AP 10th Class Physical Science Model Paper Set 8 with Solutions

Solving AP 10th Class Physical Science Model Papers Set 8 regularly is an effective strategy for time management during exams.

AP SSC Physical Science Model Paper Set 8 with Solutions

Time: 2 Hours
Max. Marks: 50

Instructions:

• The question paper consists of 4 sections and 17 questions.
• Internal choice is available only for Q.No.12 in section III and for all the questions in section IV.
• In 2 hours, 15 minutes is allotted to read the question paper.
• All answers shall be written in the answer booklet only.
• Answers shall be written neatly and legibly.

Section – I (8 × 1 = 8 Marks)

Note:
1) Answer all the questions.
2) Each question carries 1 marks.

Question 1.
Predict, Exhalation air is hotter than Inhalation air in respiration process.
Answer:
Yes. Exhaled air is typically warmer than inhaled air due to body heat warming the air during its passage through the respiratory system.

Question 2.
_______ gas is released on the reaction of zinc granules with dilute sulphuric acid.
Answer:
The gas released when zinc granules react with dilute sultunc acid is hydrogen gas (H₂).

Question 3.

Which is the strong alkaline solution amont the solutions given in the table?
Answer:
In the given table, solution C has a pH of 12, which indicates it is the strongest alkaline solution.

Question 4.
Write any one physical property of metals.
Answer:
Generally metals are good conductors of heat, except lead and mercury, which are poor conductors of heat. Metals like copper and silver are the best conductors of heat.

Question 5.
Write any one use of carbon compound.
Answer:
Ans One common use of a carbon compound is methane (CH₄), which is primarily used as a fuel for heating, cooking and electricity generation. Methane is a major component of natural gas and is valued for its efficiency and relatively lower emissions compared to other fossil fuels.

Question 6.
The least disance of distinct vision for a young adult which normal vision is about ( )
a) 25 m
b) 2.5 cm
c) 25 cm
d) 2.5 m
Answer:
c) 25 cm

Question 7.
Draw the symbol of an electric cell.
Answer:
An electric cell

Question 8.
What is the SI unit of resistance of a conductor connected in an electric circuit?
Answer:
The SI unit of resistance is the ohm, represented by the symbol Ω.

Section – II (3 × 2 = 6M)

Note:
1) Answer all the questions.
2) Each question carries 2 marks.

Question 9.
Which of the following hydrocarbons undergo addition reactions.
C₂H₆,C₃H₈,C₃H₆,C₂H₂ and CH₄
Answer:
Unsaturated hydrocarbons undergo addition reactions.
So C₃H₆ i.e., CH₃ – CH = CH₂ and C₂H₂ i.e., CH = CH will give addition reactions but C₂H₆, C₃H₈ and CH₄ will not give addition reactions because they are saturated.

Question 10.
A ray of light travelling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal? Why?
Answer:
The light ray bend towards the normal. When light travels from rarer (air) to denser (water) medi¬um it bends towards the normal.

Question 11.
Post any two questions to understand the concept of Ohm’s law.
Answer:
1. What is the relationship between voltage, current and resistance in an electrical circuit, according to Ohm’s Law?
2. If the resistance in a circuit is doubled while the voltage remains constant, what happens to the current flowing through the circuit?

Section – III (3 × 4 = 12M)

Note:
1) Answer all the questions.
2) Each question carries 4 marks.

Question 12.
Draw any one of the following diagrams.
A) Draw the ray diagrams of image formed when the object is placed infront of a biconvex lens in the following positions.
i) Beyond 2F₁
ii) At F₁
Answer:

B) Draw the diagram which shows that acid solution in water conducts electricity.
Answer:

Question 13.
Write two important uses of washing soda and baking soda each.
Answer:
Uses of washing soda
i) Washing soda is used In laundary. In other words, washing soda’ls used as a cleansing agent for domestic purposes, It is a main component of many dry soap powders.
ii) It is used in the manufacture of many useful sodium compounds like caustic soda, borax, glass and soap.

Uses of baking soda :
i) Baking soda (sodium hydrogen carbonate) is used as an antacid in medicine to remove acidity of the stomach.

ii) Baking soda is used in making powder. Baking powder contains baking soda in acid like tartaric acid. When baking powder is mixed In the dough for preparing cake or bread is heated, sodium hydrogen carbonate present It decomposes it give carbon dioxide and sodium carbonate. This carbon dioxide gas bubbles out causing the cake or bread to rise and becomes light and spongy. The tartaric acid present in baking powder reacts with sodium carbonate and neutralises it. If tartaric acid is not present in baking power, then the cake will taste bitter due to the presence of sodium carbonate in it.

