Students must start practicing the questions from CBSE Sample Papers for Class 10 Science with Solutions Set 6 are designed as per the revised syllabus.
CBSE Sample Papers for Class 10 Science Set 6 with Solutions
Time: 3 Hours
Maximum Marks: 80
General Instructions:
- This question paper consists of 39 questions in 5 sections.
- 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 consists of 20 objective type questions carrying 1 mark each.
- Section B consists of 6 Very Short questions carrying 02 marks each. Answers to these questions should be in the range of 30 to 50 words.
- Section C consists of 7 Short Answer type questions carrying 03 marks each. Answers to these questions should be in the range of 50 to 80 words.
- Section D consists of 3 Long Answer type questions carrying 05 marks each. Answer to these questions should be in the range of 80 to 120 words.
- Section E consists of 3 source-based/case-based units of assessment of 04 marks each with sub-parts.
Section – A
Select and write the most appropriate option out of the four options given for each of the questions 1 – 20.
There is no negative mark for incorrect response.
Question 1.
What is the shape of the image of a distant flag, shown below, when you focus it on a screen using a convex lens?
Answer:
Explanation: As the image is focused on the screen, then the image formed is real. And real images are inverted.
Question 2.
Dominant alleles are expressed exclusively in a heterozygote, while recessive traits are expressed only if the organism is ……………… for the recessive allele.
(a) Homozygous
(b) Heterozygous
(c) Normal
(d) None of these
Answer:
(a) Homozygous
Explanation: According to Mendel’s law of dominance, when there is a heterozygote (an organism with two different alleles for a trait), one allele will dominate and mask the presence of the other allele. Only the dominant allele will be expressed and contribute to the physical characteristics (phenotype). The recessive allele remains hidden but can still be passed on to offspring in the same way as the dominant allele. The recessive trait will only be visible in offspring who inherit two copies of this recessive allele.
Question 3.
The image formed on the retina of the human eye is :
(a) Virtual and erect
(b) Real and inverted
(c) Virtual and inverted
(d) Real and erect
Answer:
(b) Real and inverted
Explanation: The image formed on the retina of the human eye is real and inverted. On retina, the image is converted into electrical impulses and sent to the brain. The brain flips that image into a virtual and erect image.
Question 4.
Which of the following salts does not contain water of crystallisation?
(a) Blue vitriol
(b) Baking soda
(c) Washing soda
(d) Gypsum
Answer:
(b) Baking soda
Explanation:
Baking soda: It is sodium bicarbonate (NaHCO3) in anhydrous form without any water of crystallisation.
Blue Vitriol: It is hydrated salt of copper sulphate containing 5 molecules of water of crystallisation (CuSO4. 5H2O).
Washing soda: It is hydrated salt of sodium carbonate containing 10 molecules of water of crystallisation (Na2CO3.10H2O)
Gypsum: It is hydrated salt of calcium sulphate containing 2 molecules of water of crystallisation (CaSO4.2H2O
Question 5.
Identify A, B and C in the given diagram and match the labelling referred in column I and correlate with the function in column II.
Column I | Column II |
A. | (i) Special reproductive female sex cell which combines with male gamete to form the zygote. |
B. | (ii) A male gamete moves down through it towards the female gamete in the ovary. |
C. | (iii) Receiving the pollen grains from the anther of stamen during pollination. |
(a) A – (iii), B – (ii), C – (i)
(b) A – (ii), B – (i), C – (iii)
(c) A – (i), B- (ii), C -(iii)
(d) A – (iii), B – (i), C – (ii)
Answer:
(a) A – (iii), B – (ii), C – (i)
Explanation: A – Stigma. The top part of the carpel is called stigma. Stigma is for receiving the pollen grains from the anther of stamen during pollination.
B – Pollen tube. When a pollen grain falls on the stigma, it bursts open and grows a pollen tube downward through the style towards the female gamete in the ovary. A male gamete moves down the pollen tube.
C – Female gamete (ovum). It is a special reproductive female sex cell which combines with the male gamete to form a zygote.
Question 6.
A farmer wants to grow banana plants genetically similar enough to the plants already available in his field. Which one of the following methods would you suggest for this purpose?
