Friday, April 25, 2025

class 7th competancy based ques

 https://cbseacademic.nic.in/web_material/Manuals/Science_PracticeBook23_C7.pdf

Theme 1: Nutrition in Plants 

Prior Knowledge

It is recommended that you revise the following topics before you start working on these questions. 

  • Mode of nutrition in plants 

  • Photosynthesis - food-making process in plants 

  • Products of photosynthesis 

  • Autotrophs and heterotrophs


The World Before Photosynthesis


Our planet is 4.5 billion years old. The earliest undisputed evidence of life comes from around 3.4 billion years ago. At that time, our planet looked nothing like what it does today. It was hot and barren. Our atmosphere was filled with toxic gases and water vapour condensed in this thick atmosphere to form massive clouds and rained acid rain upon the Earth to create the first seas. So, how did life, as we know it now, come to be?

We have to thank a bunch of bluish-green cyanobacteria who figured out how to trap

sunlight and carbon dioxide from the atmosphere, and convert it into nutrition for

themselves. In the process of making their food, these autotrophic (creatures that make

their own food) organisms release oxygen. As there was plenty of carbon dioxide and

sunlight during this time, the population of these cyanobacteria grew so much that within a

million years, they had changed the composition of our planet’s atmosphere. The Earth's

atmosphere now had oxygen that these bacteria had given out as they produced their own

food. Plants, like the ones we see now, as well as animals, found the planet as just the right

place to grow and thrive, and so here we are.

Even today, the tiny cyanobacteria are responsible for producing most of the oxygen on our planet.


Case Study A - How did we discover Photosynthesis?

Around 1772, British scientist Joseph Priestley tried a set of experiments to study the role of air in photosynthesis. 

He used two airtight glass containers. In one container (Container A), he placed a candle and a mouse. In another (Container B), he placed a mint plant, a candle and a mouse together. He observed that the mouse died within seconds in Container A but in Container B, the mouse survived for a few minutes before its death. 

At the end of the experiment, Priestley concluded that the candle and the mouse “spend” air while the mint plant “revives” the air. 

Another scientist named Jan Ingenhousz repeated Priestley’s experiment, using a similar set-up, but also proved that plants need sunlight to “revive” the air. 

It was only by 1905 that scientists had figured out the details of photosynthesis, and the role of chlorophyll.  


Question 1 What might have been the observation made by Priestley that led him to conclude that the candle and the mouse “spent” the air in Container A? 

a. After some time, the candle and mouse heated the air in Container A. 

b. After some time, the candle did not extinguish and the mouse was active in Container A. 

c. After some time, the candle extinguished and the mouse died. 

d. It is not clear what Priestley meant by “spending” of air.

Answer : c. After some time, the candle extinguished and the mouse died


Question 2 Priestley’s conclusion that the mint plant “revived” the air was based on the observation that the state of the candle and the mouse in Container B did not change after some time. What explains this observation? 

a. The plant provided food for the mouse and fuel for the candle. 

b. The plant provided oxygen for the mouse to breathe and the candle to stay lit. 

c. The plant provided carbon dioxide and used up oxygen for respiration. 

d. The mouse respired and gave oxygen to the candle and the plant absorbed the carbon dioxide emitted by the candle to keep the air ‘pure’

Answer : b. The plant provided oxygen for the mouse to breathe and the candle to stay lit.



Question 3 Asha replicated Priestley’s experiment, but without the mouse. She placed a lit candle in Container A and a candle and a plant in Container B. She noticed that the candle in Container B lasted only a few minutes more than the candle in Container A. She had expected the Container B candle to burn until it ran out of wax. Why did it get extinguished before it ran out of wax?

Answer: 

The plant does not produce oxygen at the same rate at which the candle consumes oxygen. Therefore, the candle eventually runs out of oxygen. 
Note for the teacher: Give full credit to the student if the above reason is provided. Please add the following while discussing the answer with your students: “The burning candle produces carbon dioxide, which comes down after rising initially and extinguishes the flame.”



Question 4 : Jan Ingenhousz repeated Joseph Priestley’s experiment to see whether sunlight had a role to play in photosynthesis. Which of the following could have been a suitable design for his experiment? 

