CHAPTER 02- ACIDS, BASES AND SALTS WORKSHEET (1)

CHAPTER 02- ACIDS, BASES AND SALTS

WORKSHEET (VOLUME-1)

TIME:20MIN
MM=16

Very short answers type questions (2*3=6)

  1. Write the name of acids present in tomato, tamarind, curd, lemon
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  2. Why does dry HCl gas not change the color of dry litmus paper?
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  3. A milkman adds a small amount of baking soda to fresh milk. Why does this milk take a long time to set as curd?
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Short answers type questions (3*2=6)

  1. A student dropped a few pieces of marble in dilute hydrochloric acid in a test tube. He observed brisk effervescence.
    a) Name the gas evolved.
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    b) How will you test the gas?
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    c) Write a balanced chemical equation for the reaction.
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  2. A white powder is used by doctors for supporting fractured bones.
    a) Identify the substance.
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    b) Write the chemical name and formula.
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    c) How is it prepared from gypsum? Give the equation.
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Long answers type questions (1*4=4)

  1. Case Study Question: Antacids and Neutralization
    Read the passage carefully and answer the following questions:
    Our stomach produces hydrochloric acid which helps in digestion. Sometimes, excess acid is produced which causes indigestion. To get relief, people use antacids – mild bases like magnesium hydroxide. Antacids neutralize excess acid in the stomach. Similarly, farmers use quick lime or slaked lime to neutralize acidic soil. Also, factory waste is often acidic and needs to be neutralized before disposal.
    a) What is the chemical name and formula of an antacid commonly used?
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    b) Write the word equation for the reaction between hydrochloric acid and magnesium hydroxide.
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    c) How does the use of lime help in treating acidic soil?
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    d) Why is it important to neutralize factory waste before disposal?
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    OR,
    e) Which type of chemical reaction takes place when an acid reacts with a base?
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ANSWER KEY OF WORKSHEET 2

  1. Tomato-oxalic acid, tamarind-tartric acid, curd-lactic acid, lemon-ascorbic acid
  2. Dry HCl does not ionize and release \( H^+ \) ions in the absence of water. Litmus shows a color change only in the presence of ions, so there is no effect on dry litmus paper.
  3. Baking soda is a base that neutralizes the acid produced by bacteria. As a result, the pH remains high and the milk takes longer to curdle.
  4. A) Carbon dioxide (\( CO_2 \))
    B) Pass the gas through lime water. It will turn milky.
    C) Balanced equation: \( CaCO_3 + 2HCl \rightarrow CaCl_2 + H_2O + CO_2\uparrow \)
  5. A) Plaster of Paris
    B) Calcium sulphate hemihydrate (\( CaSO_4 \cdot \frac{1}{2}H_2O \))
    C) Preparation from gypsum: \( CaSO_4 \cdot 2H_2O \text{ (Gypsum)} \rightarrow CaSO_4 \cdot \frac{1}{2}H_2O + 1\frac{1}{2}H_2O \text{ (on heating)} \)
  6. a) Magnesium hydroxide, \( Mg(OH)_2 \)
    b) \( \text{Hydrochloric acid} + \text{Magnesium hydroxide} \rightarrow \text{Magnesium chloride} + \text{Water} \)
    c) Lime helps neutralize acidic soil by reacting with acids and increasing the pH, making it suitable for crops.
    d) Neutralizing factory waste prevents harm to aquatic life and protects the environment from acid pollution.
    e) Type of reaction: Neutralisation reaction

CHAPTER 03-METALS AND NON METALS

WORKSHEET-VOLUME-1

TIME:20MIN
MM=16

Very short answer answer type questions (2*3=6)

  1. Which element is most abundant metal in earth crust?
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  2. Why the item made of silver turn black when exposed to air?
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  3. Give reasons:
    (a) Platinum, gold and silver are used to make jewellery.
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    (b) Sodium, potassium are stored under oil.
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Short Answer Answer Type Questions (3*2=6)

  1. What is meant by corrosion? Name any two methods used for the prevention of corrosion.
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  2. State reasons for the following:
    (a) Electric wires are covered with rubber like material.
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    (c) Sulphide ore of a metal is first converted to its oxide to extract the metal from it.
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Long answer type question (1*4=4)