(a) Regeneration
(b) Budding
(c) Vegetative propagation
(d) Sexual reproduction
Answer:
(c) Vegetative propagation
Explanation: Vegetative propagation is a method of plant reproduction that involves using vegetative parts of a plant, such as stems, roots, or leaves, to create new individuals that are genetically identical to the parent plant. This method ensures that the desired characteristics of the parent plant, including fruit quality and traits, are preserved in the new plants. This method avoids the genetic variation that can occur through sexual reproduction, which involves the combination of genetic material from two different parent plants and may result in offspring with different traits.
Question 7.
An electric fuse is connected with :
(a) live wire
(b) earthing
(c) neutral wire
(d) parallel to the line wire
Answer:
(a) live wire
Explanation: An electric fuse is connected with live wire because it gets blown up when an excess current tries to pass through it to save the electrical appliances by restricting the flow of that current,
Question 8.
Characters transmitted from parents to offspring are present in:
(a) Cytoplasm
(b) ribosome
(c) Golgi bodies
(d) genes
Answer:
(d) genes
Explanation: Characters are transmitted from parents to offspring through genes. Genes are the heredity units of the body in living organisms. Chromosomes in the nucleus of a cell contain information for the inheritance of features from parents in the form of DNA (Deoxyribonucleic acid). This DNA contains genes.
Question 9.
Which of the following terms does not represent electrical power in a circuit ?
(a) I2R
(b) IR2
(c) VI
(d) V2/R
Ans.
(b) IR2
Explanation:
We know that,
P =VI …………. (i)
Where, P = Power
V = Potential difference
I = Current
V = IR ……………. (ii)
On substituting equation (ii) in equation (i), we get
P = I2R ………….. (iii)
Again, from Ohm’s law
I = \(\frac{\mathrm{V}}{\mathrm{R}}\) ……….. (iv)
On substituting equation (iv) in equation (i), we get,
P = \(\frac{\mathrm{V}^2}{\mathrm{R}}\)
Hence, option (b) IR2 does not represent power.
Question 10.
Which of the following statement is not true about Thyroxine?
(a) Thyroxine regulates the basal metabolism of our body.
(b) Iodine is an important component required for synthesis of Thyroxine.
(c) Under secretion of Thyroxine causes simple goitre.
(d) Iron is essential for synthesis of Thyroxine.
Answer:
(d) Iron is essential for synthesis of Thyroxine.
Explanation: Iodine, not iron, is required to synthesise of Thyroxine. It controls the glucose, protein, and fat metabolism of the body. The thyroid gland produces Thyroxine, which is also known as thyroid hormone.
Question 11.
Let us consider the flow of the current through a metallic wire, if the temperature of the entire system increases. What will happen from the following options ?
(a) Potential difference (V) increases
(b) Resistance (R) decreases
(c) Potential difference (V) decreases
(d) V and R remain the same
Answer:
(c) Potential difference (V) decreases
Explanation: If the temperature of the entire system increases when the current flows through a metallic wire, the power (VI) is dissipated in the form of heat. Hence the potential difference (V) decreases.
Question 12.
Which direction do afferent neurons carry nerve impulses?
(a) From the central nervous system (CNS) to muscles
(b) From the CNS to receptors
(c) From receptors to the CNS
(d) From effector organs to the CNS
Ans.
(c) From receptors to the CNS
Explanation: Afferent neurons, also known as sensory neurons, are responsible for transmitting nerve impulses from sensory receptors in the peripheral nervous system to the central nervous system (CNS). These neurons carry sensory information, such as touch, temperature, pain, and other stimuli, from various parts of the body to the CNS for processing and interpretation.
Question 13.
Which of the following properties of aluminium makes it suitable for making cooking utensils?
(i) Good thermal conductivity
(ii) Good electrical conductivity
(iii) Ductility
(iv) High melting point
(a) (i) and (ii)
(b) (i) and (iii)
(c) (ii) and (iii)
(d) (i) and (iv)
Answer:
(d) (i) and (iv)
Explanation: Good thermal conductivity and a high melting point of aluminium are the properties that makes it suitable for making cooking utensils.
Question 14.
