CBSE EXAMINATION PAPER-2023

Answer: (i) 'X' is gypsum, which is calcium sulphate dihydrate CaSO₄ . 2H₂O.
'Y' is Plaster of Paris, which is calcium sulphate hemihydrate CaSO₄ . 1/2H₂O.
(ii) 'X' (gypsum) can be reobtained from 'Y' (Plaster of Paris) by adding water to Plaster of Paris. It reacts with water and forms gypsum again which sets into a hard solid mass. This process is called hydration.

Answer:

(i) Solution M gives a red colour with a universal indicator, which means it is acidic. Acidic solutions have more hydrogen ions (H+). Hence, hydrogen ion concentration is more in solution M.

(ii) When solutions M and N are mixed, the universal indicator turns green, indicating the solution is neutral. This means the hydrogen ions from the acidic solution M have been neutralised by the hydroxide ions from the basic solution N, resulting in the formation of a neutral salt.

Answer: (i) Dendrite: Its function is to receive electrical signals or information from other neurons.
(ii) Axon: Its function is to carry electrical impulses away from the cell body towards other neurons, muscles, or glands.

Answer: Photosynthesis is the process by which plants make their food using sunlight. The first event is the absorption of light energy by chlorophyll in the leaves. Next, the light energy is converted into chemical energy. Water molecules are then split into hydrogen and oxygen using this energy. The plant takes carbon dioxide from the air. Finally, using hydrogen and carbon dioxide, the plant produces glucose (a sugar) and releases oxygen into the air.

Answer: Leaves of a plant help in excretion mainly through transpiration and guttation. Transpiration is the process where water absorbed by roots moves up and evaporates as water vapor through small pores called stomata on the leaves, removing excess water. Guttation occurs when excess water, along with dissolved waste products, is released as droplets from specialized structures on leaf edges. Thus, leaves remove waste products and excess water from plants.

Answer: In human digestion, two important enzymes that digest proteins are Pepsin and Protease. Pepsin is secreted by the gastric glands in the stomach, where it breaks down proteins into smaller peptides. Protease, on the other hand, is secreted by the pancreas and released into the small intestine to further digest proteins into amino acids. These enzymes play vital roles in protein digestion during the digestive process.

Answer: In a food chain, energy decreases at each trophic level because energy is lost as heat and used for metabolic activities. The given food chain has five trophic levels: grass (producer), grasshopper (primary consumer), frog (secondary consumer), snake (tertiary consumer), and hawk (quaternary consumer). If 50 J of energy is available to the hawk (fifth level), we can calculate the energy at the first and third levels by considering that only about 10% of energy is transferred from one level to the next. Therefore, the energy at the snake (fourth level) is 500 J, at the frog (third level) is 5000 J, and at the grass (first level) is 50000 J. Hence, the first trophic level (grass) has 50000 J and the third trophic level (frog) has 5000 J of energy. This shows that energy decreases considerably along the food chain, leading to less energy available at higher levels.

Answer: (a) A double displacement reaction is a type of chemical reaction where two compounds exchange their ions to form two new compounds. This usually takes place in aqueous solutions, and the positive ions (cations) and negative ions (anions) swap partners.

(b)(i) Example of a double displacement reaction that is also a neutralization reaction:
HCl (aq) + NaOH (aq) -> NaCl (aq) + H₂O (l)
This is a neutralization reaction because an acid (HCl) reacts with a base (NaOH) to produce a salt (NaCl) and water.

(b)(ii) Example of a double displacement reaction that is also a precipitation reaction:
BaCl₂ (aq) + Na₂SO₄ (aq) -> BaSO₄ (s) + 2NaCl (aq)
This is a precipitation reaction because BaSO₄ forms as a white solid (precipitate) which is insoluble in water. It is also a double displacement reaction as Ba²⁺ and Na⁺ ions exchange their anions SO₄^2- and Cl^-.

Answer: (a) The pH of our mouth becomes lower than 5.5 because bacteria present in the mouth break down the food particles, especially sugars, producing acids. These acids lower the pH, making the mouth acidic. (b) When baking soda (sodium hydrogen carbonate) is heated, it decomposes to form sodium carbonate (Na₂CO₃), which is the basic salt 'X'. Industrial uses of sodium carbonate include manufacturing glass and paper, and in soap and detergent production. (c) Copper sulphate crystals are blue due to water of crystallisation present in them. On heating, this water is removed as steam, and the crystals lose their blue colour, turning white because anhydrous copper sulphate is white in color.

Answer:

(i) Calculation of image distance (v):
Given: Object height h = 5 cm, Object distance u = -20 cm (since object distance is always negative for lenses), Focal length f = -18 cm (concave lens focal length is negative).
Using lens formula, 1/f = 1/v - 1/u
=> 1/v = 1/f + 1/u = 1/(-18) + 1/(-20) = -1/18 - 1/20 = - (10/180 + 9/180) = -19/180
So, v = -180/19 ≈ -9.47 cm.

(ii) Calculation of magnification (m):
Magnification m = v / u = (-9.47) / (-20) = 0.47.
Magnification is positive and less than 1, indicating the image is virtual, erect, and diminished.

