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CBSE Class 10 Science Question Paper 2023 (Series: Z1XYW/4) with Solutions

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CBSE Class 10 Science Board Question Paper 2023 (Series: Z1XYW/4) with Solutions

Time allowed: 3 hours
Maximum marks: 80

General Instructions

  • This question paper contains 39 questions. All questions are compulsory.
  • Question paper is divided into FIVE sections. Section A, B, C, D and E.
  • In Section A Question number 1 to 20 are Multiple Choice Questions (MCQs) carrying 1 mark each.
  • In Section B Question number 21 to 26 are Very Short Answer (VSA) type questions carrying 2 marks each. Answer to these Questions should be in the range of 30 to 50 words.
  • In Section C Question number 27 to 33 are Short Answer (SA) type questions carrying 3 marks each. Answer to these Questions should be in the range of 50 to 80 words.
  • In Section D Question number 34 to 36 are Long Answer (LA) type questions carrying 5 marks each. Answer to these Questions should be in the range of 80 to 120 words.
  • In Section E question number 37 to 39 are of 3 source-based/case-based units of assessment carrying 4 marks with sub-parts.
  • There is no overall choice. However, an internal choice has been provided in some sections.

SET – I Code No. 31/4/1

Section – A (Multiple Choice Questions)

Question 1.
When sodium bicarbonate reacts with dilute hydrochloric acid, the gas evolved is:
(a) Hydrogen; it gives pop sound with burning matchstick.
(b) Hydrogen; it turns lime water milky.
(c) Carbon dioxide; it turns lime water milky.
(d) Carbon dioxide; it blows off a burning matchstick with a pop sound.
Answer:
(c) When sodium bicarbonate reacts with dilute hydrochloric acid, carbon dioxide gas is evolved.

Question 2.
When aqueous solutions of potassium iodide and lead nitrate are mixed, an insoluble substance separates out. The chemical equation for the reaction involved is:
(a) KI + PbNO3 → Pbl + KNO3
(b) 2KI + Pb(NO3)2 → Pbl2 + 2KNO3
(c) KI + Pb(NO3)2 → Pbl + KNO3
(d) KI + PbNO3 → Pbl2 + KNO3
Answer:
(b) 2KI + Pb(NO3)2 → Pbl2 + 2KNO3

Question 3.
A metal ribbon ‘X’ burns in oxygen with a dazzling white flame forming a white ash ‘Y’. The correct description of X, Y and the type of reaction is:
(a) X = Ca; Y = CaO; Type of reaction = Decomposition
(b) X = Mg; Y = MgO; Type of reaction = Combination
(c) X = Al; Y = AI2O3; Type of reaction = Thermal decomposition
(d) X = Zn; Y = ZnO; Type of reaction = Endothermic
Answer:
(b) X = Mg; Y = MgO; Type of reaction = Combination

Question 4.
Acid present in tomato is:
(a) Methanoic acid
(b) Acetic acid
(c) Lactic acid
(d) Oxalic acid
Answer:
(d) Oxalic acid

Question 5.
Sodium hydroxide is termed an alkali, while Ferric hydroxide is not because:
(a) Sodium hydroxide is a strong base, while Ferric hydroxide is a weak base.
(b) Sodium hydroxide is a base which is soluble in water while Ferric hydroxide is also a base but it is not soluble in water.
(c) Sodium hydroxide is a strong base while Ferric hydroxide is a strong acid.
(d) Sodium hydroxide and Ferric hydroxide both are strong base but the solubility of Sodium hydroxide in water is comparatively higher-than that of Ferric hydroxide.
Answer:
(b) Sodium hydroxide is a base which is soluble in water while Ferric hydroxide is also a base but it is not soluble in water.

Question 6.
The name of the salt used to remove permanent hardness of water is:
(a) Sodium hydrogen carbonate (NaHCCL)
(b) Sodium chloride (NaCl)
(c) Sodium carbonate decahydrate (Na2CO3 . 10 H2O)
(d) Calcium sulphate hemihydrate (CaSO4 . \(\frac { 1 }{ 2 }\)H2O)
Answer:
(c) Sodium carbonate decahydrate (Na2CO3 . 10 H2O)

Question 7.
The electron dot structure of chlorine molecule is:

Answer:
(c)

Question 8.
Observe the given diagram and identify the process and its significance from the following options:
(a) Evaporation: maintains water contents in leaf cells.
(b) Transpiration: creates a suction force which pulls water inside the plant.
(c) Excretion: helps in excreting out waste water from the plant.
(d) Translocation: helps in transporting materials from one cell to another.

Answer:
(b) Transpiration: creates a suction force which pulls water inside the plant.

Question 9.
Opening and closing of stomata is due to:
(a) High pressure of gases inside the cells.
(b) Movement of water in and out of the guard cells.
(c) Stimulus of light in the guard cells.
(d) Diffusion of CO2 in and out of the guard cells.
Answer:
(b) Movement of water in and out of the guard cells.

Question 10.
A cross between pea plant with white flowers (vv) and pea plant with violet flowers (VV) resulted in F2 progeny in which ratio of violet (VV) and white (vv) flowers will be: -1
(a) 1 : 1
(b) 2 : 1
(c) 3 : 1
(d) 1 : 3
Answer:
(c) 3 : 1

Question 11.
In plants the role of cytokinin is:
(a) Promote cell division.
(b) Wilting of leaves.
(c) Promote the opening of stomatal pore.
(d) Hetpm the growth of stem.
Answer:
(a) Promote cell division.

Question 12.
The number of chromosomes in parents and offsprings of a particular species undergoing sexual reproduction remain constant due to:
(a) doubling of chromosomes after zygote formation.
(b) halving of chromosomes after zygote formation.
(c) doubling of chromosomes before-gamete formation.
(d) halving of chromosomes at the time of gamete formation.
Answer:
(d) halving of chromosomes at the time of gamete formation.

Question 13.
Two LED bulbs of 12 W and 6 W are connected in series. If the current through 12W bulb is 0.06 A the current through 6 W bulb will be:
(a) 0.04A
(b) 0.06A
(c) 0.08A
(d) 0.12A
Answer:
(b) Current through 6 W bulb will be 0.06 A. Since current remains constant in series.

