Category: CBSE

CBSE Class 9 Science Practical Skills – Solution, Colloids, Suspension

EXPERIMENT

AIM
To prepare

1. A true solution of common salt, sugar and alum.
2. A suspension of soil, chalk powder and fine sand in water.
3. A colloidal solution of starch in water and egg albumin in water and distinguish between these on the basis of
   • transparency
   • filtration
   • stability.

MATERIALS REQUIRED
Test tubes, beakers, water, common salt, sugar, alum, soil, chalk powder, fine sand, egg albumin, starch, funnel, glass rod, filter paper.

THEORY

1. A true solution has particle size < 10^-9 m. It is a homogeneous mixture which is clear and transparent. It passes through the filter paper easily and is stable.
2. Colloid has particle size 10^-9 m- 10^-6 m. It is heterogeneous and is translucerft. It passes through filter paper and is stable.
3. The particle size of Suspension is >10^-6 m. It is heterogeneous and opaque. Its particles do not pass through filter paper and settle under gravity when left undisturbed for sometime.

PROCEDURE
Preparation of solutions/mixtures

1. Take 8 beakers and label them as A, B, C, D, E, F, G and H and add 50 ml of water to each of these beakers.
2. Add samples of salt, sugar and alum to beakers A, B and C. Stir to make clear solution.
3. Add soil, chalk powder and fine sand to beaker D, E and F. Stir for sometime.
4. Add a small amount of starch to the water and boil the content in beaker G.
   In beaker H, add egg albumin and stir vigorously.
5. Observe the mixture and record them in the table given below.

OBSERVATION TABLE

RESULTS

1. Beaker A – Common salt forms a true solution in water.
2. Beaker B – Sugar forms a……………. in water.
3. Beaker C – Alum forms a …………….in water.
4. Beaker D – Soil forms a……………. in water.
5. Beaker E – Chalk Powder forms a……………. in water.
6. Beaker F – Fine sand forms a……………. in water.
7. Beaker G – Starch forms a……………. in water.
8. Beaker H – Egg albumin forms a …………….in water.

PRECAUTIONS

1. Distilled water should be used.
2. The components should be mixed in small amounts.
3. Fold the filter paper properly and moisten it before use.
4. Glass rod should be clean.
5. Wash your hands properly with soap after experiment.

INTERACTIVE SESSION

Examiner:
Define the term solute.
Examinee:
Solute is the dissolved component of a solution, e.g. in a solution of common salt in water, common salt is solute.

Examiner:
What does the term solvent mean?
Examinee:
Solvent is the substance in which the solute is dissolved to get a solution.
For example in a solution of sugar in water, water is the solvent.

Examiner:
Are colloids homogeneous or heterogeneous?
Examinee:
Colloids seem to be homogeneous, but they are actually heterogeneous.

Examiner:
Give two examples of colloids.
Examinee:
Blood, jelly.

Examiner:
Classify as solution, suspension and colloid.
Chalk powder in water, sulphur in water, sulphur in CS₂ , sugar in water, CuS0₄ in water, milk, cloud.
Examinee:
Solution: sulphur in CS₂, CuS0₄ in water, sugar in water.
Suspension: chalk powder in water.
Colloid: sulphur in water, milk, cloud.

Examiner:
What is Tyndall Effect?
Examinee:
The scattering of light by colloidal particles making the path of light visible is called Tyndall effect.

Examiner :
Give an example of a solution in which gas is solute and liquid is solvent.
Examinee:
Aerated drinks.

Examiner:
What is dispersed phase?
Examinee:
The solute like component or dispersed particles in the colloid form dispersed phase.

Examiner:
What is dispersion medium?
Examinee:
The medium in which dispersed phase is suspended is called the dispersion medium.

Examiner:
What are the dispersed phase and dispersion medium in milk?
Examinee:
Fat is the dispersed phase and water is the dispersion mediurfi in milk.

Examiner:
How can fat and water in milk be separated?
Examinee:
By centrifugation.

Examiner:
Why is sky blue in colour?
Examinee:
Because of scattering of light.

Examiner:
What is a gel?
Examinee:
A gel is a colloid in which dispersed phase is a liquid and dispersion medium is a solid.

Examiner:
Is the formation of a true solution from its components a physical or a chemical change?
Examinee:
It is a physical change.

Examiner:
What is a sol? .
Examinee:
A sol is a colloid in which dispersed phase is a solid and the dispersion medium is a liquid. For example paints, milk of magnesia, etc.

Examiner:
What are aerosols?
Examinee:
A colloid in which dispersed phase is either solid or liquid and dispersion medium is gas is called aerosol.

Examiner:
What is Brownian motion?
Examinee:
The zig-zag motion of colloidal particles is called Brownian motion.

Examiner:
Can colloidal particles be seen?
Colloidal particles cannot be seen by naked eye, but can be seen under an electron microscope.

NCERT LAB MANUAL QUESTIONS 

Question 1:
Why are the particles of a true solution not visible to naked eye?
Answer:
The particles of a true solution are not visible to the naked eye as they are very small in size
(less than 1 nm).

Question 2:
What is the order of the size of a particle that can be seen by naked eyes?
Answer:
The size of particles that can be seen by naked eye are of the order of 10-3 mm.

Question 3:
What different techniques of separation can be employed for the separation of components of homogeneous and heterogeneous mixtures?
Answer:
Techniques that can be used to separate the components of a homogeneous mixture are evaporation, distillation, etc., whereas techniques for the separation of the components of heterogeneous mixture are decantation, filtration, boiling, centrifugation, etc.

Question 4:
What will be your observation when a beam of light is passed through a true solution and a suspension respectively?
Answer:
A beam of light when passed through a true solution passes straight through it, whereas light is scattered when passed through a suspension.

Question 5:
What will be the effect of passing light through colloidal solution of sulphur?
Answer:
Light undergoes scattering when it is passed through a colloidal solution of sulphur.

Question 6:
What is the difference in the particle size of colloid, true solution and suspension?
Answer:
Particle size of

1.  Colloid is 1 nm-1000 nm
2. True solution is < 1 nm
3. Suspension is > 1000 nm.

Question 7:
Classify the following as a true solution, suspension or colloid:

1. Milk;
2.  CuS0₄ solution;
3. Jam;
4. Gum;
5. Soil in water
6. Sand in water.

Answer:
True Solution – (2) CuS0₄ solution.
Suspension – (4) Sand in water.
Colloid – (1) Milk, (3) Jam, (4) Gum and (5) Soil in water.

PRACTICAL BASED QUESTIONS
Multiple Choice Questions/VSA

Question 1:
Choose the correct observation about a solution of sugar in water prepared by you in the laboratory.
(a) Its particles are seen by naked eyes.
(b) Its components can be separated by filtra-tion.
(c) Solid particles settle at the bottom after sometime.
(d) It is transparent.

Question 2:
You have prepared four different mixtures in water using
(i) charcoal powder
(ii) chalk powder
(iii) slaked lime
(iv) detergent powder.
If you filter these mixtures through a filter paper, there will be no residue left after filtration in the case of
(a) charcoal powder
(b) chalk powder
(c) slaked lime
(d) detergent powder.

Question 3:
A true solution is
(a) homogeneous
(b) heterogeneous
(c) translucent
(d) opaque.

Question 4:
A solution is transparent blue. It is
(a) a suspension
(b) colloid
(c) true solution
(d) both a colloid and a true solution.

Question 5:
An example of colloid in which dispersed phase is gas and dispersion medium is solid
(a) pumice stone
(b) whipped cream
(c) gemstones
(d) jellies.

Question 6:
Which of the following is stable when allowed to stand undisturbed for sometime?
(a) Sugar solution
(b) Solution of starch in water
(c) Milk
(d) All of the above.

Question 7:
Out of the following, the only incorrect statement is
(a) no residue is left on the filter paper when a colloidal solution is filtered off.
(b) in a colloidal system dispersion medium is a gas.
(c) in a colloidal system dispersion medium is always in the liquid state.
(d) colloidal system is a heterogeneous mixture.

Question 8:
Identify the suspension in the following
(a) soap in water
(6) milk in water
(c) alcohol in water
(d) saw dust in water.

Question 9:
The statement not true for suspensions is
(a) they are transparent
(b) they are unstable
(c) they are heterogeneous
(d) they are opaque.

Question 10:
What is the correct description of a, b and c in the following figure related to the preparation of a true solution?

(a) a = glass rod, b = beaker, c = solution
(b) a = beaker, b = glass rod, c = solution
(c) a = glass rod, b = solution, c = beaker
(d) a = beaker, b = solution, c = glass rod.

Question 11:
A student takes three test tubes A, B and C containing salt solution, starch in water and suspension of sand in water. He pastes small
strips of coloured paper on one side of each test tube. He then observes the coloured paper from the other side of the test tube through the liquid one by one. The correct observation out of the following is
(a) coloured spot is visible in A, not visible in B, appear dim in C.
(b) coloured spot is not visible in A, not visible in B and not visible in C.
(c) coloured spot is not visible in A, appear dim in B, visible in C.
(d) coloured spot is clearly seen in A, appears dim in B, not visible in C.

Question 12:
The correct designation representing the particle size is

Question 13:
Which of the following is true for a mixture of (NH₄)₂C0₃ in water?

Question 14:
Which of the following is true for a mixture of clay in water?

Question 15:
A student carefully observed the colloid of starch in water, human blood and cow milk. On the basis of observations he made, certain conclusions given below. Choose the wrong conclusion.
(a) Components cannot be separated by ordinary filtration
(b) Translucent
(c) Show Tyndall effect
(d) Homogeneous and stable.

ANSWER KEY
Multiple Choice Questions/VSA

1. (d)
2. (d)
3. (a)
4. (c)
5. (a)
6. (d)
7. (c)
8. (d)
9. (a)
10. (a)
11. (d)
12. (c)
13.  (a)
14. (b)
15. (d)
16. (6)
17. (b)

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CBSE Class 9 Science Practical Skills – Osmosis in Raisins

EXPERIMENT

AIM
To determine the mass percentage of water imbibed by raisins.

MATERIALS REQUIRED
A few raisins with stalk intact, blotting paper, weight box, physical balance, petri dish.

THEORY
Raisins are dehydrated grapes. These when kept in water swell up. This happens due to imbibition of water and endosmosis.
Imbibition is the adsorption of water or any other liquid by the solid particles of a substance causing it to increase in volume without forming a solution.
The solid particles which adsorb water or any other liquid are called imbibants. The liquid which is imbibed is called imbibate. The phenomenon of imbibition occurs due to presence of hydrophilic colloids. Proteins, starch and cellulose are some examples of hydrophilic substances. Different substances have different imbibing capacity. Proteins have very high imbibing capacity, starch has less imbibing capacity and cellulose is the weakest imbiber. Because of difference in imbibing capacities, proteinaceus pea seeds swell more on imbibition than starchy wheat seeds.

PROCEDURE

1. Select a few raisins with intact stalks and weigh them.
2. Now place these raisins in petri dish filled with water and cover it.
3. Keep the set-up undisturbed overnight.
4. Take out the swollen raisins from the petri dish and place them on a dry blotting paper to soak away the extra water.
5. Weigh the swollen raisins and note their weight.
   

OBSERVATIONS AND CALCULATIONS

1. Weight of dry raisins taken =x₁ g
2. Weight of swollen raisins =x₂ g
3. Weight of water imbibed by raisins = (x₂ – x₁ ) g = x g
4. Percentage of water imbibed by raisins = \(\frac { X }{ { X }_{ 1 } }\) x 100 =………… %

RESULT
The percentage of water imbibed by raisins =……….. %

PRECAUTIONS

1. Raisins should be with intact stalks.
2. Proper care should be taken while weighing is done.
3. Sufficient water should be taken in petri dish for soaking raisins.
4. Before taking final weight, raisins should be dried with blotting paper to remove excess water.

INTERACTIVE SESSION

Examiner :
Define osmosis.
Examinee:
The movement of water from a low concentration solution to a high concentration solution through a semipermeable membrane is called osmosis.

Examiner :
What are the two types of osmosis?
Examinee:
Osmosis is of two types-

1. Exosmosis
2. Endosmosis.

Examiner :
What is exosmosis?
Examinee:
The outward movement of water from a cell through a semipermeable membrane is called exosmosis.

Examiner :
What would happen if grapes are placed in a concentrated sugar solution?
Examinee:
Exosmosis happens leading to shrinkage of grapes.

Examiner :
What is endosmosis?
Examinee:
The inward movement of water in a cell through a semipermeable membrane is called endosmosis.

Examiner :
Why do raisins swell up in water?
Examinee:
Raisins (e.g., a dried grape) swell because the water is imbibed in the raisins.

Examiner :
What is imbibition?
Examinee:
Imbibition is absorption of water by the solid particles of an adsorbent causing it to increase in volume without forming the solution.

Examiner :
State the factors that affect imbibition.
Examinee:
Factors include:

1. Temperature
2. Pressure
3. pH
4. Tonicity.

Examiner :
Name the constituents in plant cell that cause imbibition.
Examinee:
Cellulose, lignin, pectin, starch and proteins.

Examiner :
Name the first step hivolved in the germination of seeds.
Examinee:
Imbibition of water.

Examiner :
What is tonicity?
Examinee:
Tonicity is the phenomenon which deals with the external concentration.

Examiner :
Differentiate between hypertonic and hypotonic solution.
Examinee:
Hypertonic solution is an external solution whose concentration is more than the internal concentration of cell, whereas concentration is less than the internal concentration of cell of a Hypotonic solution.

Examiner :
Name the type of solution whose concentration is same as compared to the internal composition of cell.
Examinee:
Isotonic solution.

