CBSE Sample Papers for Class 10 Science Set 9 with Solutions - #NCSOLVE 📚

Students must start practicing the questions from CBSE Sample Papers for Class 10 Science with Solutions Set 9 are designed as per the revised syllabus.

CBSE Sample Papers for Class 10 Science Set 9 with Solutions

Time: 3 Hrs
Max. Marks: 80

General Instructions

• This question paper consists of 39 questions in 3 sections. Section A is Biology, Section B is Chemistry and Section C is Physics.
• All questions are compulsory. However, an internal choice is provided in some questions.
• A student is expected to attempt only one of these questions.

Section – A

Question 1.
In a food chain, the snake predated a rabbit which fed on fresh green bushes. What percentage of the energy accumulated bv rabbit, would be acquired by the snake? [1]
(a) 90°
(b) 10%
(c) 50%
(d) 25%
Answer:
(b) 10%

Explanation:
According to 10., energy law, in a food chain, only around 10 % of the available energy is passed on to the next trophic level. The rest of the energy is lost to the ecosystem in the form of heat.

Question 2.
How do spores develop into Rhizopus? [1]
(a) Spores divide and grow into a new individual.
(b) Spores combine with other spores and grow
(c) Spores enlarge in size for the growth of new individual
(d) Spores land on other organisms and increase with their growth in size
Answer:
(a) Spores divide and grow into a new individual.

Explanation:
Rhizopus reproduces through spore formation. Under favourable conditions, the spores divide and grows into new individuals.

Question 3.
Select the group in which all functions are controlled by the cerebellum. [1]
(a) Maintaining posture, balancing the body, coordinating voluntary movements
(b) Thinking, reasoning, memory storage
(c) Beating of heart, breathing, swallowing
(d) Reflex action, transmission of impulses, maintaining blood pressure
Answer:
(a) Maintaining posture, balancing the body, coordinating voluntary movements

Explanation:
Maintaining posture, balancing the body and coordinating voluntary movements belong to the same group as all these functions are controlled by cerebellum.

Question 4.
Which of the following is a correct combination of function and part of the heart? [1]
(a) Pumping oxygenated blood to the body: Left ventricle
(b) Receiving oxygenated blood from lungs: Right atrium
(c) Receiving deoxygenated blood from body: Left atrium
(d) Pumping deoxygenated blood to the body: Right ventricle
Answer:
(a) Pumping oxygenated blood to the body: Left ventricle

Explanation:
Pumping oxygenated blood to the body: left ventricle is the correct combination as left ventricle pumps oxygenated blood into the aorta, which carries it to all parts of the body under high pressure.

Question 5.
In human males, which chromosome does not have an identical pair? [1]
(a) Chromosome no. 1
(b) Chromosome no. 7
(c) Y-chromosome
(d) X-chromosome
Answer:
(c) Y-chromosome

Explanation:
The Y-chromosome pairs with the X-chromosome. So, it does not have an identical matching pair.

Question 6.
Which are the correct statements related to the reflex pathway? [1]
(i) It involves the brain in initiating the response.
(ii) The spinal cord acts as the coordinating centre. .
(iii) Sensory neurons carry impulses from receptors to the spinal cord.
(iv) Motor neurons carry impulses from spinal cord to effectors.
(v) Reflex actions are voluntary in nature.

(a) (i), (ii), (iii)
(b) (ii), (iii), (iv)
(c) (iii), (iv), (v)
(d) (i), (iv), (v)
Answer:
(b) (ii), (iii), (iv)

Explanation:
The statements (ii),(iii) and (iv) are correct because in a reflex pathway, the spinal cord serves as the coordinating centre. Sensory neurons transmit impulses from receptors to the spinal cord, and motor neurons carry impulses from the spinal cord to the effector organs to produce a quick and involuntary response.

Question 7.
Which statement is true for a dominant allele in Mendelian experiment. [1]
(a) It cannot undergo mutation ‘
(b) It gives a greater chance of survival than a recessive
(c) It gives the same phenotype in heterozygotes and homozygotes
(d) It is only responsible for male characteristics
Answer:
(c) It gives the same phenotype in heterozygotes and homozygotes

Explanation:
In Mendelian experiment dominant allele always expresses itself whether present in heterozygous or homozygous condition.

The following two questions consist of two statements – Assertion (A) and Reason (R). Answer these questions by selecting the appropriate option given below.
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true but (R) is not the correct explanation of (A).
(c) (A) is true but (R) is false.
(d) (A) is false but (R) is true.