A current flows in a wire, running between the S and N poles of a magnet lying horizontally, as shown inithe figure below:
The force on the wire due to the magnet is directed.
(a) From N to S (c)
(b) From S to N (d)
(c) Vertically downwards
(d) Vertically upwards
Answer:
(c) Vertically downwards
Explanation: Force on conductor is calculated using Fleming’s left hand rule.
Question 15.
The height of a plant is regulated by:
(a) DNA which is directly influenced by growth hormones.
(b) Genes which regulate the proteins directly.
(c) Growth hormones under the influence of the enzymes coded by a gene.
(d) Growth hormones directly under the influence of a gene.
Answer:
(c) Growth hormones under the influence of the enzymes coded by a gene.
Explanation: The height of a plant is regulated by a complex interplay of genetic factors, growth hormones, and enzyme activity.
Genes play a crucial role in determining the production and activity of growth hormones in plants. Specific genes code for enzymes that are involved in the synthesis, metabolism, and signaling pathways of growth hormones. These enzymes control the production, transportation, and breakdown of growth hormones within the plant.
While DNA carries the genetic information that influences the production of growth hormones, it is not directly influenced by growth hormones itself (option a). Genes regulate the production of proteins, including the enzymes involved in growth hormone metabolism, rather than directly regulating the proteins themselves (option b). Option d suggests that growth hormones act directly under the influence of a gene, but it is the enzymes coded by genes that mediate the effects of growth hormones.
Question 16.
Mutual induction is a process in which current is induced in the neighbouring coil if current flows in a coil. In the figure shown below:
(a) Maximum in the situation (A)
(b) Maximum in situation (B)
(c) Maximum in the situation (C)
(d) Same in all situations
Answer:
(a) Maximum in situation (A)
Explanation: As both the coils are in the same plane the induced current is found to be highest when 1 the direction of the coil is at the right angle to the magnetic field.
Question No 17 to 20 consist of two statements – Assertion (A) and Reason (R). Answer these questions 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 and 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 17.
Assertion: Diabetes mellitus is caused due to lack of insulin in the body.
Reason: Insulin helps in utilisation of glucose in the body.
Answer:
(a) Both A and R are true and R is the correct explanation of A.
Explanation: Insulin is a hormone released by the p-cells of the pancreas. It maintains the blood glucose level in the body, by its proper utilisation. So, lack of insulin causes accumulation of glucose in the blood and hence a disease called diabetes mellitus is caused.
Question 18.
Assertion: Electric current flowing through a metallic wire is directly proportional to the potential difference across its ends.
Reason: Ohm’s law expression V = IR, where R (resistance) of the wire is always varying.
Answer:
(c) A is true, but R is false.
Explanation: Ohm’s law states that the electric current flowing through a metallic wire is directly proportional to the potential difference across its two ends. The expression is written as :
V = IR
Here, R (resistance of the wire) is a constant value, then only the statement will be valid.
V ∝ I only if \(\frac{V}{I}\) = constant
Question 19.
Assertion: Lumen of the fallopian tube is lined by ciliated epithelium.
Reason: Ciliated epithelium helps to move the zygote towards the uterus for implantation.
Answer:
(a) Both A and R are true and R is the correct explanation of A
Explanation: Lumen of the Fallopian tube is lined by ciliated epithelium as cilia have a rhythmic I waving and beating motion that helps substances to travel from one place to another.
Question 20.
Assertion: Hydrogen peroxide is kept in coloured bottles.
Reason: Hydrogen peroxide is a moderately reactive metal that can react with light or heat slowly to produce water.
Answer:
(c) A is true but R is false
Explanation: Hydrogen peroxide is a highly reactive metal that can react with light or heat to produce water. It decomposes into water and oxygen in the presence of sunlight. To prevent this reaction with light and heat it is stored in coloured bottles so that light cannot pass through it.
Section – B
Question No. 21 to 26 are very short answer questions.
Question 21.
Write the essential functions performed by ozone in the upper layer of Earth’s atmosphere? How is it produced? Name the synthetic chemicals mainly responsible for the drop of the amount of ozone in the atmosphere. How can the use of these chemicals be reduced?
