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Published on: 12/05/2020
10th Standard Science English Medium Public Exam Model Question Paper June 2020
Download Tamil Nadu 10th Standard Science question papers, model tests, one-mark questions, important questions, and public exam papers in PDF format. Free study materials and answer keys for TN State Board students.
Questions + Answers key
Take MCQ Science Test1.
Draw the ultrastructure of a chloroplast and label the parts.
2.
Why copper were is used as connecting wires in the circuit?
3.
What is Sprite?
4.
How does insulin deficiency occur?
5.
6.
How is diastema formed in rabbit?
7.
What is the minimum distance needed for an echo?
8.
What is Molar volume of a gas?
9.
Classify the types of force based on their application.
10.
'Our teeth and an elephant's tusks are homologous organs'. Justify this statement. What do the analogous organs indicate?
11.
With a neat labelled diagram explain the techniques involved in gene cloning.
12.
Arrive at, systematically, the IUPAC name of the compound: CH3–CH2–CH2–OH
13.
State the universal law of gravitation and derive its mathematical expression.
14.
Sneezing, yawning etc are examples of _______
voluntary actions
involuntary actions
reflex actions
planned actions
15.
SI unit of heat is
calorie
joule
kilo calorie
kelvin
16.
In molecular biology, radioisotope are used in ________ surgical instruments.
engraving
sterilizing
sharpening
preserving
17.
The number of elements present in sixth period of modern periodic table is __________
8
18
16
32
18.
The momentum of a heavy object at rest will be
large
infinity
zero
small
19.
Which is used to edit programs?
Inkscape
script editor
stage
sprite
20.
The ‘use and disuse theory’ was proposed by __________.
Charles Darwin
Ernst Haeckel
Jean Baptiste Lamarck
Gregor Mendel
21.
Pharyngeal ganglion in leech is a part of
Excretory system
Nervous system
Reproductive system
Respiratory system
22.
A solution is a __________ mixture.
homogeneous
heterogeneous
homogeneous and heterogeneous
non homogeneous
23.
The volume occupied by 4.4 g of CO2 at S.T.P
22.4 liter
2.24 liter
0.24 liter
0.1 liter
24.
A convex lens forms a real, diminished point sized image at focus. Then the position of the object is at
focus
infinity
at 2f
between f and 2f
25.
Write a note on Obesity.
26.
Explain the uses of radio isotopes in medicine field?
27.
Explain the working of the eye.
28.
How are chromosomes classified based on the position of centromere?
29.
Program for print the word “Hello” with sound
30.
Name three improved characteristics of wheat that helped India to achieve high productivity.
31.
Why is the circulation in man referred to as double circulation?
32.
Why is Archaeopteryx considered to be a connecting link?
33.
Find the final temperature of a copper rod. Whose area of cross section changes from 10 m2 to 11 m2 due to heating. The copper rod is initially kept at 90 K. (Coefficient of superficial expansion is 0.0021 /K)
34.
A plant hormone was first discovered in Japan when rice plants were suffering from Bakanae disease caused by Gibberella fujikoroi. Based on this information answer the following questions:
a) Identify the hormone involved in this process.
b) Which property of this hormone causes the disease?
c) Give two functions of this hormone.
35.
'A' is a blue coloured crystalline salt. On heating it loses blue colour and to give 'B'. When water is added, 'B' gives back to 'A'. Identify A and B, write the equation.
36.
A solid compound ‘A’ decomposes on heating into ‘B’ and a gas ‘C’. On passing the gas ‘C’ through water, it becomes acidic. Identify A, B and C.
37.
a) Identify the bond between H and F in HF molecule.
b) What property forms the basis of identification?
c) How does the property vary in periods and in groups?
1.
2.
Copper wire is used as connecting wires because copper has very low resistivity.
3.
The characters on the background of a Scratch window are known as Sprite. Usually a cat appears as a sprite when the Scratch window is opened. The software provides facilities to make alternations in sprite.
4.
Insulin deficiency occurs due to destruction of \(\beta \)-cells of the pancreas.
5.
6.
Diastema is a gap between incisors and premolar formed due to absence of canine.
7.
The minimum distance required to hear an echo is 1/20th part of the magnitude of the velocity of sound in air.
i.e. \(\frac{1}{20} \times 344 = 16.7m\)
8.
(i) One mole (6.023 x 1023 of entities) of any gas occupies 22.4 litre or 22400 ml at S.T.P.
(ii) This volume is called as molar volume of gas.
9.
Based on the direction in which the force acts, they can be classified into two types as:
(i) Like Parallel forces
(ii) Unlike Parallel forces.
10.
(i) Our teeth and an elephant's tusks are homologous organs as both of them have the same basic structure, but different functions (we chew with our teeth, whereas elephants use their tusks to hold things).
(ii) Analogous organs indicate that even organisms having different structures can adapt by modifying these structures to perform similar functions for their survival under the given environmental conditions.
11.
Gene Cloning:
(i) The carbon copy of an individual is often called a clone. However, more appropriately, a clone means to make a genetically exact copy of an organism.
Techniques involved in gene cloning:
(i) In gene cloning, a gene or a piece of DNA fragment is inserted into a bacterial cell where DNA will be multiplied (copied) as the cell divides.