Examiner :
What do you mean by differentially permeable membrane?
Examinee:
Differentially permeable membrane allows selective passage of solutes through them.

Examiner :
What is diffusion?
Examinee:
The movement of molecules, atoms of gases, liquids and solids without any semipermeable membrane from high concentration to low concentration is diffusion.

Examiner :
How is diffusion different from osmosis?
Examinee:
In diffusion, the movement of a substance is from the place of its high concentration to an area of its low concentration that does not require a semi-permeable membrane, whereas osmosis is the movement of water from its high concentration to its low concentration that requires a semi-permeable membrane.

Examiner :
Define plasmolysis.
Examinee:
Shrinkage of protoplasm of a cell from its cell wall under the influence of a hypertonic solution is called plasmolysis.

Examiner :
What will happen to weeds if excess salts are added to weeds in lawn tennis ground?
Examinee:
The cells of weeds undergo plasmolysis and consequent death of weeds.

Examiner :
What is deplasmolysis?
Examinee:
The swelling up of a plasmolysed protoplast under the influence of hypotonic solution is called deplasmolysis.

NCERT LAB MANUAL QUESTIONS

Question 1:
Will a piece of iron also swell when it is kept in water? Justify your answer.
Answer:
No, iron does not swell but rather undergoes rusting in water.

Question 2:
Have you experienced difficulty in closing wooden doors or windows during rainy seasons? Give a suitable explanation.
Answer:
During rainy season, imbibition of water from atmosphere makes the doors/windows swell up.
Question 3:
Suggest an experiment by which the swollen raisins can be shrunk again.
Answer:
Place the swollen raisins in 10% salt solution.
Question 4:
In some plants seed coats are very hard and thick. How do they break before seeds germinate?
Answer:
The process of imbibition leads to breakdown of seed coats when soaked in water.

Question 5:
What will happen to the shape of a grape when it is placed in a viscous sugar solution?
Answer:
Grape will shrink.

Question 6:
What is the effect of temperature on the rate of imbibition?
Answer:
Imbibition increases.

PRACTICAL BASED QUESTIONS
Multiple Choice Questions/VSA

Question 1:
A student weighed 108 raisins and designated the weight as A. She then soaked them in 50 ml distilled water in a beaker. After 2 hours, she removed the raisins wiped them dry from outside and weighed again and called that weight as B. For determining the percentage of water imbibed by raisins, she should calculate as follows
(a)\(\frac { B-A }{ A }\)
(b)\(\frac { B-A }{ B }\) x 100
(c)\(\frac { B-A }{ A } x\frac { 1 }{ 100 }\)
(d)(B-A)x100

Question 2:
Raisins swell up after being placed them in a beaker containing water for sometime because i
(а) the concentration of water in the cell sap is higher than the water in the beaker.
(b) the concentration of water in the cell sap is lower than the water in the beaker.
(c) the concentration of water in the cell sap is the same as that of water in the beaker.
(d) water inside the raisins passes out of them when placed in a beaker of water.

Question 3:
Raisins selected for the experiment should
(a) have intact stalks.
(6) be swollen raisins.
(c) be without stalks
(d) none of these.

Question 4:
Hypertonic solution as compared to hypotonic solution will have
(a) less solute.
(6) same solute.
(c) more solute.
(d) equal solvent.

Question 5:
Which of the following component facilitates imbibition in a plant cell?
(a) Cellulose
(b) Pectin
(c) Lignin
(d) All of these

Question 6:
Rate of osmosis is affected by rise in temperature as
(a) osmosis increases with increase in temperature.
(b) osmosis decreases with increase in temperature.
(c) rate of osmosis remains same.
(d) none of the above.

Question 7:
The water imbibed by raisins is calculated as
(a) weight of wet raisin – weight of dry raisin.
(b) weight of dry raisin – weight of wet raisin.
(c) weight of wet raisin – weight of wet raisin.
(d) weight of wet raisin + weight of dry raisin.

Question 8:
Entry of water in a cell depends on
(a) solute concentration.
(b) tonicity.
(c) temperature.
(d) all of these

Question 9:
For the experiment, “To show imbibition of water”, raisins should be immersed into water
(a) partially.
(b) completely.
(c) both of the above.
(d) none of these.

Question 10:
While performing the experiment with raisins to determine the percentage of water imbibed by them, a student recorded the following data
Weight of water in the beaker = 50 g
Weight of raisins before soaking = 20 g
Weight of raisins after soaking for one hour = 30 g
Weight of water left in the beaker after the experiment was over = 40 g
The percentage of water imbibed by raisins is:
(a) 10
(b) 20
(c) 45
(d) 50

Question 11:
If the water imbibed by soaked raisins is 50% then the weight of raisins before soaking in water (W₁) and the weight of raisins after soaking in water (W₂) might have been
(а) W₁₌ 30 g and W₂= 45 g
(B)W₁= 30 g and W₂ = 50 g
(c)W₁= 50 g and W₂= 100 g
(d)W₁= 0.5 g and W₂= 25 g

Question 12:
A student dissolved 5 g of sugar in 100 mL of distilled water in beaker A. Then she dissolved 100 g of sugar in 100 mL of distilled water in beaker B. After that she dropped a few raisins of equal weight in each beaker. After two hours she found the raisins in A swollen and those in B shrunken. The inference drawn is that
(a) sugar concentration of raisins is lower than that of solution A and higher than that of solution B.
(b) sugar concentration of raisins is higher than that of solution A and lower than that of solution B.
(c) in B the cell membranes of raisins were damaged resulting in leaching.
(d) in A the permeability of water of the cell membranes of raisins was enhanced.

Question 13:
Each of the three beakers ‘A’, ‘B’ and ‘C’ contained 50 ml of distilled water. A student placed five raisins in each beaker. The raisins for each beaker weighed the same. The beakers were kept at room temperature. The raisins were removed from beaker. ‘A’ after 10 minutes, from beaker ‘B’ after 20 minutes and from beaker ‘C’ after one hour. On calculating the percentage of water absorbed by the raisins, it was found that
(a) maximum absorption of water by raisins was in beaker ‘C
(b) maximum absorption of water by raisins was in beaker ‘B
(c) minimum absorption of water was by raisins in beaker ‘C’.
(d) absorption of water was equal in raisins of all the three beakers.

Question 14:
A student soaked 5 raisins each of equal weight in three beakers A, B and C each containing 100 ml of distilled water and left them at room temperature. After 10 minutes he removed the raisins from beaker A, after 20 minutes from beaker B and after one hour from beaker C. He calculated the percentage absorption of water in each case. If PA, PB and Pc denote per cent
absorption, then which of the following is correct?
(o) PA = PB = Pc
(b) PA < PB > Pc
(c) PA > PB < Pc
(d) PA < PB < Pc

Question 15:
A student soaked 5 g of raisins in beaker (A) containing 25 mL of ice-chilled water and another 5 g of raisins in beaker (B) containing 25 mL of tap water at room tempe-rature. After one hour the student observed that
(a) water absorbed by raisins in beaker (A) was more than that absorbed by raisins of beaker (B).
(b) water absorbed by raisins in beaker (B) was more than that absorbed by raisins of beaker (A).
(c) the amount of water absorbed by the raisins of both beakers (A) and (B) was equal.
(d) no water was absorbed by raisins in either of the beakers (A) and (B).

ANSWER KEY
Multiple Choice Questions/VSA

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

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CBSE Previous Year Solved  Papers  Class 12 Computer Science Outside Delhi 2016

Time allowed : 3 hours                                                                                           Maximum Marks: 70

General Instructions :

1.  There are a total of 26 questions and five sections in the question paper, All questions are compulsory.
2. Section A contains question number 1 to 5, Very Short Answer type questions of one mark each.
3.  Section B contains question number 6 to 10, Short Answer type I questions of two marks each.
4.  Section C contains question number 11 to 22, Short Answer type II questions of three marks each.
5.  Section D contains question number 23, Value Based Question of four marks.
6. Section E contains question number 24 to 26, Long Answer type questions of five marks each.
7. There is no overall choice in the question paper, however, an internal choice is provided in one question of two marks, one question of three marks and all three questions of five marks. An examined is to attempt any one of the questions out of two given in the question paper with the same question number.

SET-I

SECTION-A

Question.1. (a) Out of following, find those identifiers, which cannot be used for naming Variables, Constants or Functions in a C++ program :
Total*Tax, double, case, My name, New, switch, Column 31,_Amount .
(b) Ronica Jose has started learning C++ and has typed the
following program. When she compiled the following code written by her, she discovered that she needs to include some header files to successfully compile and execute it. Write the names of those header files, which are required to be included in the code.

(c) Rewrite the following C++ code after removing any/all syntactical errors with each correction underlined.
Note : Assume all required header files are already being included in the program.

(d) Find and write the output of the following C++ program code:
Note : Assume all required header files are already included in the program.


(e) Find and Write the output of the following C++ program code:
Note : Assume all required header files are already being included in the program,



(f) Look at the following C++ code and find the possible output(s) from the options (i) to (iv) following it. Also, write the maximum and the minimum values that can be assigned to the variable PICKER.
Note : Assume all the required header files are already being included in the code.
The function random (n) generates an integer between 0 and n-1.

Question.2. (a) Write any four important characteristics of Object Oriented Programming ? Give example of any one of the characteristics using C++.
(b)Observe the following C++ code and answer the questions (i) and (ii). Assume all necessary files are included :



(i) Which specific concept of object oriented programming out of the following illustrated by Member Function 1 and Member Function 2 combined together ? [1]
• Data Encapsulation
• Polymorphism
• Inheritance
• Data Hiding
(ii)How many times the message “Book Discarded!” will be displayed after executing the above C++ code? Out of line 1 to Line 9, which line is responsible to display the message “Book Discarded!”
(c) Write the definition of a class CITY in C++ with following description:

(d) Answer the questions (i) to (iv) based on the following :


(i) Which type of Inheritance out of the following is illustrated in the above example ?
– Single Level Inheritance
– Multi Level Inheritance
– Multiple Inheritance
(ii) Write the names of all the data members, which are directly accessible from the member functions of class SALEPOINT.
(iii) Write the names of all the member functions, which are directly accessible by an object of class SALEPOINT.
(iv) What will be order of execution of the constructors, when an objects of class SALEPOINT is declared?
Answer : (a) Four characteristics of Object Oriented Programming are :
(a) Encapsulation
(b) Abstraction
(c) Inheritance
(d) Polymorphism Examples:


In the above program, by declaring the class we are using the concept of encapsulation and in the main method while calling the functions using the object of the class is abstraction.
(b) (i) Polymorphism
(ii) 2 times, once for each object when the destructor will be invoked

Question.3. (a) Write the definition of a function Fix Salary (float Salary [ ], int N) in C++, which should modify each element of the array Salary having N elements, as per the following in the memory along the row with each of its element rules :

(b) R[ 10] [50] is a two dimensional array, which is stored occupying 8 bytes, find the address of the element R[5] [15], if the element R[8][10] is stored at the memory location 45,000.
(c) Write the definition of a member function DELETE () for a class QUEUE in C++, to remove a product from a dynamically allocated Queue of products considering the following code is already written as a written as a part of the program.

(d) Write definition for a function DISPMID (int A[ ] [5], int R, int C) in C++ to display the elements of middle row and middle column from a two dimensional array A having R number of rows and C number of columns. For example, if the content of array is as follows :

The function should display the following as output:
103 901 921 802 601 ‘
516 921 609
(e) Convert the following Infix expression to its equivalent Post fix expression, showing the stack contents for each step of conversion.
P/(Q-R)*S+T

Question.4. (a) Write function definition of DISP3CHAR( ) in C++ to read the content of a text file KIDINME.TXT, and display all those words, which has three characters in it. Example:
If the content of the file KIDINME.TXT is as follows :

(b) Write a definition for function ONOFFER() in C++ to read each object of a binary file TOYS.DAY, find and display details of those toys, which has status as “ON OFFER”. Assume that the file TOYS.DAT is created with the help of objects of class, TOYS, which is defined below:

(c) Find the output of the following C++ code considering that the binary file CLIENT. DAT exists on the hard disk with a data of 1000 clients.

SECTION-B
(Only For Candidates, Who Opted for Python)

Question.1. (a) Out of the following, find those identifiers, which cannot be used for naming Variables or Functions in a Python program:
Total*Tax, While, class, switch,
3rd Row, finally, Column 31,_Total
(b) Name the Python Library modules which need to be
imported to invoke the following functions. [1]
(i) sqrt ()
(ii) dump ()
(c) Rewrite the following code in Python after removing all syntax error(s).
Underline each correction done in the code.
for Name in [Ramesh, Suraj, Priya]
IF Name [0] = ‘S’ :
Print (Name)
(d) Find and write the output of the following Python code :

(e) Find and write the output of the following Python code :


(f) What are the possible Outcome, (s) executed from the following code ? Also specify the maximum and minimum values that can be assigned to variable PICKER.

Question.2. (a) What is the difference between Multilevel and Multiple inheritance ? Give suitable examples to illustrate both.
(b)What will be the output of the following Python code considering the following set of inputs ?


(c) Write a Class CITY in Python with following specifications.

(d) How do we implement abstract method in Python ? Give an example for the same.
(e) What is the significance of super () method ? Give and example for the same.
Answer : (a) Difference between Multi level and multiple inheritance
• In Multiple Inheritance we can inherit more than one class in the same class where as in Multi-Level Inheritance one class can inherit only one base class and the derived class can become base class of some other class

(b) Enter a number : AMAR
Re-enter an integer 2
Enter a number : THREE
Re-enter an integer 4
Enter a number: A123
Enter a number : 1200
try part is used to execute the statement and try to read an integer value and except part is executing every time when the program is reading a non-integer value and redirecting to the beginning of the loop.