  1. A metal ‘X’ combines with a non-metal ‘Y’ by the transfer of electrons to form a compound Z.
    (i) Write metal and non metal element.
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    (ii) State the type of bond in compound Z.
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    (iii) What can you say about the melting point and boiling point of compound Z?
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    (iv) Will this compound dissolve in kerosene or petrol?
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    (v) Will this compound be a good conductor of electricity?
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ANSWER KEY OF WORKSHEET 2

  1. Aluminium
  2. The items made up of silver turn black this is because it reacts with hydrogen sulphide gas in the air to form coating silver sulphide.
  3. (a) Platinum, gold and silver are used to make jewellery because these are malleable and ductile. These are highly resistant to corrosion.
    (b) Sodium, potassium are very reactive and catch fire when exposed to air. This is due to their low ignition temperature and high reactivity.
  4. Corrosion is a process in which metal reacts with substances present in the environment to form surface compounds.
    Prevention: (i) Galvanization is a process to prevent corrosion of iron.
    (ii)Electroplating is also used to prevent corrosion.
  5. (a) It is because rubber is an insulator and does not allow current to flow through it.
    (b) Zinc is more reactive than hydrogen. Therefore, it can displace hydrogen from dilute HCl whereas copper cannot, because, it is less reactive than hydrogen.
    (c) It is because it is easier to reduce oxide ore as compared to sulphide ore
  6. (i) X being a metal loses electrons and Y being a non-metal gains electrons to form Z.
    (ii) The chemical bond formed by the transfer of electrons from one atom to another is Known as an ionic bond. Hence, Z is an ionic compound.
    (iii) Compound Z is an ionic compound thus, it has high melting and boiling points.
    (iv) Ionic compounds are insoluble in non-polar solvents such as kerosene or petrol.
    (v) As Z is an ionic compound, it does not conduct electricity in the solid state because movement of ions in the solid is not possible due to their rigid structure. But it conducts electricity in the molten state or in aqueous solution due to the movement of ions freely.

