PLEASE SOLVE THESE QUESTIONS AFTER YOU HAVE REVISED THE CHAPTER.
Sense organ answers
Q1 Solution:
(i) Hypermetropia.
(ii) Macula / fovca / yellow spot.
(iii) Eustachian tube.
(iv) Auditory nerve.
(v) Vitreous humour.
(vi) Ciliary muscles.
(vii) Pupil.
(viii) Rhodopsin.
(ix) Iris.
(x) Stapes.
(xi) Accomodation.
(xii) Cones.
Q2 Solution:
(i) False. Rods are photoreceptor cells that are sensitive to
dim light.
(ii) False. Deafness is caused due to rupturing of tympanum.
(iii) False. Rods are receptor cells in retina of eye sensitive
to dim light.
(iv) False. Myopia defect of eye is caused due to eyeball being
elongated.
Q3 Solution:
(i) Auditory canal, tympanum, ear ossicles, oval window,
cochlea.
(ii) Conjunctiva, cornea, aqueous humour, pupil, lens, vitreous
humour, yellow spot.
Q4 auditary nerve
Q5 Solution:
(i) Parts of inner ear. Malleus.
(ii) Disease of eye. Xerophthalmia.
Q6 Solution:
Choroids forms nutritive layer of eye. Contains iris. It
prevents reflection of light inside the eye.
Q7 Solution:
(i) Side of head, collects sound waves.
(ii) Middle ear, increases the magnitude of vibrations.
(iii) Axis of eye in retina, more clear and distinct image is
formed.
Q8 hearing
Q9 Solution:
(i) It serves to adjust the pressure inside the middle ear to
the change in the atmospheric pressure for normal hearing so that the tympanum
is not affected by high pressure.
(ii) The radial and circular muscles of the iris contract or
relax and hence the size of the pupil is adjusted.
(iii) It transmits and increase the magnitude of vibrations to
inner ear.
Q10 eye
Q11. Solution:
(i) Myopia : – It is also called short
sightedness.
Symptom – Near vision is clear and distant
vision is blurred.
Cause – Eye ball becomes too longer i.e. eye
lens is too convex and hence focal length is shortened. The image of distant
object is formed in front of the retina.
(ii) Tympanum : – Location – It is located
between external and middle ear.
Function – Sound waves from outside cause
tympanum to vibrate which are ultimately transmitted across the middle ear and
produce sound.
(iii) Yellow spot : – Location – At the back
of the eye almost at the centre of the horizontal axis of the eye ball.
Importance –Place the best focus of the rays
of light that enters and a very clear image is seen in the normal eye.
Q12 Solution:
(i)
Choriod
|
Sclerotic
|
Absorb light and prevent total
internal reflection that may cause blurring of image.
|
Provides and maintains shape of
eye ball.
|
(ii)
Dynamic
balance
|
Static
balance
|
The sensory cells in cristae of
internal ear are concerned with balance while body is in motion.
|
The sensory cells in ear which are
located in Maculae of vestibule are responsible for static balance in human
beings by detecting change in the head or body at rest. Concerned with positional
balance with respect to gravity.
|
(iii)
Rod
cells
|
Cone
cells
|
Rhodopsin
|
Iodopsin
|
(iv)
Myopia
|
Hypermetropia
|
Lens
too convex
|
Flattened
condition of lens
|
Q13 Solution:
Power of accommodation is the ability of eye to see objects
equally clear from various distance. Muscle – Ciliary muscles.
Q14 Solution:
(i) Present on the lateral aspects
of eye. They secrete tears which contains lysozymes which kills bacteria.
(ii)
Yellow
spot
|
Blind
spot
|
1.
Area of best vision.
|
1.
Area of no vision.
|
2.
Contains maximum cone cells.
|
2.
Contains no rod and cones.
|
(iii) Presbyopia.
(iv) Real, inverted and diminished.
(v) Rod and cone cells.
Q16 Solution:
(i) 1 – Stapes, 2 – Semicircular canal, 3 – Auditory nerve, 4 –
Cochlea,
5 – Eustachian tube, 6 – Ear drum or Tympanum, 7 – Malleus, 8 –
Pinna.