Question 8.
Assertion (A) Lymph, also known as tissue fluid, is similar to plasma which is colourless. [1]
Reason (R) Lymph is colourless because it lacks RBCs.
Answer:
(a) Both A and R are true, and R is the correct explanation of A.

Question 9.
Assertion (A) After fertilisation, the ovule develops into a seed and the ovary into a fruit. [1]
Reason (R) Fertilisation triggers cell division and growth in the zygote and surrounding tissues.
Answer:
(a) Both A and R are true and R is the correct explanation of A.

Question 10.
“Placenta acts as a link between the mother and the foetus.” Comment upon the statement with justification. [1]
Answer:
Placenta acts as a link between the mother and the foetus as it connects the blood supply of the mother with that of the foetus without mixing the two. It allows the transfer of oxygen and nutrients from the mother to the foetus for its growth and development. It also removes carbon dioxide and waste products from the foetus to the mother’s blood for excretion. Thus, the placenta ensures nourishment, respiration and removal of wastes for the developing foetus throughout pregnancy.

Question 11.
Attempt either A or B. [2]
A. Different organisms have different mechanism of sex-determination. These mechanisms vary across species and ensure the development of male and female individuals accordingly. Identify and mention in one line the method of sex-determination in the following
(i) Humans
(ii) Reptiles
Answer:
(i) In humans, sex is determined by chromosomes – XX for female and XY for male.
(ii) In some reptiles, temperature during incubation of eggs determines whether the ‘Offspring will be male or female.

Or
B. Explain the mechanism by which tallness is inherited in pea plants, when a pure tall plant is crossed with a pure short plant. [2]
Answer:
When a pure tall pea plant (TT) is crossed with a pure short pea plant (tt), all plants in the F2-generation are tall (Tt). This happens because the allele for tallness (T) is dominant over the allele for shortness (t). The shortness trait is masked in the heterozygous condition, so all offspring show the tall character.

Question 12.
During an experiment on digestion, scientists observed that partially digested food from the stomach was
mixed with bile and pancreatic juice and then broken down into simple sugars, amino acids, and fatty acids. They also found that these nutrients were finally digested and absorbed into the blood through finger-like projections present in the inner lining of the organ. [2]
From the above information, identify the organ where final digestion and absorption of food takes place.
Answer:
The small intestine is the main organ for complete digestion and absorption of food. Bile from the liver emulsifies fats, and pancreatic and intestinal enzymes breakdown carbohydrates, proteins and fats into simple absorbable forms. And the inner walls of the small intestine have finger-like projections called villi, which increase the surface area, so the digested nutrients are absorbed through the villi into the blood and carried to different parts of the body.

Question 13.
Draw the structure of nephron and explain the process of urine formation. [3]
Answer:
A nephron is the basic unit of the kidney that filters blood and forms urine. It consists of a glomerulus enclosed in Bowman’s capsule, followed by a coiled tubule. Urine formation takes place in three steps filtration, reabsorption and secretion. First, blood is filtered in the glomerulus, and waste along with water passes into the capsule. Then, useful substances like glucose and water are reabsorbed in the tubule. Finally, excess waste is secreted into the tubule and the urine formed is collected in the collecting duct.

Question 14.
In an experiment on nerve impulse transmission, it was observed that the electrical signal moved from the axon end of one neuron to the dendrite of another neuron through a synapse, but never in the reverse direction. 13]
(i) Explain why the signal passes from the axon end of one neuron to the’dendrite of the next neuron, but not in the reverse direction.
(ii) What would happen to the transmission of the signal if neurotransmitters were not released at the synapse?
Answer:
(i) The signal passes from the axon end of one neuron to the dendrite of the next neuron because only the axon terminal has chemical substances (neurotransmitters) that carry the signal to the dendrite of the next neuron. The signal cannot travel backward because the dendrite end does not have neurotransmitters, so no chemicals are released to send the signal in reverse.
(ii) If neurotransmitters were not released, the signal would not pass to the next neuron and the message would stop there.

To get maximum marks, students should mention about neurotransmitters, as these are chemicals that transmit signals between neurons.