Question 14.
The correct pattern of magnetic field lines of the field produced by a current carrying circular loop is:

Answer:
(c)

Question 15.
The resistance of a resistor is reduced to half of its initial value. If other parameters of the electrical circuit remain unaltered, the amount of heat produced in the resistor will become:
(a) four times
(b) two times
(c) half
(d) one fourth
Answer:
(c) As we know, H = I2Rt …(i)
If R’ = \(\frac { 1 }{ 2 }\)R
Now, H’ = I2R’t = I2 × \(\frac { 1 }{ 2 }\)Rt = \(\frac { 1 }{ 2 }\)I2Rt
∴ H’ = \(\frac { 1 }{ 2 }\)H …[From (i)

Question 16.
An alpha particle enters a uniform magnetic field as shown. The direction of force experienced by the alpha particle is:

(a) towards right
(b) towards left
(c) into the page
(d) out of the page
Answer:
(d) out of the page

Q. No. 17 to 20 are Assertion-Reasoning based questions.
These consist of two statements – Assertion (A) and Reason (R). Answer these questions selecting the appropriate option given below:
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of (A).
(c) Assertion (A) is true, but Reason (R) is false.
(d) Assettion (A) is false, but Reason (R) is true.

Question 17.
Assertion (A): Reaction of Quicklime with water is an exothermic reaction.
Reason (R): Quicklime reacts vigorously with water releasing a large amount of heat.
Answer:
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of (A).

Question 18.
Assertion (A): In humans, if gene (B) is responsible for black eyes and gene (b) is responsible for brown eyes, then the colour of eyes of the progeny having gene combination Bb, bb or BB will be black only.
Reason (R): The black colour ofrthp_eyes is a dominant trait.
Answer:
(d) Assettion (A) is false, but Reason (R) is true.

Question 19.
Assertion (A): The inner walls of the small intestine have finger like projections called villi which are rich in blood.
Reason (R): These villi have a large surface area to help the small intestine in completing the digestion of food.
Answer:
(c) Assertion (A) is true, but Reason (R) is false.

Question 20.
Assertion (A): A current carrying straight conductor experiences a force when placed perpendicular to the direction of magnetic field.
Reason (R): The net charge on a current carrying conductor is always zero.
Answer:
(c) Assertion (A) is true, but Reason (R) is false.

Section – B (Very Short Answer Questions)

Question 21.
(a) A student took a small amount of copper oxide in a conical flask and added dilute hydrochloric acid to it with constant stirring. He observed a change in colour of the solution.
(i) Write the name of the compound formed and its colour.
(ii) Write a balanced chemical equation for the reaction involved.
OR
(b) The industrial process used for the manufacture of caustic soda involves electrolysis of an aqueous solution of compound ‘X’. In this process, two gases ‘Y’ and ‘Z’ are liberated. ‘Y’ is liberated at cathode and ‘Z’, which is liberated at anode, on treatment with dry slaked lime forms a compound ‘B’. Name X, Y, Z and B.
Answer:
(a) (i) When copper oxide reacts with dil. hydrochloric acid then copper chloride and water will be formed. In this reaction the black colour of copper oxide changes to bluish green colour of copper chloride.

OR
(b) Compound X ⇒ Sodium Chloride [NaCl] or (Brine)
Compound Y ⇒ Hydrogen gas [H2]
Compound Z ⇒ Chlorine gas [Cl2]
B ⇒ Calcium Oxychloride
or

Question 22.
(a) Name the part of brain which is responsible for the following actions:
(i) Maintaining posture and balance
(ii) Beating of heart
(iii) Thinking
(iv) Blood pressure
OR
(b) Where are auxins synthesized in a plant? Which organ of the plant shows:
(i) Positive phototropism
(ii) Negative geotropism
(iii) Positive hydrotropism
Answer:
(a) (i) Maintaining posture and balance = Cerebellum
(ii) Beating of heart = Medulla
(iii) Thinking = Cerebrum
(iv) Blood pressure = Medulla
OR
(b) Auxin is synthesized in young shoots, buds, young leaves and roots where growth occurs by cell elongation and is stimulated by presence of sunlight.
(i) Positive phototropism → Plant stem
(ii) Negative geotropism → Plant stem
(iii) Positive hydrotropism → Plant roots

Question 23.
Write one specific function each of the following organs in relation with excretion in human beings:
(i) Renal Artery
(ii) Urethra
(iii) Glomerulus
(iv) Tubular part of nephron
Answer:
(i) Renal Artery. The Renal Artery brings in the dirty blood containing waste substances (nitrogenous waste like urea etc.) into the kidneys.

(ii) Urethra. The urine collected in the bladder is passed out from the body through the urethra.

(iii) Glomerulus. One end of the glomerulus is attached to renal artery which brings the dirty blood containing the urea waste into it. The glomerulus filters this blood. During filtration, the substances like glucose, amino acids, salts, water and urea etc. present in the blood of glomerulus pass into Bowman’s capsule.

(iv) Tubular part of nephron. When the filtrate containing useful substances as well as the waste substances passes through the tubule, then the useful substances like glucose, all amino acids, most salts and most water etc. are reabsorbed into the blood through blood capillaries surrounding the tubule. Only the waste substances urea, some unwanted salts and excess water remain behind in the tubule in the form of urine.

Question 24.
Two green plants are kept separately in oxygen free containers, one in the dark and other in sunlight. It was observed that plant kept in dark could not survive longer. Give reason for this observation.
Answer:
The plant which is kept in oxygen-free container in continuous sunlight will live longer because it will be able to produce oxygen required for its cellular respiration by the process of photosynthesis.
The plant which is also kept in oxygen-free container, but in the dark will not photo- synthesize and hence it will finally die due to non-availability of oxygen for its respiration.

Question 25.
(a) Observe the following diagram and answer the questions following it:

(i) Identify the defect of vision shown.
(ii) List its two causes.
(iii) Name the type of lens used for the correction of this defect.
OR
(b) The colour of clear sky from the earth appears blue but from the space it appears black. Why?
Answer:
(a) (i) Defect of vision shown is Myopia.
(ii) Causes:
1. Due to the elongation of the eye ball.
2. Due to decrease in the focal length of the eye lens and hence the eye lens becomes more convergent.
(iii) Concave lens of suitable focal length is used for the correction of Myopia.
OR
(b) Sky appears blue. The molecules of air and other fine particles in the atmosphere have a size smaller than the wavelength of visible light. So these particles scatter more effectively the light rays of shorter wavelength at the blue end than light of longer wavelengths at the red end. When the scattered blue light enters our eyes, it gives us the feeling of a blue sky.

Colour of Sky to an Astronaut. There is no atmosphere containing air in space to scatter sunlight. As there is no scattering of light in space, the scattered light does not reach the eyes and the sky appears dark (black) instead of blue to An astronaut in the outer space.

Question 26.
Use of several pesticides which results in excessive accumulation of pesticides in rivers or ponds, is a matter of deep concern. Justify this statement.
Answer:
At present time, water pollution by pesticides is a critical issue, especially in range of extensive agriculture where leaching of these harmful chemicals in water bodies like rivers or ponds cause toxic effects on aquatic animals and human health. The toxic substances may run off into streams and ground water. It may cause the death of aquatic animals. Some harmful chemicals like pesticides, when absorbed by the plants through soil and water, get transferred from first trophic to the last trophic level of the food chain.