Notes-1. Answer all the questions. (3 v 8 – 24 M)
2. Each question carries 8 marks.
3. Each question has internal choice.

Question 14.

Materials medium	Air	Ice	Ruby	Benzene
Refractive Index	1.0003	1.31	1.71	1.50

Observe the table and answer the following questions.
i) Which material medium light travels faster?
ii) In which material medium the speed of light is least?
iii) In which material medium the speed of light is least
iv) Calculate the speed of light in Benzene? (Speed of light in vacuum is 3 × 10⁸ msl^-1).
Answer:
i) The material medium with the lower refractive index allows light to travel faster.
From the table, Air has the lowest refractive index (1.0003).
Therefore, light travels faster in Air.

ii) The material medium with the highest refractive index slows down light the most.
From the table, Ruby has the highest refractive index (1.71).
Therefore, the speed of light is least in Ruby.

iii) Answer same a point ii.

iv) We know that refractive index (n) = speed of light in vacuum (c) / speed of light in medium (v) Therefore, speed of light in medium (v) = c / n
Given, refractive index of benzene (n) = 1.50, speed of light in vacuum (c) = 3 × 10⁸ m/s So, speed of light in benzene (v) = (3 × 10⁸ m/s) / 1.50 = 2 × 10⁸ m/s.
Therefore, the speed of light in Benzene is 2 × 10⁸ m/s.

Section – IV (3 × 8 = 24 M)

Note:
1) Answer all the questions.
2) Each question carries 8 marks.
3) Each question has an internal choice.

Question 15.
A) Explain the following.
i) Twinkling of stars
ii) Formation of Rainbow
Answer:
i) Twinkling of stars : Light emitted by star passes through the atmosphere of the earth before reaching our eyes. The atmosphere of the earth is not uniform but consists of many layers of different densities.

The layers close to the surface of the earth are denser. As we go higher the density of layers and refractive index decreases progressively. As the light from a star enters the upper-most layer of the atmosphere, it bends towards the normal as it enters the next layer. This process continues till the light enters our eyes. So due to refraction of light, the apparent position of the star is different from the actual position of the star.

Moreover, the different layers of the atmosphere are mobile and the temperature and the density of layers of atmosphere changes continuously. Hence the apparent position of the star changes continuously. This change in the apparent position of the star continuously leads to the twinkling of a star.

ii) Rainbow formation : A rainbow is a natural spectrum appearing in the sky after rain shower. It is caused by dispersion of sunlight by tiny water droplets, present in the atmosphere. The water droplets act like small prism. They refract and disperse the incident sunlight, then reflect it internally and finally refract it again.
Due to dispersion of light and internal reflection different colours appear.

B) Deduce the expression for the equivalent resistance of three resistors connected in parallel in an Electric Circuit.
Answer:
Aim : To prove that when a number of resistors are connected in parallel then the total current in the circuit is divided into different resistors while the potential difference remains the same.
Materials Required : Three resistors, plug key, an ammeter, battery and voltmeter.

Procedure :

1. Make a parallel combination, XY, of three resistors having resistances R₁, R₂, and R₃, respectively.
2. Connect it with a battery, a plug key and an ammeter. Also, connect a voltmeter in parallel with the combination of resistors.
3. Plug the key and note the ammeter reading.
4. Let the current be I. Also take the voltmeter reading. It gives the potential V across the combination.
5. The potential differences across each resistor are also V. This can be checked by connecting the
6. Take out the plug from the key. Remove the ammeter and voltmeter from the circuit. Insert the ammeter in series with the resistor R₁ Note the ammeter reading l₁
7. Similarly, measure the currents through R₂ and R₃. Let these be l₂ and l₃ respectively.

Observation :

1. The total current I is equal to the sum of the current through each branch of the combination,
   1 = I₁ + l₂ + l₃.
2. The potential difference across each resistor is same.

Conclusion :
when a number of resistors are connected in parallel then the total current in the circuit is divided among different resistors while the potential difference remains the same.