  1. Place a lit candle in Container A. Place a plant and a lit candle in Container B. Keep both containers in sunlight. 

  2. Place a lit candle in Container A and keep it in a dark place. Place a plant and a lit candle in Container B and keep it under the Sun. 

  3. Place a plant and a lit candle in each container. Keep one container in a dark room and another in sunlight. 

  4.  Place a plant and a lit candle in each container. Keep both containers first in the dark and then in sunlight.

Answer: c. Place a plant and a lit candle in each container. Keep one container in a dark room and

another in sunlight


Question 5

Based on your answer to Question 4, predict what the outcome of Jan Ingenhousz’s

version of the experiment was.

Answer: The outcome was that the candle in the container, which was in the sunlight, stayed lit for longer because sunlight was needed for photosynthesis and the plant produced oxygen, which was needed for the wick to burn.


Question 6
The scientist , whi discovered oxygen is 
(a) Priestley
(b) Ingenhousz
(c) Engelman
(d)Van Niel 

Question 7 
Sunlight is essential for plants was concluded by 
(a) Joseph Priestley
(b) Jan Ingenhousz
(c) van Niel
(d) none of these

Case Study B - Parasites 

Several plants do not have chlorophyll (green pigment) and cannot perform photosynthesis. They obtain nutrition from a host plant and are called parasites. A parasite gives no benefit to its host. Instead, some parasites cause extreme damage to their hosts. 

Witchweed or Striga is one such parasite. It is a small plant with pretty flowers. It needs a host plant in order to germinate and grow into a young plant. Once it grows, it can create its own food. However, as a sapling, it attaches itself to the roots of its host plant and uses all its nutrition. In the process, the host plant either cannot grow fully or dies completely. Striga especially affects farmers in Asia, Africa, America and Australia as they attach themselves to crops like sugarcane, rice and corn. They can reduce the yield or destroy the entire crop.


Question 1: 

i. Which of the plants shown in Fig 1.6 are parasites?

Answer:. (i) Bird’s Nest Orchid


ii. What factors about the plant did you consider while answering Part i.?

a. Colour b. Shape c. Type d. Size

Answer a. Colour


Question2 : 

A group of scientists in Kenya are trying to help farmers whose crops are affected by Striga/witchweed. They have found 3 chemical solutions that prevent Striga seeds from germinating in the roots of the host plant. They are called Nij1, GR24 and MP1. 

To test the effect of the solutions, they apply the solutions to a collection of four groups of Striga infested rice plant pots. They observe the plants to see whether they are effective in controlling the growth of the parasite or not. Here are their results:


Answer the following questions based on the above results: 

I. Look at Graph (a). Apart from the three solutions Nij1, GR24 and MP1, why did the experimenters also look at the impact of just water?

Answer: The experimenters have included water to see how many Strigas would grow in a pot if no other solution is applied to the rice plant. That way the scientists can infer if adding the other chemical solutions has any impact on the germination of Striga plants


ii. As a farmer, which solution should I use to prevent Striga growth?

a. Water b. Nij1 c. GR24 d. MP1

Answer d. MP1 Explanation: MP1 is most effective in controlling the growth of the parasitic plants and it also enables the rice plant to reach its optimal height.


iii.Which of the following shows that Striga has a negative effect on the growth of its host - the rice plant? 

a. Graph (a) b. Graph (b)

c. Both Graph (a) and Graph (b) d. No such conclusion can be drawn from either of 

the two graphs. 


Answer bGraph (b)

Case Study C - Photosynthesis - CO2 

In this activity, one leaf is exposed to potassium hydroxide, which will absorb carbon dioxide from the surrounding air. It is then exposed to light and tested for starch (carbohydrate). The experiment is designed to test whether carbon dioxide is needed for the plant to produce starch/carbohydrates.

Question 1 Adil was surprised to see that the plant grew by a few cm even when it was kept in the dark. He concluded that photosynthesis was possible even in the absence of light. Rehana countered his conclusion by giving the following statements: 

Assertion(A): In the dark, plants use up the food stored in the form of starch to grow. 

Reason(R): Food cannot be prepared in the absence of sunlight. 

What is your opinion about this debate between the two? 


a. Adil is right. 

B. Adil is wrong and Rehana gave the correct explanation for the observed growth. 

c. Both Adil and Rehana made incorrect conclusions. 