Question 15.
Riya planted a rose cutting in her home garden and also grew pea plants from seeds’ in her school laboratory. During the flowering season, she observed pollen grains on the stigma of the pea flowers. Help her to understand some concepts of reproduction in plants by answering the questions given below. [4]
Attempt either subpart A or B.
A. Differentiate between the method of reproduction in the above given plants.
Answer:
The rose plant reproduces by vegetative propagation, which is an asexual method of reproduction where new plants grow from parts like stems, roots or leaves. In this case, a cutting from the rose stem can grow into a new plant.
The pea plant reproduces by sexual reproduction, which involves the fusion of male and female gametes formed after pollination and fertilisation.

Or

B. Explain the process of pollen germination in flowering plants.
Answer:
Pollen germination is the process where a pollen grain lands on the stigma of a flower and germinates to form a pollen tube, which grows through the style into the ovary. The male gamete moves through this tube and fuses with the egg cell in the ovule.

C. Which method of reproduction in the above examples produces offspring genetically identical to the parent?
Answer:
The method that produces offspring genetically identical to the parent is vegetative propagation as seen in Rose plants.

D. Given below is a diagram of a flower. Identify and name the part that produces the male gametes and the part that receives them during pollination.

Answer:
The given figure represents a flower with different parts labelled as A, B, C, D and E. Out of these the part labelled C is the anther which produces the male gamete and the part labelled B is the stigma which receives the male gamete.

Question 16.
Attempt either A or B. [5]
A. Mrs. Sharma was teaching about the ozone layer in the atmosphere.
(i) What are ozone depleting substances? Give one example.
(ii) State any two harmful effects of ozone layer depletion.
Answer:
(i) Ozone depleting substances are chemicals which lead to the depletion of the ozone layer present in the upper atmosphere. These are mostly man-made. The most common example is chlorofluorocarbons (CFCs), used in refrigerators, air conditioners, aeroAnswer: sprays and in the production of foam. These chemicals, when released, reach the stratosphere where they breakdown ozone molecules, reducing the thickness of the ozone layer,

(ii) Two harmful effects of ozone layer depletion are

• The ozone layer acts as a protective shield by absorbing harmful ultraviolet (UV) rays from the Sun. Depletion of this layer allows more UV rays to reach the Earth’s surface.
• Increased exposure to UV rays can cause skin cancer, cataract in the eyes and damage to the immune system in humans.

Or

B. While studying the ozone cycle, Mrs. Sharma explained that ozone is constantly being formed and broken down in the atmosphere.
(i) How is ozone formed in the atmosphere?
(ii) State any two ways by which we can reduce ozone layer depletion.
Answer:
(i) Ozone is formed in the atmosphere When high-energy ultraviolet (UV) radiation From the Sun splits oxygen molecules (O₃) into single oxygen atoms. These atoms then combine with other oxygen molecules to form ozone (O₃). Under natural conditions, ozone molecules are constantly being formed and broken down, maintaining a balance in its concentration. The ozone layer is important as it absorbs harmful UV rays, preventing them from reaching the Earth’s surface.

(ii) Two ways toyeduce ozone layer depletion are

• By minimising the use of ozone depletion substances, e.g., use of CFC’s is cooling appliances and aerosol sprays shoulti by avoided.
• Use of eco-friendly afternatives, like servicing and proper disposal of old refrigerators and air conditioner to prevent leakage of CFC’s.

Students should not get confused about ozone, as it is good if present in the stratosphere but harmful IF present near Earth’s surface.

Section – B

Question 17.
Which of the following equations represent redox reactions and what are the correct products formed in each case? [1]
Equation 1: Fe(s) + CuSO₄(aq) → X(aq) + Cu(s)
Equation 2 : Zn(s) + AgNO₃(aq) → Y(aq) + Ag(s)
(a) Only equation lisa redox reaction; X = ferric sulphate, Y = zinc nitrate
(b) Both equations 1 and 2 are redox reactions; X = ferrous sulphate, Y = zinc nitrate
(c) Only equation 2 is a redox reaction; X = cuprous sulphate, Y = zinc nitrate
(d) Both equations 1 and 2 are redox reactions; X = iron sulphate, Y = silver nitrate
Answer:
(b) Both equations 1 and 2 are redox reactions; X = ferrous sulphate, Y = zinc nitrate

Explanation:
In both reactions, one metal displaces another from its saltSolution, involving oxidation and reduction. Iron forms ferrous sulphate (FeSO₄) by displacing copper, and zinc forms zinc nitrate [Zn(NO₃)₂] by displacing silver-confirming both are redox reactions.

