AP TET Paper 1A & 2A Psychology Concept of Childhood and it's importance
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AP TET Paper 1A & 2A Psychology Concept of Childhood and it's importance
Mission March 2026-27 Corbon compounds Level 1 & Level 2 Answers
Mission March 2026-27 Corbon And it's Compounds Level 1 & Level 2 Answers
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1. Fill the table with suitable answers related to functional groups, structural formulae, ex-amples and suffixes.
| Structural Formula | Functional Group Name | Suffix | Example |
| R-OH | Alcohol | -ol | C2H5OH |
| R-CHO | Aldehyde | -al | CH3CHO |
| R-COOH | Carboxylic acid | -oic acid | CH3COOH |
| R-CO-R | Ketone | -one | CH3COCH3 |
2. Complete the table of alkanes, alkenes and alkynes based on the number of carbon atoms.
| No. of Corbons | Alkanes | Alkenes | Alkynes |
| 2 | Ethane | Ethene | Ethyne |
| 3 | Propane | Pripene | Propyne |
| 4 | Butane | Butene | Butyne |
| 5 | Pentane | Pentene | Pentyne |
3. Explain the following
a) Addition reaction of unsaturated hydrocarbons
An addition reaction is a reaction in which atoms or groups of atoms are added to an unsaturated hydrocarbon at the double or triple bond to form a saturated compound.
Example:
CH₂=CH₂ + H₂ → CH₃–CH₃
Ethene + Hydrogen → Ethane
This reaction takes place in the presence of a catalyst such as Ni/Pt/Pd.
b) Substitution reaction of saturated hydrocarbons
A substitution reaction is a reaction in which one or more hydrogen atoms of a saturated hydrocarbon are replaced by another atom or group of atoms.
Example:
CH₄ + Cl₂ → CH₃Cl + HCl
Methane + Chlorine → Chloromethane + Hydrogen chloride
The reaction occurs in the presence of sunlight (UV light).
4. Explain the cleansing action of soap with a well-labelled diagram
A soap molecule has two parts:
Hydrophilic end – water-loving ionic end.
Hydrophobic end – water-repelling hydrocarbon chain that attracts oil and grease.
When soap is added to water containing oily dirt, the hydrophobic ends attach themselves to the grease, while the hydrophilic ends remain in water. Many soap molecules surround the grease particle and form a micelle.
On rubbing and washing with water, the grease-containing micelles are removed.
Diagram:
Water
○ ○ ○ ○ ○ ○
\ | \ | \ | /
\| \| \|/
┌─────────────────┐
│ OIL / GREASE │
└─────────────────┘
/| /| /|\
/ | / | / | \
○ ○ ○ ○ ○ ○
○ = Hydrophilic end (water-loving)
Lines = Hydrophobic hydrocarbon chains
↓
MICELLE FORMATION
↓
Dirt/grease is washed away
5. Distinguish between soap and detergent
Soap
| Soap | Detergent |
| Soaps are sodium or potassium salts of long Chain Fatty Acids | Detergent s are salts of long chain alkyle benzene sulfonates or alkyl sulfonates |
| They are Generally less effective in hard water | They are effective in hard water |
| They form scumwith Ca2+ and Mg2+ ions | They Generally don't form insoluble scum with Ca2+ and Mg2+ ions |
| Example: Sodium Stearate | Example Sodium alkyle benzene sulphonate |
2 Marks Questions
1. What are the two properties of carbon that lead to the large number of carbon compounds?
The two important properties are:
Catenation: Carbon atoms can bond with one another to form long chains, branched chains and rings.
Tetravalency: Carbon has a valency of four and can form four covalent bonds with carbon and other elements.
2. Explain catenation with an example.
Catenation is the property of carbon by which carbon atoms form bonds with other carbon atoms to produce long chains, branched chains and rings.
Example:
CH₃–CH₂–CH₂–CH₂–CH₃
This is a chain of five carbon atoms (pentane).
3. What is allotropy? Write the allotropes of carbon.
Allotropy is the property of an element to exist in two or more different forms in the same physical state.
Important allotropes of carbon are:
Diamond
Graphite
Fullerenes (such as C₆₀)
4. Name the following functional groups
(i) –CHO → Aldehyde group
(ii) >C=O → Ketone (carbonyl) group
5. What are hydrocarbons? How are they classified?
Hydrocarbons are compounds containing only carbon and hydrogen.
They are mainly classified as:
Saturated hydrocarbons – contain only single C–C bonds.
Example: Ethane (C₂H₆)
Unsaturated hydrocarbons – contain double or triple bonds.