(iii) Comparison of magnification for concave and convex lenses when both form virtual images:
A concave lens always forms virtual, erect, and smaller images, so magnification is positive but less than 1.
A convex lens forms virtual images only when the object is placed between the lens and its focal point; in this case, the magnification is positive and greater than 1 indicating virtual, erect, and magnified images.
Therefore, the magnification of a convex lens is greater than 1 for virtual images, whereas for a concave lens it is less than 1.

Answer:

A convex lens can form two types of magnified images depending on the position of the object relative to the focal length (10 cm) of the lens.

Case (i): Real, inverted and magnified image
For this type of image, the object should be placed between the focal length (10 cm) and twice the focal length (20 cm) from the lens. When the object distance is between 10 cm and 20 cm, the lens forms a real, inverted, and magnified image on the other side of the lens beyond 20 cm.

Case (ii): Virtual, erect and magnified image
For this case, the object must be placed closer to the lens than the focal length, that is within 10 cm of the lens. When the object is inside the focal length, the lens produces a virtual, erect, and magnified image on the same side as the object.

Ray Diagrams:
In the first case, rays converge on the opposite side forming a real image while in the second, rays appear to diverge from a point on the same side, creating a virtual image. The image formed in the first case can be obtained on a screen, but the virtual image in the second case cannot be projected on a screen.

Answer:

(a)

(i) The retina is the light-sensitive layer at the back of the eye that captures the light rays focused by the lens and converts them into electrical signals. These signals are then sent to the brain through the optic nerve, allowing us to see images.

(ii) The pupil is the opening in the center of the iris that controls the amount of light entering the eye. It adjusts in size — becoming small in bright light and large in dim light — to protect the eye and improve vision.

(b)

The ciliary muscles play a crucial role in focusing objects at different distances, a process called accommodation. When we look at nearby objects, the ciliary muscles contract, causing the crystalline lens to become thicker and more curved. This increases the lens's focusing power, allowing the eye to clearly see close objects. Conversely, when looking at distant objects, the ciliary muscles relax, making the lens thinner and less curved, which helps focus light from far objects onto the retina correctly. Thus, the ciliary muscles adjust the lens shape to ensure sharp vision at varying distances.

Answer:

(i) Effects on the displacement of the conductor:

1. If the current through the conductor is increased, the force acting on the conductor increases. Hence, the displacement of the conductor towards left will increase because the force on the conductor in a magnetic field is directly proportional to the current.

2. If the horse-shoe magnet is replaced by a stronger magnet, the magnetic field strength increases. Since the force on the conductor is also proportional to the magnetic field strength, the conductor will experience a greater force and thus its displacement towards left will increase.

3. If the direction of the current through the conductor is reversed, the direction of the force experienced by the conductor will also reverse. Therefore, the conductor will be displaced towards the right side instead of left.

(ii) Name and state the rule:

The rule is Fleming's Left-Hand Rule. It states that if you align your left hand such that the first finger points in the direction of the magnetic field (from North to South), the second finger points in the direction of the current (from positive to negative), then the thumb will point in the direction of the force or motion experienced by the conductor in the magnetic field.

Answer:

The magnetic field produced around a vertical current-carrying straight conductor passing through a horizontal cardboard consists of concentric circles around the conductor. The conductor is represented as a dot (if the current is coming out of the cardboard) or a cross (if the current is going into the cardboard). The direction of the magnetic field lines can be determined using the Right-Hand Thumb Rule. According to this rule, if you point the thumb of your right hand in the direction of the current, then the curl of your fingers around the conductor shows the direction of the magnetic field lines. For example, if the current flows upward (out of the cardboard), the magnetic field lines will be in the anticlockwise direction around the conductor. The key here is that magnetic field lines form closed loops around the current-carrying conductor.

Answer: Ozone in the higher levels of the atmosphere, also called the ozone layer, is formed when ultraviolet (UV) light from the Sun breaks oxygen molecules (O2) into single oxygen atoms. These single oxygen atoms then combine with oxygen molecules to form ozone (O3). Although ozone is poisonous at ground level, in the ozone layer it plays a crucial role by absorbing harmful UV radiation from the Sun. Damage to this ozone layer is a cause of concern because it reduces the protection against UV rays. Increased UV radiation reaching Earth's surface can cause serious health issues, such as skin cancer and cataracts in humans, and harm animals and plants. Therefore, preserving the ozone layer is essential to protect life on Earth.

Answer:

(i) Identification of compounds:

Compound ‘X’ with molecular formula C₂H₆O is ethanol (CH₃CH₂OH), a neutral organic compound. On reacting with acidified potassium dichromate (K₂Cr₂O₇), ethanol is oxidized to compound ‘Y’, which is ethanoic acid (CH₃COOH), an acidic compound. When ethanol (X) reacts with ethanoic acid (Y) in the presence of concentrated sulphuric acid (H₂SO₄) and heat, an esterification reaction occurs producing ‘Z’, which is ethyl ethanoate (CH₃COOCH₂CH₃), a sweet smelling ester.