Answer:
Ozone layer absorbs most of the harmful ultraviolet radiations from the sun to the earth. It is formed high up in the atmosphere by the action of ultraviolet radiation on oxygen gas. Chlorofluorocarbons are the synthetic chemicals responsible for the drop of amount of ozone in the atmosphere.
The use of these chemicals can be reduced by:
1. Replacement of chloroflurocarbons with hydrochlorofluorocarbons because it breaks down more quickly.
2. Safe disposal of old appliances such as refrigerators and freezers.
Question 22.
Answer the following questions:
(i) Define the power of a lens.
(ii) Name the lens that has:
(a) Negative power.
(b) Positive power.
Answer:
(i) The ability of a lens to converge the rays of light falling on it is called the power of the lens,
(ii) (a) The focal length of a concave lens is negative, so its power is negative.
(b) The focal length of a convex lens is positive, so its power is positive.
Question 23.
Draw a diagram of the human excretory system and label the following terms given below. Aorta, Vena cava, Urinary bladder and Urethra.
OR
List two vital functions of the kidneys.
Answer:
Two vital functions of kidneys are:
(1) They help in the removal of waste products that are harmful to the body. It is called excretion.
(2) They maintain the proper level of salt and water concentration in the body. It is called osmoregulation.
Question 24.
The diagram shows one layer of carbon atoms in graphite.
(i) Identify the type of bonding in graphite.
(ii) Which property of graphite makes it suitable for use as a dry lubricant? Explain.
Answer:
(i) Between the carbon atoms in each layer covalent bond is found and Van Der Waal’s forces are found between the layers of carbon atoms in graphite.
(ii) Graphite is soft which makes it suitable for use as a dry lubricant. Between the carbon atoms in each layer covalent bond is found and weak Van Der Waals forces are found between the layers of carbon atoms in graphite. So, the layers can slide over one another making it a relatively soft substance.
Question 25.
Guinea pig having black colour when crossed with guinea pig having the same colour produced 80 offspring, out of which 60 were black and 20 were white. Now find out:
(i) What is the possible genotype of the guinea pigs?
(ii) Which trait is dominant and which trait is recessive?
(iii) What is this cross called as and what is its phenotypic ratio?
OR
Why do all the gametes formed in the human females have an X-chromosome?
Answer:
(i) The possible genotype of the guinea pigs is Bb x Bb.
(ii) Black colour is dominant and white colour is recessive.
(iii) This is an example of a monohybrid cross and its phenotypic ratio is 3 : 1.
OR
The sex chromosome in human female is homomorphic i.e., they contain the same chromosome XX. During meiosis process at the time of gamete formation all egg cells will get one copy of X chromosome, hence all the gametes formed in human females have an X-chromosome.
Question 26.
State the various characteristics of chemical reactions.
Answer:
Some of the characteristics of chemical reactions are:
- Formation of a precipitate
- Change in colour
- Change in temperature
- Change in the state.
- Evolution of a gas
Section – C
Question No. 27 to 33 are short answer questions.
Question 27.
B1, B2 and B3 are three identical bulbs connected, as shown in the figure. Ammeters A1, A2 and A3 are connected as shown. When all the bulbs glow, the current of 3 A is recorded by ammeter A.
(i) What happens to the glow of the other two bulbs when bulb B1 gets fused?
(ii) What happens to the reading of A1, A2, A3 and A when the bulb B2 gets fused?
(iii) How much power is dissipated in the circuit when all the bulbs glow together?
Answer:
Resistance of combination of three bulbs in parallel,
Req = \(\frac{V}{I}=\frac{4.5}{3}\) = 1.5Ω
If R is the resistance of each wire, then
\(\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{\mathrm{R}}+\frac{1}{\mathrm{R}}+\frac{1}{\mathrm{R}}\)
⇒ \(\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{3}{\mathrm{R}}\)
⇒ R = 3Req
⇒ R = 3 × 1.5 = 4.5Ω
Current in each bulb, I = \(\frac{\mathrm{V}}{\mathrm{R}}=\frac{4.5 \mathrm{~V}}{4.5 \Omega}\) = 1A
(i) When bulb B1 gets fused, the currents in B2 and B3 remain same I2 = I3 = 1A, as the voltage across them remains the same so, their glow remains unaffected.