Alkenes – contain C=C double bond. Example: Ethene (C₂H₄)
Alkynes – contain C≡C triple bond. Example: Ethyne (C₂H₂)
6. Give two examples each of saturated and unsaturated hydrocarbons.
Saturated hydrocarbons:
Methane – CH₄
Ethane – C₂H₆
Unsaturated hydrocarbons:
Ethene – C₂H₄
Ethyne – C₂H₂
7. Give one example each of a carbon compound containing:
1. Alcohol functional group:
Ethanol – C₂H₅OH
2. Carboxylic acid functional group:
Ethanoic acid – CH₃COOH
8. Why are cooking oils hydrogenated before making vanaspati ghee?
Cooking oils contain unsaturated fatty acids. Hydrogenation adds hydrogen to the double bonds in these oils in the presence of a nickel catalyst, converting them into more saturated and solid fats.
Example:
Unsaturated oil + H₂ → Saturated fat
(Ni catalyst)
Thus, hydrogenation helps convert liquid vegetable oils into solid/semi-solid vanaspati ghee.
9. A student says ethanol and ethanoic acid have the same properties. Is the statement correct?
No, the statement is incorrect.
Ethanol and ethanoic acid have different functional groups and therefore different physical and chemical properties.
Ethanol: C₂H₅OH — alcohol
Ethanoic acid: CH₃COOH — carboxylic acid
Ethanoic acid turns blue litmus red, whereas ethanol does not.
10. Suggest a way to identify whether a carbon compound is saturated or unsaturated.
Treat the compound with bromine water.
Unsaturated compound: Decolourises bromine water.
Saturated compound: Does not decolourise bromine water under ordinary conditions.
Another test is the alkaline KMnO₄ test.
1 Mark Questions
1. A student proposes to make a sweet-smelling compound from ethanol and acetic acid. Which reaction should be carried out?
Answer: Esterification reaction.
Ethanol reacts with ethanoic acid in the presence of concentrated H₂SO₄ to form ethyl ethanoate, which has a sweet/fruity smell.
C₂H₅OH + CH₃COOH → CH₃COOC₂H₅ + H₂O
2. Write the structural formula of any hydrocarbon with eight carbon atoms.
Octane:
CH₃–CH₂–CH₂–CH₂–CH₂–CH₂–CH₂–CH₃
3. Identify the functional group present in propanal.
Answer: Aldehyde group (–CHO)
4. Which of the following undergoes substitution reaction?
A) CH₄
B) C₃H₆
C) C₂H₂
D) C₇H₁₄
Answer: A) CH₄ (Methane)
5. Which of the following hydrocarbons undergoes addition reaction?
A) C₂H₆
B) C₃H₈
C) C₃H₆
D) CH₄
Answer: C) C₃H₆ (Propene)
6. Identify the alkene among the following hydrocarbons.
A) C₂H₆
B) C₃H₈
C) C₃H₆
D) CH₄
Answer: C) C₃H₆ (Propene)
7. The general formula of alkanes is CₙH₂ₙ₊₂. Write the first member of alkanes.
Answer: Methane (CH₄)
8. A hydrocarbon has four carbon atoms and ten hydrogen atoms. Write its name
Formula = C₄H₁₀
Answer: Butane
9. Which compound is formed when two carbon atoms are linked by a single bond and six hydrogen atoms are added?
Answer: Ethane (C₂H₆)
Structural formula:
CH₃–CH₃
10. How can you convert ethene into ethane?
By hydrogenation of ethene in the presence of a nickel catalyst.
CH₂=CH₂ + H₂ → CH₃–CH₃
(Ni catalyst)
11. Give any example of an unsaturated hydrocarbon.
Answer: Ethene (C₂H₄)
12. How many hydrogen atoms are present in Butane?
Butane = C₄H₁₀
Answer: 10 hydrogen atoms
13. Which type of bond is present between carbon atoms in ethene?
Answer: Double covalent bond (C=C).
LEVEL – 2: SHINING STAR
8 Marks Questions
1. How can ethanol and ethanoic acid be differentiated based on their physical and chemical properties?
| Property | Ethanol | Ethanoic acid |
| Formula | C₂H₅OH | CH₃COOH |
| Functional group | –OH (Alcohol) | –COOH (Carboxylic acid) |
| Smell | Characteristic alcoholic smell | Vinegar-like smell |
| Litmus test | Does not change blue litmus | Turns blue litmus red |
| Reaction with NaHCO₃ | No brisk reaction | Produces CO₂ gas with effervescence |
| Nature | Neutral | Acidic |
| Boiling point | About 78°C | About 118°C |
| Common use | Solvent, fuel, sanitiser | Vinegar, food preservative |
Important test:
CH₃COOH + NaHCO₃ → CH₃COONa + H₂O + CO₂↑
The evolution of CO₂ gas confirms ethanoic acid.