Question.3. (a) What will be the status of the following list after the ‘ First, Second and Third pass of the selection sort method used for arranging the following elements in descending order ?
Note : Show the status of all the elements after each pass very clearly underlining the changes.
12, 14, -54, 64, 90, 24
(b)For a given list of values in descending order, write a L method in Python to search for a value with the help of Binary search method. The method should return position of the value and should return —1 if the value ; not present in the list.
(c) Write Insert (City) and Delete (City) methods in Python to add City and Remove City considering them to act as Insert and Delete operations of the data structure Queue. 
(d) Write a method in Python to find and display the prime numbers between 2 to N. Pass N as argument to the method.
(e) Evaluate the following post fix notation of expression. Show status of stack after every operation.
12, 2, /, 34, 20, -, +, 5, +

Question.4. (a) Write a statement in Python to perform the following operations:
• To open a text file “MYPET.TXT” in write mode
• To open a text file “MYPET.TXT” in read mode
(b) Write a method in Python to write multiple line of text contents into a text file daynote.txt line.
(c) Consider the following definition of class Employee, write a method in Python to search and display the content in a pickled file emp.dat, where Empno is matching with
‘A0005’.

SECTION-C
[For all the candidates]

Question.5. (a) Observe the following PARTICIPANTS and EVENTS’ tables carefully and write the name of the RDBMS operations which will be used to produce the output as shown in RESULT ? Also, find the Degree and Cardinality of the result.

(b) Write SQL queries for (i) to (iv) and find outputs for SQL queries (v) to (viii), which are based on the tables.


Note : • Km is Kilometers travelled
• NOP is number of passengers travelled in vehicle
(i) To display CNO, CNAME, TRAVELDATE from the table TRAVEL in descending order of CNO.
(ii) To display the CNAME of all the customers from the table TRAVEL who are traveling by vehicle with code V01 orV02.
(iii) To display the CNO and CNAME of those customers
from the table TRAVEL who travelled between ‘2015-12-31’ and ‘2015-05-01’. .
(iv) To display all the details from table TRAVEL for the customers, who have travel distance more than 120 KM in ascending order of NOP.
(v) Select Count (*), Vcode From Travel Group By Vcode Having Count (*)>1;
(vi) Select Distinct Vcode From Travel;
(vii) select a.vcode, cname, vehicletype from travel a, vehicle b
where a.vcode=b.vcode and km<90;
(viii) Select Cname, Km* Perkm From Travel A, Vehicle B Where A. Vcode=B.Vcode And A. Vcode = ‘V05′;

Question.6. (a) Verify the following using Boolean Laws.
X’ + Y Z = X’. Y’.Z’ + X’. Y.Z’ + X’ Y. Z + X’. Y’.Z + X. Y’.Z
(b) Write the Boolean Expression for the result of the Logic Circuit as shown below:

(c) Derive a Canonical SOP expression for a Boolean function G, represented by the following truth table :

(d) Reduce the following Boolean Expression to its simplest form using K-Map :

Question.7. (a) Differentiate between PAN and LAN types of networks.
(b) Which protocol helps us to transfer files to and from a remote computer ?
(c) Write two advantages of 3G over 2G Mobile Telecommunication Technologies in terms of speed and services ?
(d) Write two characteristics of Web 2.0.
(e) What is the basic difference between Computer Worm and Trojan Horse ?
(f) Categorise the following under Client side and Server Side script category ?
(i) Java Script
(ii) ASP
(iii) VB Script
(iv) JSP
(g) Intelligent Hub India is a knowledge Community aimed to uplift the standard of skills and knowledge in the society. It is planning to set up its training centres in multiple towns and villages pan India with its head offices in the nearest cities. They have created a model of their network with a city, a town and 3 villages as follows.
As a network consultant, you have to suggest the best network related solutions for their issues/problems raised in (i) to (iv), keeping in mind the distances between various locations and other given parameters.


Note : In villages, there are community centres, in which one room has been given training center to this organization to install computers.
The organization has got financial support from the government and top IT companies.
(i) Suggest the most appropriate location of the SERVER in the YHUB (out of the 4 locations), to get the best and effective connectivity. Justify your answer.
(ii) Suggest the best wired medium and draw the cable layout (location to location) to efficiently connect various locations with the YHUB.
(iii) Which hardware device will you suggest to connect all the computers within each location of YHUB?
(iv) Which service/protocol will be most helpful to conduct live interactions of Experts from Head office and people at YHUB locations ?
Answer : (a) PAN or Personal Area Network is limited to few meters where a desktop is connected with a Laptop or Mobile phone or Tab etc for personal use where as LAN or Local Area Network is spread within a building connecting all the computers and other peripherals through cables.
(b) FTP – File Transfer Protocol ‘
(c) Advantages of 3G over 2G Mobile Telecommunications are :
(i) Very high speed network
(ii) Customer will get wireless broadband
(iii) Customer can see video or satellite based programs
(iv) All the facilities can be used at same time
(d) Two characteristics of Web 2.0
(i) Folksonomy : Traditional Web like Yahoo and DMOZ uses a pre-defined classification of information like category and sub category on the other hand Web 2.0 without sticking to the existing framework of classification allows user to create free arrangement of information.
(ii) Rich User experience : Traditional web built with HTML and CSS, CGI and had been offered as a static page on the other hand Web 2.0 uses AJAX (Asynchronous JavaScript + XML) presenting dynamic and rich user experience to the users.
(e) Basic difference between Computer Worm and Trojan Horse is there in their functionality. Worm is a kind of program that replicate itself from one system other another without any aid of host file where as Trojan Horse a program that claims to be something useful but actually harmful to the system and also transfer the systems secret to the sender.
(f) (i) Java Script – Server side script
(ii) ASP – Client side script
(iii) VB Script – Server side script
(iv) JSP – Client side script
(g) (i) Server must be placed at YTOWN as maximum computers are there thus decreasing the cabling cost and increasing the efficiency of the computers in the network.
(ii) The following structure would be the most economic way to connect all the villages with YTOWN with reasonable high speed fibre optic cables for VILLI and VILL2 and radio wave transmission for V1LL3 to connect to YTOWN.

(iii) Switch would be needed in all the buildings as more than 10 computers are there in the buildings.
(iv) Video conferencing would be the best option.

CBSE Previous Year Solved  Papers  Class 12 Computer Science Delhi 2014

Time allowed : 3 hours                                                                                           Maximum Marks: 70

General Instructions :

1.  There are a total of 26 questions and five sections in the question paper, All questions are compulsory.
2. Section A contains question number 1 to 5, Very Short Answer type questions of one mark each.
3.  Section B contains question number 6 to 10, Short Answer type I questions of two marks each.
4.  Section C contains question number 11 to 22, Short Answer type II questions of three marks each.
5.  Section D contains question number 23, Value Based Question of four marks.
6. Section E contains question number 24 to 26, Long Answer type questions of five marks each.
7. There is no overall choice in the question paper, however, an internal choice is provided in one question of two marks, one question of three marks and all three questions of five marks. An examined is to attempt any one of the questions out of two given in the question paper with the same question number.

SET I

Question.1. (a) What is the difference between actual parameter and formal parameter ? Give a suitable example to illustrate using a C++ code.
Answer:

(b) Observe the following C++ code and write the name(s) of the header file(s), which will be essentially required to run it in a C++ compiler:

(c) Rewrite the following C++ code after removing all the r syntax error (s), if present in the code. Make sure that you underline each correction done by you in the code.
Important Note:
— Assume that all the required header files are already included, which are essential to run this code.
— The corrections made by you do not change the logic of the program.

(d) Obtain the output of the following C++ program as . expected to appear on the screen after its execution.
Important Note:
—All the desired header files are already included in the code, which are required to run the code.

(e) Obtain the output of the following C++ program, which will appear on the screen after its execution.
Important Note:
— All the desired header files are already included in the code, which are required to run the code.


Answer:
OUTPUT:
A@ 1
0
A@2
75
A @ 3 120

(f) Read the following C++ code carefully and find out, which out of the given option (i) to (iv) are the expected correct output(s) of it. Also, write the maximum and minimum value that can be assigned to the variable Start used in the code:

Answer: (i) 200 #150# Maximum value=3 Minimum value=2

Question.2. (a) Write 4 characteristics of a constructor function used in a class.
Answer: 4 characteristics of a constructor class are:-

1. It is a function of a class.
2. It has the name same that of its class name.
3. It does not have any return type.
4. It is called automatically whenever an object is created.

(b) Answer the question (i) and (ii) after going through the following class:

(i) Which of the function out of Function 1,2,3,4 or 5 will get executed when the Statement 1 is executed in the above code?
(ii) Write a statement to declare a new object G with reference to already existing object H using Function 3.
Answer:
(i) Function 1 will be executed when the Statement 1 is executed
(ii) Health G(H);

(c) Define a class CABS in C++ with the following g specification:
Data Members

•  CNo — to store Cab No
• Type — to store a character ‘A’, ‘B’ or ‘C’ as City Type
• PKM – to store per kilo metre charges
• Dist – to store Distance travelled (in km)

Member Functions

• A constructor function to initialize Type as ‘A’ and CNo as‘1111’
• A function ChargesO to assign PKM as per the 0 following table:
  
• A function Register() to allow administration to enter the values for CNo and Type. Also, this function should call Charges!) to assign PKM Charges.
• A function ShowCab() to allow user to enter the value of Distance and display CNo, Type, PKM, PKM * Distance (as Amount) on screen.

(d) Consider the following C++ code and answer the question from (i) to (iv):


(i) Which type of Inheritance is shown in the above example?
(ii) Write the names of those member functions, which are directly accessed from the objects of class Applicant.
(iii) Write the name of those data members, which can be directly accessible from the member functions of class Applicant.
(iv) Is it possible to directly call function Display O of class Campus
from an object of class Dept? (Answer as YES or NO).
Answer:
(i) Multilevel inheritance is shown in the above example.
(ii) Member functions: void Enroll ();
void View();
void Enter();’
void Show();
(iii) Data Members :
long RegNo;
char Name [20];
double Bugdget;
(iv) No, it is not possible because Display () function of Campus becomes private for the object of Dept class.

Question.3. (a) Write code for a function odd Even(in ts[], int N) in C++, to add 5 in all the odd values and 10 in all the even values of the array S.

(b) An array T[25] [20] is stored along the row in the memory with each element requiring 2 bytes of storage. If the base address of array is 42000, find out the location of T [10] [15]. Also, find the total number of elements present in this array.
Answer:
B=42000
W=2
T[I] [J] = B + W [ N (I -1,) + (J + J,) ]
T[10] [15] = 42000 + 2[ 20 (10) + (15) ]
= 42000 + 2[ 200 +15]
= 42000 + 2[ 215]
= 42000 + 430 = 42430
Location ofT[10][15] = 42430
Total number of elements present in this array = 25 * 20 = 500

(c) Write user-defined function Sum Last 3(int A[] [4], int N, int M) in C++ to to find and display the sum of all the values, which are ending with 3 (ie., units place is 3). For example if the content of array is:

The output should be 49.

(d) Evaluate the following post fix expression. Show the status of Stack after execution of each operation separately: F, T, NOT, AND, F, OR, T, AND
Answer:

(e) Write a function POPBOOKO in C++ to perform delete operation from a Dynamic Stack, which contains Bno and Tide. Consider the following definition of NODE, while writing your C++code.

Question.4. (a) Fill in the blanks marked as as the Statement 1 and the Statement 2, in be program segment given below the appropriate functions for the required task.

(b) Write the function EU Count() in C++, which should read character of a text file IMP.TXT, should count and display the occurrence of alphabets E and U (including small case e and u too)
EXAMPLE:
If the file content is as follows :
Updated information is simplified by official websites.
The EU Count () function should display the output as E: 4 U: 1

(c) Assume the class GAMES as declared below, write a functions in C++ to read the objects GAMES from binary file GAMES.DAT and display those details of those GAMES, which are meant for children of Age Range “8 to 13”.

Question.5. (a) Explain the concept of Union between two tables, with the help of appropriate example.


NOTE
Answer the question (b) and (c) on the basis of the following tables STORE and ITEM

(b) Write the SQL query (1 to 4)
1. To display I Name and Price of all the Items in ascending
order of their Price.
2. To display SNo and S Name of all Stores located in Cp.
3. To display Minimum and Maximum price of each I Name ’ from the table Item.
4. To display I Name, Price of all the Items and their respective S Name where they are available.
Answer:
(1) select I Name, Price from ITEM order by Price
(2) select SNo, S Name from STORE where Area=’CP’;
(3) select MIN(Price), MAX(Price) from ITEM group by I Name;
(4) select I Name, Price, S Name from STORE, ITEM where STORE. SNo= ITEM. SNo;

(c) Write the output of the following SQL command (1 to 4)
1. SELECT DISTINCT I NAME FROM ITEM WHERE , PRICE>=5000;
2. SELECT AREA, COUNT(*), FROM STORE GROUP BY AREA;
3. SELECT COUNT (DISTINCT AREA) FROM STORE;
4. SELECT I NAME, PRICE * 0.05 DISCOUNT FROM ITEM WHERE SNO IN (‘S02’, ‘S03’);

Question.6 (a) Name the law shown below and verify it using a truth table.
A+B.C=(A+B).(A+C)

(b) Obtain the Boolean Expression for the logic shown below:

(c) Write the Sum of Product form of the function F.(P,Q,R) for the following truth representation F :

(d) Obtain the minimal form for the following Boolean expression using Karnaugh’s Map.