Question 16.
A) Explain the following with an example.
i) Chemical combination
ii) Chemical decomposition
iii) Chemical displacement
iv) Chemical double displacement
Answer:
i) Chemical Combination: The reaction in which two or more substances combine to form a new substance is called chemical combination.
Eg : i) Burning of coal

ii) Chemical Decomposition The reaction in which a single compound breaks down to pro¬duce into two of more products is called chemical decomposition.
Eg :

iii) Chemical Displacement: The reaction in which an atom or a group of atoms in a molecule is displaced by another atom is called chemical displacement.
Eg :
CuSO₄(aq) + Zn(s) → ZnSO₄(aq) + Cu(s)
CuSO₄(aq) + Fe(s) → FeSO₄(aq) + Cu(s)

iv) Chemical Double Displacement : The reactions in which there is an exchange of ions be¬tween the reactants are called chemical double displacement.
Eg :
i) Na₂SO₄(aq) + BaCl₂(aq) → BaSO₄(s) + 2NaCl (aq)
ii) AgNO₃(aq) + NaCl (aq) → AgCl₅ + NaNO₃(aq)

(OR)
B) Explain the cleaning action of soap.
Answer:
Soap molecules possess two distinct ends : one end is hydrophilic (water-soluble), while the other end is hydrophobic (water-repellent). When soap is present at the surface, the hydrophobic tail of the soap is insoluble in water, causing the soap molecules to arrange themselves along the water surface. The ionic end of the soap molecule remains in contact with the water, while the hydrocarbon “tail” extends out of the water.

When soap is introduced into water, the molecules adopt a specific orientation that keeps the hydrophobic end shielded from the water. This orientation is achieved by the formation of clusters, or micelles, wherein the hydrophobic tails are nestled within the interior of the cluster, while the ionic ends reside on the surface. This micelle formation enables soap to effectively clean, as the oily dirt is captured within the core of the micelles. The micelles remain dispersed in the solution as a colloidal suspension and do not precipitate due to repulsion between ions. Consequently, the dirt suspended in the micelles can be easily rinsed away.

Question 17.
A) Explain the procedure to show that compass needle is deflected on passing an electric current
through a metallic conductor (Oersted’s experiment).
Answer:
Aim : To show that the direction of a magnetic field due to a current- carrying straight conductor depends upon the direction of the current flowing through it.

Materials Required : Battery, copper wire, plug key and compass.

Procedure :

1. Take a long straight copper wire, two or three cells of 1.5 V each, and a plug key. Connect all of them in series as shown in the figure (a).
2. Place the straight wire parallel to and over a compass needle.
3. Plug the key in the circuit.
4. Observe the direction of deflection of the north pole of the needle. If the current flow from north to south as shown in figure (a), the north pole of the compass would move towards the east.
5. Replace the cell connections in the circuit as shown in figure (b). this would result in the change of the direction of current through the copper wire, that is from the south to the north.
6. Observe the change in the direction of deflection of the needle. We will see that now the needle moves in the opposite direction, that is towards the west. It means that the direction of the magnetic field produced by the electric current is also reversed.

Observation :
When the direction of current through a copper wire changes, the direction of deflection of the compass needle also changes.

Conclusion :

1. The direction of a magnetic field due to a current-carrying straight conductor depends upon the direction of the current flowing through it.
2. If the direction of current in the conductor is reversed then the direction of a magnetic field is also reversed.

(OR)
B) Explain the experimental procedure to investigate the conductions under which iron rusts.
Answer:
Rusting is a process in which raddish brown coating of hydrated ferric oxide is formed on the surface of iron.
Aim : To show that air and water are necessary for rusting.
Materials Required : Three boiling tubes, iron nails, cork, anhydrous calcium chloride, oil.

Procedure :

1. Take three test tubes and place clean iron nails in each of them.
2. Label these test tubes as A, B and C.
3. Pour water in test tube A and cork it.
4. Pour boiled distilled water in test tube B, add about 1 ml of oil and cork it. The oil will float on water and prevent the air from dissolving in the water.
5. Put some anhydrous calcium chloride in test tube C and cork it. Anhydrous calcium chloride will absorb the moisture, if any, from the air.
6. Leave the test tubes for a few days and observe.

Observations :

1. Iron nails rust in test tube A but they do not rust in test tubes B and C.
2. In the test tube A, the nails are exposed to both air and water.
3. In the test tube B, the nails are exposed to only water.
4. The nails in test tube C are exposed to dry air.

Conclusion : From this, we can conclude that iron gets rusted in the presence of air and water.