(ii) When bulb B2 gets fused, the current in B2 becomes zero and currents in B1 and B3 remains 1 A.
Total current I = I1 + I2 + I3
= 1 + 0 + 1 = 2A
Current in ammeter A1 = 1 A
Current in ammeter A2 = 0
Current in ammeter A3 = 1 A
Current in ammeter A = 2 A
(iii) When all the bulbs are connected in the circuit:
Power dissipated, P = \(\frac{\mathrm{V}^2}{\mathrm{R}_{\mathrm{eq}}}\)
⇒ P = \(\frac{(4.5)^2}{1.5}\) = 13.5 W
Question 28.
Obtain an expression for the magnification of an image formed by a concave mirror.
Answer:
Consider the formation of the image A’B’ of an object AB by a concave mirror. As shown in given
\(\frac{\mathrm{A}^{\prime} \mathrm{B}^{\prime}}{\mathrm{AB}}=\frac{\mathrm{PB}^{\prime}}{\mathrm{PB}}\)
As per sign convention followed, PB = – u, PB’ = – υ, AB = size of the object = + h and A’B’ = size of the
image – h’.
Hence, we have
\(\frac{-h^{\prime}}{h}=\frac{-v}{-u}\)
or \(\frac{h^{\prime}}{h}=\frac{-v}{u}\)
Thus, by definition of magnification of image, we have
Magnification, m = \(\frac{h^{\prime}}{h}=-\frac{v}{u}\).
Question 29.
(i) Define activity series of metals. Arrange the metals gold, copper, iron, and magnesium in order of their increasing reactivity.
(ii) What will you observe when:
(a) Some zinc pieces are put in a copper sulphate solution.
(b) Some silver pieces are put into a green coloured ferrous sulphate solution.
Answer:
(i) The series of metals in which metals are arranged in decreasing order of their reactivity is called activity series of metals. The increasing order of reactivity will be:
Au < Cu < Fe < Mg.
(ii) (a) When zinc pieces are put into blue colour copper sulphate solution. Zinc displaces copper from copper sulphate solution. In this process, blue copper sulphur solution turns into colourless.
(b) When some silver pieces are put into green coloured ferrous sulphate solution, there will be no reaction because Ag is less reactive than iron.
Ag(s) + FeSO4(aq) → No reaction
The reaction will not take place because Ag is less reactive than iron.
Question 30.
How do Mendel’s experiment show that traits are inherited independently?
OR
The given box diagram represents the ratio of females to males or the sex ratio in our country for 10 decades (1901 to 2001). Answer the following questions in light of your knowledge of sex determination and the data presented in the box diagram.
(i) What does the bar diagram show?
(ii) As per scientific knowledge regarding sex determination, what should be the sex ratio or the male to female ratio at a given point of time?
(iii) Assign one reason for the trend showing deviation from the expected sex ratio.
(iv) Suggest a way for which such a trend can be stopped.
Answer:
Mendel’s experiment show that:
(i) When a cross was made between a tall pea plant with round seeds and a short pea plant wrinkled seeds, the F1 progeny plants are all tall with round seeds. This indicates that tall and round seeds are the dominant traits.
(ii) When the F1 plants are self-pollinated and the F2 progeny show that some progencies tall with round seeds and some progenies were short plants with wrinkled seeds, which are the traits visible in the F2 generation.
(iii) There were also some progenies with new combinations like tall plants with wrinkled seed short plants with round seeds.
OR
(i) Bar diagram shows the proportion of females in the population over a decade.
(ii) 1 : 1 should be the sex ratio or the male-to-female ratio at a given point of time.
(iii) Female foeticide is the main reason for this trend showing deviation from the expected sex ratio.
(iv) Banning sex tests of unborn babies; increasing awareness and education will help to stop such trends.
Question 31.
Define:
(i) Food chain
(ii) Food web
(iii) Biological magnification.
Answer:
(i) Food chain: The series of organisms taking part at various trophic levels for the flow of energy is called a food chain.
(ii) Food web: The food web represents many interlinked food chains as each organism is generally eaten by two or more other kinds of organisms. It cannot be represented by a straight line.