Question.7. (a) Write one characteristic each for 2G and 3G Mobile Technologies.
Answer:
Characteristic of 2G mobile technologies is that it is used to make only voice calls.
Characteristic of 3G mobile technologies is that it is used for voice calls as well as video calls.

(b) What is the difference between Video Conferencing and Chat ?
Answer: In Video Conferencing, both the communicators can see each other while talking, whereas, in Chat, we can not see each other.

(c) Expand the following:

• GPRS
• CDMA

Answer:
GPRS—General Packet Radio Service
CDMA—Code Division Multiple Access

(d) Which type of network (out of LAN, PAN and MAN) is formed, when you connect two mobiles using Bluetooth to transfer a picture file. 
Answer: PAN (Personal Area Network)

(e) Trine Tech Corporation (TTC) is a professional consultancy company. The company is planning to set up their new offices in India with its hub at Hyderabad. As a network adviser, you have to understand their requirement and suggest them the best available solutions. Their queries are mentioned as (i) to (iv) below.

(i) What will be the most appropriate block, where TTC should plan to install their services?
(ii) Draw a block to block cable layout to connect all the buildings in the most appropriate manner far efficient communication.
(iii) What will be the possible connectivity out of the following, you will suggest to connect the new set up of offices in Hyderabad with its London based office.

• Satellite
• Infrared
• Ethernet Cable

(iv) Which of the following device will be suggested by you to connect each computer in each of the buildings ?

• Switch
• Modem
• Gateway

Answer:
(i) Finance block is appropriate to install server.
(ii) CABLE LAYOUT:

(iii) Satellite Link ” .
(iv) Switch

(f) Write the name of any two popular Open Source Software, which are used as operating system.
Answer: Linux and Unix are two Open Source operating system.

(g) Write any two important characteristics of Cloud Computing.
Answer: Two characteristic of Cloud Computing are:-

1. It is controlled by entity and restricted to their authorized user.
2. It is delivered through internet 24 x 7.

CBSE Class 9 Science Practical Skills – Plant and Animal Tissues

EXPERIMENT (a)

AIM
To identify parenchyma and sclerenchyma tissues in plants, from prepared slides and to draw their labelled diagrams.

MATERIALS REQUIRED
Prepared slides of parenchyma, collenchyma and sclerenchyma, compound microscope.

You can also Download tissues class 9 notes PDF to help you to revise complete Syllabus and score more marks in your examinations.

THEORY
A group of cells that are similar in structure and work together to achieve a particular function forms a tissue. The main types of plant tissues include-

PROCEDURE

1. Observe the prepared slides of all the plant tissues one by one.
2. First focus the slide at low power and then observe it at high power.
3. Study the characters and draw the diagrams in your notebook.

OBSERVATIONS

IDENTIFYING FEATURES
I. Parenchyma

1. Cells of Parenchyma tissue are isodiametric.
2. Intercellular spaces are present in between the cells.
3. Parenchymatous cells possess large central vacuole and peripheral cytoplasm with a nucleus.
4. These are generally present in the soft parts of plants like leaves, roots, flowers, etc.
5. The important functions of parenchymatous cells are storage, photosynthesis, etc.
   

II. CoIIenchyma

1. Collenchymatous cells are somewhat oval to elongated.
2. Each cell possesses large central vacuole and peripheral cytoplasm with prominent nucleus.
3. Thickenings are present at corners of cells. Thickening comprise of cellulose and pectin.
4. Intercellular spaces are absent.
5. Collenchymatous cells are commonly present below the epidermis in petiole, leaves and stems. Its
6. main function is to provide mechanical strength.
   

III. Sclerenchyma

1. Cells of sclerenchyma tissue are dead with highly thickened walls.
2. Thickenings consist of lignin.
3. There are two types of sclerenchyma cells:
   • fibres which are elongated cells with tapering ends and
   • sclereids (also called stone cells), which are roughly isodiametric cells with narrow cavities.
4. Sclerenchymatous cells have pits which act as connections with adjacent cells.
5. The main function of sclerenchyma is to provide support and mechanical strength to the plant.
   

PRECAUTIONS

1. Handle the microscope carefully.
2. Always focus the slide first at low power (10 x) and then at high power (40x).

EXPERIMENT (b)

AIM
To identify striped muscle fibres and nerve cells in animals, from prepared slides and to draw their labelled diagrams.

MATERIALS REQUIRED
Prepared slides of non-striated, striated and cardiac muscle fibres and nerve cells, compound microscope.

THEORY
A tissue is a group of cells that are similar in structure, origin and function.

PROCEDURE

1. Clean the microscope and adjust the light in microscope.
2. Place the prepared slide of tissue on the stage of microscope and focus first at low power and then at high power.
3. Observe the structure of tissues and draw the diagram of tissues as seen under microscope.

OBSERVATIONS

I. Striated Muscles or Striped or Skeletal Muscles

1. Striated muscle cells are cylindrical, elongated and enclosed in a membrane called sarcolemma.
2. These muscle cells are multinucleated.
3. Striated muscles show presence of light and dark bands which gives it striped appearance.
4. These muscles are present attached to the skeleton of the body.
5. They are voluntary muscles (i.e., work according to our will).
   

II. Non-striated Muscles or Unstriped or Smooth Muscles

1. The cells are spindle-shaped.
2. Nucleus is centrally located.
3. These muscles do not show striations {i.e., no light and dark bands).
4. Non-striated muscles are involuntary in nature.
5. They are commonly found in alimentary canals and blood vessels.
   

III.Cardiac Muscles

1. Cardiac muscle cells are long, branched and uninucleate.
2. These show the presence of intercalated discs.
3. They show alternate light and dark bands.
4. These are involuntary muscles and responsible for rhythmic contraction and relaxation of heart.
5. Cardiac muscles are present only in walls of heart.
   

NERVE CELL

1. Nerve cell comprises of a cell body or cyton with a single nucleus and cytoplasm.
2. Many small cytoplasmic projections arise from cyton which are called dendrons. The dendrons divide further to form dendrites.
3. A long cytoplasmic projection arising from cell body is called axon.
4. A myelin sheath is present over the axon in some nerve fibres, these are called myelinated nerve fibre and when myelin sheath is absent these are called non-myelinated nerve fibres.
5. Myelin sheath is not continuous. There are gaps along the entire length. Each gap is called nodes of Ranvier.
6. Nerve cells help in conduction of nerve impulse.
   

PRECAUTIONS

1. Handle the microscope carefully.
2. Always focus the slide first at low power and then at high power.

INTERACTIVE SESSION

Examiner :
What are meristematic tissues?
Examinee:
The cells which retain the property of division throughout their lives are called meristematic tissues. These remain undifferentiated.

Examiner :
What is differentiation?
Examinee:
Differentiation is a process by which cells lose its ability to divide and are assigned specialised functions.

Examiner :
What are permanent tissues? Name the types of permanent tissues.
Examinee:
Permanent tissues are the tissues which lose their capacity of division and perform specialised functions.The types are – Simple and Complex tissues.

Examiner :
What are simple tissues? Give examples of simple tissues.
Examinee:
Simple tissue is a group of cells having similar structure and function. Parenchyma, Collenchyma and Sclerenchyma are the examples of simple tissues.

Examiner :
What are complex tissues?
Examinee:
Complex tissue is composed of different types of cells to perform a common function.

Examiner :
Why are xylem and phloem called complex tissues?
Examinee:
Xylem and phloem are called complex tissues because these comprise of various types of cells. Xylem includes tracheids, vessels, xylem fibres and xylem parenchyma. Phloem includes sieve tubes, companion cells, phloem parenchyma and phloem fibres.

Examiner :
Name a simple plant tissue which is dead at maturity.
Examinee:
Sclerenchyma.

Examiner :
Give one structural difference between parenchyma and collenchyma.
Examinee:
Parenchyma are thin walled cells with no thickenings whereas collenchyma cells show thickenings at their corners.

Examiner :
What is the composition of thickenings of collenchyma cells?
Examinee:
Cellulose and pectin.

Examiner :
Name the tissue whose presence in the pulp of pear makes it full of grit.
Examinee:
Sclereids of sclerenchyma.

Examiner :
Name a living tissue and a dead tissue which provides mechanical support in plants.
Examinee:
Living tissue is collenchyma and dead tissue is sclerenchyma. *

Examiner :
What is the function of parenchyma?
Examinee:
Main function of parenchyma is storage of food material.

Examiner :
Name the modification of parenchyma cells which contain chlorophyll.
Examinee:
Chlorenchyma.

Examiner :
What is histology?
Examinee:
Histology is branch of biology dealing with study of tissues.

Examiner :
Which type of muscle cells show presence of intercalated discs?
Cardiac muscles.

Examiner :
How would you differentiate between a slide of striated and cardiac muscles?
Examinee:
Striated muscles are long and unbranched whereas cardiac muscles are short and branched. Striated muscles lack intercalated discs whereas intercalated discs are present in cardiac muscles.

Examiner :
Name the components of light and dark bands of striated muscle cells.
Examinee:
Light bands consist of actin proteins and dark bands of myosin proteins.

Examiner :
Which is the longest cell in animal body?
Examinee:
Nerve cell.

Examiner :
How does nerve impulse travel from one neuron to another?
Examinee:
Nerve impulse travels from one neuron to another through a synapse.

Examiner :
What is a synapse?
Examinee:
Synapse is a functional gap between axon of one neuron and dendron of another.

Examiner :
From where does axon and dendron arise in nerve cell?
Examinee:
Axon and dendron arise from cyton (or cell body) of the nerve cell.

Examiner :
Name the type of muscle cells which do not get fatigued and rhythmically contracts and relaxes throughout life.
Examinee:
Cardiac muscles.

Examiner :
What is the shape of non-striated muscle fibres?
Examinee:
Spindle-shaped.

Examiner :
What is the outer membrane surrounding muscle cell called?
Examinee:
Sarcolemma.

Examiner :
What is the cytoplasm of muscle cells called as?
Examinee:
Sarcoplasm

NCERT LAB MANUAL QUESTIONS
EXPERIMENT (a)

Question 1:
In the transverse section of stem which tissue occupies larger space—parenchyma or sclerenchyma?
Answer:
Parenchyma

Question 2:
Which tissue, when matured, has dead cells?
Answer:
Sclerenchyma

Question 3:
Mention the main function of sclerenchyma tissue?
Answer:
To provide mechanical support and strength.

Question 4:
You can bite fruits like guava, grapes, banana etc, but not a piece of wood. Why?
Answer:
Wood is hard due to thickenings present in the cells, mainly made up of dead cells. It comprises of thickened secondary xylem.

EXPERIMENT (b)

Question 1:
What are the features of striated muscle fibre? Where do we find these in our body?
Answer:

1. Multinucleated cells
2. Show striations i.e., light and dark bands.
3. Unbranched, cylindrical.

Location of striated muscles – In fore limbs and hind limbs attached to skeletal bones.

Question 2:
Mention the function of skeletal muscles in our body. –
Answer:
Main function is to help in movement and locomotion.

Question 3:
What are the features observed in a neuron?
Answer:

1. Three main parts – dendrites, cell body, axon.
2. For transmission of nerve impulse.
3. Unidirectional flow of impulse.

PRACTICAL BASED QUESTIONS
Multiple Choice Questions/VSA

Question 1:
The cell appear elongated, tapering at ends as observed under a microscope. It is
(a) striped muscle.
(b) sclerenchyma fibre.
(c) nerve cell.
(d) parenchyma.

Question 2:
A student identified the following figure as striped muscles because of

(a) light and dark striations and is uninucleated.
(b) light and dark striations and is multinucleated.
(c) has no striations and is uninucleated.
(d) has no striations and is multinucleated.

Question 3:
In the figure of neuron, X can be identified as

(a) denton
(b) axon
(c) dendrite
(d) cell body

Question 4:
Smooth muscle fibres are
(a) spindle shaped, unbranched, non-striated, multinucleate and involuntary.
(b) spindle shaped, unbranched, unstriated, uninucleate and involuntary.
(c) cylindrical, striated unbranched, multinucleate and voluntary.
(d) cylindrical, unbranched, non-striated, multinucleate and involuntary.

Question 5:
If your teacher asked you to select the slide showing parenchyma cells from a slide box containing a few unlabelled slides. Which of the following would help you to correctly identify the slide?
(a) Structure of the cells as seen under low power of the compound microscope
(b) Observing the slide with the help of a simple microscope
(c) A colour of the material on the slide
(d) None of these

Question 6:
You are viewing a prepared slide of striated muscle fibres from cockroach leg. When you focus the microscope, the striations appear pale and indistinct. To make the striations clearly visible, you would
(a) slowly close the diaphragm to reduce the light.
(b) remove the mirror to cut out light.
(c) change the eyepiece to increase magnification.
(d) replace the objective to decrease magnification.

Question 7:
On observing a permanent slide under the microscope, a student found the structure without cell wall. It had light and dark bands. The slide could be of

(a) nerve cell.
(b) striated muscle.
(c) parenchyma.
(d) unstriated muscle.

Question 8:
You are shown two slides of plant tissues- Parenchyma and sclerenchyma. You can identify the sclerenchyma by
(a) location of nucleus.
(b) thickness of cell wall.
(c) presence of sarcolemama
(d) position of vacuoles.

Question 9:
The undifferentiated, actively dividing cells come under the category of
(a) parenchyma.
(b) collenchyma.
(c) meristematic
(d) permanent.