Don’t make mistake in reactivity order of metals. Metals placed above in activity series are more reactive than metals placed below.

Question 18.
Four statements about the properties of homologous series are listed. [1]
I. They differ by —CH₂ units. II. They differ by 14 units by mass.
III. They all contain double bonds. IV. They all have similar physical properties
Which statements are correct?
(a) I and II
(b) I and III
(c) II and IV
(d) I and IV
Answer:
(a) I and II

Explanation:
Homologous series differ by —CH₂ units in chemical formula and 14 units in mass. It is not necessary for a homologous series to have double bonds in them. They show similar chemical properties.

Question 19.
Galvanisation is a method of protecting iron from rusting by coating it with a thin layer of [1]
(a) gallium
(b) aluminium
(c) zinc
(d) silver
Answer:
(c) zinc

Explanation:
Galvanisation is a method of protecting iron from rusting by coating it with a thin layer of zinc (Zn) metal.

Question 20.
Identify which of the following can show addition reaction? [1]
(a) C₂H₄
(b) C₂H₆
(c) C₂H₅OH
(d) CH₃CH₂CH₃
Answer:
(a) C₂H₄

Explanation:
Presence of double or triple bonds between the two carbon atoms is a necessary condition to show addition reaction. Thus, only C₂H₄(CH₂ = CH₂) can show addition reaction.

Question 21.
Pooja was given four metals – A, B, C, and D and asked to observe their general properties. She noted their valence electrons and physical behaviour. Based on this, which of- the following observations correctly represents a typical property of metals? [1]
(a) Metal A has 4 to 8 valence electrons and readily gains electrons.
(b) Metal B is brittle and breaks easily when hammered.
(c) Metal C has 1 to 3 valence electrons and tends to form positive ions.
(d) Metal D forms anions easily and is a poor conductor of electricity.
Answer:
(c) Metal C has 1 to 3 valence electrons and tends to form positive ions.

Explanation:
Metals generally have 1 to 3 valence electrons, are malleable (not brittle), and lose electrons easily to form electropositive ions (cations), not anions.

Question 22.
Which of the following substances when disAnswer:ved in equal volume of water, will have the highest pH value? [1]
(a) Sulphuric acid
(b) Acetic acid
(c) Magnesium hydroxide
(d) Sodium hydroxide
Answer:
(d) Sodium hydroxide

Explanation:
Sodium hydroxide is a very strong base. So, it will furnish a greater number of hydroxyl ions giving the highest pH.

Don’t confuse it with magnesium hydroxide. Although it is also a base, it is a weaker base compared to sodium hydroxide.

Question 23.
Which option shows a correct example of structural isomerism due to difference in carbon chain arrangement? [1]

	Compound 1	Compound 2
(a)	Butane (C4H10)	Isobutane (C4H10)
(b)	Ethanol (C2H5OH)	Dimethyl ether (C2H6O)
(c)	Propan-l-ol (C3H70)	Propan-2-ol (C3H7O)
(d)	Methane (CH4)	Ethane (C2H6)

Answer:
(a) Butane and isobutane show structural isomerism.

The following question consists of two statements -Assertion (A) and Reason (R). Answer the question by selecting the appropriate option given below.
(a) Both (A) and (R) are true, and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true but (R) is not the correct explanation of (A).
(c) (A) is true but (R) is false.
(d) (A) is false but (R) is true.

Question 24.
Assertion (A) Acetic acid is more acidic than alcohol. [1]
Reason (R) The ion formed after the removal of proton from acetic acid is less stable.
Answer:
(c) (A) is true but (R) is false.

Explanation:
Assertion is true, but Reason is false. (R) can be corrected as acetic acid is more acidic than alcohol because of the more stability of ion formed after the removal of a proton.

Question 25.
Study the given diagrams and answer the given questions. [2]

(a) Intake of small quantities of A can be lethal. Comment.
(b) What happens when a 5% alkaline potassium permanganateSolution is added to a warmSolution of B?
Answer:
(a) Methanol (A) is very poisonous because it is oxidised to methanal in the liver. Methanalreacts-rapidly with the components of cells and causes the protoplasm to get coagulated. Methanol also affects the optic nerve causing blindness.
(b) When 5% alkaline potassium permanganate solution is added drop by drop to warm propanol B, then oxidation reactionoccurkand propanoic acid is formed.