(ii) Function of Part 2 – It brings about the dynamic
equilibrium which help to maintain body position in response to movement.
Function of Part – 7 – The hammer or Malleus is fastened to the
ear drum whose vibrations are transmitted through these three bones (malleus,
incus and stapes) to the membrane across the oval window. These bones act as a
lever and increases the magnitude or power of vibrations.
(iii) As the pin or any sharp object may injure the ear drum.
Hence it is harmful to use pin or any sharp object and must be avoided.
(iv) Ear vax has insect repellent properties and prevent their
entry in the auditory canal. It also check the entry of dust particles in the
ear. Thus it lubricates and protects the ear drum.
Q17. Solution:
I. 1 – Cochlea, 2 – Auditory nerve, 3 – Semicircular canals, 4
– Ear ossicles,
5 – Auditory canal, 6 – Pinna, 7 – Ear drum, 8 – Oval window or
fenestra ovalis,
9 – Eustachian tube, 10 – Malleus or hammer.
II. (i) Ear drum.
(ii) Cochlea.
(iii) Semicircular canal.
(iv) Auditory nerve.
(v) Eustachian tube.
Q18. Solution:
(i) Ossicles.
(ii) (A) Cochlea,
(B) Semi-circular canals,
(C) Vestibular apparatus.
(iii) (A) Hearing,
(B) Dynamic balancing.
(iv) Organ of Corti.
Biology - Human
Population QUESTIONS COMPILED BY SOUVIC JATI
Question .1. Name the following :
(i) The surgical method of sterilization in the human
female.
(ii) A statistical study of human population of a
region .
(iii) The number of persons per square kilometer at any
one given time.
Question .2. Define the following :
(i) Birth rate.
(ii) Death rate.
Question .3. Rewrite the wrong statement by inserting a word/words.
Death rate is the number of deaths of a given population per year.
Question .4. Mention two reasons for increase in population in India.
Question .5. Answer the following :
(i) What are the age restriction for marriage by law
for boys and girls in India ?
(ii) Name two surgical techniques that can be used to
prevent pregnancy.
(iii) Give two advantage of small family.
Question .6. The need to limit the size of the family is more vital today than ever before. Give two reasons.
Question .7. Why resources cannot keep pace with the rising population ? Give two examples.
Populations answers
compiled by souvic jati
Q1Solution:
(i) Tubectomy.
(ii) Demography.
(iii) Population density.
Q2 Solution:
(i) Number of births per thousand people of
a given population per year in a particular area is called the birth rate.
(ii) The number of deaths per thousand of
people of a given population per year in a particular area is called the death
rate.
Q3 Solution:
Death rate is the number of deaths per 1000
people of a given population per year.
Q4 Solution:
(i) Illiteracy.
(ii) Poverty.
Q5 Solution:
(i) For boys – 21 years. For girls – 18
years.
(ii) For males – Vasectomy. For females –
Tubectomy.
(iii) 1. Parents can provide better
education to their children.
2. There is chances of good health of the mother and child.
2. There is chances of good health of the mother and child.
Q6 Solution:
(i) Due to population growth, the country
has to spend more on food, clothing and shelter (housing).
(ii) More population produces more wastes
causing pollution and ecological imbalance.
Q7 Solution:
(i) Population grows by geometric
progression but food production increases by arithmetic progression.
(ii) Indiscriminate cutting of trees for
agriculture is one of the serious consequences of overpopulation.
QUESTIONS Photosynthesis
COMPILED BY SOUVIC SIR
Question .1. Name the following :
(i) The structure where photosynthesis takes place.
(ii) The part of the chloroplast where the dark reaction of
photosynthesis takes place.
(iii) Plants that prepare their own food from basic raw
materials.
(iv) The energy currency of the cell.
(v) The process by which green plants prepare their own food.
(vi) The product of photosynthesis.
(vii) A plant that does not perform photosynthesis.