Question 10:
A student prepared a slide of thigh muscles of cockroach. Which type of muscles he would observe, when slide is viewed under the microscope?
(a) Striated muscles
(b) Non-striated muscles
(c) Both of these
(d) None of these

Question 11:
The stain used for staining the muscle slide is
(a) safranin.
(b) acetocarmine.
(c) iodine.
(d) methylene blue.

Question 12:
Which of the following shows the presence of intercellular spaces?
(a) Collenchyma
(b) Parenchyma
(c) Sclerenchyma
(d) All of these

Question 13:
Which of the following shows collenchyma?

(a) A
(b) B
(c) C
(d) None of these

Question 14:
Observe the location and shape of the nuclei in the four drawings of the striated muscle fibres given. The drawings that most resembles the slide of striated muscle fibre under the microscope is

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

Question 15:
Aditi observed following observations while looking into a permanent slide.

1. Cells are long and cylindrical.
2. Light and dark bands are present giving striated appearance

It could be a
(a) slide of striated muscle fibre.
(b) slide of smooth muscle fibre.
(c) slide of neuron.
(d) slide of parenchyma cells.

Question 16:
Identify the tissues in given diagrams and choose the correct sequence.

Question 17:
A figure depicting parts of a neuron is given below. The correct identification of the labels 1, 2, 3
,

(a) dendrite, cytoplasm, Nissl’s granules, nerve fibre.
(b) cilia, endoplasmic reticulum, nucleoli, nerve fibre.
(c) dendron, cell body, Nissl’s granules, axon.
(d) dendrite, cyton, nucleus, axon.

Question 18:
Given below is the diagram showing the structure of parenchyma cell.

Which markings are wrong?
(a) Cell membrane and nucleus
(b) Intercellular space and cytoplasm
(c) Nucleus and cell wall
(d) Cell wall and cell membrane

Question 19:
The following diagram gives the structure of nerve cell. Identify the part labelled as A.

(a) Axon
(b) Dendrites
(c) Node of Ranvier
(d) Myelin sheath

ANSWER KEY
Multiple Choice Questions/VSA

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

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In this article, we are providing The Book that Saved the Earth Extra Questions and Answers PDF Class 10 English Footprints Without Feet CBSE, Extra Questions for Class 10 English Footprints Without Feet was designed by subject expert teachers.

The Book that Saved the Earth Extra Questions and Answers Class 10 English Footprints Without Feet

Extract Based Questions [3 Marks each]

Read the following extracts and answer the questions that follow.
Question 1.
Noodle : (bowing) O Great and Mighty Think-Tank, most powerful and intelligent creature in the whole universe, what are your orders?
Think-Tank : (peevishly) You left out part of my salutation, Apprentice Noodle. Go over the whole thing again.
(a) Who is Noodle?
(b) What does the extract reveal about Think-Tank?
(c) Find a word from the extract which means ‘feeling or showing nutation’.
(d) What is the opposite of ‘powerful’?
Answer:
(a) Noodle is the apprentice to Think-Tank, the ruler of Mars.
(b) The extract reveals Think-Tank to be a vain and egoistic creature.
(c) ‘Peevishly’ from the extract means ‘feeling or showing invitation’.
(d) It opposite is ‘weak’.

Question 2.
Iota : I can’t figure it out, Captain. (holding up a book) I’ve counted two thousand of these peculiar items.
This place must be some sort of storage barn. [CBSE 2014]
(a) What could Iota not figure out?
(b) Where was Iota?
(c) Find the antonym of the word ‘ordinary’ in the extract.
(d) What part of speech is ‘peculiar’ as used in the extract?
Answer:
(a) Iota could not figure out what the book were because he had never seen a book before.
(b) Iota was at Centerville Public Library on Earth.
(c) The antonym of ordinary is peculiar.
(d) It is an adjective.

Question 3.
Think-Tank : It can’t be worth much, but go ahead. Give us your trifling bit of data.
(a) Whom was Think-Tank addressing?
(b) What was the data?
(c) Find the word ih the extract which is a synonym of ‘insignificant’.
(d) What part of speech is ‘It’ as used in the extract?
Answer:
(a) Think-Tank was addressing his apprentice, Noodle.
(b) The data was that the earthlings did not eat those ‘sandwiches’. They used them for communication.
(c) The synonym of‘insignificant’ in the extract is ‘trifling’.
(d) ‘It’ is a pronoun.

Question 4.
Historian : (chuckling) And that’s how one dusty old book of nursery rhymes saved the world from a Martian invasion.
(a) What was the name of the book?
(b) Why did the Historian chuckle?
(c) Which word in the extract is a synonym of ‘Conquer’?
(d) What part of speech is ‘And’ as used in the extract?
Answer:
(a) ‘Mother Goose’ was the name of the book.
(b) The historian chuckled at the idea that the misinterpretation of a book saved the Earth.
(c) ‘Invasion’ from the extract is the synonym of‘conquer’.
(d) ‘And’ is a conjunction.

Short Answer Type Questions [2 Marks each]

Question 1.
Why did Think-Tank wish to contact the space probe?
Answer:
Think-Tank wanted to contact the probe to find out what the crew of the probe had observed about Earth.

Question 2.
What difficulty do the crew of the space probe face on Earth?
Answer:
The space probe lands in a public library. The crew is shocked because the place looked very strange as they had no idea of books. They think the library to be some kind of storage barn.

Question 3.
How does Think-Tank explain the books to his probe crew?
Answer:
Think-Tank told his crew that the books were sandwiches. He even ordered his crew to eat them. Then, he called them communication devices and later he said that they were meant for watching.

Question 4.
Why do the space probe crew take vitamins?
Answer:
Think-Tank wanted the probe crew to find out what was in the books. So, the probe crew take vitamins to increase their intelligence. They are able to read the books after taking those vitamins.

Question 5.
What is Think-Tank’s interpretation of words like ‘shells’, ‘silver bells’ and ‘garden’ from the nursery rhyme? What is its impact?
Answer:
Think-Tank interprets the given words to mean that the earthlings grew shells and silver in their gardens. This caused him to withhold their planned invasion of Earth as he felt that Earth was more advanced than Mars.

Question 6.
Think-Tank thinks that Earthlings are after him. What is the reason? [CBSE 2015]
Answer:
Think-Tank crew read about Humpty-Dumpty’s fall in the rhyme and showed the picture to Think-Tank. Think-Tank’s head resembled that of Humpty-Dumpty. This makes Think-Tank feel that earthlings were planning to invade Mars and capture him.

Question 7.
Mention any two striking qualities of Noodle.
Answer:
The apprentice of Think-Tank, Noodle, comes out as an interesting figure in the story. He is very polite and smart. He corrects the mistakes of his ruler very humbly by giving indirect suggestions. Also, he is very wise and well informed.

Question 8.
Elaborate the escape that Think-Tank plans for.
Answer:
Think-Tank orders his entire fleet to evacuate Mars. He gets a capsule for himself. They decide to move to the distant star Alpha Centauri to escape invasion by earthlings.

Question 9.
Describe Think-Tank in your our words.
Answer:
Think-Tank is the ruler of the planet Mars. His head is huge and egg shaped. He wears a robe decorated with stars and circles. He is very proud of his intelligence and might.

Question 10.
How did one old book of nursery rhymes save the Earth from a Martian invasion? [CBSE 2011]
or
How did a book change Think-Tank’s opinion about the Earthlings? [CBSE 2014]
Answer:
This book is totally misunderstood by the Martians.
The Martians thought that earthlings were planning to invade Mars. So, they cancelled their invasion of Earth and went back. Think-Tank totally misunderstands the book. Phrases like ‘shell’, ‘silver’, ‘garden’ gave him a false idea that earthlings grew silver and weapons. He starts thinking that earthlings are very advanced technologically.

Long Answer (Value Based) Type Questions [8 Marks each]

Question 1.
Noodle hails Think-Tank as the most powerful and intelligent creature in the whole universe. Do you agree? Write your opinion of Think-Tank citing instances from the given text.
Answer:
Noodle is an trainee under Think-Tank. He bows before him and hails his ruler as the most powerful and intelligent creature in the universe. However, Think-Tank is not really intelligent. He misinterprets the nursery rhymes completely. ‘Silver’, ‘shells’ and ‘garden’ for him mean that the earthlings are growing metals in farms.
He thinks books to be sandwiches and wants the crew to eat them. When he is read out the nursery rhyme which says, ‘Humpty-Dumpty had a great fall’ he is terrified and plans his escape from Mars Thus, Think-Tank in nothing more than a show off and a pompous fool.

Question 2.
Mars and Earth become friends. How?
Answer:
Think-Tank evacuates Mars and the martians move to Alpha Centauri. They fear an invasion by Earth. In the twenty-fifty century, the two planets become friends. The wise and wonderful Noodle replaces Think-Tank. Earthlings tell Martians the difference between sandwiches and books.
They even taught them how to read. At the same time, they have created a model library for Martians at Marsopolis. Still there is one thing that Martians have never learnt to do. They have never tried to read the nursery rhymes of ‘Mother Goose’.

Question 3.
Compare and contrast the characters of Noodle and Think-Tank. [CBSE 2012]
Answer:
Noodle and Think-Tank have contrasting characters. Think-Tank likes to be called the ‘Great and the Mighty’. He orders and commands. He is the ruler of Mars, but has no intelligence.
He thinks books to be sandwiches. He is wrong about everything. He demands that the crew obey him. He likes to pass on other’s idea as his own. He is a coward who simply boasts about his power.
Noodle, his apprentice, is very clever and wise. He corrects the mistakes of the ruler (Think-Tank) very gently. He never takes credit for his ideas. But he offers them to the ruler very gently. Think-Tank makes them as his own.

Question 4.
The play, The Book That Saved the Earth’ conveys the message that misunderstanding of cultural differences between various races can cause confusion and conflict. Based on your reading of the play, write how such confusion and conflicts can be checked so that peace and harmony is maintained. [CBSE 2015]
Answer:
The confusion in the mind of Think-Tank occurred due to the literal interpretation of the nursery rhymes in the book-‘Mother Goose’. To ensure that conflicts are checked so that peace and harmony are maintained, cultural differences should be sorted out by initially sending mature and wise people from one culture to the other to establish contact before deciding to wage war.
Think-Tank had sent an initial probe, but the crew members of the probe were not sufficiently intelligent or mature to understand Earth’s culture. That is why they literally interpreted the nursery rhymes and caused misunderstanding in the mind of Think-Tank.

For More Resources

• NCERT Solutions for Class 10 English
• CBSE Class 10 English First Flight Chapter Wise Question Bank
• CBSE Class 10 English Footprints Without Feet Chapter Wise Question Bank
• CBSE Class 10 English First Flight Extra Questions
• CBSE Class 10 English Footprints Without Feet Extra Questions
• CBSE Revision Notes for Class 10 English First Flight
• CBSE Revision Notes for Class 10 English Footprints Without Feet

CBSE Class 9 Science Practical Skills – Slide of Onion Peel and Cheek Cells

EXPERIMENT (a)

AIM
To prepare stained temporary mount of onion peel cells and to record observations and draw labelled diagrams.

MATERIALS REQUIRED
Onion, plain slides, coverslip, watch glass, needles, forceps, brush, blade, safranin, blotting paper, glycerine and compound microscope.

THEORY
Onion is a multicellular plant. Like other plant cells, the cell of onion peel consists of a cell wall, cell membrane, cytoplasm, a large vacuole and a nucleus. The nucleus lies at the periphery of cytoplasm and vacuole is located in the centre. Presence of large vacuoles and cell wall confirms that cells of onion peel are plant cells.

PROCEDURE

1. Take a piece of onion and bend it to remove the transparent membranous structure called onion epidermal peel. With help of forcep remove the peel from its inner side.
2. Place the peel in water in a watch glass.
3. Add a few drops of stain safranin, to the watch glass containing the peel for staining.
4. Now, wash the leaf peel with water and transfer it on to a clean slide with the help of brush.
5. Remove extra water from the slide surrounding the peel with the help of blotting paper.
6. To this slide, add a drop of glycerine over the peel and place the coverslip in a manner to avoid entry of air bubbles.
7. Soak away the extra glycerine with blotting paper.
8. Examine slide under the microscope.
   

OBSERVATIONS

1. A large number of rectangular cells with distinct cell wall can be observed.
2. Cytoplasm is seen as thin layer of deep coloured substance on inner surface of cell wall.
3. A big central vacuole is present in the cell.
4. A deeply stained round body called nucleus is seen in each cell.
   

RESULT

1. The epidermal peel of onion comprises of rectangular shaped cells. Each cell comprises of a nucleus, a central vacuole, thin layer of cytoplasm and cell wall.
2. As cell walls and large prominent vacuole are present in each cell, the cells placed under observation are plant cells.

PRECAUTIONS

1. Always take a clean slide and hold it by its edges to avoid making the slide dirty.
2. Peel should be properly stain. Avoid under-staining or excessive staining of the peel.
3. Always transfer the peel with the help of brush.
4. Mounting of the peel should be done in centre of slide.
5. Avoid folding of the leaf peel.
6. Remove extra glycerine with the help of blotting paper.
7. Avoid entry of air bubbles while placing the cover slip.

EXPERIMENT (b)

AIM
To prepare stained temporary mount of human cheek cells and to record observations and draw labelled diagrams.

MATERIALS REQUIRED
Slide, coverslip, needle, blotting paper, tooth pick, methylene blue, watch glass, compound microscope.

THEORY
Animal cells lack cell wall and prominent vacuole. In these cells a semi-permeable membrane called cell membrane surrounds the cytoplasm. In comparison to plant cell, in animal cell cytoplasm is dense and occupies larger space. Vacuoles are absent or very small in size. In animal cells, the position of nucleus is usually central.Cheek cells are part of squamous epithelium. They are thin and flat cells. They found arranged like pavement tiles, i.e., arranged end to end forming a delicate lining.