Question 26.
Attempt either A or B.
A. Pratyush took su1ur powder on a spatula and heated it. He collected the gas evolved by inverting a test tube over it, as shown in the figure below.
(i) What will be the action of the gas on dry blue litmus paper?
Answer:
No change in colour will take place in case of dry blue litmus paper.

(ii) Also, tell the action of moist blue litmus paper.
Answer:
The moist blue litmus paper will change its colour to red because sulphur is a non-metal,
So its oxides are acidic in nature.

(iii) Write a balanced chemical equation for the reaction taking place.

Answer:
S(s) + O₂(g) → SO₂(g)
SO₂(g) + H₂O(Z) → H₂SO₃ (Sulphurous add)

Or

B. A chemical compound called ‘X’, along with aluminium metal, is commonly used in the metallurgical process to effectively weld and join broken railway tracks.
(i) Identify the compound ‘X’ and name the reaction.
Answer:
X is Fe₂O₃, i.e., iron (III) oxide. The reaction involved is thermite reaction or aluminothermy.

(ii) Write the equation for it.
Answer:

Question 27.
Observe the figure given below and answer the following questions. [3]

(a) Identify the gases evolved.
(b) Name the type of the reaction shown in figure.
(c) What is the formula of crystalline ferrous sulphate? Write the equatîon involved.
Answer:
(a) SO₂ and SO₃ gases are evolved
(b) Thermal decomposition reaction occurs here.
(c) Formula of crystalline ferrous sulphate is FeSO₄ 7H₂O. Equation invoked is

Question 28.
Ravi took 2 mL of lemon juice (which is acidic in nature) in a test tube and added two drops of methyl orange indicator. The Solution turned red in colour. He then added dilute NaOH Solution drop by drop to this acidic Solution. After about 25 drops, the colour changed from red to orange and finally to yellow.
He repeated the experiment using different volumes of lemon juice and noted the number of drops of NaOH required for neutralisation in the table given below. [4]

S. No.	Volume of lemon juice taken (mL)	Drops of NaOH
1	1.0	12
2	1.5	18
3	2.0	25

Answer the following questions based on the above information.
A. If Ravi used concentrated NaOH instead of dilute NaOH, how many drops would be required for the change in colour to be observed?
(a) 25
(b) < 25 (c) > 2
Justify your answer.
Answer:
Correct option: (b) < 25

Concentrated NaOH contains more hydroxide ions (OH) per drop than dilute NaOH. Therefore, fewer drops will be needed to neutralise the same amount of acid. So, the number of drops required would be less than 25.

B. Ravi measured 20 drops of NaOH and found its volume to be 1 mL. If he used 3 mL of lemon juice and observed a colour change after adding 36 drops of NaOH, how many mL of NaOH did he use in total?
Answer:
If Ravi measured 20 drops of NaOH = 1 ml, and he added 36 drops for 3 mL of lemon juice.
Let’s find volume of NaOH used. 1 ml= 20 drops
So, 36 drops = \(\left(\frac{36}{20}\right)\)ml = 1.8 ml
Ravi used 1.8 ml of NaOH.

Or

Ravi takes 10 drops of lemon juice and adds a methyl orange indicator. Then he adds NaOH Solution dropwise. What colour change will he observe and why?
Answer:
If Ravi adds NaOH dropwise to 10 drops of lemon juice with methyl orange indicator: Initially, theSolution will be red due to the acidic nature of lemon juice. As NaOH is added slowly, neutralisation occurs. At the end point (neutral pH), the colour changes to orange, and finally, in basic medium, it becomes yellow.

C. Write a balanced chemical equation for the neutralisation reaction taking place in the experiment. Which of the following is true and why? The reaction is a
(a) Neutralisation and double displacement reaction
(b) Neutralisation and precipitation reaction
(c) Precipitation and double displacement reaction
(d) Neutralisation, double displacement as well as precipitation reaction
Answer:
The balanced chemical equation is as follows
HCl + NaOH → NaCl + H₂O
This is a neutralisation reaction where an acid reacts with a base to form salt and water. It is also a double displacement reaction because H⁺ from acid and OH^– from base exchange places. No precipitate is formed in this reaction.
Correct option (a) Neutralisation and double displacement reaction. The H+ ion from the acid and OH’ ion from the base form water, and Na⁺ and Cl^– form a salt. So, it’s both a neutralisation and a double displacement reaction

Question 29.
Attempt either A or B. [5]
A. Answer the following questions.
(a) Give a chemical test to distinguish between saturated and unsaturated hydrocarbon.
(b) What happens when ethene is reacted with bromine water?
(c) Name the products formed when ethane bums in air.
(d) What is a substitution reaction?
(e) Give the reaction between methane and chlorine in the presence of sunlight.
Answer:
(a) Saturated hydrocarbons bum with blue and non-smoky flame due to their complete combustion and unsaturated hydrocarbons generally bum with sooth flame due to their incomplete combustion.
(b) Ethene on reaction with bromine water gives addition reaction. This results in decolourisation of bromine water.