A brief outline of the basic steps involved in gene cloning are:
(i) Isolation of desired DNA fragment by using restriction enzyme
(ii) Insertion of the DNA fragment into a suitable vector (Plasmid) to make rDNA
(iii) Transfer of rDNA into bacterial host cell (Transformation)
(iv) Selection and multiplication of recombinant host cell to get a clone
(v) Expression of cloned gene in host cell.
(vi) Using this strategy several enzymes, hormones and vaccines can be produced.
12.
Step 1: The parent chain consists of 3 carbon atoms. The root word is 'Prop'.
Step 2: There are single bonds between the carbon atoms of the chain. So, the primary suffix is 'ane'
Step 3: Since, the compound contains - OH group, it is an alcohol. The carbon chain is numbered from the end which is closest to -OH group. (Rule 3)
Step 4: The locant number of –OH group is 1 and thus the secondary suffix is ‘1-ol’.
The name of the compound is Prop + ane + (1-ol) = Propan-1-ol
Note: Terminal ‘e’ of ‘ane’ is removed as per Rule 5
13.
Statement:
Universal law of gravitation states that, 'every particle of matter in this universe attracts every other particle with a force. This force is directly proportional to the product of their masses and inversely proportional to the square of the distance between centers of these masses. The direction of the force acts along the line joining the masses'.
Deviation: Force between the masses is always attractive and it does not depend on the medium where they are placed.

Let m1 and m2 be the masses of two bodies A and B placed at r meter apart in space
Force \(\mathrm{F} \propto \mathrm{m}_{1} \times \mathrm{m}_{2}\)
\(\mathrm{F} \propto 1 / r^{2}\)
On combining the above two expressions,
\(\mathrm{F} \propto \frac{\mathrm{m}_{1} \times \mathrm{m}_{2}}{\mathrm{r}^{2}} \)
\(F=\frac{G m_{1} m_{2}}{r^{2}}\)
Where G is the universal gravitational constant.
Its value in SI unit is \(6.674 \times 10^{-11} \mathrm{~N} \mathrm{~m}^{2} \mathrm{~kg}^{-2}\).
14.
(c)
reflex actions
15.
(b)
joule
16.
(b)
sterilizing
17.
(d)
32
18.
(c)
zero
19.
(b)
script editor
20.
(c)
Jean Baptiste Lamarck
21.
(b)
Nervous system
22.
(a)
homogeneous
23.
(b)
2.24 liter
24.
(b)
infinity
25.
(i) Obesity is the state in which there is an accumulation of excess body fat with an abnormal increase in body weight.
(ii) Obesity occurs if intake of calories is more than the expenditure of energy.
(iii) Overweight and obesity are conditions where the body weight is greater than the mean standard weight for age and height of an individual.
(iv) Body Mass Index (BMI) is an estimate of body fat and health risk. BMI = Weight (kg) / Height (m)2
Causes and risk factors :
(i) Obesity is due to genetic factors, physical inactivity, eating habits (overeating) and endocrine factors.
(ii) Obesity is a positive risk factor in development of hypertension, diabetes, gall bladder disease, coronary heart disease, and arthritis.
Prevention and Control of Obesity :
(i) Diet Management: Low calorie, normal protein, vitamins and mineral, restricted carbohydrate and fat, high fiber diet can prevent overweight.
(ii) Physical exercise: A low calorie diet accompanied by moderate exercise will be effective in causing weight loss. Meditation, yoga and physical activity can also reduce stress related to overeating.
26.
Medical applications of radioisotopes can be divided into two parts
i) Diagnosis ii) Therapy Radio.
Isotopes are used as tracers to diagnose the nature of blood circulatory disorders, defects of bone metabolism, to locate tumors, etc. Some of the radio isotopes which are used as tracers are: hydrogen, carbon, nitrogen, sulfur, etc.
(i) Radio sodium (Na24) is used for the effective functioning of heart.
(ii) Radio - Iodine (I131) is used to cure the goiter.
(iii) Radio iron is (Fe59) is used to diagnose anemia and also provide treatment for the same.
(iv) Radio phosphorous (P32) is used in the treatment of skin diseases.
(v) Radio cobalt (Co60) and radio gold (Au79) are used in the treatment of skin cancer.
(vi) Radiations are used to sterilize the surgical devices as they can kill the germs and microbes.
27.