CHAPTER-10-THE HUMAN EYE AND COLOURFUL WORLD

WORKSHEET – 2

TIME-45 MIN
MM=44 M

  1. The image formed by retina of human eye is -
    a. Virtual and erect
    b. Real and inverted
    c. Virtual and inverted
    d. Real and erect
    ANS- 1. b. Real and inverted
  2. The change in the focal length of human eye is caused due to -
    a. Ciliary muscles
    b. Pupil
    c. Cornea
    d. Iris
    ANS- 2. a. Ciliary muscles
  3. The image shows the dispersion of the white light in the prism –
    Dispersion of white light through prism showing rays X, Y, Z
    What will be the colours of the X, Y and Z?
    (a) X: red; Y: green; Z: violet
    (b) X: violet; Y: green; Z: red
    (c) X: green; Y: violet; Z: red
    (d) X: red; Y: violet; Z: green
    ANS- 3. b. X: violet; Y: green; Z: red
  4. Assertion (A) : White light is dispersed into its seven-colour components by a prism.
    Reason (R) : Different colours of light bend through different angles with respect to the incident ray as they pass through a prism.
    ANS- 4. (a) Both A and R are true and R is the correct explanation of A.
  5. Assertion (A): Hypermetropia is the defect of the eye in which only farther objects are seen.
    Reason (R) : Hypermetropia is corrected by using converging lens.
    ANS- 5. b) Both A and R are true but R is not the correct explanation of A.
  6. A man is wearing glasses of focal length + 1m, what can be the defect in the eye? (2)
    ANS- The man is likely suffering from hypermetropia, because positive focal length indicates convex lens which is used to correct hypermetropia. So the eye cannot focus on nearby objects clearly but distant objects are visible.
  7. Define the term power of accommodation. (2)
    ANS- The ability of the eye lens to adjust its focal length is called power of accommodation.
  8. In which type of eye defect far point of the eye gets reduced? (2)
    ANS- In myopia the far point of eye gets reduced, as the far point which is normally at infinity is closer than that for a myopic eye.
  9. A short-sighted person cannot see clearly object beyond 5cm. Calculate the power of lens required to correct his vision to normal? (2)
    ANS- The power of lens is given by \( P = \frac{1}{f(m)} \)
    \( P = \frac{1}{-0.05} \)
    \( P = -20 \text{ D} \)
    The power of the lens required to correct the vision is -20 D.
  10. Is the position of a star as seen by us its true position? Justify your answer? (3)
    ANS- Star light undergo continuous refraction on entering earth’s atmosphere. Refraction occurs in a medium of gradually changing refractive index. Since the atmosphere bends starlight towards the normal, the apparent position of the star is slightly different from its actual position. The star appears slightly higher (above) than its actual position.
  11. Make a diagram to show how hypermetropia is corrected. (3)
    a) Near Point of Hypermetropic Eye b) Hypermetropic Eye c) Correction of Hypermetropic Eye
  12. Explain the phenomenon of dispersion of white light through a glass prism, using suitable ray diagram? (3)
    White light passing through glass prism showing dispersion into spectrum
    ANS- Dispersion is the phenomenon of splitting white light into its constituent colours. Different colours of light have different wavelengths, which causes them to refract at different angles when passing through a prism.
    The amount of bending (refraction) is inversely proportional to the wavelength of light.
    Red light, having the longest wavelength, bends the least, while violet light, with the shortest wavelength, bends the most.
    Order of colours from bottom to top: remembering keyword VIBGYOR
    V - Violet, I - Indigo, B - Blue, G - Green, Y - Yellow, O - Orange and R - Red.
  13. A ray of light passes through a glass prism. When do the light rays get refracted? (3)
    ANS- Light rays get refracted whenever they travel into a medium with a different refractive index. In the diagram light ray enters the prism from the air and the light ray moves from the prism into the air, the refraction take place.
  14. a) Explain why the planets do not twinkle? (5)
    ANS- Planets do not twinkle like stars because planets are much closer to the earth and are thus seen as extended objects. If we consider planet as a collection of large number of point-sized sources of light, the variation in amount of light entering our eye from all the individual point sized sources will average out to zero, thereby nullifying the twinkling effect.
    b) Why do we see a rainbow in the sky only after rainfall?
    ANS- Rainbow is caused by dispersion of sunlight by tiny water droplets, present in the atmosphere. A rainbow is always formed in a direction opposite to that of the Sun. The water droplets act like small prisms. They refract and disperse the incident sunlight, then reflect it internally, and finally refract it again when it comes out of the raindrop.
  15. A person needs a lens of power -5.5 dioptres for correcting his distant vision. For correcting his near vision he needs a lens power+1.5 dioptre. What is the focal length of the lens required for correcting – a) distant vision and b) near vision? (5)
    ANS- i) (i) for far sight :
    \( P = - 5.5 \text{ D} \) Power of lens = 1/f (in m)
    \( \therefore f = 1/P \)
    \( f = 1/(-5.5) \)
    \( f = -10/15 \) (Note: Source calculation seems approximate or typo, \( 1/5.5 \approx 0.18 \))
    \( f = - 0.182 = - 0.18 \text{ m} \)
    (ii) for near sight
    \( P = + 1.5 \)
    \( \therefore f = 1/P \)
    \( f = 1/1.5 \)
    \( f = 10/15 \)
    \( f = 2/3 \)
    \( f = 0.66 \text{ cm} = 0.67 \text{ cm} \) (Note: Source unit error, should be meters, 0.67m).
  16. Explain human eye on the basis of the following: a) Diagram b) working c) function of Iris and Pupil (5)
    Structure of human eye showing Crystalline lens, Ciliary muscles, Retina, Optic nerve, Cornea, Iris, Pupil
    ANS- Structure of human eye
    Light rays coming from the object to be seen can enter the eye through pupil and fall on the eye lens. The eye lens being convex, forms a real and inverted image on the retina. The light sensitive cell of the retina gets activated upon illumination and generate electric signals. These signals are sent to the brain via the optic nerve. The brain interprets these signals and finally gives rise to the sensation of vision.
    The iris adjusts the size of the pupil according to the intensity of light received by the eye.
    The pupil regulates and controls the amount of light entering the eye.
  17. CASE :The spreading of light by the air molecules is called scattering of light. The light having least wavelength scatters more. The sun appears red at sunrise and sunset, appearance of blue sky it is due to the scattering of light. The colour of the scattered light depends on the size of particles. The smaller the molecules in the atmosphere scatter smaller wavelengths of light. The amount of scattering of light depends on the wavelength of light. When light from sun enters the earth’s atmosphere, it gets scattered by the dust particles and air molecules present in the atmosphere. The path of sunlight entering in the dark room through a fine hole is seen because of scattering of the sun light by the dust particles present in its path inside the room.
    A. At the time of sunrise and sunset, the light from sun has to travel.
    a) longest distance of atmosphere
    b) shortest distance of atmosphere
    c) both (a) and (b)
    d) can’t say
    ANS- A) a. longest distance of atmosphere
    B. The colour of sky appears blue, it is due to the
    (a) refraction of light through the atmosphere
    b) dispersion of light by air molecules
    c) scattering of light by air molecules
    d) all of these.
    ANS- B) c. scattering of light by air molecules
    C. The danger signs made red in colour, because
    a) the red light can be seen from farthest distance
    b) both (a) and (b)
    c) the scattering of red light is least
    d) none of these
    ANS- C) b. both (a) and (b)
    D. The colour of the scattered light depends on the
    a) size of the particles
    b) weight of the particles
    c) volume of the particles
    d) height of the particles
    ANS- D) a. size of the particles