As these chemicals are non-degradable, their concentration in the bodies of living organisms at each trophic level progressively increases. This increase in the concentration of harmful chemicals in the body of living organisms at each trophic level of a food chain is called biological magnification. The level of concentration of chemicals is maximum for human beings as they are at the highest trophic level.

Section – C (Short Answer Questions)

Question 27.
(i) While electrolysing water before passing the current some drops of an acid are added. Why? Name the gases liberated at cathode and anode. Write the relationship between the volume of gas collected at anode and the volume of gas collected at cathode.
(ii) What is observed when silver chloride is exposed to sunlight? Give the type of reaction involved.
Answer:
(a) (i) While electrolysing water before passing the current some drops of an acid are added to make water a good conductor of electricity.
Hydrogen gas (H2) is liberated at cathode and Oxygen gas (O2) is liberated at anode. The ratio of volume of O2 (gas collected at anode) and the volume of H2 (gas collected at cathode) is 1 : 2 by volume.

(ii) When silver chloride is exposed to sunlight, it decomposes to form silver metal and chlorine gas.

This is a decomposition reaction which is brought about by light energy.

Question 28.
(i) Suggest a safe procedure of diluting,a strong concentrated acid.
(ii) Name the salt formed when sulphuric acid is added to sodium hydroxide and write its pH.
(iii) Dry HCl gas does not change the colour of dry blue litmus paper. Why?
Answer:
(i) The dilution of a concentrated acid should always be done by adding concentrated acid to water gradually with stirring and not by adding water to concentrated acid because when a concentrated acid is added to water for preparing a dilute acid, then the heat is evolved gradually, and easily absorbed by the large amount of water.

(ii) H2SO4 + 2NaOH → Na2SO4 + 2H2O
When sulphuric acid is added to sodium hydroxide sodium sulphate (Na2SO4) is formed. The pH of Na2SO4 is 7.

(iii) Dry HCl gas does not change the colour of dry blue litmus paper because in the absence of water dry HCl will not form hydrogen ions and hence will not show its acidic behaviour.

Question 29.
(a) (i) How does Paramecium obtain its food?
(ii) List the role of each of the following in our digestive system:
(a) Hydrochloric acid
(b) Trypsin
(c) Muscular walls of stomach
(d) Salivary amylase
OR
(b) (i) What is double circulation?
(ii) Why is the separation of the right side and the left side of the heart useful? How does it help birds and mammals?
Answer:
(a) (i) Paramecium is a unicellular organism which lives in water. Paramecium uses its hair like structures called cilia to sweep the food particles from water and put them into its mouth.

(ii) (a) • Hydrochloric acid makes the medium of gastric juice acidic so that the enzyme pepsin can digest the protein properly.
• Hydrochloric acid kills any bacteria which may enter the stomach with food.
(b) Trypsin. Enzyme Trypsin digests the protein.
(c) Muscular walls of stomach churn the food in the stomach for about three hours.
(d) Salivary amylase breaks down (digests) the starch present in the food into sugar.
OR
(i) Double circulation. The blood travels twice through the heart in one complete cycle of the body which called double circulation. It involves two circulations: (i) Pulmonary circulation. The pathway of the blood from the heart to the lungs and back to the heart is called pulmonary circulation. It is a small circulation. Deoxygenated blood in the right ventricle flows into the vascular system of the lungs, becomes oxygenated and returns to the heart’s left atrium through pulmonary veins.

(ii) Systemic circulation. The pathway of the blood from the heart to the : rest of the body and back to the heart is called systemic circulation. It is a large circulation. Left ventricle sends the blood into the aorta. Aorta divides into arteries, arterioles and capillaries and supplies oxygenated blood to various parts of the body. From there the deoxygenated blood is collected by venules, which join to form veins and finally vena cava and pours blood back into the right atrium.

(ii) (a) Humans have a four chambered heart which consists of two atria and two ventricles. In a four chambered heart, the left side and right side of the heart are completely separated to prevent the oxygenated blood from mixing with deoxygenated blood. Such a separation allows a highly efficient supply of oxygen to the body cells which is necessary for producing a lot of energy. This energy is useful for a warm blooded animal (like humans) which has high energy needs to maintain body temperature.

(b) All the animals having four chambered hearts have double circulation in which the blood passes through the heart ‘twice’ in one complete cycle of the body. This ensures the separation of oxygenated blood from deoxygenated blood.

Question 30.
(a) Define the following terms in the context of a diverging mirror:
(i) Principal focus
(ii) Focal length
Draw a labelled ray diagram to illustrate your answer.
OR
(b) An object of height 10 cm is placed 25 cm away from the optical centre of a converging lens of focal length 15 cm. Calculate the image distance and height of the image formed.
Answer:
(a) (i) The Principal focus of a concave mirror (diverging mirror) is a point on its principal axis from which a beam of light rays, initially parallel to the axis appears to diverge after being reflected from the convex mirror.

A concave mirror has virtual focus as it is situated behind the mirror.
(ii) The focal length of a convex mirror is equal to half of its radius of curvature.

OR
(b) Given. Object height, h1 = +10 cm;
Object distance, u = – 25 cm
Focal length, f = +15 cm;
Image distance, v = ?
Image height, h2 = ?
According to lens formula, \(\frac { 1 }{ v }\) – \(\frac { 1 }{ u }\) = \(\frac { 1 }{ f }\)
⇒ \(\frac { 1 }{ v }\) – \(\frac { 1 }{ -25 }\) = \(\frac { 1 }{ 15 }\)
⇒ \(\frac { 1 }{ v }\) + \(\frac { 1 }{ 25 }\) = \(\frac { 1 }{ 15 }\)
⇒ \(\frac { 1 }{ v }\) = \(\frac { 1 }{ 15 }\) – \(\frac { 1 }{ 25 }\) = \(\frac { 5-3 }{ 75 }\) = \(\frac { 2 }{ 75 }\)
⇒ v = \(\frac { 75 }{ 2 }\) or +37.5 cm
∴ Image is 37.5 cm away from the converging lens, i.e., convex lens and +ve sign of v shows that the image is real.

Now; \(\frac{h_2}{h_1}\) = \(\frac{v}{u}\)
⇒ \(\frac{h_2}{+10}\) = \(\frac{75}{2}\) × \(\frac{1}{-25}\)
⇒ h2 = –\(\frac{75 \times 10}{50}\)
∴ h2 = -15 cm
∴ Height of the image is 15 cm and -ve sign shows that the image is inverted.

