depressant as it forms-zinc complex, Na₂[Zn(CN)₄] on the surface of ZnS thereby preventing it from forming froth.

Question.11. Complete the following equations :
(i) Ag + PCl₅
(ii) CaF₂ + H₂SO₄
Answer: (i) 2Ag + PCl₅ —> 2AgCl + PCl₃
(ii) CaF₂ +H₂SO₄ —> CaSO₄ + 2HF

Question.12. Draw the structures of the following:
(i) XeF₄
(ii) HClO₄

Question.13. (i) Write the type of magnetism observed when the magnetic moment are oppositely aligned and cancel out each other.
(ii) Which stoichiometric defect does not change the density of the crystal?
Answer: (i) Antiferromagnetism observed when the magnetic moment are oppositely aligned and cameal out each other.
(ii) Frenkel defect does not change the density of the crystal.

Question.14. Define the following terms :
Fuel cell
(ii) Limiting molar conductivity

Answer: (i) Fuel cell is a device that converts chemical energy from combustion of a fuel into electric energy through a chemical reaction.
(ii) Molar conductivity of electrilyte at infinite dilution or when consentration approaches zero is‘called limiting molar conductivity. It is expressed as

Question.19. Define the following terms :
(i) Glycosidic linkage
(ii) Invert sugar
(iii) Oligo saccharides
Answer : (i) The linkage between two monosaceharide through oxygen atom in an oligosaccharide ora polysaccharide is known as glycosidic linkage.
(ii) Sucrose is called invert sugar as it produces equimolar quantities of glucose and fructose on hydrolysis it gives an equimolar mixture of D – (+)- glucose, and D -fructose, which is laevo rotatory. This change of specific rotation from dextrorotation to laevo-rotation is called inversion-1 of sugar and the mixture is obtained is called invert sugar.
(iii) Carbohydrate which .on hydrolysis give two to ten molecules of monosaccharides are called oligosaccharide e.g. sucrose.

SET III

Note: Except for the following questions, all the remaining question have been asked in previous sets.

Question.1. What are the dispersed phase and dispersion medium in milk?
Answer : Dispersed phase is oil or fat and dispersion medium is water.

Question.2. Name the method used for refining of copper metal.
Answer: Electrolytic refining.

Question.3. Why does NH₃ act as a Lewis base?
Answer : Due to the presence of lone pair on nitrogen in NH₃, it forms coordinate bonds with Lewis acids and acts as Lewis base.

Question.5. Which of the following is a fibre? Nylon, Neoprene, PVC.
Answer: Nylon

Question.6. Write the products of hydrolysis of lactose.
Answer:

Question.8. Write the structure of 2-hydroxybenzoic acid.

Question.9. Complete the following equations :
(i) Cu + 2H₂SO₄ (Cone.) —> CuSO₄+ SO₂+ 2H₂O
(ii) XeF₂ + H₂O
Answer:
(i) Cu + cone 2H₂SO₄ —-> CuSO₄ + SO₂+ 2H₂0
(ii) 2XeF₂ + 2H₂0 —–> 2Xe + 4HF + O₂

Question.10. Draw the structure of the following:
(i) XeO₃ 
(ii) H₂SO₄
Answer:

Question.11. Write the name of monomers used for getting the following polymers:
(i) Teflon (ii) Buna-N
Answer: (i) Tetrafluoroethene
(ii) Butadiene and acrylonitrile

Question.13. (i) Write the type of magnetism observed when the magnetic moment are aligned in parallel and anti-parallel directions in unequal numbers.
(ii) Which stoichiometric defect decreases the density of the crystal?
Answer: (i) Ferrimagnetism.
(ii) Schottky defect decreases the density of the crystal.

Question.14. Define the following terms :
(i) Molar conductivity 
(ii) Secondary batteries
Answer : (i) Molar conductivity is defined as the conducting power of all the ions produce by dissolving one more of an electrolyte in solution.

(ii) Secondary batteries are those batteries which can be recharged by passing electric current through them and hence can be used over again e.g. Lead storage battery.

Question.17. Write the principle behind the froth floatation process. What is the role of collectors in this process?
Answer: Froth floatation method has been in use for removing gangue from sulphide ores. In this process, a suspension of the powdered ore is made with water.
Collectors (e.g. pine oils, fatty acids, etc.) enhance non-wettability of the mineral particles and used to skim the froth off the surface.

Question.23. Define the following terms :
(i) Nucleotide
(ii) Anomers
(iii) Essential amino acids
Answer: (i) A Nucleotide contain all the three basic components of nucleic acid i.e., apentose sugar, a nitrogeneous base and a phosphoric acid. When nucleoside is linked to phosphoric acid at 5′ position of sugar moiety, we get a nucleotide, (ii) The carbohydrate which differ in configuration at the glycosidic carbon ( i.e., C1 in aldoses and in ketoses) are called anomers. e.g. α -D-(+) glucose and β-D-(+) glucose, (iii) Those amino acids which cannot be synthesised by the body and need to be consumed through diet. eg. Valine.