PROCEDURE

1. Rinse your mouth to remove any food particles.
2. With the help of the toothpick scrap the inner side of the cheek to get some epithelial cells.
3. Place the scrappings in a drop of water over a clean slide.
4. Spread the scrappings with the help of needle.
5. Add a drop of methylene blue over the slide.
6. Wait for 2 minutes and then remove the extra stain with blotting paper.
7. Add a drop of glycerine on the slide.
8. With the help of needle, gently place the coverslip over the slide to avoid entry of air bubbles.
9. Observe the slide under the microscope.

OBSERVATIONS

1. Large number of flat polygonal cells with irregular boundaries attached edge to edge are seen.
2. The cells have thin membrane called cell membrane which encloses jelly like substance called
   

RESULT
Cells under observation do not have cell wall and large prominent vacuole. So, cells of specimen on the slide are animal cells.

PRECAUTIONS

1. Scrap the cheek cells gently with help of clean toothpick to avoid injury.
2. Always rinse the mouth before scrapping the tissue so as to remove any food particles from mouth.
3. Place coverslip gently to avoid entry of air bubbles.
4. Remove extra glycerine with blotting paper.
5. Spread the scrappings so that cells are properly distributed.

INTERACTIVE SESSION

Examiner :
Name the basic structural and functional unit of life.
Examinee:
Cell is the basic unit of life.

Examiner :
State the difference between plant cell and animal cell with respect to outer membranes.
Examinee:
Plant cell possesses a cell wall along with a cell membrane. Cell wall provides shape, protects and supports the plant cell. In contrast to this, animal cells lack cell wall. Cell membrane is the outermost covering in animal cells.

Examiner :
What is the cell wall in plants made up of?
Examinee:
Cell wall is made up of cellulose.

Examiner :
Do both plant and animal cells possess cell membrane?
Examinee:
Yes.

Examiner :
What is the significance of having a cell membrane around a living cell?
Examinee:
Cell membrane around a cell acts as a boundary membrane and encloses the protoplasm. Cell membrane is selectively permeable and helps in transport of materials in and out of cell.

Examiner :
How are plant cells different from animal cells?
Examinee:

Examiner :
Name any four cell organelles found both in plant cell and animal cell.
Examinee:
These include nucleus, endoplasmic reticulum, ribosomes and mitochondria.

Examiner :
Name the two cell organelles which apart from nucleus contain DNA
Examinee:
Mitochondria and chloroplast.

Examiner :
Name the stain used for staining the human cheek cell during the experiment.
Examinee:
Methylene blue.

Examiner :
Even after staining of slide with methylene blue, certain cell organelles like mitochondria are not visible in the microscope. Why?
Examinee:
All the cell organelles do not get stained with methylene blue, so they cannot be seen under the microscope.

Examiner :
Why is glycerine used for mounting of the onion cell or human cheek cell?
Examinee:
Glycerine is hygroscopic in nature and hence, does not allow the cells to dry up.

Examiner :
Why is teasing or spreading the cells important in case of cheek cell slide?
Examinee:
Teasing is important as it helps in proper dispersion of cells.

Examiner :
Why is it essential to place the coverslip gently to avoid entry of air bubbles?
Examinee:
Entry of air bubbles on the slide should be avoided as these would mask the structure to be observed.

Examiner :
Name the stain used for staining of onion peel
Examinee:
Safranin.

Examiner :
Compare the shapes of onion peel cells with human cheek cells.
Examinee:
Onion peel cells are rectangular in shape, having distinct cell walls whereas human cheek cells are flat polygonal cells with irregular boundaries.

Examiner :
Name the cell organelle found only in cheek cells (animal cells) but absent in onion peel cells.
Examinee:
Centrosome.

Examiner :
What is the function of centrosome?
Examinee:
Centrosome helps in spindle fibre formation during cell division.

Examiner :
What is the function of nucleus of a cell?
Examinee:
Nucleus controls all the activities of cell.

Examiner :
Why should you rinse your mouth before scrapping the epithelial cells from inner side of cheek? Rinsing helps in removal of any food particles left in the mouth and which may come with the cells. Why is staining done before mounting?
Examinee:
Rinising helps in removal of any food particless left in the mouth and which may come with the cells.

Examiner :
Why is atinin done before mounting?
Examinee:
Staining provides colour to the cell, hence making it visible under microscope.

NCERT LAB MANUAL QUESTIONS
EXPERIMENT NO. 14 (a)

Question 1:
What is the size of nucleus in relation to size of cell?
Answer:
The size of nucleus is very small.

Question 2:
Name the stained parts of the cell.
Answer:
Cell wall, nucleus, cytoplasm.

Question 3:
Pick the odd one out.

1. Plastids
2. Large vacuoles
3. Cell wall
4. Centrioles.

Answer:
(d) Centrioles (as these are present in animal cells and absent in plant cells.)

Question 4:
Which of the following is a correctly labelled cell of an onion peel?

Answer:
(b) is correctly labelled

EXPERIMENT NO. 14 (b)

Question 1:
Arrange the following steps in correct sequence-
(i) Putting a drop of glycerine on the cheek cells on a slide.
(ii) Scrapping the inner side of cheek,
(iii) Adding methylene blue stain, and
(iv) Placing the cover slip over the material.
(a) (i), (ii), (iii), (iv)
(b) (ii), (i), (iv), (iii)
(c) (iv), (ii), (iii), (i)
(d) (ii), (iii), (i), (iv)
Answer:
(d) (ii), (iii), (i), (iv)

Question 2:
Which one of the following is absent in animal cells?
(a) Cell membrane
(b) Nucleolus
(c) Cell wall
(d) Cytoplasm
Answer:
(c) Cell wall

Question 3:
Cheek epithelial cells are an example of
(a) squamous epithelial cells.
(b) cuboidal epithelial cells.
(c) columnar epithelial cells.
(d) all of these.
Answer:
(a) Squamous epithelial cells.

Question 4:
Why are cheek epithelial cells always moist?
Answer:
Due to presence of saliva.

Question 5:
Name two structures which you would see in cheek cells if you were using a very high magnifying power of microscope.
Answer:
Nucleus, Cytoplasm.

PRACTICAL BASED QUESTIONS
Multiple Choice Questions/VSA

Question 1:
Which of the following components are seen in a slide of human cheek cell when observed under a microscope?
(a) Cell membrane, protoplasm, nucleus
(b) Cell wall, cytoplasm, nucleus
(c) Cell wall, protoplasm, nucleus
(d) Cell membrane, cytoplasm, nucleus

Question 2:
One end of the onion peel when focussed under microscope, the other end of the peel became out of focus. The reason is that the
(a) light coming from diaphragm is not uniform.
(b) material is overstained.
(c) material is not spread uniformly.
(d) material is understained.

Question 3:
The extra stain from the slide if present is removed by
(a) drying stain under sunlight.
(b) evaporation by heating.
(c) keeping slide under running water.
(d) soaking with filter paper.

Question 4:
To prepare a mount of human cheek cell, the sample is collected from
(a) outer side of cheek with a blade.
(b) inner side of cheek wfrh a blade.
(c) outer side of cheek with a toothpick.
(d) inner side of cheek with a toothpick.

Question 5:
A student used a red stain for mounting a peel of onion. This corresponds to Pie stain
(a) Acetocarmine.
(b) Safranin.
(c) Methylene.
(d) Iodine.

Question 6:
Pooja draws a cell with cell wall and presence of large vacuole. The cell belongs to which type of cells
(a) animal cell.
(b) bacterial cell.
(c) plant cell.
id) none of these.

Question 7:
You are viewing a prepared slide of temporary mount of cells under high magnification microscope. You would observe
(a) more number of cells and size is large.
(b) less number of cells and size of cells is large.
(c) less number of cells and small in size.
(d) more number of cells and small in size.

Question 8:
Roshini is making a temporary mount of onion peel. What precautions should be taken to avoid the entry of air bubbles on the slide?
(a) Coverslip should be dropped on to the peel from a certain height.
(b) Peel should be allowed to fold over itself once or twice.
(c) Coverslip should be gently dropped over the peel.
(d) Peel should be immersed in stain for over an hour.

Question 9:
Given below are four steps for preparing a temporary mount of human cheek cells.
(i) Taking the scrap from inner side of the cheek and spreading it on a clean slide.
(ii) Putting a drop of glycerine on the material.
(iii) Adding two or three drops of methylene blue.
(iv) Rinsing the mouth with fresh water and disinfectant solution.
The correct sequence of these steps is:
(a) (i) – (ii) – (iii) – (iv)
(b) (iv) – (i) – (iii) – (ii)
(c) (iv) – (i) – (ii) – (iii)
(d) (i) – (iii) – (ii) – (iv)

Question 10:
A compound microscope is generally used in the following manner:
(a) Objective of lOx and eye-piece of lOx
(b) Objective of 45x and eye-piece of lOx
(c) Objective of lOx and eye-piece of 15x
(d) Objective of 45x and eye-piece of 15x
The microscope is used in its high power in
(a) d
(b) b, d
(c) c, d
(d) b

Question 11:
Neha observed a slide of human cheek cells under a microscope in its (i) low magnifying power, (ii) high magnifying power settings. In the first setting, she must have observed
(a) fewer cells in a darker field of view.
(b) more cells in a brighter field of view.
(c) more cells in a darker field of view.
(d) fewer cells in a brighter field of view.

Question 12:
Diagrams of cells of an onion peel were labelled by four students as given below. The correctly labelled diagram is

(a) nucleus, cytoplasm,
(b) nucleus, vacuole.
(c) vacuole, nucleus.
(d) cytoplasm, vacuole.

Question 13:
The correct labelling of A and B is

(a) nucles, crtoplasm
(b)  nucles, vacuole
(c) vacuole ,   nucles,
(d) crtoplasm ,   vacuole

Question 14:
Following four slides were given for observation under microscope. The correct identification of the slides is

Question 15:
When an onion peel cell i solution, it will look like

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

Question 16:
Which of the following is not seen while studying the slide of squamous epithelium?
(a) Cell membrane
(b) Vacuole
(c) Cell wall
(d) Nucleus

Question 17:
Where does protein synthesis in a cell occur?
(a) mitochondria.
(b) ribosomes.
(c) plastids.
(d) golgi apparatus.

Multiple Choice Questions/VSA

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

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CBSE Class 10 Science Practical Skills – Dicot Seed

BASIC BUILDING CONCEPTS
The ovule after fertilisation develops into seed. A scar called hilum is present at the outer surface of seed. This is the point where seed stalk remains attached to seed. A small pore called micropyle is also present near the hilum. Each seed has an outer covering called seed coat. The seed coat encloses an embryo which is differentiated into plumule, radicle and cotyledons. Plumule is the upper terminal part of embryo which elongates and develops into future shoot. The lower end of embryo is radicle. Radicle develops into future root. In a seed, one (in monocots) or two (in dicots) cotyledons may be present. The cotyledons are also known as seed leaves. The part of embryonic axis between the plumule and point of attachment of cotyledons is called epicotyl. Similarly, the part of embryonic axis between the radicle and point of attachment of cotyledons is called hypocotyl.

AIM
To identify the different parts of an embryo of a dicot seed (pea, gram or red kidney bean).

MATERIALS REQUIRED
Seeds of bengal gram/red kidney bean, forceps, cloth, magnifying glass, petridish, water.

THEORY

PROCEDURE

1. Soak a few bean seeds and keep them overnight.
2. Drain the excess water.
3. Cover the seeds with a moist cloth and keep them aside for a day.
4. Peel off the seed coat carefully.
5. Using your thumbs or the forceps, open the bean seed so that it falls into two halves.
6. With the help of the magnifying glass, locate the parts of the bean embryo carefully.
7. Draw a sketch of the inside of the dicot seed that you examined. Label all its parts.
   

OBSERVATIONS

1. Bean seed is kidney-shaped with a concave and convex side.
2. The concave side which is slightly darker has a scar called hilum.
3. Adjacent to hilum a small pore known as micropyle is present.
4. The seed is covered by a seed coat.
5. The embryo consists of two large, white and kidney-shaped cotyledons.
6. The cotyledons are attached laterally to the curved embryonal axis.
7. Rod shaped and slightly bulgy lower end of embryonal axis which lies towards micropylar end is called radicle.
8. Plumule is the upper end of embryonal axis.
9. The part of embryo axis between radicle and attachment of cotyledon leaves is called hypocotyl.
10. The part of embryo axis between plumule and attachment of cotyledon leaves is called epicotyl.

RESULT
The embryo of dicot seeds consists of three main parts – plumule, radicle and two cotyledons.

PRECAUTIONS

1. The cloth which is used to cover seeds should be moist and should not get dry.
2. Seeds should be open carefully so that the embryo does not get damaged.

INTERACTIVE SESSION

Question 1.
What is the difference between a dry seed and the soaked seed?
Answer:
Dry seed has hard and wrinkled seed coat, soaked seed has soft seed coat which can be removed easily.

Question 2.
Why should we soak the seed overnight?
Answer:
Soaking helps the cotyledons to absorb water and the embryo becomes active.

Question 3.
Why should we keep the seed covered with a moist cloth for another day?
Answer:
So that the baby plant grows up a little more.

Question 4.
What is the function of cotyledons?
Answer:
Cotyledons contain food for the baby plant.

Question 5.
What do you understand by dicot seed?
Answer:
A seed which contains two cotyledons.