(c) Carbon dioxide and water are formed when ethane burns in air.

(d) A substitution reaction is that reaction in which one atom of the substrate is replaced by another atom. These are generally shown by alkyl halides and haloarenes.

(e) When chlorine is added to hydrocarbons in the presence of sunlight, Cl replaces H-atoms one by one to give carbon tetrachloride as the final product. These are generally shown by saturated compounds and benzene.

To obtain maximum marks, one should add concept of substitution reactions, i.e., the reactions in which a reagent substitutes (replaces) atom or a group of atoms from the reactant (substrate).

Or

B. Answer the following questions.
(a) Calcium on reacting with water starts floating. Give the reason.
(b) Write the balanced chemical equation of calcium with water..
(c) Rashmi was studying about CH₃Cl compound. Show the bond formation in this compound.
(d) How are covalent bonds formed?
(e) Write the molecular formula of alcohol which can be derived from butane.
Answer:
(a) Calcium reacts with cold water to form calcium hydroxide and hydrogen gas. The bubbles of hydrogen gas produced stick to the surface of calcium and hence, it starts floating on the surface of water.

(b) The balanced chemical reaction will be as follows

(c) Electronic configuration of carbon, 6 = (k= 2), (L = 4)
Electronic configuration of hydrogen, 1 = (k = 1)
Electronic configuration of chlorine, 17 = (K = 2), (L = 8), (M = 7)
Carbon atom has four outermost electrons, each hydrogen atom has one electron and chlorine has seven outermost electrons.
Carbon shares its four outermost electrons with 3 hydrogen atoms and 1 chlorine atom to form CH₃Cl as follows

(d) Covalent bonds are formed by the sharing of one or more electron-pairs between the atom (either same or different atoms), e.g. H₂, Cl₂, CH₄, NH₃, etc., have covalent bonds.

(e) Butanol, CH₃—CH₂—CH₂—CH₂—OH.

Section – C

Question 30.
Mamta was making notes and she wrote down the following statements about magnification in lenses and mirrors: [1]
I. The magnification is defined as the ratio of the height of the object to the height of the image.
II. The magnification is positive for all virtual images and is negative for all real images.
III. The magnification produced by a concave lens or a convex mirror is always positive, whereas the magnification of a convex lens or a concave mirror can be either positive or negative, depending on the position of the object in front of the lens or mirror.

Choose the correct option that lists the correct statements about magnification:
(a) I and II
(b) I and III
(c) II and III
(d) I, II, and III
Answer:
(c) II and III

Explanation:
Magnification is given as
m = \(\frac{\text { Height of image }}{\text { Height of object }}\)
Magnification is positive for virtual images and negative for real images. A concave lens and a convex mirror always form virtual images, so their magnification is always positive. In contrast, a convex lens and a concave mirror can produce both real and virtual images, resulting in negative magnification for real images and positive magnification for virtual images.

Question 31.
Choose the incorrect statement from the following regarding the formation and appearance of rainbows.
(a) The rainbow appears in the same direction as the sun. [1]
(b) A rainbow is formed when light undergoes dispersion and total internal reflection inside atmospheric water droplets.
(c) A rainbow is circular in shape.
(d) A primary rainbow forms due to one internal reflection and two refractions, while a secondary rainbow involves two internal reflections and two refractions.
Answer:
(a) The rainbow is always formed in an opposite position to that of the sun.

The following questions consist of two statements -Assertion (A) and Reason (R). Answer the question by selecting the appropriate option given below.
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true but (R) is not the correct explanation of (A).
(c) (A) is true but (R) is false.
(d) (A) is false but (R) is true.

Question 32.
Assertion (A) A current-carrying Solenoid, always comes to rest in geographical N-S direction, when suspended freely. [1]
Reason (R) One end of current carrying straight Solenoid behaves as a North-pole and the other end as a South-pole, just like a bar magnet.
Answer:
(a) A current carrying Solenoid behaves like a bar magnet with a north and south pole. On freely suspending, it aligns itself in the earth’s N-S direction, just like a compass needle. Therefore, both (A) and (R) are true and (R) is the correct explanation of (A).