Question 31.
The power of a lens is +4D. Find the focal length of this lens. An object is placed at a distance of 50 cm from the optical centre of this lens. State the nature and magnification of the image formed by the lens and also draw a ray diagram to justify your answer.
Answer:
We have, P = +4D
Now, f = \(\frac{1}{P}\)m or \(\frac{100}{P}\)cm = \(\frac{100}{+4}\) = +25 cm
+Ve sign of f shows that the lens is a convex lens.

Here, f = +25 cm;
u = -50 cm;
v = ?
m = ?
⇒ \(\frac { 1 }{ v }\) – \(\frac { 1 }{ -50 }\) = \(\frac { 1 }{ 25 }\)
⇒ \(\frac { 1 }{ v }\) + \(\frac { 1 }{ 50 }\) = \(\frac { 1 }{ 25 }\)
⇒ \(\frac { 1 }{ v }\) = \(\frac { 1 }{ 25 }\) – \(\frac { 1 }{ 50 }\) = \(\frac { 2-1 }{ 50 }\) = \(\frac { 1 }{ 50 }\)
∴ v = +50 cm
Now, magnification m = \(\frac { v }{ u }\) = \(\frac{+50}{-50}\) = -1
Hence, image is formed at 50 cm from the optical centre of the lens but on the other side of the object, therefore the image is real.
Here, m = -1 shows that size of the image is same as the object and ve sign of magnification shows that the image is inverted.
Therefore, A real, inverted, same size of the object image is formed.

Question 32.
(a) (i) Why is an alternating current (AC) considered to be advantageous over direct current (DC) for the long distance transmission of electric power?
(ii) How is the type of current used in household supply different from the one given by a battery of dry cells?
(iii) How does an electric fuse prevent the electric circuit and the appliances from a possible damage due to short circuiting or overloading.
OR
(b) For the current carrying solenoid as shown, draw magnetic field lines and give reason to explain that out of the three points A, B and C, at which point the field strength is . maximum and at which point it is minimum?

Answer:
(a) (i) Alternating Current (AC) is considered to be advantageous over Direct Current (DC) because Alternating Current (AC) can be transmitted over long distances without much loss of electrical energy.
(ii) The current used in household supply is Alternating Current (AC) whereas the current given by a battery of dry cells is Direct Current (DC).

Alternating Current (AC) Direct Current (DC)
(i) If the current reverses its direction after equal intervals of time, it is called alternating current. If the current flows in one direction only, it is called a direct current.
(ii) The magnitude and direction of current change continuously at definite intervals of time. The magnitude and direction of flow of current remain the same.
(iii) The magnitude of AC becomes Zero after a regular interval of time. The magnitude of DC does not become zero with the passage of time.

(iii) Overloading or short circuiting can be highly damaging to electrical appliances and buildings. So, fuse of proper rating must be used to avoid such damage. A fuse-wire will melt before the temperature of the heated circuit wire becomes too high and causes the circuit to break.
OR
(b) The magnetic field produced is directly proportional to the current passing in the wire and inversely proportional to the distance of the point from the current carrying conductor. Therefore as the distance of a point from the current carrying conductor increases, the strength of

the magnetic field becomes weaker. Out of the three given points A, B and C, Point A has the maximum magnetic field strength as it is at minimum distance from the current carrying conductor. Point C has the minimum magnetic field strength as it is at maximum distance from the current carrying conductor.

Question 33.
Write one difference between biodegradable and non-biodegradable wastes. List two impacts of each type of the accumulated waste on environment if not disposed off properly.
Answer:

Biodegradable wastes Non-biodegradable wastes
(i) Wastes which can be broken down into non-poisonous substances in nature in due course of time by the action of microorganisms are called biodegradable wastes. Wastes which cannot be broken down into non-poisonous substances in nature are called non-biodegradable wastes.
(ii) Examples: Sewage, peals of fruits and vegetables, etc. Examples: Plastic and glass, DDT etc.

Impact of biodegradable waste on the environment:
1. Biodegradable wastes are decomposed by the action of micro-organisms and may produce foul smell during the decomposition process.
2. Waste decomposition in landfills produces harmful methane gas which is a green house gas (100 to 120 times more powerful than CO2) that contributes to Global warming.

Impact of non-biodegradable waste on the environment:
1. Non-biodegradable waste does not decompose. These accumulate in the environment and cause biomagnification. As a result of this accumulation, they pollute soil and water.
2. Non-biodegradable wastes such as pesticides deplete soil fertility and so reduce the crop yield.
3. If these wastes are burnt, it causes air pollution.

Section – D (Long Answer Questions)

Question 34.
(a) (i) Draw the structure of the following compounds:
(a) Butanoic acid
(b) Chloropentane
(ii) How are structure (i) and structure (ii) given below related to one another? Give reason to justify your answer.

Draw one more possible structure for above case.
(iii) Differentiate between saturated and unsaturated carbon compounds on the basis of their general formula.
OR
(i) What happens when a small piece of sodium is dropped in ethanol? Write the equation for this reaction.
(ii) Why is glacial acetic acid called so?
(iii) What happens when ethanol is heated at 443 K in the presence of cone. H2SO4? Write the role of cone. H2SO4 in this case.
(iv) Write an equation showing saponification.
Answer:

The two given compound structures (i) and (ii) are the two isomers of C6H14 because both compound have same molecular formula but different structure formula.

(iii)

Saturated Hydrocarbon Unsaturated Hydrocarbon
General Formula CnH2n+2 (a) General formula of unsaturated hydrocarbon having double bond is CnH2n
(b) General formula of unsaturated hydrocarbon having triple bond is CnH2n-2

(b) (i) When a small piece of sodium is dropped in ethanol, then ethanol reacts with sodium to form sodium ethoxide and hydrogen gas. The reaction used as a test for ethanol is

(ii) When pure ethanoic acid (CH3COOH) is cooled, it freezes to form a colourless, ice-like solid which looks like a glacier. Therefore pure ethanoic acid is called glacial acetic acid.

(iii) When ethanol is heated at 443 k in the presence of conc. H2SO4, it gets dehydrated to form ethane.

The conc. H2SO4 acts as a dehydrating agent which removes water molecule from the ethanol molecule.

(iv) The process of making soap by the hydrolysis of fats and oils with alkalis is called saponification.

The alkaline hydrolysis of esters using alkali like sodium hydroxide is known as saponification.

Question 35.
(i) Name and explain the two modes of asexual reproduction observed in hydra.
(ii) What is vegetative propagation? List two advantages of using this technique.
Answer:
(i) Mode of asexual reproduction in Hydra:
1. Budding. Hydra reproduces by budding which is an asexual type of reproduction.