(iii) Biological magnification: The accumulation of harmful chemicals at each trophic level with the help of various biotic and abiotic factors is called biological magnification.
Question 32.
(i) Define the term alloy and amalgam. Name the alloy used for welding electric wires together. What are its constituents?
(ii) Name the constituents of the following alloys:
(a) Brass
(b) Stainless steel
(c) Bronze
Answer:
(i) An alloy is a homogeneous mixture of a metal with a metal or a non-metal. Amalgam is an alloy of a metal with mercury.
The alloy used for welding electric wires together is Solder. Its constituents are lead and Tin.
(ii) (a) Brass: Brass alloy is made of zinc (Zn) and copper (Cu).
(b) Stainless steel: It contains chromium (12.14%), molybdenum (0.21 %), nickel (less than 2%), and carbon (about 0.11%).
(c) Bronze: It is an alloy consisting primarily of copper, commonly with about 12% tin and often with the addition of other metals (such as aluminium, manganese, nickel or zinc) and sometimes non-metals or metalloids such as arsenic, phosphorus or silicon.
Question 33.
State one characteristic of each of the chemical reactions which takes place when:
(i) Dilute hydrochloric acid is added to sodium carbonate.
(ii) Lemon juice is added gradually to the potassium permanganate solution.
(iii) Dilute sulphuric acid is added to the barium chloride solution.
Answer:
(i) Carbon dioxide gas is released.
(ii) Colour changes from purple to colourless
(iii) White precipitate of barium sulphate is formed
Section – D
Questioh No. 34 to 36 are long answer questions.
Question 34.
Explain the nutrition process in Amoeba.
OR
(i) Explain the statement “Bile does not contain any enzyme but it is essential for digestion”.
(ii) Explain the process of breathing in human beings.
Answer:
Amoeba follows holozoic nutrition.
It involves the following steps:
- Ingestion – Amoeba engulfs the food by using its temporary finger-like projections called pseudopodia. This process is called ingestion.
- Digestion – The food which is taken inside the amoeba forms a food vacuole. Many enzymes are secreted into the food vacuole and the complex food molecule is converted into a simple and diffusible form.
- Absorption and assimilation – The digested food is absorbed by the cell by the diffusion process.
- Egestion – The undigested residue which remains in the vacuole is expelled.
OR
(i) Though bile juice secreted by liver has no enzymes in it is still essential for digestion because:
- It makes the medium alkaline in the small intestine so that intestinal juice can perform its functions because the food from stomach which enters the small intestine is acidic in nature.
- It emulsifies fats so that enzymes like lipase can act upon fats to convert them into fatty acids and glycerol.
(ii) Breathing process occurs by inhalation and exhalation.
1. Inhalation: It is the process by which oxygen is taken in through nostrils. During inhalation process, the ribs move upwards and outwards due to contraction of intercostal muscles. The diaphragm is lowered so that the volume of thoracic cavity increases. As a result air is forced inside the lungs through nostrils.
2. Exhalation: It is the process by which carbon dioxide is exhaled out from lungs through nostrils. During this process the ribs moves inwards and diaphragm comes back to its original position. The volume of thoracic cavity decreases so air is forced out through the lungs.
Question 35.
(i) Observe the given figure and answer the following:
(a) Identify the gas evolved.
(b) How will you test the presence of gas evolved?
(c) What is the nature of the gas evolved?
(d) What will happen to the lime water when the gas evolved is passed through it? Give chemical equation.
(e) What happens to the lime water when this gas is passed for a longer period of time?
(ii) Account for the following:
(a) State the relation between the hydrogen ion concentration of an aqueous solution and its pH.
(b) An aqueous solution has a pH value of 7.0. Is this solution acidic, basic or neutral?
(c) Which has a higher pH, 1 M HCl or 1 M NaOH solution?
(d) Tooth enamel is one of the hardest substances in our body. How does it undergo damage due to eating chocolates and sweets? What should we do to prevent it?
(e) How do H+ ions exist in water?
OR
Answer the following questions:
(i) What colour will be produced when we put a drop of distilled water on pH paper? What is the pH of distilled water?
(ii) The pH of a solution is less than 7. What does it indicate? What is the pH of 1 M HCl solution?