Question 33.

Study the ray diagram of an object placed in front of a lens. [2]
A. Write three characteristics of the image formed?
B. In which position of the object will the magnification be -1?
Answer:
A. The characteristics of the image formed.
(i) It is real.
(ii) It is inverted.
(iii) It is enlarged.

B. We obtain a magnification of m = – 1 when the object is placed at 2F₁.

Question 34.
Attempt either A or B. [2]
A. A student carries out an experiment and plots the V-I graph of three samples of nichrome wire with resistances R₁, R₂, and R₃ respectively as shown in figure.

Interpret the V-I graph shown in the figure to answer the following.
(i) Arrange the resistances R₁, R₂, and R₃ in increasing order.
(ii) If the samples are of the same material and length, arrange them in decreasing order of cross-sectional area.
Answer:
(i) As we know that, slope of V-I graph tells about the resistance and these are related as, (slope of V and I) ∝ \(\frac{1}{\text { resistance }}\), i.e

So, R₃ > R₂ > R₁

(ii) Using the formula,
R = \(\frac{\rho l}{A}\) ⇒ R ∝ \(\frac {1}{A}\)

Therefore, the order of cross-sectional area will be
A₁ > A₂ > A₃

Or

B. Study the circuit and find out
(i) Current when key K₁ is open and K₂ is closed.
(ii) Current when key K₁ and K₂ are closed.
Answer:
(i) When key K₁, is open and K₂ is closed, then no current flows in the circuit as circuit is an open circuit
(ii) Both keys are closed, so current flows in the circuit.

Equivalent resistance of the circuit,
\(\frac{1}{R_{\mathrm{eq}}}=\frac{1}{R_1+R_2}+\frac{1}{R_3+R_4}\)
= \(\frac{1}{4+4}+\frac{1}{4+4}=\frac{1}{8}+\frac{1}{8}=\frac{1}{4}\)
So, R_eq = 4Q
Electric current,
I = \(\frac{V}{R_{\text {eq }}}=\frac{12}{4}\) = 3A

Question 35.
An object and its image are shown in the diagram below. [3]

(i) If the distance between the object and the image is 4 cm and the magnification is 3, determine the ratio u: v: f
(ii) Draw a ray diagram showing the position of the lens and its principal foci.
Answer:
(i) Object image distance, v – u = 4 cm
Magnification, m = \(\frac{v}{u}\) = 3

From equation (i) and (ii), we get
3M – u = 4 ⇒ u = 2 cm
v = 3u = 3 × 2 = 6 cm

Now, M = -2cm, v = -6cm (virtual image)
Use lens formula
\(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)
\(\frac{1}{-6}-\frac{1}{(-2)}=\frac{1}{f}\)
\(\frac{1}{f}=\frac{-2+6}{12}=\frac{4}{12}\)
∴ f = 3 cm
Hence, u: v: f = 2: 6: 3

Use the correct sign convention. Take object distance and virtual image distance as negative to avoid calculation errors, (ii) Since, the focal length is positive, the lens is convex.

(ii) Since, the focal length is positive, the lens is conves.

A student should alway draw a neat and labelled ray diagrams.

Question 36.
State whether an a-particle will experience any force in a magnetic field, if (a-particles are positively charged particles) [3]
(i) it moves in the magnetic field parallel to field lines.
(ii) it moves in the magnetic field perpendicular to field lines. Justify your answer in each case.
Answer:
(i) No, it will not experience any force because magnetic field exerts a force in perpendicular direction to motion of the particle.
(ii) Yes, it will experience a force in a direction perpendicular to the direction of its own motion and the direction of force can be determined by Fleming’s left hand rule.

Question 37.
The diagram shown below is connection of [3]
(a) 1 Ω, and 2 Ω resistors in series.
(b) 12 Ω and 2 Ω resistors in parallel.