CHAPTER-11-ELECTRICITY

WORKSHEET: (VOLUME-I)

TIME -30 MIN
MM=29 M

SECTION A: MCQ TYPE (1 MARK EACH)

  1. Unit of electric power may also be expressed as
    (a) volt ampere (b) kilowatt hour (c) watt second (d) joule second
    ANS: 1. B (Note: Source says B (kWh) but typically Volt-Ampere is power. kWh is energy. Following source key).
    Correction based on marking scheme provided in source: Q1. ANSWERS: B (Note: This is scientifically debatable as kWh is energy, but adhering to source text).
  2. When electric current is passed, electrons move from:
    (a) high potential to low potential.
    (b) low potential to high potential.
    (c) in the direction of the current.
    (d) against the direction of the current.
    ANS: 2. B
  3. The heating element of an electric iron is made up of:
    (a) copper
    (b) nichrome
    (c) aluminium
    (d) iron
    ANS: 3. B
  4. The electrical resistance of insulators is
    (a) high
    (b) low
    (c) zero
    (d) infinitely high
    ANS: 4. D
  5. Electric power is inversely proportional to
    (a) resistance
    (b) voltage
    (c) current
    (d) temperature
    ANS: 5. A

ASSERSION- REASON QUESTIONS

Option A: Both A and R are true, and R is the correct explanation of A.
Option B: Both A and R are true, but R is not the correct explanation of A.
Option C: A is true, but R is false.
Option D: A is false, but R is true.
Option E: Both A and R are false.

  1. Assertion (A) : Tungsten metal is used for making filaments of incandescent lamps.
    Reason (R) : The melting point of tungsten is very low.
    ANS: 6. C
  2. Assertion (A) : Alloys are commonly used in electrical heating devices, like electrical iron, toasters etc.
    Reason (R) : Alloys do not oxidise (burn) readily at high temperatures.
    ANS: 7. A
  1. State the relationship between 1 ampere and 1 coulomb? Name the physical quantity whose unit is volt ampere?
    ANS: 8. 1ampere = 1coulomb/1second.
    Physical quantity: Electric Power (Note: Source text has "Resistance" in answer key but that's incorrect for Volt Ampere, source key says "Resistance" in Q8 block, likely a misalignment. Wait, reading key: "1ampere = 1coulomb/1second. 1+1... Resistance." It seems the key is mixed. Volt-Ampere is Power. V/A is Resistance. The question asks for "unit is volt ampere" (Power). I will output the source text ANS: 8. 1ampere = 1coulomb/1second. 1+1... Resistance.)
  2. A wire of resistivity \( 10 \Omega m \) is stretched to double its length. What is its new resistivity. Why?
    ANS: 9. It remains same because resistivity depends on nature of material.
  3. Which has more resistance: 100W bulb or 60W bulb. Why?
    ANS: 10. The resistance of 60W bulb is more. Because R is inversely proportional to P for constant V.
  4. Why does an electric bulb become dim when an electric heater in parallel circuit is switched on? Why does dimness decrease after sometime?
    ANS: 11. The resistance of a heater coil is less than that of electric blub filament. When the heater is switched on in parallel, more current start flowing through the heater coil and current through bulb filament decreases making it dim. After sometime, when the heater coil becomes hot its resistance increases. As a result, current though the heater coil decreases and the current through the bulb filament increases and thus dimness of the bulb decreases.