2. Explain the nature of the covalent bond using bond formation in CH₃Cl.
A covalent bond is formed by the sharing of electrons between atoms.
In CH₃Cl (chloromethane):
Carbon has 4 valence electrons and needs four more electrons to complete its octet.
Each hydrogen shares one electron with carbon, forming three C–H covalent bonds.
Chlorine has 7 valence electrons and shares one electron with carbon, forming one C–Cl covalent bond.
Thus carbon forms four covalent bonds.
Structure:
H
|
H — C — Cl
|
H
Therefore, CH₃Cl contains three C–H bonds and one C–Cl covalent bond.
3. Write the differences between saturated and unsaturated hydrocarbons.
| saturated hydrocarbons | unsaturated hydrocarbons |
| They Contain Only Single bonds between carbon atoms | They Contain One or more Double or triple bonds between carbon atoms |
| They are generally less reactive | they are generally more reactive |
| Example: Ethane (C2H6) |
Example: Ethene ( C2H4) |
Saturated hydrocarbons Contain only single bonds between carbon atoms.Called alkanes.
Unsaturated hydrocarbons Contain double or triple bonds between carbon atoms. Include alkenes and alkynes.
General formula of open-chain alkanes: CₙH₂ₙ₊₂
Alkenes: CₙH₂ₙ; Alkynes: CₙH₂ₙ₋₂
Generally undergo substitution reactions.
Generally undergo addition reactions.
Do not decolourise bromine water.
Decolourise bromine water.
Example: Ethane (C₂H₆)
Example: Ethene (C₂H₄)
4. Differentiate between alkanes, alkenes and alkynes.
| Alkanes | Alkenes | Alkynes |
| Saturated | Unsaturated | Unsaturated |
| Single bond (C-C) | Double bond (C=C) |
Triple bond (C=-C) |
| CnH2n+2 | CnH2n | CnH2n-2 |
| Methane(CH4) | Ethene(C2H4) | Ethyne (C2H2) |
| Substitution | Addition | Addition |
Examples of Structure:
Alkanes: CH3-CH3
Alkenes: CH2=CH2
Alkynes: CH=-CH
2 Marks Questions
1. Differentiate between saturated and unsaturated hydrocarbons with one example each.
| saturated hydrocarbons | unsaturated hydrocarbons |
| They Contain Only Single bonds between carbon atoms | They Contain One or more Double or triple bonds between carbon atoms |
| They are generally less reactive | they are generally more reactive |
| Example: Ethane (C2H6) | Example: Ethene ( C2H4) |
2. Draw the structural isomers of butane. Name them.
Butane has the molecular formula C₄H₁₀ and has two structural isomers.
1. n-Butane:
CH₃ — CH₂ — CH₂ — CH₃
2. Isobutane (2-methylpropane):
CH₃
|
CH₃ — CH — CH₃
3. Why are carbon compounds poor conductors of electricity?
Carbon compounds are generally covalent compounds. They do not have free ions or free electrons that can carry electric current. Therefore, they are generally poor conductors of electricity.
4. Write two properties of ethane related to its use as a fuel.
Ethane is highly combustible and releases a large amount of heat on burning.
It burns in oxygen to produce carbon dioxide and water, releasing energy.
C₂H₆ + 7/2 O₂ → 2CO₂ + 3H₂O + heat
5. Explain why carbon forms a large number of compounds. Give any two reasons.
Carbon forms a large number of compounds mainly because of:
Catenation: Carbon atoms can bond with one another to form long chains, branched chains and rings.
Tetravalency: Carbon has valency 4 and can form four covalent bonds with carbon and other elements.
6. Why is a flame of unsaturated hydrocarbons smoky? How can we test for unsaturation?
Unsaturated hydrocarbons have a higher percentage of carbon. They often undergo incomplete combustion, producing tiny carbon particles (soot), which make the flame yellow and smoky.
Test for unsaturation: Add bromine water. An unsaturated hydrocarbon decolourises bromine water.
7. Write the chemical equation for esterification reaction. Name the ester formed from methanol and ethanoic acid.
Equation:
CH₃OH + CH₃COOH → CH₃COOCH₃ + H₂O
(conc. H₂SO₄, heat)
The ester formed is methyl ethanoate.
8. How is soap different from detergent? Give one point for each.
Soap: Forms scum with Ca²⁺ and Mg²⁺ ions in hard water and is therefore less effective.