D. Adil is right and Rehana's assertion is incorrect though her reason is right.


b. Adil is wrong and Rehana gave the correct explanation for the observed growth.


Question 2 Rehana performed the experiment described in Case Study C herself. After making the leaves colourless, when she put iodine, she observed that none of the leaves turned blue-black. What can be the possible source of error? 

  1. The leaves were not de-starched properly initially, when the plant was placed in the dark room.  

  2. The leaf chosen as Control could not receive proper sunlight. 

  3. The leaves were not boiled in water for a sufficient period of time to kill the cells. 

  4. There was a hole in the Test bag through which CO2 (carbon dioxide) could enter


Question 3 The experiment described in Case Study C was repeated two more times, without any error. In one of the repetitions, the plant was kept in the dark room for 24 hours and in the other one, the plant was kept in the dark room for 120 hours. The same process was followed afterwards to test if CO2 (carbon dioxide) is required to produce starch. The test results are shown in Table 1.1. 

What accounts for the starch detected in the Test leaf of the plant that was kept in the dark room for 24 hours? 

  1. Starch was already present in the Test leaf before the leaf was enclosed in the potassium hydroxide bag. 

  2. Starch was prepared by the Test leaf in Step 2 in the absence of CO2 2. (carbon dioxide).

  1. 1 only b. 2 only

  2. c. Both 1 and 2 d. An error made in the later stages of the activity.


Question 4. CO2 concentration in air is indicated in terms of parts per million or ppm. 100 ppm would mean that 100 particles of CO2 are present in 1,000,000 air particles. The graph in Fig. 1.8 shows the effect of CO2 levels on the plant growth rate. 

I. On the basis of the activity given in Case Study C and Fig. 1.8, which of the following inferences are correct? Select all that apply.

  1. CO2  is necessary for photosynthesis. 

  2. Plant growth increases with increase in atmospheric CO2 , only up to a certain level 

  3. A continuous increase in atmospheric CO2 after a certain level can have a negative effect on plant growth. \

  4. CO2  is not necessary for photosynthesis but it enhances plant growth. 

  5. Answer a. CO2 is necessary for photosynthesis.b. Plants growth increases with increase in atmospheric CO2, only up to a certain level.c. A continuous increase in atmospheric CO2after a certain level can have a negativeeffect on plant growth.

Ii. Over the past 60 years, CO2 in the Earth’s atmosphere has increased by nearly 100 ppm. It stands at 420 ppm at present. How do you think it might have affected plant life on the planet?

Answer: The rate of photosynthesis has increased for plants on Earth


1. Photosynthesis is an important process for life on earth because

(a) it is the primary source of all food on earth

(b) it is responsible for the release of oxygen

(c) it is the only natural process responsible for the utilisation of sunlight

(d) All of the above


2. The main purpose of photosynthesis is to

(a) consume CO2

(b) produce ATP

(c) convert light energy into Chemical energy

(d) produce starch


3. The process of digestion in our digestive system involves

(a) conversation of complex substances into simple form

(b) absorption of monomers by the body

(c) conversion of monomers into polymers

(d) absorption of water and food


4. Digestion in the digestive system of humans is accomplished by

(a) mechanical and Chemical processes

(b) chemical processes only

(c) mechanical processes only

(d) none of the above


5. A baby boy aged two years is admitted to play school and passed through a dental checkup . The dentist observed that the boy had twenty teeth. Which teeth were absent.

(a) Incisors

(b) Canines

(c) Premolars

(d) Molars


6. The primary dentition in human differs from permanent dentition not having one of the following type of teeth.

(a) Canine

(b) Premolars

(c) Molar

(d) incisor



7. Write the dental formula of human being.
Ans

8. How many deciduous teeth are present in humans?
(a) 22
(b) 24
(c) 20
(d) 18

9.The hard chewing surface of teeth is made up of :
(a) Dentine
(b) enamel
(c) teeth
(d) bone

10. During intake of food , the entry of food into the glottis (opening of windpipe) is prevented by
(a) Glottis itself
(b) air present in windpipe
(c) annular rings of pharynx
(d) epiglottis