During this type of reproduction, a bulb like projection arises from the parent body which is known as bud. The bud may be unicellular or multicellular formed by mitotic division of its cells. This bud then grows gradually to form a small hydra and finally, the tiny new hydra detaches itself from the body of the parent hydra and lives as a separate organism. Thus, the parent hydra produces a new hydra.

2. Regeneration. Hydra can reproduce by the asexual mode of regeneration. If the body of hydra gets cut into a number of pieces, then each body piece can regenerate into a complete hydra by growing all the missing parts. The regeneration of hydra from its cut body part occurs by the process of growth and development. The cells of cut body part divide rapidly to make ball of cells. The cells then become specialised to form different types of tissues which again form various organs and body parts.

(ii) Vegetative propagation is the mode of asexual reproduction in plants in which new plants are obtained from the parts of old plants like stem, roots and leaves, without the help of any reproductive organs.
Advantages of Vegetative propagation:
(a) Plants grown by vegetative propagation grow much faster than those grown from seeds.
(b) A large number of plants can be produced by the method of vegetative propagation.
(c) The plants grown by vegetative propagation usually need less attention in their early years than the plants grown from seeds.
(d) Seedless plants can also be grown from this method.

Question 36.
(i) How is electric current related to the potential difference across the terminals of a conductor? Draw a labelled circuit diagram to verify this relationship.
(ii) Why should an ammeter have low resistance?

(iii) Two V-I graphs A and B for series and parallel combinations of two resistors are as shown. Giving reason state which graph shows (a) series, (b) parallel combination of the resistors.
Answer:
(i) Ohm’s law derives a relation between electric current and potential difference across the terminals of a conductor.
Ohm’s law states that the electric current, through a conductor, is directly proportional to the potential difference across its two ends when, other physical conditions like temperature, etc., remain constant.
V ∝ I or \(\frac { V }{ I }\) = Constant = R or V = IR
Thus, the ratio V : I is a constant. This constant is called the resistance (R) of the conductor. Circuit diagram for Ohm’s law:

(ii) Since the entire current passes through the ammeter, therefore, an ammeter should have low resistance so that it may not change the value of the currer flowing in the circuit.

(iii) Resistance of combination in ‘A’ is greater than the resistance of combination in ‘B’. Therefore, Graph A shows series combination as resultant resistance in series is given by
R = R1 + R2

Whereas Graph B shows parallel combination as resultant resistance in parallel combination is given by
\(\frac{1}{R}\) = \(\frac{1}{R_1}\) + \(\frac{1}{R_2}\)

Section – E (Source Based/Case Based Questions)

Question 37.
The melting points and boiling points of some ionic compounds are given below:

Compound Melting Point (K) Boiling Point (K)
NaCl 1074 1686
LiCl 887 1600
CaCl2 1045 1900
CaO 2850 3120
MgCl2 981 1685

These compounds are termed ionic because they are formed by the transfer of electrons from a metal to a non-metal. The electron transfer in such compounds is controlled by the electronic configuration of the elements involved. Every element tends to attain a completely filled valence shell of its nearest noble gas or a stable octet.
(i) Show the electron transfer in the formation of magnesium chloride.
(ii) List two properties of ionic compounds other than their high melting and boiling points.
(iii) (A) While forming an ionic compound say sodium chloride how does sodium atom attain its stable configuration?
OR
(iii) (B) Give reasons:
(i) Why do ionic compounds in the solid state not conduct electricity?
(ii) What happens at the cathode when electricity is passed through an aqueous solution of sodium chloride?
Answer:
(i) Mg has two electrons in its outermost shell (atomic no. of Mg is 12 so electronic configuration is 2, 8, 2).
Whereas Cl has seven electrons in its outermost shell (atomic no. of Cl is 17, so electronic configuration is 2, 8, 7).

(ii) 1. Ionic compounds conduct electricity when dissolved in water or when melted.

2. Ionic compounds are usually soluble in water but insoluble in organic solvents like ether, acetone etc.
(iii) (A) Atomic no. of Na = 11;
Electronic configuration = 2, 8, 1
For attaining stable electronic configuration either it has to lose 1 electron or it gain 7 electrons. So, losing one electron is easier than gaining 7 electrons.
Thus, Na loses its one electron and attains one positive charge and gives it to Cl which needs one electron to attain stable configuration and thus Cl acquires one negative charge.

OR
(iii) (B) (i) Ionic compounds are made up of ions but they do not conduct electricity in their solid state because in the solid ionic compound, the ions are held together in fixed positions by strong electrostatic forces and can not move freely.
(ii) When electricity is passed through an aqueous solution of sodium chloride, sodium ions move towards cathode (which is a negatively charged electrode).

Question 38.
The most obvious outcome of the reproductive process is the generation of individuals of similar design, but in sexual reproduction they may not be exactly alike. The resemblances as well as differences are marked. The rules of heredity determine the process by which traits and characteristics are reliably inherited. Many experiments have been dbne to study the rules of inheritance.
(i) Why an offspring of human being is not a true copy of his parents in sexual reproduction?
(ii) While performing experiments on inheritance in plants, what is the difference between F1 and F2 generation?
(iii) (A) Why do we say that variations are useful for the survival of a species over time?
Or, (B) Study Mendel’s cross between two plants with a pair of contrasting characters.

He observed 4 types of combinations in F2 generation. Which of these were new combinations? Why do new features which are not present in the parents, appear in F2 generation?
Answer:
(i) An offspring of human being is not a true copy of his parents in sexual reproduction because it inherits half of its genetic material from each parent. During the formation of gametes, the genetic material undergoes a process of recombination known as crossing over.
(ii)

F1 Generation F2 Generation
It is the first filial generation obtained by crossing two different parents. It is the second filial generation obtained by self crossing the F1 generation.

(iii) (A) Variation is necessary for the survival of a species as variation makes species more adapted to survive and grow in the changing environmental conditions. The variant species helps the species overcome the adversities of extreme conditions in their surrounding environment. Therefore, they can survive better and reproduce to pass the traits to the offspring.
(B)

In the inheritance of more than one pair of traits in a cross simultaneously, the factors responsible for each pair of traits are distributed independently to the gametes. It means that though the two pairs of original characters (shape and colour of the seed) combine in F1 generation but they are separate and behave independently in subsequent generations which is known as “Law of Independent Assortment.”

Question 39.
The ability of a medium to refract light is expressed in terms of its optical density. Optical density has a definite connotation. It is not the same as mass density. On comparing two media, the one with the large refractive index is optically.denser medium than the other. The other medium with a lower refractive index is optically rarer. Also the speed of light through a given medium is inversely proportional to its optical density.
(i) Determine the speed of light in diamond if the refractive index of diamond with respect to vacuum is 2.42. Speed of light in vacuum is 3 × 108 m/s.