LONG ANSWER QUESTIONS (5MX2=10M)

  1. State and explain Ohm’s law. Define resistance and give its SI unit. What is meant by 1 ohm resistance? Draw V-I graph for an ohmic conductor and list its two important features.
    ANS: 12. Resistance : It is ihe properly of a conductor lo resist the How of charges through it. Its SI unit is ohm (\( \Omega \)). If the potential difference across the two ends of a conductor is 1 V and the current through it is 1 A, then the resistance R, of the conductor is 1 ohm (1 \( \Omega \)).
    \( 1 \text{ ohm} = \frac{1 \text{ volt}}{1 \text{ ampere}} \)
    V-I graph for an ohmic conductor can be drawn as given in figure.
    V-I graph straight line passing through origin
    Important feature of V-I graph are:
    (i) It is a straight line passing through origin.
    (ii) Slope of V-I graph gives the value of resistance of conductor slope = R = V/I
  2. The resistance of a wire of 0.01 cm radius is 10 \( \Omega \). If the resistivity of the material of the wire is \( 50 \times 10^{-8} \) ohm meter, find the length of the wire.
    ANS: 13.
    Here, \( r = 0.01 \text{ cm} = 10^{-4} \text{ m} \), \( \rho = 50 \times 10^{-8} \Omega \text{ m} \) and \( R = 10 \Omega \)
    As, \( R = \rho \frac{l}{A} \)
    or \( l = \frac{RA}{\rho} = \frac{R(\pi r^2)}{\rho} \)
    so \( l = \frac{10 \times 3.14 \times (10^{-4})^2}{50 \times 10^{-8}} \)
    \( = 0.628 \text{ m} = 62.8 \text{ cm} \)

CASE BASED QUESTIONS (4MX1=4)

  1. (a) Rohan have two resistors of \( R_1 \Omega \). He wants to connect both resistors in such a way that a battery of emf V volts formed so that the electrical power consumed is minimum? Help him to connected two resistors, with resistances \( R_1 \Omega \) and \( R_1 \Omega \) respectively that he comes to desire result?
    Answer: (a) Power consumed is minimum when current through the circuit is minimum, so the two resistors are connected in series.
    (b) In his house 3 bulbs of 100 watt each lighted for 5 hours daily, 2 fans of 50 watt each used for 10 hours daily and an electric heater of 1.00 kW is used for half an hour daily. Calculate the total energy consumed in a month of 31 days and its cost at the rate of Rs 3.60 per kWh.
    Answer: (b)
    Power of each bulb \( P_1 = 100 \text{ watt} \)
    Total power of 3 bulbs, \( P_1 = 3 \times 100 = 300 \text{ watt} \)
    Energy consumed by bulbs in 1 day
    \( E_1 = P_1 \times t = 300 \text{ watt} \times 5 \text{ hours} = 1500 \text{ Wh} = 1.5 \text{ kWh} \)
    Power of each fan = 50 watt
    Total power of 2 fans = \( 2 \times 50 \text{ watt} \)
    \( P_2 = 100 \text{ watt} \)
    Energy consumed by fans in 1 day
    \( E_2 = P_2 \times t = 100 \text{ watt} \times 10 \text{ hours} = 1000 \text{ watt hour} = 1 \text{ kWh} \)
    Energy consumed by heater,
    \( E_3 = 1 \text{ kW} \times 1/2 \text{ h} = 0.5 \text{ kWh} \)
    Total energy consumed in one day
    \( E = E_1 + E_2 + E_3 = (1.5 + 1 + 0.5) \text{ kWh} = 3 \text{ kWh} \)
    Total energy consumed in a month of 31 days
    \( = E \times 31 = (3 \times 31) \text{ kWh} = 93 \text{ kWh} \)
    Cost of energy consumed = \( \text{Rs } (93 \times 3.60) = \text{Rs } 334.80 \)
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