11. Length of oesophagus is:
(a) 25 cm
(b) 55 cm
(c) 33 cm
(d) 45 cm

12. Salivary glands pour their secretion into
(a) Stomach
(b) blood
(c) buccal cavity
(d) intestine

13. Which is the largest gland of human body?
(a) Liver
(b) Gastric gland
(c) Pancreas
(d) intestine

14. Bile juice is stored in which organ of human body?
(a) Gall bladder
(b) Liver
(c) kidney
(d) Pancreas

15. Which food component gets 30% hydrolysed in mouth?
(a) Starch
(b) Protein
(c) Fats
(d) Nucleic acids

16. The secretion released into the small intestine are :
(a) bile and pancreatic juice
(b) succus entericus only
(c) pancreatic juice, bile and intestinal juice
(d) pancreatic juice and intestinal juice

17.

CLASS 7TH TEST

 NUTRITION IN PLANTS 


https://forms.gle/WB5cjhhZCor9G8QF8

Saturday, February 8, 2025

VI A

 Annual syllabus 2025

1. Body movements

2. Living Organisms and their surroundings 

3. Light, shadows and reflection

4. Electricity and circuits

5. Fun with magnets

6. Separation of substances

7. 



Sunday, December 29, 2024

The Living Organisms and their Surroundings class 6th

 Assignment 

Lesson :The Living Organisms and their Surroundings 


Why is reproduction important for organisms? 

Fill in the blanks: 

(a) Saline water, hot air and sand are ................ components of a habitat. 

(b) The habitat of plants and animals that live in ................ is called the aquatic habitat.

(c) ................ enable a plant or an animal to live in its surroundings. 

(d) Plants and animals that live on land are said to live in ................ habitats

Paheli has a rose plant in her garden. How can she increase the number of rose plants in the garden? 

Why do desert snakes burrow deep into the sand during the day? 

Write the adaptation in aquatic plants due to which 

(a) submerged leaves can bend in the flowing water. 

(b) leaves can float on the surface of water. . 

Mention one adaptation present in the following animals: 

(a) In camels to keep their bodies away from the heat of sand. 

(b) In frogs to enable them to swim. 

(c) In dolphins and whales to breathe in air when they swim near the surface of water.

Some desert plants have very small leaves whereas some others have only spines. How does this benefit the plants? 

What are the specific features present in a deer that helps it to detect the presence of predators like lion? 

Read the features of plants given below: (a) Thick waxy stem (b) Short roots

(c) Cone shaped plants (d) Sloping branches (e) Small or spine-like leaves (f) Hollow stem 

Choose the type of plant for every feature given in a, b, c, d, e and f from the list given below: Aquatic plant, Desert plant, Mountainous plant

What are the adaptive features of a lion that helps it in hunting?

Answers 

1.Reproduction is very important for living organisms as it helps them to produce young ones of their own kind and maintains the continuity of life.

2.(a) abiotic ( non-living)

(b) water

(c) Adaptations

(d) terrestrial

3.Paheli can increase the number of rose plants in the garden by planting stem cuttings of rose plant. It is a common method of plant propagation in which 20-30 cm long pieces of one year old stems are cut and their lower ends are planted in the soil.

4. As the deeper layers of sand are cooler, so desert snakes burrow deep into the sand to stay away from heat of the desert during day time.

5.(a) Submerged leaves are long narrow and ribbon – like which can easily bend in flowing water and withstand water currents without getting damaged, e.g., tape grass.

(b) Stems/stalks of floating leaves are long, hollow and light which help the leaves to float over water surface.

6.(a) Long legs of camel help to keep the upper part of the body away from the heat of the sand on the ground.

(b) Webbed feet enable frog to swim in water.

(c) Blow holes present in dolphins and whales help them to breathe in air when they swim near the surface of water

7.Small leaves and spines are modifications of leaves, which are adaptations of plants to live in desert habitat. In these modifications the surface of leaf lamina is reduced thereby reducing water loss by transpiration.

8.Specific features present in deer that help them to detect the presence of predators like lion are:

(i) Long ears to hear and recognise movement of predators.

(ii) Eyes on the sides of its head which allow it to look in all directions.

9.(a) Desert plant

(b) Aquatic plant

(c) Mountainous plant

(d) Mountainous plant

(e) Desert plant

(f) Aquatic plant

10. Adaptive features of a lion that help it in hunting are as follows :

Brown body colour helps it to hide in dry grassland for avoiding detection by its prey.