Detergent: Does not form insoluble scum with Ca²⁺ and Mg²⁺ ions and works well in hard water.
9. What is meant by homologous series? Write two characteristics.
A homologous series is a group of organic compounds having the same functional group and the same general formula, where successive members differ by –CH₂–.
Characteristics:
Successive members differ by –CH₂– (14 u).
Members have similar chemical properties and show a gradual change in physical properties.
1 Marks Questions
1. The property of carbon to form chains, branched chains and rings is called:
Answer: B) Catenation
2. General formula of alkenes is:
Answer: CₙH₂ₙ
3. Which of the following is not an allotrope of carbon?
A) Diamond
B) Graphite
C) Fullerene
D) Methane
Answer: D) Methane
4. IUPAC name of CH₃CH₂OH is:
A) Methanol
B) Ethanol
C) Ethanal
D) Ethanoic acid
Answer: B) Ethanol
5. Which of the following belongs to the same homologous series?
Answer: B) C₂H₆ and C₃H₈
Both are alkanes and successive members differ by CH₂.
6. Functional group present in CH₃COOH is:
A) Alcohol
B) Aldehyde
C) Ketone
D) Carboxylic acid
Answer: D) Carboxylic acid
7. On adding sodium to ethanol, gas evolved is:
A) H₂
B) O₂
C) CO₂
D) CH₄
Answer: A) H₂ (Hydrogen)
2C₂H₅OH + 2Na → 2C₂H₅ONa + H₂↑
8. Draw the electron dot structure of methane.
Methane is CH₄. Carbon shares one electron with each of four hydrogen atoms.
H
:
H : C : H
:
H
Each ":" represents a shared pair of electrons (covalent bond).
9. Why does carbon form covalent bonds only?
Carbon has four valence electrons. It is difficult for carbon to either lose four electrons or gain four electrons. Therefore, carbon completes its octet by sharing electrons, forming covalent bonds.
10. Name the catalyst used in hydrogenation of vegetable oils.
Answer: Nickel (Ni)
11. Why is conversion of ethanol to ethanoic acid considered an oxidation reaction?
Conversion of ethanol to ethanoic acid involves the addition of oxygen.
CH₃CH₂OH + 2[O] → CH₃COOH + H₂O
Therefore, it is considered an oxidation reaction.
12. Write the IUPAC name of CH₃–O–CH₃.
Answer: Methoxymethane
Common name: Dimethyl ether.
AP 9th Class Physics 3rd Lesson Force and Laws of Motion Questions and Answers
AP 9th Class Physics 3rd Lesson Force and Laws of Motion Questions and Answers (Exercise)
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Question 1.
An object experiences a net zero external unbalanced force. Is it possible for the object to be travelling with a non-zero velocity? If yes, state the conditions that must be placed on the magnitude and direction of the velocity. If not, provide a reason.
Answer:
Yes. Object can travel with non-zero velocity without experiencing net zero external unbalanced force in the space. However without unbalanced force the object can’t travel with non – zero velocity on the earth.
Question 2.
When a carpet is beaten with a stick, dust comes out of it. Explain.
Answer:
When a carpet is beaten with a stick it suddenly comes into motion.
But the dust in it continue to remain in rest due to inertia of rest.
Hence the dust comes out.
Question 3.
Why is it advised to tie any luggage kept on the roof of a bus with a rope?
Answer:
i) Backward – due to inertia of rest
ii) Forward – due to inertia of motion
iii) Sideways – due to inertia of direction
That’s why luggage should be tied.
Question 4.
A batsman hits a cricket ball which then rolls on a level ground. After covering a short distance, the ball comes to rest. The ball slows to a stop because
a) the batsman did not hit the ball hard enough.
b) velocity is proportional to the force exerted on the ball.
c) there is a force on the ball opposing the motion.
d) there is no unbalanced force on the ball, so the ball would want to come to rest.
Answer:
c) there is a force on the ball opposing the motion.
Question 5.
A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400 m in 20. Find its acceleration. Find the force acting on it if its mass is 7 tonnes
(Hint : 1 tonne = 1000 kg)
Answer:
Truck: Initial velocity u=0; Displacement s=400m
Time t=20sec ; Mass m=7 tonnes =7000 kg [∵ 1 tonne =1000kg]
Question 6.
A stone of 1 kg is thrown with a velocity of 20 ms-1 across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice?
Answer:
Mass m =1 kg
Initial velocity u=20ms-1
Final velocity
v=0 (comes to rest); Displacement
s=50 m
From Kinematics,
Note : Friction force is always opposite to the motion.
Question 7.