(ii) Refractive indices of glass, water and carbon disulphide are 1.5, 1.33 and 1.62 respectively. If a ray of light is incident in these media at the same angle (say θ), then write the increasing order of the angle of refraction in these media.

(iii) (A) The speed of light in glass is 2 × 108 m/s and in water is 2.25 × 108 m/s.
1. Which one of the two is optically denser and why?
2. A ray of light is incident normally at the water-glass interface when it enters a thick glass container filled with water. What will happen to the path of the ray after entering the glass? Give reason.
OR
(B) The absolute refractive indices of water and glass are 4/3 and 3/2 respectively. If the speed of light in glass is 2 × 108 m/s, find the speed of light in (i) vacuum and (ii) water.
Answer:
(i) Given. Refractive index of diamond w.r.t. vacuum, n2 = 2.42
Speed of light in vacuum = 3 × 108 m/s
Let speed of light in vacuum be v1.
Speed of light in diamond = v2
As we know, n2 = \(\frac{\text { Speed of light in vaccum }\left(v_1\right)}{\text { Speed of light in diamond }\left(v_2\right)}\)
⇒ 2.42 = \(\frac{3 \times 10^8}{v_2}\)
⇒ v2 = \(\frac{3 \times 10^8}{2.42}\) m/s
∴ v2 = 1.24 × 108 m/s

(ii) ng = 1.5 ; nw = 1.33 and \(n_{c s_2}\) = 1.62
Since higher the refractive index of a substance, more it will change the direction (or bending towards the mirror) of a beam of light passing through it.
Increasing order of angle of refraction in media:
Carbon disulphide (1.62) < Glass (1.5) < Water (1.33)

(iii) (A) 1. Speed of light in glass, Vg = 2 × 108 m/s
Speed of light in water, Vw = 2.25 × 108 m/s
Glass is optically denser than water because speed of light decreases as the optical density of a medium increases.

2. When a ray of light is incident normally then it goes straight (or normally) in the other miedium (i.e., glass) because the incident ray goes along the normal to the surface, the angle of incidence in this case is zero (0) and the angle of refraction is also zero (0).
OR
(iii) (B) Given. Absolute refractive index of water, = vnw = \(\frac { 4 }{ 3 }\)
Absolute refractive index of glass, = vng = \(\frac { 3 }{ 2 }\)
Speed of light in glass, Vg = 2 × 108m/s
Speed of light in vacuum, Vl = ?;
Speed of light in water, Vw = ?
Now, vng = \(\frac{\mathrm{V}_l}{\mathrm{~V}_g}\)
⇒ \(\frac { 3 }{ 2 }\) = \(\frac{\mathrm{V}_l}{2 \times 10^8}\)
⇒ Vl = \(\frac { 3 }{ 2 }\) × 2 × 108 = 3 × 108 m/s
Hence Speed of light in Vaccum, Vl = 3 × 108 m/s.
Now, vng = \(\frac{\mathrm{V}_l}{\mathrm{~V}_w}\)
⇒ \(\frac { 4 }{ 3 }\) = \(\frac{3 \times 10^8}{\mathrm{~V}_w}\)
∴ Vw = 3 × 108 × \(\frac { 3 }{ 4 }\) = 2.25 × 108 m/s
Therefore speed of light in water, Vw is 2.25 × 108 m/s

SET – II Code No. 31/4/2
Except for the following questions, all the remaining questions have been asked in Set I.

Question 9.
During adolescence, reproductive phase starts and:
(a) general growth rate begins to slow down.
(b) height becomes less.
(c) the body weight is reduced.
(d) hair growth decreases.
Answer:
(a) general growth rate begins to slow down.

Question 10.
Which pair of sex chromosomes will determine a male?
(a) XO
(b) XX
(c) XY
(d) YY
Answer:
(c) XY

Question 11.
One of the events that does not occur during photosynthesis is:
(a) Chlorophyll absorbs solar energy.
(b) Carbon dioxide is released during the process.
(c) Oxygen is released during the process.
(d) Carbon dioxide is absorbed during the process.
Answer:
(b) Carbon dioxide is released during the process.

Question 21.
(i) State the essential function performed by ozone at the higher levels of the atmosphere.
(ii) Why was there a sharp drop in the amount of ozone in the atmosphere in 1980s.
Answer:
(i) Ozone layer is very important for the existence of life on earth because it absorbs most of the harmful ultraviolet radiations coming from the Sun and prevents them from reaching the earth.

(ii) The depletion of ozone layer is due to the use of chemicals called chlorofluorocarbons (CFCs). CFCs are the chemicals which are widely used in refrigeration (refrigerators and air conditioners) as a coolant; in fire extinguishers and in aerosol sprayers. When CFCs is released into the air, it reacts with ozone gas present in the ozone layer and destroys it gradually. Due to this, the ozone layer in the upper atmosphere has become thinner, allowing more UV rays to pass through it to the earth.

Question 24.
Give two reasons, why bile juice is considered to be an important secretion of liver in the process of digestion?
Answer:
Use of bile juice.
(i) Bile makes the acidic food coming from the stomach alkaline, so that pancreatic enzymes can act on it.
(ii) Bile salts break the fats present in the food into small globules making it easy for the enzymes to act and digest them.

Question 25.
Name the hormone secreted in scary situations by animals. Write any three responses which enable the animal body to deal with it.
Answer:
Adrenaline hormone is secreted by animals in large amounts when the person is frightened or excited.
When adrenaline is secreted in large amounts –

  • It prepares the body for action.
  • It speeds up heartbeat and breathing.
  • It raises blood pressure and allows more glucose (carbohydrate) to go into the blood to give a lot of energy quickly to fight or flight.
  • Adrenal glands are often called glands of emergency.

Question 29.
List two differences in the characteristic properties of the virtual images formed by the two types of spherical lenses (concave and convex). How are these characteristics of the two lenses used in the correction of the two common defects of vision namely myopia and hypermetropia?
Answer:

Convex lens Concave lens
A convex lens produces an enlarged virtual image when an object is within the focus of a convex lens. A concave lens produces a diminished image for all positions of the object.

– When a concave lens of suitable power is placed in front of the myopic eye, the parallel rays of light coming from the distant object (at infinity) are first diverged by concave lens. Due to this the concave lens forms a virtual image of the distant object at the far point of the myopic eye. Now this virtual image of the distant object is easily focused by the eye lens to form an image on the retina.
– When a convex lens of suitable power is placed in front of the hypermetropic eye, then the diverging rays of light coming from the nearby object are first converged by the convex lens. Due to this, the convex lens forms a virtual image of the nearby object (which is lying at 25 cm) at the near point of the hypermetropic eye, which can be easily focused by the eye-lens to the image on the retina.