 Eyes placed in front allow it to know the exact location and movements of its prey.

Strong paws, sharp teeth and long claws help it to catch and kill the prey.


Wednesday, December 11, 2024

Structure of atom Lesson 4 class 9th Chemistry

 Chapter 4 class 9th notes 

Introduction : 

  • Till now we have learnt that matter is made up of atom and molecules

  • Atom of same element are identical 

  • Atom of different elements are different .

  • But the ques arises in what sense these are different?

  • THis was a major challenge to scientists to give answer .

  • This mystery was solved towards the end of 19th century by scientists like J.J Thomson , Goldstein, Rutherford, Chadwick , Bohr and many more. 

  • With their researches , it was established that atom is not as simple as was predicted by Dalton . 

  • It is made up of a large number of particles out of which the most important are Electrons , protons and Neutrons 


Charged particles  in matter 

The electric nature of matter was indicated quite early in the form of fictional electricity. It was produced by rubbing two articles together such as glass rod with silk or fur . As a result of rubbing , they got electrically charged . As matter is constituted by atoms , this suggested the presence of charge in an atom. 


Discovery of electrons (e-) and cathode rays : 

First identified by JJ Thomson.

He performed this experiment in discharge tube . It is a long cylindrical tube sealed at both the ends and is fitted with two electrodes. Two electrodes are connected to a source of high voltage called a battery.

The electrodes connected to the negative terminal of the battery is called cathode .

The electrodes connected to the positive terminal of the battery is called anode .

The tube is connected to a vacuum pump so as to reduce the pressure inside the tube to an extent.

Fluorescent material is used on opposite side of cathode .

Observations :

(i) A current of very high voltage was passed through discharge tube under normal pressure of one atmosphere. 

No current passed as air is a bad conductor of electricity.

(ii) The pressure inside the discharge tube was slowly reduced to about 1 mm by working the vacuum pump and then current is passed .

Electricity start flowing and and light was emitted inside the tube. 

(iii) The emossion of light stopped when the pressure inside the tube was further reduced to about 0.01 mm of mercury. 

The wall of tube opposite to cathode glow with greenish light called fluorescence. 

Conclusion : 

Florescence is due to bombardment of the walls of tube by the rays emitted from the cathode . these were known as cathode rays .


Thomson repeat the same experiment by replacing  gases . He got same results.


Since these rays of particles moves towardspositivecharges electrode , so it itself carry negative charge. These negatrive charged particles are called electrons




Discovery of protons(p+) and anode rays : 

First identified by E. Goldstein.

He performed this experiment in modified discharge tube.It is a long cylindrical tube sealed at both the ends and is fitted with two electrodes. Two electrodes are connected to a source of high voltage called a battery.

The electrodes connected to the negative terminal of the battery is called cathode .

He used perforated electrode as cathode and placed it in centre and not at end. 

The electrodes connected to the positive terminal of the battery is called anode .

The tube is connected to a vacuum pump so as to reduce the pressure inside the tube to an extent.

Fluorescent material is used on opposite side of anode 

He used hydrogen gas first for his experiment.


Observation : 

Red colour glow on the side opposite to anode .

Conclusion : 

Florescence is due to bombardment of the walls of tube by the rays emitted from the anode. these were known as anode rays .

These rays are also called canal rays as these passed through canal of cathode. 


Thomson repeat the same experiment by replacing  gases . He got different results.


Since these rays of particles moves towards negative charges electrode , so it itself carry positive charge. These positive charged particles are called protons


Discovery of Neutrons (n) : 

Rutherford found that except for hydrogen atom , the atomic mass of no other element could be explained by electrons and protons only. . So Rutherford predicted the presence of some neutral particles present in the nucleus of an atom.


These neutral particles were discovered by Chadwick in 1931 , when bombarded a thin foil of beryllium metal with the fast moving alpha particles .

These particles have mass similar to proton but carry no charge.  




How does the atom look like?

Thomson` Model of an atom : 

According to him 

(i)An atom consists of a positively charged sphere and the electrons are embedded in it. 

(ii) The negative and positive charges are equal in magnitude. 