Question 6.
What is the function of seed coat?
Answer:
Seed coat protects the cotyledons and the embryo during the dormant period.

Question 7.
What would happen if the seed covered with moist cloth is kept in the fridge? Why?
Answer:
If the seeds are kept in the fridge, the temperature required for germination is not available, so the embryo would not be able to grow.

Question 8.
What are the two main parts of the embryo?
Answer:
Plumule which grows into a shoot and the radicle which grows into a root.

Question 9.
What does the radicle grow into?
Answer:
The radicle grows to develop into the root system.

Question 10.
What conditions are required for the seed to grow into a complete plant?
Answer:
The conditions required for the seed to grow into a plant are

• Water
• Oxygen
• Right temperature
• Soil.

PRACTICAL BASED QUESTIONS

Multiple Choice Questions /VSA (1 Mark)

Question 1.
Seeds are produced within
(a) root
(b) stem
(c) fruit
(d) leaf

Question 2.
The development process in which a mature seed develops into a seedling is called
(a) transpiration
(b) germination
(c) photosynthesis
(d) dormancy

Question 3.
The protective covering of a seed is
(a) hilum
(b) stem
(c) seed coat
(d) micropyle

Question 4.
The scar where the seed was attached to the fruit is the
(a) micropyle
(b) hilum
(c) seed coat
(d) stem

Question 5.
Which of the following forms a tiny root?
(a) Radicle
(b) Epicotyl
(c) Hypocotyl
(d) Cotyledon

Question 6.
The baby shoot is formed by the
(a) hilum
(b) radicle
(c) hypocotyl
(d) plumule

Question 7.
Seed coat protects
(a) hypocotyl, hilum
(b) cotyledons, embryo
(c) micropyle, epicotyl
(d) cotyledons, micropyle

Question 8.
Water enters the seed through the
(a) seed coat
(b) hilum
(c) embryo
(d) micropyle

Question 9.
The part of the embryonic axis located above the point of attachment of cotyledons is called
(a) hypocotyl
(b) micropyle
(c) hilum
(d) epicotyl

Question 10.
The part of the embryo between root and cotyledons is called
(a) hypocotyl
(b) epicotyl
(c) plumule
(d) micropyle

Question 11.
The number of cotyledons present in pea, gram, j kidney bean is
(a) one
(b) two
(c) three
(d) either one or two

Question 12.
The food for the baby plant is stored in
(a) seed coat
(b) cotyledons
(c) micropyle
(d) embryo

Question 13.
Embryo consists of
(a) only plumule
(b) only radicle
(c) plumule and radicle both
(d) hilum

Question 14.
After a seed is germinated, it will develop and grow into a young plant. Arrange the following to show the correct order of events taking place during seed germination.
A. root appears
B. seed coat ruptures
C. seedling
D. shoot appears
(a) A, B, D, C
(b) B, A, D, C
(c) B, D, A, C
(d) C, A, B, D

Question 15.
The seed is produced by the fusion of
(a) micropyle and ovum
(b) pollen grains and hypocotyl
(c) pollen grains and ovum
(d) ova and ovum

Question 16.

What is A?
(a) Hilum
(b) Seed coat
(c) Plumule
(d) Micropyle

Question 17.
A student draws the diagram of a dicot seed to show the different parts of its embryo. He has marked its parts as A, B, C and D:

The correct labelling of these parts respectively is
(a) plumule, radicle, cotyledon, seed coat
(b) plumule, radicle, seed coat, cotyledon
(c) radicle, plumule, cotyledon, seed coat
(d) radicle, plumule, seed coat, cotyledon

Question 18.
Following are some differences between dicot and monocot seeds:

Of the above mentioned differences, which are correct?
(a) I and II only
(b) II and III only
(c) I and III only
(d) All I, II and III

ANSWER KEY

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

Short Answer Questions (2 Marks)

Question 1.
What are the various parts of an embryo of a dicot seed?
Answer:
The embryo is differentiated into three parts— plumule, radicle and cotyledons. Plumule is upper terminal part of embryo which elongates and develops into future shoot. The lower end of embryo is radicle. Radicle develops into future root. Seed leaves are called cotyledons.

Question 2.
What are cotyledons? What is their function in a seed? How is bean seed different from maize seed on the basis of number of cotyledons?
Answer:
Cotyledons are the part of embryo which provide nourishment to developing embryo during germination. Cotyledons contain food for the baby plant. Bean seed has two cotyledons while maize seed has only one cotyledon.

Question 3.
What do you understand by the terms: hypocotyl and epicotyl?
Answer:
The portion of embryonal axis which lies below the cotyledon in a dicot embryo is called hypocotyl. It terminates with radicle.
Epicotyl in the portion of embryonal axis which lies above the cotyledons in a dicot embryo. It terminates with plumule.

Question 4.
Mention the different steps of germination of seed in correct order.
Answer:

1. absorption of water by seeds.
2. rupture of seed coat.
3. emergence of radicle.
4. emergence of plumule.
5. development of radicle into baby root.
6. development qf plumule into shoot.

Question 5.
Differentiate between plumule and radicle.
Answer:

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CBSE Class 9 Science Practical Skills – Plant Kingdom

EXPERIMENT

AIM
To study the characteristics of SpirogyraJAgaricus, Moss/Fem, Pinus (either with male or female cone) and an Angiospermic plant. Draw and give two identifying features of groups they belong to.

MATERIALS REQUIRED
Permanent slide of Spirogyra, specimens of Agaricus, moss, fern, Pinus (male cone or female cone), an angiospermic plant and compound microscope.

THEORY
All the plants are placed in kingdom Plantae, plants are multicellular, eukaryotes and autotrophic. They include algae, mosses, ferns, gymnosperms and angiosperms. They have rigid cellulose cell wall.

PROCEDURE

1. Observe the permanent slide of Spirogyra under low power of microscope and record your observation. Draw a well labelled diagram.
2. Observe the given specimens carefully. Note the characteristics of each specimen. Draw well labelled diagram of each specimen.

OBSERVATIONS
1. Spirogyra (Pond silk)
Identifying Features

1. The plant body is filamentous.
2. Its filaments are long, multicellular and unbranched.
3. Each filament comprises of long, cylindrical cells which are arranged end to end. Cells are generally longer than being broader.
4. Each cell has two layered cell wall. The inner layer is made up of cellulose and outer of pectin. Pectin absorbs water to form slimy, mucilagenous sheath.
   
5. Cytoplasm is present in the form of strands which enclose a large nucleus.
6. Chloroplasts are flat, spiral and ribbon-shaped. It contains starch containing pyrenoids, placed equidistant from each other.
7. A large central vacuole is present in each cell.

2. Agaricus (Mushroom)
Identifying Features

1. It is the fruiting body of Agaricus.
2. It is macroscopic, umbrella-shaped and fleshy.
3. It can be divided into a fleshy, stalk known as the stipe and an umbrella-like cap borne on its top, known as the pileus.
4. A ring-like structure called annulus remains attached at the base of stipe.
5. On the lower side of pileus are present a number of vertical plate like structures called gills.
6. The gills bear spores.
   

3. MOSS (Funaria)
Identifying Features

1. Moss is small plant, usually 2-3 cm in height.
2. The main plant body is gametophyte which consists of short axis with spirally arranged minute green leaves.
3. Plant bears a number of thread-like rhizoids which perform the function of root.
4. Plant is monoecious (bears both male and female sex organs).
5. The female sex organ called archegonia is present at the apex of main axis. A lateral branch called male shoot bears antheridia at its tip.
6. The female branch of older plants bears sporophyte which comprises of foot, seta and capsule.
   

4. FERN (Dryopteris)
Identifying Features

1. Plant body is sporophyte which is differentiated into root, stem and leaf.
2. Stem is short, stout and underground which is known as rhizome.
3. Roots are adventitious (fibrous) growing profusely from the rhizome.
   
4. From the adventitious buds on the rhizome arise large compound leaves. Leaflets are small which are arranged on either side of rachis.
5. Young leaves are circulate (rolled from the apex downwards).
6. The stem and the petiole are covered with numerous brownish scales known as ramenta.
7. The mature leaves on ventral surface bear sori. Such leaves are called sporophylls. Sori bear sporangia that contain spores.

5. PINUS (Chir)
Identifying Features

1. Pinus is a tall erect evergreen tree, usually found on slopes of hills.
2. Plant body is differentiated into root, stem and leaves. Plant shows well developed tap root system.
3. Plant bear two types of shoots – long shoots and dwarf shoots. Long shoots are branches of unlimited growth and dwarf shoots are branches of limited growth.
4. Leaves are of two kinds-
   • long needle like foliage leaves borne on dwarf shoots. They are photosynthetic.
   • small brown scaly leaves borne on both kinds of branches.
5. Pinus is monoecious and distinct male and female cones present on different branches of the same tree.
6. Seeds produced are not enclosed in fruits,
   i. e., naked seeds present.
   

Male Cone

1. Male cone are borne in clusters on long branches. It consists of many microsporophylls which are arranged spirally around the axis.
2. Male cone is ovoid and dark brown in colour.
3. On ventral surface each microsporophyll bears two microsporangia.
4. Each microsporangia bears a number of winged microspores.
   

Female Cone

1. Female cones arise in the axil of scale leaves on the long branches.
   Each cone consists of a short axis around which a number of megasporophylls are spirally arranged.
2. Each megasporophyll bears two ovules on its upper surface.
   
3. Size of female cone varies with age.
   • In first year, the cones are very small.
   • In second year, the cones are large and woody and microsporophylls are compactly arranged.
   • In third year, microsporophylls elongate and separate out from one another.
     

6. ANGIOSPERMIC PIJANT [Mustard (Brassica campestris)]
Identifying Features

1. Plant body is differentiated into root system and shoot system.
2. Root system is usually under the ground. It comprises of main root associated with many lateral branches. The aerial part of the plant constitutes shoot system. It consists of stem, leaves, flower, fruits and seeds.
3. Stem is soft green which is differentiated into nodes and intemodes.
4. Leaves are present at the nodes.
5. Flowers are reproductive organs. It has four parts – calyx, corolla, androecium and gynoecium.
6. Seeds produced are enclosed within fruits.
   

INTERACTIVE SESSION

Examiner :
Define taxonomy.
Examinee:
The branch of science dealing with identification, classification and nomenclature of different organisms.

Examiner :
What are thallophytes? Give examples.
Examinee:
The plants in which plant body is thallus like, i.e., not differentiated into stem, root and leaves
are called thallophytes. e.g., Ulothrix.

Examiner :
What is the study of algae called as?
Examinee:
Phycology.

Examiner :
Give any two important characteristics of algae.
Examinee:
Algae are autotrophic and contain chlorophyll. Plant body is thalloid and cells have cell wall made up of cellulose.

Examiner :
What is peculiar characteristic of chloroplast in Spirogyra?
Examinee:
Chloroplast in Spirogyra are ribbon-shaped with pyrenoid bodies present at equidistant.

Examiner :
What is the function of pyrenoid bodies?
Examinee:
To store starch of the cell.

Examiner :
Why is Spirogyra slimy to touch?
Examinee:
In Spirogyra, the outer cell wall is made up of pectin which absorbs water and gives slimy appearance.

Examiner :
What are fungi?
Examinee:
Fungi are non-green heterotrophic organisms with thalloid body.

Examiner :
Give an example of unicellular fungi.
Examinee:
Yeast.

Examiner :
What is the cell wall of yeast made up of?
Examinee:
Chitin.

Examiner :
Mention various ways of reproduction in fungi.
Examinee:
Budding, sporulation, conjugation, etc.

Examiner :
What is alternation of generations?
Examinee:
The life cycle of plants has two phases – gametophytic phase and sporophytic phase. These two phases alternate in life cycle. This is called alternation of generations.

Examiner :
Why are bryophytes called amphibians of plant kingdom?
Examinee:
Bryophytes require water for the fertilisation of male and female gametes.

Examiner :
How do leaves of mosses differ from those of angiosperms?
Examinee:
Leaves in mosses are small, sessile and spirally arranged around the axis. Leaves of angiosperms may have stalk, are broad and possess a number of veins.

Examiner :
WTiat is dominant phase in mosses?
Examinee:
Gametophytic phase.

Examiner :
Name the group in which vascular tissues in its proper form first appeared.
Examinee:
Pteridophytes.

Examiner :
Why are pteridophytes considered to be more advanced than bryophytes?
Examinee:
Pteridophytes has plant body differentiated into leaves, stem and roots. They also possess vascular system.

Examiner :
What is the group containing naked seeds called?
Examinee:
Gymnosperms.

Examiner :
Why are seeds of gymnosperms called naked seeds?
Examinee:
Because the seeds in gymnosperms are not enclosed in fruits.

Examiner :
What do you mean by monoecious plant?
Examinee:
In monoecious plant, both male and female sex organs are present on the same plant.

Examiner :
Name the two categories in which angiosperms are divided.
Examinee:
Dicotyledons and Monocotyledons.

Examiner :
What are phanerogams?
Examinee:
Phanerogams include plants in which body is differentiated into stem, root and leaves and produce seeds.

Examiner :
Name the group of angiosperms which show reticulate veina^ion in leaves.
Examinee:
Dicots. .

Examiner :
How do gymnosperms and angiosperms differ with respect to seeds?
Examinee:
Seeds in gymnosperms are naked whereas in angiosperms they are enclosed in fruits.

Examiner :
Name four whorls of an angiospermic flower.
Examinee:
Four whorls are calyx, corolla, androecium and gynoecium.

NCERT LAB MANUAL QUESTIONS

Question 1:
Why is a spirogyra plant green in colour?
Answer:
Due to presence of green pigment, chlorophyll.