Question 30.
(a) An object is kept at a distance of 1 m from a lens of power +2D:
(i) Identify the type of lens.
(ii) Calculate its focal length and distance of the image formed.
OR
(b) Define the following terms in the context of a diverging lens:
(i) Principal focus,
(ii) Focal length.
Draw a labelled ray diagram to illustrate your answer.
Answer:
(a) Given u = 1 m or -100 cm; P = +2D
(i) +Ve sign of P shows that the lens is a convex lens.
(ii) As we know, f = \(\frac { 1 }{ P }\) or \(\frac { 1 }{ P }\) × 100 cm
⇒ f = \(\frac { 1 }{ 2 }\) × 100 cm = +50 cm
According to lens formula, \(\frac { 1 }{ v }\) – \(\frac { 1 }{ u }\) = \(\frac { 1 }{ f }\)
⇒ \(\frac { 1 }{ v }\) – \(\frac { 1 }{ -100 }\) = \(\frac { 1 }{ +50 }\)
⇒ \(\frac { 1 }{ v }\) + \(\frac { 1 }{ 100 }\) = \(\frac { 1 }{ 50 }\)
⇒ \(\frac { 1 }{ v }\) = \(\frac { 1 }{ 50 }\) – \(\frac { 1 }{ 100 }\)
⇒ \(\frac { 1 }{ v }\) = \(\frac { 2 – 1 }{ 100 }\) = \(\frac { 1 }{ 100 }\)
∴ v = 100 cm or 1 m
OR
(b) (i) The principal focus of a concave lens (diverging lens) is a point on its principal axis from which light rays, originally parallel to the Principal axis, appear to diverge after passing through the concave lens. A concave lens has a virtual focus.

(ii) The distance between the optical centre of the concave lens and the Principal focus is called the focal length. The focal length of a concave lens is always negative (according to sign convention).

Question 32.
(i) Why is acidified water considered to be a good conductor of electricity?
(ii) Write a chemical equation showing the ionic products formed on dissolving potassium hydroxide in water.
(iii) Care must be taken while diluting concentrated nitric acid with water. Why?
Answer:
(i) When an acid is dissolved in water, then acid dissociates and produces Hydrogen ions into the water. The presence of free ions that can move through the medium makes the acidified water a good conductor of electricity.
(ii) Potassium hydroxide which is a base when dissolves in water, it produces potassium ion and hydroxide ions.

(iii) The dilution of a concentrated acid like conc. nitric acid should always be done by adding concentrated acid to water gradually with stirring and not by adding water to concentrated acid because:
(a) When a concentrated acid is added to water for dilution, then the heat is evolved (because the process of mixing the concentrated acid with water is highly exothermic) gradually, and easily absorbed by the large amount of water.
(b) When water is added to concentrated acid, then a large amount of heat is evolved at once. This heat changes some of the water to steam explosively which can splash the acid on our face or clothes and cause acid burns.

Question 36.
(a) (i) It is observed that covalent compounds are bad conductors of electricity. Give reason.
(ii) Carbon can neither form C4+ cation nor C4- anion. Why?
(iii) Draw electron dot structure of Ethanol.
(iv) Identify hetero atom(s) in the following compounds:

OR
(b) (i) What are soaps? Explain the mechanism of cleaning action of soap with the help of a labelled diagram.
(ii) Detergents are better than soaps. Justify.
Answer:
(a) (i) Covalent compounds are bad conductor of electricity because they do not contain ions.
(ii) Carbon atoms have 4 electrons in their outermost shell, so it needs to gain or lose 4 electrons to attain noble gas configuration. If carbon gains four electrons forming C4+ anion, it would be difficult for the nucleus with six protons to hold ten electrons. If carbon loses 4 electrons forming C4+ cation, it would require a large amount of energy to remove four electrons from its outermost shell. Thus, carbon can neither form C4+ cation nor C4- anion.

OR
(b)
(i) A soap is the sodium salt (or potassium salt) of a long chain carboxylic acid (fatty acid) which has cleaning properties in water.

Action of soap in removing an oily spot from a piece of cloth. Soaps are molecules in which the two ends have differing properties, one is hydrophilic, that is, it dissolves in water, while the other end is hydrophobic, that is, it dissolves in hydrocarbons. When soap is at the surface of water, the hydrophobic ‘tail’ of soap will not be soluble in water and the soap will align along the surface of water with the ionic end in water and the hydrocarbon ‘tail’ protruding out of water. Inside water, these molecules have a unique orientation that keeps the hydrocarbon portion out of the water.

This is achieved by forming clusters of molecules in which the hydrophobic tails are in the interior of the cluster and the ionic ends are on the surface of the cluster. This formation is called a micelle. Soap in the form of a micelle is able to clean, since the oily dirt will be collected in the centre of the micelle. The micelles stay in solution as a colloid and will not come together to precipitate because of ion-ion repulsion. Thus, the dirt suspended in the micelles is also easily rinsed away.

(ii) Detergents are better than soaps because of the following reasons:
(1) Detergents can be used even with hard water whereas soaps are not suitable for use with hard water.
(2) Detergents have a stronger cleansing action than soaps.
(3) Detergents are more soluble in water than soaps.

SET – III Code No. 31/4/3
Except for the following questions, all the remaining questions have been asked in Set I & IL

Question 9.
Water in the root enters due to:
(a) the function of the root to absorb water.
(b) difference in the concentration of ions between the root and the soil.
(c) excess water present in the soil.
(d) diffusion of water in the roots.
Answer:
(d) diffusion of water in the roots.

Question 10.
Which one of the given statements is incorrect:
(a) DNA has the complete information for a particular characteristic.
(b) DNA is the molecule responsible for the inheritance of characters from parents to offsprings.
(c) Change in information will produce a different protein.
(d) Characteristics will remain the same even if protein changes.
Answer:
(d) Characteristics will remain the same even if protein changes.

Question 11.
Sensory nerve of a reflex arc carries information from the receptor cells to the:
(a) spinal cord
(b) brain
(c) muscles of the effector organ
(d) bones of the receptor organ
Answer:
(a) spinal cord

Question 15.
An alpha particle enters a uniform magnetic field as shown. The direction of motion of the alpha particle is:

(a) towards right
(b) towards left
(c) into the page
(d) out of the page
Answer:
(c) into the page

Question 21.
Name a plant hormone responsible for bending of a shoot of a plant when it is exposed to unidirectional light. How does it promote phototropism?
Answer:
‘Auxin’ hormone is responsible for bending of a shoot of a plant when it is exposed to unidirectional light. Auxin hormone is secreted in the growing plant which detects light. It is present in the tip of the growing stem.