(iii) What is the pH of solution A’ which liberates CO2 gas with a carbonate salt? Give the reason?
(iv) What is the pH of solution ‘B’ which liberates NH3 gas with an ammonium salt? Give reason?
(v) How do you increase or decrease the pH of pure water?
Answer:
(i) (a) Carbon dioxide gas is evolved.
(b) Carbon dioxide gas is a non-supporter of combustion so when a burning splinter is brought near this gas, it gets extinguished.
(c) This gas is acidic in nature as it turns blue litmus red.
(d) Lime water will turn milky.
Ca(OH)2 + CO2 → CaCO3 + H2O
(e) When this gas is passed through lime water for a longer period, the milky lime water disappears.
CO2 + CaCO3 + H2O → Ca(HCO3)2
(ii) (a) Hydrogen ion concentration of an aqueous solution is inversely proportional to its pH.
(b) The solution is neutral.
(c) 1 M NaOH solution has higher pH because bases have higher pH values than acids.
(d) Tooth enamel gets corroded slowly when pH in the mouth is below 5.5. Acid is produced in mouth due to the degradation of food which is partially hydrolysed by saliva. But if excess acid is produced, it causes tooth decay. It can be prevented by using tooth paste which is generally basic.
(e) H+ ions exist in water as H3O+ ions.
OR
(i) The colour produced will be green. Its pH is 7.
(ii) It indicates that the solution is acidic in nature. The pH of 1M HCL is zero.
(iii) The pH of solution ‘A’ is lesser than 7. Carbonates salts react with acids (A) to liberate CO2 gas.
(iv) The pH of solution ‘B’ is lesser than 7 because ‘B’ is an alkali as it liberates NH3 gas.
(v) By adding few drops of alkali to pure water, it’s pH increases; and by adding a few drops of an acid decreases the pH of pure water.
Question 36.
(i) “A convex lens can form a magnified erect as well as magnified inverted image of an object placed in front of it.” Draw a ray diagram to justify this statement stating the position of the object with respect to the lens in each case.
(ii) An object of height 4 cm is placed at 20 cm from a concave lens of focal length 10 cm. Use lens formula to determine the position of the image formed.
OR
(i) To construct a ray diagram, we use two rays which are so chosen that it is easy to know their directions after reflection from the mirror. List two such rays and state the path of these rays after reflection in case of concave mirrors. Use these two rays and draw a ray diagram to locate the image of an object placed between the pole and the focus of a concave mirror.
(ii) A concave mirror produces three times magnified image on a screen. If the object is placed 20 cm in front of the mirror, how far is the screen from the object?
Answer:
(i) Case I: Image formed is magnified and erect when the object is placed between the optical centre and focus on a convex lens.
Case II: Image formed is magnified and inverted when the object is placed between F and 2F of a convex lens.
(ii) Given: u = – 20 cm, f = – 10 cm.
Using lens formula,
\(\frac{1}{f}=\frac{1}{υ}-\frac{1}{u}\)
⇒ \(\frac{1}{υ}=\frac{1}{f}+\frac{1}{u}\)
⇒ \(\frac{1}{υ}=-\frac{1}{10}-\frac{1}{20}\)
⇒ \(\frac{1}{υ}=\frac{-2-1}{20}=\frac{-3}{20}\)
⇒ υ = \(\frac{-20}{3}\) cm
OR
(i) The following rays of light are usually used to locate the images formed by a spherical mirror : The incident ray passing through the centre of curvature: In this case, light after reflecting from the spherical mirror moves back along the same path.
The ray incident obliquely to the principal axis: In this case, the incident ray will be reflected back by the reflecting surface of the spherical mirror obliquely, making equal angles with the principal axis.
When the object is placed between the pole and the focus of the concave mirror, an image is formed behind the mirror and is virtual, erect and magnified.
(ii) Given, Magnification, m = – 3 (As the image is real), Object distance, u = – 20 cm
We have, m = –\(\frac{υ}{u}\)
⇒ – 3 = – \(\left(\frac{υ}{-20}\right)\)
⇒ υ = – 60 cm
The image is located at 60 cm in front of the mirror.
Thus, the screen is at 40 cm from the object.