(i) The transparent layer cornea bends the light rays through pupil located at the centre part of the Iris.
(ii) The adjusted light passes through the eye lens. Eye lens is convex in nature. So, the light rays from the objects are converged and a real and inverted image is formed on retina.
(iii) Then, retina passes the received real and inverted image to the brain through optical nerves. Finally, the brain senses it as erect image.
28.
Based on the position of centromere, the chromosomes are classified as Telocentric, Acrocentric, Submetacentric andMetacentric.
(i) Telocentric - The centromere is found on the proximal end. They are rod shaped chromosomes.
(ii) Acrocentric - The centromere is found at the one end with a short arm and long arm. They are also rod - shaped chromosomes.
(iii) Submetacentric - The centromere is found near the centre of the chromosome. Thus forming two unequal arms. They are J

(iv) Metacentric - The centromere occurs in the centre of the chromosome and form two equal arms. They are V shaped chromosomes.
29.

1. Click events in script option

3. Click Looks in script option. Drag “say” to script area.

4. Type “Hello “ word in say tab.

5.Click sounds in script option. Drag play sound to script area. Choose the hello sound from the audio file.

6. From File menu choose the Save option.
7. Click the green flag at the top right corner of the stage window to run the program


30.
Semi dwarf nature high yield, disease resistance, early maturity.
31.
(i) When the blood circulates twice through the heart in one complete cycle it is called double circulation.
(ii) In double circulation oxygenated blood does not mix with deoxygenated blood.
32.
(i) Archaeopteryx is considered to be a connecting link between reptiles and birds.
(ii) It had wings with feathers like a bird. It has long tail, clawed digits and conical teeth like a reptile.
33.
Area of copper rod, Ao = 10 m2
Changes of Area of cross section,
Initial temperature \(\Delta \mathrm{A} =11 -10 =1 \mathrm{~m}^{2} \)
\(\mathrm{~T}_{1} =90 \mathrm{~K} \)
\(a_{\mathrm{A}} =0.0021 / \mathrm{K} \)
\(\mathrm{T}_{2} =? \)
\(\frac{\Delta A}{A_{0}} =a_{\mathrm{A}} \Delta \mathrm{T} \)
\(\frac{1 }{10 } =0.0021\left[\mathrm{~T}_{2}-90\right] \)
\(0.1 =0.0021\left[\mathrm{~T}_{2}-90\right]=\frac{0.1}{0.0021}+90=\mathrm{T}_{2} \)
\(\mathrm{~T}_{2} =137.61 \mathrm{~K}\)
So the final temperature of a copper rod is 137.61 K
34.
a) Gibberellin
b) The active substance was identified as Gibberellic acid, which caused this disease.
c) i) Application of gibberellins on plants stimulate extraordinary elongation of internode. e.g. Corn and Pea.
ii) Treatment of rosette plants with gibberellin induces sudden shoot elongation followed by flowering. This is called bolting.
35.
(i) 'A' is copper sulphate pentahydrate (CuSO4.5H2O).
(ii) 'B' is Anhydrous copper sulphate (CuSO4)
36.
(i) On passing ' C ' through water it becomes acidic.
(ii) Therefore the gas ' C ' must be a non-metal oxide \(\left(\mathrm{CO}_{2}\right)\).
(iii) So a solid compound must be a calcium carbonate.
(iv) It decomposes into calcium oxide and carbon dioxide. (C)
\(\mathrm{CaCO}_{3(\mathrm{~g})} \rightarrow \mathrm{CaO}_{(\mathrm{S})}+\mathrm{CO}_{2(\mathrm{~g})} \uparrow\\ \quad \mathrm{A} \quad \quad \quad \quad \mathrm{B} \quad \quad \quad \quad \mathrm{C}\)
| A | CaCo3 | Calcium carbonate |
| B | CaO | Calcium oxide |
| C | CO2 | Carbon di oxide |
37.
(a) Ionic bond.
(b) Electronegativity property.
(c) (i) Along the period, from left to right in the periodic table, the electronegativity increases, because of the increase in the nuclear charge which in turn attracts the electrons more strongly.
(ii) On moving down a group, the electronegativity of the element decreases because of the increased number of valence shells.
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Tamilnadu Stateboard 10th Standard Subjects
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