(viii) A gas released by green plants during photosynthesis.
(ix) The raw materials needed for photosynthesis.
(x) The site of light reaction.
(xi) The substance from which oxygen is evolved during
photosynthesis.
Question .2. State true or false :
(i) Photosynthesis occurs in all the cells of a plant.
(ii) Stomata is stimulated by light.
(iii) Grana helps in diffusion of gases.
(iv) Photosynthesis results in the loss of dry weight of the
plant.
(v) The unit of light absorbed by the chlorophyll during photosynthesis
is the proton.
(vi) Photosynthesis stops to occur at a temperature above 35 ºC.
(vii) Plants can be destarched by keeping in dark for 48 hours.
Question .3. Choose the correct alternative :
(i) A plant is kept in a dark cupboard for about 48 hours before
conducting any experiment on photosynthesis to (remove chlorophyll from leaves,
remove starch from the plant , ensure that no photosynthesis occurs, ensure
that leaves are free from starch).
(ii) The site of light reaction in the cells of a leaf is
(grana, cytoplasm, stroma).
(iii) The chemical substance used to test the presence of starch
in the cell of a leaf is (cobalt chloride paper, iodine solution, Benedict’s
solution).
(iv) The specific function of light energy in the process of
photosynthesis is to (reduce carbon dioxide, synthesise glucose, activate
chlorophyll, split water).
(v) If the rate of respiration becomes more than the rate of
photosynthesis plants will (continue to live but will not be able to store
food, be killed instantly, grow more vigorously because more energy will be
available, stop growing and gradually die of starvation).
(vi) Which one of these reaction occurs during photosynthesis ?
(carbon dioxide is reduced and water is oxidized, water is reduced and carbon
dioxide is oxidized, carbon dioxide and water both are oxidized, carbon dioxide
and water both are reduced).
Question .4. Give exact location and function of :
(i) Stoma,
(ii) Thylakoids,
(iii) Guard cell,
(iii) Grana.
Question .5. (i) Explain the following terms : Photolysis.
(ii) What is meant by photolysis of water ?
Question .6. Answer the following :
Question .6. Answer the following :
(i) Name the two phases of photosynthesis.
(ii) What is the role of light in this process ?
(iii) Describe any three adaptation in a green leaf for photosynthesis.
(iv) Describe an experiment to show that light is necessary for
photosynthesis.
Question .7. Define :
(i) Photophosphorylation.
Question .8. Re-write the correct form of statement by inserting a suitable word/words at right place. Do not delete any word in the statement :
Question .8. Re-write the correct form of statement by inserting a suitable word/words at right place. Do not delete any word in the statement :
(i) Destarching a plant means removing the starch from the
plant.
(ii) The splitting of water molecules into hydrogen and hydroxyl
ions is termed photolysis.
Question .9. Re-write the terms in correct order in a logical
sequence :
(i) Water molecules, oxygen, grana, hydrogen and hydroxyl ions,
photons.
Question .10. Complete the following process to show how the oxygen in the air reaches a mesophyl cell of the leaf.
Oxygen in air → −− → −− → Mesophyl cell.
Question .11. Fill in the blanks : Chloroplasts and _______.
Question .12. Distinguish between the following :
(i) Light reaction and Dark reaction.
(ii) Autotrophs and Heterotrophs.
(iii) Stroma and Grana.
(iv) Photosynthesis and Respiration.
(v) Chloroplast and chlorophyll.
Question .13. Enumerate the steps involved in testing a green
leaf for the presence of starch.
Question .14. Write the full form of NADP and ATP.
Question .15. A healthy Croton plant bearing variegated leaves was kept in a dark cupboard to destarch it, after which it was placed in sunlight for few hours. One of the leaves was then plucked and an outline of the leaf marking the green and non-green regions was drawn. The leaf was then tested for starch. Using the above information, answer the following questions:
(i) State the aim of the above experiment.
(ii) Name the chemical used for testing the presence of starch.
(iii) Why is the leaf boiled in water and alcohol before testing
for the presence of starch ?