Section – E
Question No. 37 to 39 are case – based/data -based questions with 2 to 3 short sub – parts. Internal choice is provided in one of these sub-parts.
Question 37.
Mass can neither be created nor destroyed in a chemical reaction. That is, the total mass of the elements present in the products of a chemical reaction has to be equal to the total mass of the elements present in the reactants. In other words, the number of atoms of each element remains the same, before and after a chemical reaction. Hence, we need to balance a skeletal chemical equation.
To make a chemical equation more informative, the physical states of the reactants and products are mentioned along with their chemical formulae. The gaseous, liquid, aqueous and solid states of reactants and products are represented by the notations (g), (l), (aq) and (s), respectively. Aqueous (aq) is written if the reactant or product is present as a solution in water.
Usually, physical states are not included in a chemical equation unless it is necessary to specify them. Sometimes the reaction conditions, such as temperature, pressure, catalyst, etc., for the reaction are indicated above and/or below the arrow in the equation. For example,
(a) Balance the chemical equation for a given reaction with the proper physical state.
Fe + H2O → Fe3O4 + H2
(b) Which law is demonstrated above?
OR
Write a balanced chemical equation for reaction of Barium chloride and Aluminium sulphate to give Barium sulphate and aluminium chloride.
Answer:
(a) The balanced chemical equation for given reaction is;
3Fe(s) + 4H2O(g) → Fe3O4(s) + 4H2(g)
(b) Law of Conservation of Mass is demonstrated in the above equation.
OR
Question 38.
Stomata are tiny pores present on the surface of the leaves. Massive amounts of gaseous exchange takes place in the leaves through these pores for photosynthesis. But it is important to note here that exchange of gases occurs across the surface of stems, roots and leaves as well. Since large amounts of water can also be lost through these stomata, the plant closes these pores when it does not need carbon dioxide for photosynthesis. The opening and closing of the pore is a function of the guard cells. The guard cells swell when water flows into them, causing the stomatal pore to open. Similarly, the pore closes if the guard cells shrink.
1. Take two healthy potted plants which are nearly the same size.
2. Keep them in a dark room for three days.
3. Now, place each plant on separate glass plates. Place a watch-glass containing potassium hydroxide by the side of one of the plants.
4. The potassium hydroxide is used to absorb carbon dioxide.
5. Cover both plants with separate bell-jars.
6. Use Vaseline to seal the bottom of the jars to the glass plates so that the set-up is air-tight.
7. Keep the plants in sunlight for about two hours.
8. Pluck a leaf from each plant and check for the presence of starch as in the above activity.
(a) What can be concluded from the activity?
(b) What is the function of stomata?
(c) How do the guard cells control the opening and closing of stomata?
OR
In the above activity, do both the leaves show the same amount of starch?
Answer:
(a) It shows that the amount of carbon dioxide affects the process and outcome of photosynthesis.
(b) Massive amounts of gaseous exchange takes place in the leaves through these pores for the purpose of photosynthesis.
(c) The guard cells swell when water flows into them, causing the stomatal pore to open. Similarly the pore closes if the guard cells shrink.
OR
No, both the leaves show the presence of a different amount of starch in the given activity.
Question 39.
The given table shows that the resistivity of an alloy is generally higher than that of its constituent metals. Alloys do not oxidise (bum) readily at high temperatures. For this reason, they are commonly used in electrical heating devices, like electric iron, toasters etc. Tungsten is used almost exclusively for filaments of electric bulbs, whereas copper and aluminium are generally used for electrical transmission lines.
(a) How much current will an electric bulb draw from a 200 V source, if the resistance of the bulb filament is 1500 Ω?
(b) Why are alloys used in heating devices?
(c) Which alloy is used exclusively for filaments of electric bulbs?
OR
(d) Write the composition of Nichrome.
Answer:
(a) Given that V = 200V; R = 1500 Ω.
Therefore, current I = 200 V/1500 Ω = 0.13 A.
(b) Alloys do not oxidise (bum) readily at high temperatures. For this reason, they are commonly used in electrical heating devices, like electric iron, toasters etc.
(c) Tungsten
OR
(d) Nichrome is an alloy of Ni, Cr, Mn, and Fe.