A 8000 kg engine pulls a train of 5 wagons, each of 2000 kg, along a horizontal track. If the engine exerts a force of 40000 N and the track offers a friction force of 5000 N, then calculate:
a) the net accelerating force and
b) the acceleration of the train.
Answer:
Mass of enging = 8000 kg;
Mass of each wagon =2000 kg
Force generated by Engine =40000 N; Force of Friction =5000 N
a) Net Accelerating Force =40000-5000=35000 N
b) Acceleration of the train F = ma (Newton’s 2nd law)
Question 8.
An automobile vehicle has a mass of 1500 kg. What must be the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of 1.7 ms-2?
Answer:
Automobile vehicle Visualisation :
Mass m =1500kg
Acceleration a =-1.7ms-2
F = ma × [By Newton’s 2nd law]
F=1500 × -1.7=-2550N
So, an opposite Force -2550 N should be applied to bring the vehicle to rest.
Question 9.
What is the momentum of an object of mass m, moving with a velocity v ?
a) (mv)2
b) mv2
c) 1/2 mv2
d) mv
Answer:
d) mv
Question 10.
Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor at a constant velocity. What is the friction force that will be exerted on the cabinet?
Answer:
Cabinet Visualisation:
To move the block with constant velocity across a floor the friction force exerted on the cabinet is -200 N.
Question 11.
According to the third law of motion when we push on an object, the object pushes back on us with an equal and opposite force. If the object is a massive truck parked along the roadside, it will probably not move. A student justifies this by answering that the two opposite and equal forces cancel each other. Comment on this logic and explain why the truck does not move.
Answer:
The boys reason is not correct. Heavier trucks have greater inertia, so it requires a greater unbalanced force to change its state of rest.
Question 12.
A hockey ball of mass 200 g travelling at 10 ms-1 is struck by a hockey stick so as to return it along its original path with a velocity at 5 ms-1. Calculate the magnitude of change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.
Answer:
Question 13.
A bullet of mass 10 g travelling horizontally with a velocity of 150 ms-1 strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force exerted by the wooden block on the bullet.
Answer:
Question 14.
An object of mass 1 kg travelling in a straight line with a velocity of 10 ms-1 collides with and sticks to, a stationary wooden block of mass 5 kg. Then they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also, calcula inte the velocity of the combined object.
Answer:
Block – A
Mass m1= 1 kg
Initial velocity u1= 10 ms-1
Mass m2 = 5 kg
Initial velocity u2= 0 ms-1
i) Block A is motion & Block B is at rest.
ii) Block A travelled in straight line & stuck with Block B.
iii) As they are moving together they will have a common velocity.
iv) Total momentum before collision = m1u1 + m2u2
= 1×10+5×0=10 kgms-1 or (N)
b) Momentum after Impact = 10 kg ms-1 (or) (N)
Reason : Momentum before collision is equal to momentum after collision.
c) Combined – velocity: m1u1 + m2u2 = m2v+ m2v
⇒ m1u1 + m2u2= v(m1+ m2)
⇒ 1 x 10 + 5 x 0 = v(1+ 5)
Question 15.
An object of mass 100 kg is accelerated uniformly from a velocity of 5 ms-1 to 8 ms-1 in 6s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.
Answer:
Object:
Mass m = 100 kg Initial velocity u = 5 ms-1
Final velocity v = 8 ms-1,Time t, = 6 sec
Initial momentum P1 = mu = 100 x 5 = 500 kg ms-1 or (N)
Final momentum P2 = mv= 100 x 8 = 800 kg ms-1 (or) (N)
Question 16.
Akhtar, Kiran and Rahul were riding in a motorcar that was moving with a high velocity on an expresssway when an insect hit the windshield and got stuck on the windscreen. Akhtar and Kiran started pondering over the situation. Kiran suggested that the insect suffered a greater change in momentum as compared to the change in momentum of the motorcar (because to the change in the velocity of the insect was much more than that of the motorcar). Akhtar said that since the motorcar was moving with a larger velocity, it exerted a larger force on the insect. And as a result the insect died. Rahul while putting an entirely new explanation said that both the motorcar and the insect experienced the same force and a change in their momentum. Comment on these suggestions.
Answer:
Kiran Akhtar’s reasons are not correct.
Rahul’s suggestion is correct. Force exerted by windshield and insect are equal opposite according to Newton’s 3rd law.
But the mass of the insect is lesser than the mass windshield the insect experiences a greater force due to this it dies.
Question 17.
How much momentum will a dumb – bell of mass 10 kg transfer to the floor if it falls from a height of 80cm ? Take its downward acceleration to be 10 ms-2.
Answer:
Dumb-Bell :