CBSE Sample Paper for Class 9 English Communicative

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CBSE Class 9 Science Practical Skills – Features of Monocot and Dicot Plants

AIM
To study the external features of root, stem, leaf and flowers of monocot and dicot plants.

MATERIALS REQUIRED
A pea plant and an onion plant with root, stem, leaf and flowers, two paper sheets, simple dissecting microscope, a hand lens, slide, cover slip, razor blade, a few seeds of gram, pea, maize, petri dish and water.

THEORY
Monocots
Characteristics of monocots are as follows:

1. These plants have single cotyledon in their seeds.
2. Their leaves are narrower and have parallel venation.
3. They have fibrous roots.
4. They have trimerous flowers. Trimerous flowers bear different structures in 3’s or in multiples of 3.

Dicots
Characteristics of dicots are as follows:

1. These plants have two cotyledons in their seeds.
2. Their leaves are broad and have reticulate venation.
3. They have tap roots.
4. They have pentamerous flowers. Pentamerous flowers bear different structures in 5’s or in multiples of

PROCEDURE
Root
1. Wash the roots of both the plants and spread them on a newspaper.
2. Carefully observe the nature of roots of both plants.

Leaf
3. Take one leaf of each plant and carefully observe their shape.
4. With the help of a hand lens observe the pattern of veins.

Flower
5. Take one flower of each plant.
6. Count the number of sepals, petals and stamens in each.
7. Now, carefully remove the carpel from each flower and cut its transverse section with a sharp razor blade.
8. Put a drop of water in the centre of a slide and with the help of brush transfer the section of ovary on the slide.
9.   Cover the slide with a cover slip.
10. Observe this slide under the dissecting microscope.
11.  Count the number of carpels in each ovary.

Seeds
12. Soak both types of seeds in water in separate petri dishes overnight.
13. Remove the seed coat, try to split the seed and count the number of cotyledons in each.

OBSERVATIONS
Root
Pea root: The root system consists of one main root called primary root. From the primary root several lateral roots arise.
Onion root: All the roots are similar and originating from the base of the stem.
Leaf
Pea leaf: The leaves are broad. A thick vein called mid vein is present in the middle of the leaf. From this thick vein many veins are emerging which form a net-like design on both sides of midrib.
Onion leaf: The leaves are narrow. All the veins are parallel to one another and almost similar in thickness.
Flower
Pea flower: Flower bears five seals, petals and ten stamens. Pistil consists of style, stigma and ovary. Ovary consists of one carpel.
Some bead like structures called ovules are attached to the wall of ovary.
Onion flower: Flower bears six tepals and six stamens. Pistil consists of style, stigma and ovary. Ovary consists of three carpels. Some bead like ovules are attached to the ovary wall in each carpel.

Seeds
Gram and pea seeds: Each seed of gram and pea has two cotyledons.
Maize seeds: Each seed of maize consists of only one cotyledon.

RESULT

1. The study reveals that dicot plants have tap root system, broad leaves with reticulate venation, five or in multiple of five sepals, petals and stamens in their flowers whereas monocots have fibrous root, narrow leaves with parallel venation, three or in multiple of three sepals, petals and stamens in their flowers.
2. The dicot seeds such as gram, pea have two cotyledons while monocot seeds such as maize, wheat, onion have only one cotyledon.

PRECAUTIONS

1. We should handle the blade carefully.
2. The transverse section of ovary should be kept dipped in water otherwise it may dry.

INTERACTIVE SESSION

Examiner :
Which plants have you taken for the experiment?
Examinee:
Pea and onion

Examiner :
Are both these plants angiosperms?
Examinee:
Yes

Examiner :
Why?
Examinee:
Because they both bear flowers and their seeds are enclosed in fruit.

Examiner :
Which root system is present in pea plant?
Examinee:
Tap root system

Examiner :
What are the identification features of tap root system?
Examinee:
In tap root system, one main root is present. From this secondary roots, and from secondary roots tertiary roots arise. In other words, we can say in tap root system, roots are differentiated into primary, secondary and tertiary roots.

Examiner :
Can you categorise the plants into monocot and dicot by looking their leaves?
Examinee:
Yes, dicot plants have reticulate venation while monocot plants have parallel venation.

Examiner :
What is venation?
Examinee:
Arrangement of veins in a leaf is called venation.

Examiner :
Name two plants in which venation is parallel and two plants in which venation is reticulate.
Examinee:
Parallel venation — Wheat and rice.
Reticulate venation – China rose and mustard.