(iv) What change is seen on the leaf after the starch test ?
(v) Give the chemical equation to represent the process of
syarch formation in plants.
Question .16. A candidate in order to study the importance of certain factors in photosynthesis, took a potted plant and kept in the dark for over 24 hours. Then in the early hours of the morning she covered one of the leaves with black paper in the centre only. She placed the potted plant in the sunlight for a few hours and then tested the leaf which was covered with black paper for starch.
(i) What aspect of photosynthesis was being investigated ?
(ii) Is there any control in this experiment ? If so state the
same.
(iii) Why was the plant kept in the dark before the experiment ?
(iv) Describe step by step how the candidate proceeded to test
the leaf for the presence of starch.
Question .17. Give biological reasons :
Question .17. Give biological reasons :
(i) A tiger owes its existence to chlorophyll.
(ii) Oxygen given out during photosynthesis comes from water.
Question .18. (i) Fill in the blank in the following equations for two chemical reactions :
Question .18. (i) Fill in the blank in the following equations for two chemical reactions :
1. ______ + 12H2O → C H O + ________ + ______.
2. _______ + 6O2 → 6 ______ +
6 _______ + Energy.
(ii) Name the two process represented by the above two
reactions.
(iii) Which one of these two reactions is the support of all
life on earth.
Q.19. Draw a neat labeled diagram of the experimental set up to
show that green plants give out oxygen during photosynthesis.
SOLUTION TO
PHOTOSYNTHESIS QUESTIONS.
1. Solution:
(i) Thylakoid/grana.
(ii) Stroma.
(iii) Autotrophs.
(iv) ATP.
(v) Photosynthesis.
(vi) Glucose, water and oxygen.
(vii) Mushroom/Fungus.
(viii) Oxygen.
(ix) Carbon dioxide and water.
(x) Grana of chloroplast.
(xi) Water.
2. Solution:
(i) F
(ii) T
(iii) F
(iv) F
(v) F
(vi) T
(vii) T.
3. Solution:
(i) Ensure that leaves are free from starch.
(ii) Grana.
(iii) Iodine solution.
(iv) Activate chlorophyll.
(v) Stop growing and gradually die of starvation.
(vi) Carbon dioxide is reduced and water is oxidized.
4. Solution:
(i) Stoma : Location – Epidermis of leaves.
Function – Exchange of gases for photosynthesis/respiration/loss of water as
water vapour during transpiration.
(ii) Thylakoids : Location – Granum of chloroplast.
Function – Help in traping sun-light (photosynthesis).
(iii) Guard cell : Location – Around stomata
mostly under the surface of a dorsiventral leaf. Function – Allow diffusion of
gases.
(iv) Grana : Location – Inside chloroplast.
Function – Help in trapping sunlight.
5. Solution:
(i) Splitting of water by light in photosynthesis.
(ii) Splitting of water into hydrogen and hydroxyl ion in presence
of sunlight.
6. Solution:
(i) 1. Light reaction (Photolysis of water).
2. Dark reaction (CO2reduction).
(ii) Light has energy giving photons which activate and ionize
chlorophyll to release electrons. The electrons (split) ionize water molecules.
This process is called photolysis.
The H+ reduces CO2 to
form crbohydrates.
(iii) 1. Large surface area : The surface area of the leaves is
very large due to which maximum light energy is absorbed from the
sunlight.
2. Leaf arrangement : The arrangement of leaves on the stem are in
such a way that maximum surface area is exposed to sunlight.
3. Chloroplasts : Chloroplasts containing chlorophyll are
concentrated in the upper surface of the leaf which helps in trapping maximum
radiant energy from the sunlight quickly.
(iv) Aim : To demonstrate that light is
necessary for photosynthesis.
Requirements : A destarched potted plant, Gangon’s light screen.
Procedure :
1. Gangon’s light screen is fixed to one of the leaves of the
potted plant (destchared) as shown in figure below.