(ii) Chemical equations:

1. Oxidation of ethanol (X) to ethanoic acid (Y):
CH₃CH₂OH + 2[O] → CH₃COOH + H₂O
(where [O] represents the oxygen from acidified K₂Cr₂O₇)

2. Esterification reaction of ethanol (X) with ethanoic acid (Y) in presence of conc. H₂SO₄ to form ethyl ethanoate (Z):
CH₃CH₂OH + CH₃COOH ⇌ CH₃COOCH₂CH₃ + H₂O

(iii) Roles of reagents:

1. Acidified K₂Cr₂O₇ acts as an oxidizing agent in the oxidation of ethanol (X) to ethanoic acid (Y). It provides the necessary oxygen atoms for oxidation under acidic conditions.

2. Concentrated H₂SO₄ acts as a dehydrating agent and catalyst in the esterification reaction between ethanol and ethanoic acid, promoting the formation of the ester (Z) by removing water and shifting equilibrium towards ester formation.

(iv) Reaction of ‘Z’ with alkali:

When the ester (ethyl ethanoate, Z) reacts with an alkali such as sodium hydroxide (NaOH), a saponification reaction occurs, producing an alcohol and a salt of the acid (sodium ethanoate).

Answer:

(i) Ethanol to Ethene:

Ethanol when heated with concentrated sulfuric acid (H₂SO₄) at around 170°C undergoes dehydration to form ethene and water. The reaction removes water (H₂O) from ethanol.

(ii) Ethene to Ethane:

Ethene is converted to ethane by hydrogenation, which involves addition of hydrogen (H₂) in the presence of a catalyst such as nickel (Ni) at elevated temperature and pressure.

(iii) Ethane to Chloroethane:

Ethane reacts with chlorine (Cl₂) in the presence of sunlight (UV light) to undergo substitution and form chloroethane and hydrogen chloride. This is a free radical substitution reaction.

(iv) Ethanol to Ethanoic acid:

When ethanol is oxidized by an oxidizing agent like potassium dichromate (K₂Cr₂O₇) in acidic medium (dilute sulfuric acid), ethanol is first converted to acetaldehyde and then further oxidized to ethanoic acid (acetic acid).

(v) Ethanoic acid to Ethyl ethanoate:

Ethanoic acid reacts with ethanol in the presence of an acid catalyst (usually concentrated sulfuric acid) under reflux to form ethyl ethanoate through an esterification reaction. This reaction is reversible.

Answer:

(i) In human males, the testes are located in the scrotum, which is outside the body. They are positioned outside the abdominal cavity because sperm production requires a temperature slightly lower than the normal body temperature. The two main functions of the testes are: to produce sperm cells, which are male reproductive cells, and to produce the male hormone testosterone that controls the development of male secondary sexual characteristics like hair growth and muscle development.

(ii) Changes in the female body after ovulation:

(1) If the egg is fertilized, it travels to the uterus and gets implanted in the uterine lining. This leads to the formation of an embryo and the female body begins to produce hormones to maintain the pregnancy.

(2) If the egg is not fertilized, it does not get implanted. The uterine lining breaks down and is shed along with the unfertilized egg as menstruation.

(iii) To prevent pregnancy through surgical methods, in males, vasectomy is performed where the sperm ducts are cut and tied to stop the passage of sperm. In females, tubectomy is done where the fallopian tubes are cut and tied to prevent the egg from traveling to the uterus, thus preventing fertilization.

Answer:

(i) (1) When leaves of Bryophyllum fall on the soil, small plantlets develop from the edges of the leaves. These plantlets grow into new plants. This is an example of vegetative reproduction where new plants grow from parts of the parent plant.

(2) When Planaria is cut into many pieces, each piece can regenerate and grow into a complete new Planaria. This is called regeneration, a type of asexual reproduction.

(3) When the sporangia of Rhizopus mature, they burst open and release numerous spores into the environment. These spores germinate and grow into new Rhizopus fungi. This is reproduction by spores.

Therefore, Bryophyllum reproduces by vegetative propagation, Planaria by regeneration, and Rhizopus by spore formation.

(ii) After fertilisation in a flower, several changes occur: The zygote formed develops into an embryo, the ovule transforms into a seed containing the embryo, and the ovary changes into a fruit which protects the seed and aids in its dispersal. Other floral parts like petals, sepals, and stamens usually wither and fall off as their functions are complete.

Answer: (a) Ohm's Law states that the potential difference V across a conductor is directly proportional to the current I flowing through it, provided the temperature remains constant. Mathematically, V = IR, where R is the resistance.
(b) The physical quantity determined by the slope of the V-I curve is the resistance (R). Resistance is defined as the ratio of voltage V to current I, R = V / I. To find R from the graph, take two points on the line, calculate the change in voltage and current, then divide. For example, if V = 6 volts and I = 2 amperes from the graph, then R = 6 / 2 = 3 ohms (Ω).
(c) Relationship between 1 kWh and 1 joule: 1 kilowatt hour (kWh) is the energy consumed by a 1000 watt device running for 1 hour. Since 1 hour = 3600 seconds, 1 kWh = 1000 watts * 3600 seconds = 3,600,000 joules. Therefore, 1 kWh = 3.6 * 10⁶ joules.