So, the atom as a whole is electrically neutral

Limitations: 

  • He could not offer any experimental support to his atomic model.

  • He could not explain the results of the experiments carried by Rutherfod and other scientists at a later stage.

Rutherford model of atom: 

 In this experiment, fast moving alpha (α)-particles were made to fall on a thin gold foil. 

• He selected a gold foil of about 1000 atoms thick. 

• the fast-moving α-particles was bombarded on this gold foil. 

Observations 

(i) Most of the fast moving α-particles passed straight through the gold foil. 

(ii) Some of the α-particles were deflected by the foil by small angles. 

(iii) Surprisingly one out of every 12000 particles appeared to rebound

Conclusion: 

 (i) Most of the space inside the atom is empty because most of the α-particles passed through the gold foil without getting deflected. 

(ii) Very few particles were deflected from their path, indicating that the positive charge of the atom occupies very little space. 

(iii) A very small fraction of α-particles were deflected by 1800,indicating that all the positive charge and mass of the gold atom were concentrated in a very small volume within the atom


Rutherford nuclear model of an atom(features):

(i) There is a positively charged centre in an atom called the nucleus. Nearly all the mass of an atom resides in the nucleus. 

(ii) The electrons revolve around the nucleus in circular paths. 

(iii) The size of the nucleus is very small as compared to the size of the atom


Drawbacks of Rutherford’s model of the atom:

Maxwell Questioned that:  Any particle in a circular orbit would undergo acceleration. During acceleration, charged particles would radiate energy. Thus, the revolving electron would lose energy and finally fall into the nucleus. If this were so, the atom should be highly unstable and hence matter would not exist in the form that we know.

So he could not explain the stability of atom .


BOHR’S MODEL OF ATOM:

(i)Only certain special orbits known as discrete orbits of electrons, are allowed inside the atom. 

(ii) While revolving in discrete orbits the electrons do not radiate energy

These orbits or shells are called energy levels. Energy levels in an atom are shown.

These orbits or shells are represented by the letters K,L,M,N,… or the numbers, n=1,2,3,4,….

How are Electrons Distributed in Different Orbits (Shells)? in Different Orbits (Shells)?

The distribution of electrons into different orbits of an atom was suggested by Bohr and Bury. The following rules are followed for writing the number of electrons in different energy levels or shells: 

(i) The maximum number of electrons present in a shell is given by the formula 2n2, where ‘n’ is the orbit number or energy level Hence the maximum number of electrons in different shells are as follows: 

first orbit or K-shell will be = 2 × 12 = 2, 

second orbit or L-shell will be = 2 × 22 = 8,

third orbit or M-shell will be = 2 × 32 = 18, 

fourth orbit or N-shell will be = 2 × 42 = 32, and so on. 

(ii) The maximum number of electrons that can be accommodated in the outermost orbit is 8. (iii) Electrons are not accommodated in a given shell, unless the inner shells are filled. That is, the shells are filled in a step-wise manner.


Atomic structure of the first eighteen elements is shown schematically: 

Valency : The combining capacity of an element. It can be 1,2,3 and 4 .

If outermost shell (valence shell) contain 1,2,3 or 4 electrons then the valency is same .

If outermost shell (valence shell) contain 5,6,7 or 8 electrons then the valency will be 8 minus the no. of valence electrons.

ATOMIC NUMBER(Z):

The atomic number is defined as the total number of protons present in the nucleus of an atom

Or 

The number of electrons present in neutral atom. 

Atomic Number = no of protons=no of electrons in neutral atom.


 Mass number(A)::

The sum of the total number of protons and neutrons present in the nucleus of an atom.

Notation for an atom

Isotopes :the atoms of the same element, having the same atomic number but different mass numbers.

Eg.hydrogen protium deuterium, and tritium

Eg. (i) carbon, (ii) chlorine, and , etc


Applications 

(i) An isotope of uranium is used as a fuel in nuclear reactors. 

(ii) An isotope of cobalt is used in the treatment of cancer. 

(iii) An isotope of iodine is used in the treatment of goitre.


The average atomic mass of chlorine atom,

ISOBARS: Atoms of different elements with different atomic numbers, which have the same mass number, are known as isobars.

Eg :calcium atomic no 20 and mass no 40 

Argon atomic no 18 and mass no 40