2. The plant is then fixed in light at least for 6 hours.
3. The starch test is performed.
Fig. Ans. 20 (iv)/page – 38 [evergreen]
Observation : We observe that the portion of screen that allows
the light to fall on leaf gives positive starch test and the portion covered
with light screen gives negative starch test.
Inference : Light is
necessary for photosynthesis.
7. Solution:
(i) Conversion of low grade energy ADP into high potential
chemical energy ATP is called photophosphorylation.
8. Solution:
(i) Destarching the plant means removing the starch from the
leaves of plant.
(ii) The splitting of water molecules into hydrogen and hydroxyl
ions by light is termed photolysis.
9. Solution:
(i) Photons, grana, water molecules, hydrogen and hydroxyl ions,
oxygen.
10. Solution:
Oxygen in air → stoma → substomatal space → Mesophyl cell.
11. Solution:
Photosynthesis.
12. Solution:
(i)
Light Reaction
|
Dark Reaction
|
1. Presence of light
is must.
|
1. It does not require
light.
|
2. It occurs in the
grana of chloroplast.
|
2. It occurs in the
stroma of chloroplast.
|
3. Its product are ATP
and NADPH2.
|
3. Its product are
organic compounds.
|
(ii)
Autotrophs
|
Heterotrophs
|
1. They can produce
their own food.
|
1. They can prepare
their own food.
|
2. They contain
chlorophyll.
|
2. They do not contain
chlorophyll.
|
(iii)
Stroma of Chloroplast
|
Grana of Chloroplast
|
1. Here dark reaction
takes place.
|
1. Here light reaction
takes place.
|
2. It is a non-green
granular matrix.
|
2. It is a green
flattened sac-like structure.
|
(iv)
Photosynthesis
|
Respiration
|
1. It is an anabolic
process.
|
1. It is a catabolic
process.
|
2. It takes place only
in chlorophyll bearing cells.
|
2. It takes place in
all cells.
|
3. Oxygen and glucose
released as end-products.
|
3. Carbon dioxide and
water released as end-product.
|
4. It takes place only
in presence of light.
|
4. Light is not
required.
|
(v)
Chloroplast
|
Chlorophyll
|
1. It is an organelle
of the cell.
|
1. It is the green
pigment in chloroplast.
|
2. It is living.
|
2. It is non-living.
|
13. Solution:
Step I – The leaf
is dipped in boiling water for a minute to kill the cells.
Step II – The leaf
is boiled in methylated spirit to remove chlorophyll.
Step III – The leaf
is put in normal water to make it soft.
Step IV – Iodine
solution is put which turns starch blue black.
14. Solution:
NADP – Nicotinamide
adenine dinucleotide phosphate.
ATP – Adenosine
tri phosphate.
15. Solution:
(i) To prove that chlorophyll is necessary for photosynthesis.
(ii) Iodine solution.
(iii) Cells are killed by boiling and chlorophyll is dissolved in
alcohol.
(iv) The green parts turn blue black and non-green parts turn
brown.
(v)
16. Solution:
(i) To show that sunlight is necessary for photosynthesis.
(ii) Yes, the uncovered portion of the experimental leaf is the
control.
(iii) To destarch the leaf.
(iv) 1. The leaf is boiled in alcohol to remove chlorophyll.
2. The leaf is washed in water to make it soft.
3. The iodine solution is added to the leaf.
The uncovered portion shows blue-black colour and the portion
covered shows brown colour. This indicate that sunlight is necessary for
photosynthesis.
17. Solution:
(i) Green plants synthesizes food by photosynthesis using
chlorophyll. Tiger feed on animal which feed on green plants. Thus they obtain
their food indirectly from green plants. That is existence of tiger depends
upon chlorophyll.
(ii) Due to photolysis of water oxygen is released during
photosynthesis.
18. Solution:
(i)
(ii) Photosynthesis and respiration.
(iii) Photosynthesis.
19.
Sir , can you post the answers for Absorption By Roots?
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yess pls post answers of absorption by roots
DeleteSir, can you post circulatory system, excretory system, reproductive System
ReplyDeletesir can you post ans for absorption by roots chapter pleaseeeeeeeeee
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