Question 2:
Name two functions that are common to roots and rhizoids.
Answer:

1. Attachment to substratum.
2. Absorption of water and minerals.

Question 3:
What do the cones in pine represent?
Answer:
Reproductive parts (male and female sporophylls).

PRACTICAL BASED QUESTIONS
Multiple Choice Questions/VSA

Question 1:
Chloroplast in Spirogyra is
(а) spirally arranged and ribbon-shaped with pyrenoids
(b) spirally arranged without pyrenoids
(c) circular
(d) cup-shaped

Question 2:
While locating the stem of fern, the students find that it is
(a) underground rhizome
(b) coiled like spring
(c) totally absent
(d) branched and filamentous

Question 3:
The structure by which the nucleus in Spirogyra cell is held?
(a) Cell wall
(b) Cytoplasmic strands
(c) Cell membrane
(d) Pyrenoids

Question 4:
Needle-shaped structure in Pinus plant is
(a) leaf
(b) shoot
(c) stem
(d) reproductive part

Question 5:
Some students want to prepare a temporary mount of Spirogyra. Where should they search for a fresh specimen?
(a) In a pond of a salty water
(b) In a pond of stagnant dirty water
(c) In a stream of running fresh water
(d) In a stream of running salty water

Question 6:
Which of the character is identifying feature of gymnosperms?
(а) Naked seeds
(б) Rhizoids present
(c) Chlorophyll absent
(d) Does not possess vascular bundle.

Question 7:
Protein body associated with storage of starch grains in Spirogyra is called
(a) leucoplast
(b) amyloplast
(c) pyrenoid
(d) chloroplast

Question 8:
To which group do the plants belong that have embryo, vascular tissue, seeds but lack fruits?
(a) Gymnosperms
(b) Angiosperms
(c) Pteridophytes
(d) Bryophytes

Question 9:
Mosses are
(a) monoecious
(b) dioecious
(c) both (a) and (b)
(d) none of these

Question 10:
Which group is called amphibians of plant kingdom?
(a) Algae
(b) Bryophytes
(c) Gymnosperms
(d) Angiosperms

Question 11:
Megasporophyll in female cones of gymnosperms contain ovules on its upper surface.
(a) one
(b) three
(c) two
(d) no ovule

Question 12:
In Angiosperms, the female reproductive part is represented by
(a) calyx
(b) corolla
(c) androecium
(d) gynoecium

Question 13:
Angiosperms differ from gymnosperms in having
(a) seeds
(b) fruits
(c) leaves
(d) vasculature

Question 14:
Megasporophylls and microsporophylls are associated with which of the following plants?
(a) Neem
(b) Pinus
(c) Agaricus
(d) Moss

Question 15:
Figures of two plants are given below. Observe them carefully and select the option which correctly gives their identification and names of the groups to which they belong

(а) (i) Moss: Bryophyta       (ii) Fern: Pteridophyta
(b) (i) Pine: Gymnosperm   (ii) Leafy plant: Angiosperm
(c) Both (i) and (ii) are mosses and belong to bryophyta
(d) Both (i) and (ii) are ferns and belong to pteridophyta

Question 16:
Funaria differs from Dryopteris in
(a) bearing seeds
(b) having vascular bundle
(c) having a dependent sporophyte
(d) having an independent sporophyte

ANSWER KEY
Multiple Choice Questions/VSA

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

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In this article, we are providing The Sermon at Benares Extra Questions and Answers PDF Class 10 English First Flight CBSE, Extra Questions for Class 10 English was designed by subject expert teachers.

The Sermon at Benares Extra Questions and Answers Class 10 English First Flight

 Extract Based Questions [3 Marks each]

Read the following extracts carefully and answer the questions that follow
Question 1.
GAUTAMA Buddha (563 B.C.- 483 B.C.) began life as a prince named Siddhartha Gautama, in northern India. At twelve, he was sent away for schooling in the Hindu sacred scriptures and four years later he returned home to marry a princess. They had a son and lived for ten years as befitted royalty.
(a) Who was Gautama Buddha?
(b) What did Gautama Buddha study?
(c) Find the exact word from the extract whii means ‘people of royal family’. ‘
(d) When did Gautama Buddha marry?
Answer:
(a) Gautama Buddha was a prince born in North India.
(b) Gautama Buddha studied Hindu Sacred Scriptures.
(c) ‘Royalty’ from the extract means ‘people of royal family’.
(d) Gautam Buddha married after his schooling at the age of sixteen years.

Question 2.
At about the age of twenty-five, the Prince, heretofore shielded from the sufferings of the world, while out hunting chanced upon a sick man, then an aged man, then a funeral procession, and finally a monk begging for alms. These sights so moved him that he at once went out into the world to seek enlightenment concerning the sorrows he had witnessed.
(a) Why did Gautama Buddha go out to seek enlightenment?
(b) What did Gautama Buddha see while hunting?
(c) Find the exact word from the extract which means ‘a state of high spiritual knowledge’.
(d) Describe the thought of Gautam Buddha while he encountered with the different stages of life.
Answer:
(a) Gautama Buddha was highly moved by seeing the sufferings of man. So he went out to seek enlightenment.
(b) Gautama Buddha chanced upon a sick man, then an aged man, then a funeral procession, and finally a monk begging for alms while hunting.
(c) ‘enlightenment’ from the extract means ‘a state of high spiritual knowledge’.
(d) Gautam Buddha was so upset and worried while encountering the different stages of life as a sickman, an aged man, a funeral and a monk begging for alms.

Question 3.
He wandered for seven years and finally sat down under a peepal tree, where he vowed to stay until enlightenment came. Enlightened after seven days, he renamed the tree the Bodhi Tree and began to teach and to share his new understandings. At that point he became known as the Buddha.
(a) What was the name of the peepal tree under which Buddha sat?
(b) What did Buddha do while sitting under the tree?
(c) Find the exact word from the extract which means ‘solemnly promise to do a specified thing’.
(d) After how many days Gautam Buddha got enlightenment?
Answer:
(a) The name of the peepal tree under which Buddha sat was Bodhi tree.
(b) Buddha began to teach and share his new understandings while sitting under the tree.
(c) ‘Vowed’ from the extract means ‘solemnly promise to do a specific thing’.
(d) Gautam Buddha got enlightenment after seven days of his vow that he would sit until enlightenment came.

Question 4.
The Buddha preached his first sermon at the city of Benares, most holy of the dipping places on the River Ganges; that sermon has been preserved and is given here. It reflects the Buddha’s wisdom about one inscrutable kind of suffering.      [CBSE 2014]
(a) Where did Buddha preach his first sermon?
(b) What does the sermon preached by Buddha reflect?
(c) ‘Find the exact word from the extract which means impossible to understand’.
(d) How Benares is described in the lesson?
Answer:
(a) Buddha preached his first sermon at Benares.
(b) It reflects Buddha’s wisdom about one inscrutable kind of suffering.
(c) ‘Inscrutable’ from the extract means ‘impossible to understand’.
(d) Benares is described in the lesson as the most holy of the dipping places on the river Ganges.

Question 5.
Poor Kisa Gotami now went from house to house, and the people pitied her and said, “Here is mustard-seed; take it!” But when she asked, “Did a son or daughter, a father or mother, die in your family?” they answered her, ‘Alas! the living are few, but the dead are many. Do not remind us of our deepest grief.” And there was no house but some beloved one had died in it.
(a) Why did Kisa Gotami travel from house to house?
(b) What did Kisa Gotami ask before taking the mustard seeds?
(c) Find the exact word from the extract which means ‘dearly loved’.
(d) What did the people reply Kisa Gotami when she asked about the death of a family member?
Answer:
(a) Kisa Gotami travelled from house to house to collect mustard seeds.
(b) Kisa Gotami asked everyone before taking the mustard seeds whether a family member had ever died in their family or not.
(c) ‘beloved’ from the extract means ‘dearly loved’.
(d) The people replied Kisa Gotami that there was no house but some beloved one had died in it and she should not remind them of their deepest grief.

Question 6.
“Not from weeping nor from grieving will anyone obtain peace of mind; on the contrary, his pain will be the greater and his body will suffer. He will make himself sick and pale, yet the dead are not saved by his lamentation. He who seeks peace should draw out the arrow of lamentation, and complaint, and grief.”
(a) Who said these lines?
(b) How can a person seek peace?
(c) Find the exact word from the text which means ‘to express sadness about something’.
(d) What effect does the weeping or grieving have on us?
Answer:
(a) Buddha said these lines.
(b) The person who seeks peace should draw out the arrow of lamentation, complaint and grief.
(c) ‘lamentation’ from the extract means ‘to express sadness about something.
(d) Weeping or grieving makes our pain greater than before. The person will make himself sick and pale.

Short Answer Type Questions [2 Marks each]

Question 1.
Why did Prince Siddhartha leave the palace and become a beggar?     [CBSE 2012]
Answer:
Once Prince Siddhartha, while hunting saw a sick man, then an aged man, then a funeral procession and finally a monk begging for alms. Looking at this, he left the palace and became a beggar to search for enlightenment.

Question 2.
What do you know about the early life of Buddha?
Answer:
Gautama Buddha was born in a royal family. His childhood name was Siddhartha. At the age of 12, he was sent away for schooling in Hindu sacred scriptures and four years later he got married to a princess.

Question 3.
Where did Buddha preach his first sermon?
Answer:
Gautama Buddha preached his first sermon at the city of Benares, which is regarded as the holiest of the bathing places on the river Ganges.

Question 4.
How did Kisa Gotami realise that life and death is a process?       [CBSE 2016]
Answer:
Kisa Gotami went from house to house but was unable to find one house where nobody had died. She was tired and hopeless and sat down at the wayside watching the lights of the city as they flickered up and were extinguished again. She realised that similar to the city lights human lives also flicker up for some time and are extinguished again.

Question 5.
What was the effect of the sufferings of the world on Buddha?
Answer:
At the age of 25, while hunting, one day Buddha saw a sick man, then an aged man, then a funeral procession and finally a monk begging for alms. These moved him so much that he went out into the world to seek enlightenment.

Question 6.
According to Kisa Gotami what is the greatestgrief of life?       [CBSE 2014]
Answer:
According to Kisa Gotami, the greatest grief in life is the death of one’s loved ones and one’s inability to stop them from dying. Therefore, instead of lamenting on it, the wise should not grieve. Weeping will only increase the pain and disturb the peace of mind of a person

Question 7.
Why was Kisa Gotami sad? What did she do in her hour of grief?
Answer:
Kisa Gotami was sad over the death of her only son. In the hour of grief, she went door to door in order to find medicine for her son that could bring him to life.

Question 8.
What did the Buddha do after he had attained enlightenment?
Answer:
When Buddha Attained enlightenment, he started preaching and telling people about life and its meaning. He spread his preachings far and wide so that people

Long Answer (Value Based) Type Questions [8 Marks each]

Question 1.
Life is full of trials and tribulations. Kisa Gotami also passes through a period of grief in her life. How does she behave in those circumstances?      [CBSE 2013]
Answer:
After the death of Kisa Gotami’s only child, she became very sad. She carried her dead child to her neighbours in order to get medicine to bring him to life. Her neighbours thought that she had gone insane as she was unable to accept the fact that. It was then that someone suggested her to meet Gautama Buddha. When she met Gautama Buddha he gave her an exercise to do.She was asked to collect mustard seeds from a house where no one has ever died. She went from one house to another but was unable to find a single house where no one has died. This way she realised that death is a part of life and anyone who is born is bound to die one day. Thus, Buddha changed her understanding of death by this exercise. could come to terms with the truth.

Question 2.
Personal losses are a part and parcel of life. Instead of wailing on them, we should move on in life. This message of Gautama Buddha has become more relevant in modern times. Do you agree ? Why /why not?   [CBSE2015]
Answer:
Yes, I agree with the message that Gautama Buddha has given about life. In the modem times, people have a lot to explore and move with the world at the same pace. If people don’t understand the practicality of life, they will be under stress which would in turn affects, their personal and professional lives. People need to understand that everyone who is born will have to die one day. There is no use being sad or crying over the loss. People should remain calm and composed in such situations. They should face the truth and move on in life.

Question 3.
What did Buddha say about death and suffering?
Answer:
After enlightenment, Buddha started to spread his teachings about life, truth and the likes of it. He told that death and suffering are the part and parcel of life. None can avoid this truth. One has to meet one’s destined end one day. Whoever has come to the world, will die one day. In the hour of grief, one must remain calm and composed so that grief doesn’t overcome one. People who are wise, never complain or lament over their loss. They accept the truth and be blessed with it. So, the wisdom lies in the fact that people should not get distressed with pain, suffering and death.

For More Resources

• NCERT Solutions for Class 10 English
• CBSE Class 10 English First Flight Chapter Wise Question Bank
• CBSE Class 10 English Footprints Without Feet Chapter Wise Question Bank
• CBSE Class 10 English First Flight Extra Questions
• CBSE Class 10 English Footprints Without Feet Extra Questions
• CBSE Revision Notes for Class 10 English First Flight
• CBSE Revision Notes for Class 10 English Footprints Without Feet

NCERT Solutions for Class 10 English

• A Letter to God Question Answer
• Nelson Mandela: Long Walk to Freedom Question Answer
• Two Stories about Flying Question Answer
• From the Diary of Anne Frank Question Answer
• The Hundred Dresses – I Question Answer
• The Hundred Dresses – II Question Answer
• Glimpses of India Question Answer
• Mijbil the Otter Question Answer
• Madam Rides the Bus Question Answer
• The Sermon at Benares Question Answer
• The Proposal Question Answer