Examiner :
What are pentamerous flowers?
Examinee:
Flowers having parts in multiples of five (five, ten, etc.) are called pentamerous flowers.

Examiner :
In which type of plants pentamerous flowers are present?
Examinee:
Dicot plants

Examiner :
What is seed made up of?
Examinee:
Seed is made up of seed coat and embryo.

Examiner :
How can we identify whether a seed belong to a monocot or dicot plant?
Examinee:
By counting the number of cotyledons present in seed we can identify whether a seed belongs to a monocot plant or dicot plant.

Examiner :
How many cotyledons are present in a dicot seed?
Examinee:
Two

Examiner :
What is the role of cotyledon in the seed?
Examinee:
The cotyledons are the “seed leaves” produced by the embryo. They serve to absorb nutrients packaged in the seed, until the seedling is able to produce its first true leaves and begin photosynthesis

NCERT LAB MANUAL QUESTIONS

Question 1:
How do we differentiate between fibrous root system and tap root system?
Answer:
In the fibrous root system, roots originate from the base of the stem and all the roots are alike whereas in tap root system, roots are differentiated into primary, secondary, tertiary roots.

Question 2:
A plant have leaves with reticulate venation and floral parts consisting of 5 sepals, 5 petals, 5 stamens, and 5 carpels. In which group of angiosperms would you place this plant? Give reasons.
Answer:
Leaves with reticulate venation, and pentamerous flowers are characteristic of dicot plants. So, the plant mentioned in the question is a dicot.

Question 3:
In a plant, name two features which you would examine to categorise it into a monocot or dicot plant.
Answer:
Two features that we can examine to categorise a plant into monocot and dicot are

1. number of cotyledons in a seed, and
2. leaf venation.

Question 4:
Do all the flower have all the floral parts? Explore.
Answer:
A typical flower has sepals, petals, stamens and carpels, such type of flowers are called complete flower. e.g., mustard, china rose, pea, etc. But some flowers may lacking any one or more than one part that are found in the complete flower. Such flowers are called incomplete flower e.g., sunflower, rice, wheat, etc.

Multiple Choice Questions/VSA

Question 1:
Which of the following is the characteristic of angiosperms?
(a) Presence of flower
(b) Presence of seeds enclosed in fruit
(c) Both (a) and (b)
(d) None of these

Question 2:
Fibrous root originates from
(a) radicle
(b) base of stem
(c) seed
(d) any part of plant

Question 3:
Inside the ovary tiny dot-like structures present are
(a) placenta
(b) ovule
(c) anther
(d) cotyledon

Question 4:
The individual unit of an androecium is
(a) ovule
(b) ovary
(c) stamen
(d) petals

Question 5:
The four whorls of a flower starting from outside are
(a) corolla, calyx, androecium, gynoecium
(b) gynoecium, corolla, calyx, androecium
(c) corolla, androecium, gynoecium, calyx
(d) calyx, corolla, androecium, gynoecium

Question 6:
Plant is a dicot if
(a) leaves have reticulate venation
(b) leaves have parallel venation
(c) root system is fibrous
(d) flower is trimerous

Question 7:
The following pair has dicot seeds.
(a) Pea and onion
(b) Pea and gram
(c) Maize and pea
(d) Onion and maize

Question 8:
Gynoecium is
(a) male reproductive system
(b) female reproductive system
(c) cotyledons
(d) root system

Question 9:
Which of the following is not a component of carpel?
(a) Anther
(b) Stigma
(c) Style
(d) Ovary

Question 10:
Tepals are the individual units of
(a) calyx
(b) corolla
(c) perianth
(d) gynoecium

Question 11:
The terminal part of the carpel which receives pollen grains is called
(a) anther
(b) stigma
(c) ovary
(d) style

Question 12:
In the following figure, X is

(a) placenta
(6) ovule
(c) style
(d) stigma

Question 13:
Select the odd one out.

(a) A
(6) B
(c) C
(d) D

Question 14:
In the diagram given below, what is X?

(a) Micropyle
(b) Hilum
(c) Stigma
(d) Cotyledon

Question 15:
Which type of root is found in a pea plant? How this root is different from the root of onion?
(a) 5 sepals + 3 petals + 5 stamens
(b) 3 sepals + 5 petals + 3 stamens
(c) 3 sepals + 3 petals + 5 stamens
(d) 3 sepals + 3 petals + 9 stamens

ANSWER KEY

1. (c)
2. (b)
3. (b)
4. (c)
5. (d)
6. (a)
7. (b)
8. (b)
9. (a)
10. (c)
11. (b)
12. (a)
13. (d)
14. (b)
15. (d)

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