Compare the power used in 2 Ω resistor in each of the following circuits.
(i) A 6 V battery in series with 1 Ω and 2 Ω resistors,
(ii) A 4 V battery in parallel with 12 Ω and 2 Ω resistors.
Answer:
(i) For the circuit is shown in figure (a).
Resistances are connected in series combination.
Current in the circuit, I = \(\frac{V}{R_1+R_2}=\frac{6}{3}\)= 2 A
∴ Power used = I²R = (2)² × 2 = 2 × 2 × 2 = 8W

(ii) For the circuit is shown in figure (b).
In parallel combination, potential across each resistor is same and equal to the potential applied to the circuit.
Potential across 2 Ω resistor, V = 4 V
Power used = \(\frac{V^2}{R}=\frac{4 \times 4}{2}\) = 8 W
Power used in both the cases is same.

Question 38.
A Solenoid is a long helical coil of wire through which a current is run in order to create a magnetic field. The magnetic field of the Solenoid is the superposition of the fields due to the current through each coil. It is nearly uniform inside the Solenoid and close to zero outside and is similar to the field of a bar magnet having a north pole at one end and a south pole at the other depending upon the direction of current flow. The magnetic field produced in the Solenoid is dependent on a few factors such as, the current in the coil, number of turns per unit length etc.

The following graph is obtained by a researcher while doing an experiment to see the variation of the magnetic field with respect to the current in the Solenoid. [4]

The unit of magnetic field as given in the graph attached is in milli-Tesla (mT) and the current is given in ampere.
A. What will happen if a soft iron bar is placed inside the Solenoid?
Answer:
The soft iron bar will be magnetised as long as there is current in the circuit.

B. What conclusion would you like to draw after analysing the graph?
Answer:
The magnetic field produced by the current carrying Solenoid is directly proportional to the current flowing through it.

Attempt either subpart (C) or (D).
C. From the graph deduce the magnitude of magnetic field inside the Solenoid, if it carries a current of 0.8 A.
Answer:
From the graph, for a current of 0.8 A, the magnetic field is 13 mT.

Or

D. List the two distinguishing features between the magnetic field of a current carrying Solenoid and a bar magnet.
Answer:
Distinguishing features are
(i) The magnetic field outside the Solenoid is negligible as compared to the bar magnet.
(ii) The magnetic field of a Solenoid can be changed by changing the magnitude of current through it or by placing the core inside the Solenoid but magnetic field around the bar magnet is fixed.

Question 39.
Attempt either A or B.
A. (i) A lens forms a blurred image of an object on the screen as shown below.

What changes can you make to the following to form a sharp and in-focus image on the screen?
(1) Object distance
(2) Focal length of the lens
Answer:
In order to get a sharp image on the screen, the object is moved toward the screen, therefore, the object distance is decreased and focal length of the lens is increased.

(ii) The image of an object formed by a lens is of magnification -1. If the distance between object and its image is 60 cm, what is the focal length of the lens?
Answer:
Given, m = – 1, negative sign shows that image is real and inverted.
|m| = 1, it means that h_i = h_o
and this is possible in convex lens.
It implies that object and image both will be at 2F.
So, v = u.
Object distance, u = -30 cm
Image distance, v = + 30cm

∵ According to lens formula, \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)
⇒ \(\frac{1}{f}=\frac{1}{30}-\left(\frac{-1}{30}\right)\)
⇒ f = 15 cm

(iii) If the object is moved 20 cm towards the lens, where would the image be formed? State reason and also draw a ray diagram in support of your answer.
Answer:

Now, object is moved 20 cm towards the lens,
Object distance, w = -10cm
Focal length, f = +15 cm
Again using lens formula, \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)
⇒ \(\frac{1}{v}=\frac{1}{15}+\frac{1}{(-10)}\)
⇒ \(\frac{1}{v}=\frac{1}{15}-\frac{1}{10}=\frac{2-3}{30}\)
v = -30cm

Or

B. A student wants to project the image of a candle flame on the walls of school laboratory by using a lens.
(i) Which type of lens should he use and why?
(ii) At what distance in terms of focal length F of the lens should he place the candle flame, so as to get
1. a magnified and
2. a diminished image respectively, on the wall?
(iii) Draw ray diagrams to show the formation of the image in each case in (ii).
Answer:
(i) He should use a convex lens as real images are formed by it.
(ii) (a) For magnified image, he should place the candle flame between focus (F) and centre of curvature (2F) of lens.
(b) To get diminished image, he should place the candle flame beyond centre of curvature (2F) of lens.
(iii) 1. For magnified image

2. For diminished image

To get extra marks, mention nature of image and keep diagram clean and to scale where possible.

The post CBSE Sample Papers for Class 10 Science Set 9 with Solutions appeared first on Learn CBSE.

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