Plants appear to bend towards light. As more auxin is produced on the shaded side than on the lighted side of the stem, so stem grows faster on the shaded side than on the lighted side.

Question 24.
What is the other name of ’tissue fluid’? Write its two functions.
Answer:
Tissue fluid is also called lymph.
Functions of lymph:
(i) Lymph protects the body by killing the germs drained out of the body tissue with the help of lymphocytes, and by making antibodies.
(ii) Lymph carries digested fat (which is a large molecule and could not be absorbed by blood stream) for the nutritive process.
(iii) Lymph helps in removing the waste products like fragments of dead cells etc.

Question 25.
“Although gardens are created by man but they are considered to be an ecosystem.” Justify this statement.
Answer:
An ecosystem is a self-contained unit of living things (plants, animals and decomposers) and their non-living environment (soil, air, water). An ecosystem needs only the input of sunlight energy for its functioning.
A garden although man-made is a self sufficient or independent unit. It contains all the components of the ecosystem. In this ecosystem, producers (plants and trees of the garden) trap the solar energy and then provide the basic food or energy for all other life in the garden like monkeys, squirrels, birds etc. When the producers and consumers die, the decomposers present in the garden act on their dead bodies to return the various elements back to the nutrient pool.

Question 27.
Consider the following salts:
(i) YCl
(ii) NH4X
(iii) ZCO3
(a) What would be the pH of the salt solution if in YCl, Y is sodium? Give reason for your answer.
(b) If in salt NH4X, X is nitrate, then its solution will give what colour with universal indicator? Why?
(c) What would be the change in colour in blue litmus solution if ZCO3 is added to it and Z is potassium?
Answer:
(a) In YCl, if Y is sodium then the salt in NaCl (sodium chloride) is formed by sodium hydroxide (NaOH) which is a strong base. HCl which is also a strong acid, therefore pH of NaCl (aq) is 7 . As the salts of strong acid and strong base gives neutral solutions.

(b) If in salt NH4 X , X is nitrate, then the salt is NH4NO3, which is the salt of strong acid (HNO3) and weak base (NH4OH). Thus, solution of NH4NO3 is acidic having pH less than 7. Solution of NH4HO3 will give orange colour with universal indicator.

(c) If in salt ZCO3, Z is potassium then the salt is K2CO3 which is the salt of weak acid (H2CO3 ) and strong base (KOH). Thus, solution of K2CO3 is basic having pH more than 7. If blue litmus solution is added to K2CO3 it remains blue. No colour change will be observed.

Question 29.
The magnification produced when an object is placed at a distance of 20 cm from a spherical mirror is +1/2. Where should the object be placed to reduce the magnification to +1/3?
Answer:
1. When the object is placed at a distance of 20 cm from a spherical mirror.
Given. Magnification, m = \(\frac { +1 }{ 2 }\). As we know, m = \(\frac{h_2}{h_1}\)
⇒ \(\frac{h_2}{h_1}\) = +\(\frac { +1 }{ 2 }\)
⇒ h2 = +\(\frac { 1 }{ 2 }\)h1
+ve sign of h2 shows that image is erect and diminished.
Also we have, u = -20 cm
⇒ \(\frac { -v }{ u }\) = + \(\frac { 1 }{ 2 }\)
⇒ \(\frac { -v }{ -20 }\) = + \(\frac { 1 }{ 2 }\)
∴ v = \(\frac { 1 }{ 2 }\) × 20 = + 10 cm
Now, \(\frac { 1 }{ f }\) = \(\frac { 1 }{ v }\) + \(\frac { 1 }{ u }\)
⇒ \(\frac { 1 }{ f }\) = \(\frac { 1 }{ 10 }\) + \(\frac { 1 }{ -20 }\)
⇒ \(\frac { 1 }{ f }\) = \(\frac { 1 }{ 10 }\) – \(\frac { 1 }{ 20 }\)
⇒ \(\frac { 1 }{ f }\) = \(\frac { 2 – 1 }{ 10 }\) + \(\frac { 1 }{ 20 }\)
∴ f = +20 cm
+ve sign of f shows that spherical mirror is a convex mirror.

2. When an object be placed to reduce the magnification, m to + \(\frac { 1 }{ 3 }\)
⇒ \(\frac { -v }{ u }\) = \(\frac { 1 }{ 3 }\) …[m = \(\frac { -v }{ u }\)
⇒ v = –\(\frac { 3 }{ u }\)
As we know, \(\frac { 1 }{ v }\) + \(\frac { 1 }{ u }\) = \(\frac { 1 }{ f }\)
⇒ \(\frac{1}{\frac{-u}{3}}\) + \(\frac { 1 }{ u }\) = \(\frac { 1 }{ 20 }\)
⇒ – \(\frac { 3 }{ n }\) + \(\frac { 1 }{ u }\) = \(\frac { 1 }{ 20 }\)
⇒ \(\frac{-3+1}{u}\) = \(\frac { 1 }{ 20 }\)
⇒ \(\frac{-2}{u}\) = \(\frac { 1 }{ 20 }\)
⇒ – 2 × 20 = u
∴ u = -40 cm
Thus, the object should be placed at 40 cm in front of the convex mirror so as to reduce the magnification to + 1/3.

Question 33.
(i) Define the term dispersion of white light. State the colour which bends (a) the most,
(b) the least while passing through a glass prism.
Draw a diagram to show the dispersion of white light.
OR
(ii) What is a rainbow? Draw a labelled diagram to show its formation.
Answer:
1. The splitting up of white light into seven colours on passing through a transparent medium like glass prism is called dispersion of light.
2. Since the violet colour has the minimum speed in glass prism, therefore violet colour deviates the most while passing through a glass prism.
3. Since the red colour has the maximum speed in glass prism, so the red colour deviates the least, while passing through glass prism.

OR
(b) The rainbow is an arch of seven colours visible in the sky which is produced by the dispersion of sunlight by raindrops present in the atmosphere. The raindrops in the atmosphere act like many small prisms. As white sunlight enters and leaves these raindrops, the white light splits into an arch of seven colours of the rainbow.

Question 36.
(a) Write the chemical equation for the following:
(i) Combustion of methane
(ii) Oxidation of ethanol
(iii) Hydrogenation of ethene
(iv) Esterification Reaction
(v) Saponification Reaction
OR
(b) (i) Draw two structural isomers of butane.
(ii) Draw the structures of propanol and propanone.
(iii) Name the third homologue of:
(a) alcohols
(b) aldehydes
(iv) Name the following:

(v) Show the covalent bond formation in nitrogen molecule.
Answer:


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