Vardaan Learning Institute - Mock Test 3 Solutions

1. (c) \(Ethanol + CO_2 + Energy\)
In yeast (anaerobic respiration), glucose breaks down into Ethanol, Carbon dioxide, and a small amount of energy.

2. (b) Synapse
The synapse is the junction between two neurons where the electrical impulse triggers the release of neurotransmitters (chemical signals).

3. (b) Wing of a bird and forelimb of a human
Homologous organs have the same basic structural design and origin but perform different functions. Both have a humerus, radius, ulna, etc.

4. (b) Convert organic material to inorganic forms
Decomposers break down complex organic matter from dead organisms into simple inorganic substances that go back into the soil.

5. (c) Condom
Barrier methods like condoms prevent fluid exchange, thus protecting against STDs. Other methods only prevent pregnancy.

6. (a) Both A and R are true and R is the correct explanation of A.
Energy flows from Sun -> Producers -> Consumers. It cannot flow backwards from consumers to producers or producers to Sun.

7. (a) Both A and R are true and R is the correct explanation of A.
Traits are inherited independently because the genes for them are located on separate chromosomes and segregate independently during gamete formation (Law of Independent Assortment).

8. (a) Auxin: Promotes cell elongation, root formation, and phototropism (bending towards light).
(b) Cytokinin: Promotes cell division, delays ageing of leaves, and promotes opening of stomata.

9. Importance: DNA copying ensures that the body design and traits are passed to the next generation, maintaining the species identity.
Variation: Variations (errors in copying) allow organisms to adapt to changing environments (e.g., heat resistance). While an individual variant might die, the population survives due to these variations.

10. Aerobic: Occurs in presence of oxygen, complete breakdown of glucose, end products are \(CO_2\) and \(H_2O\), releases more energy (38 ATP).
Anaerobic: Occurs in absence of oxygen, incomplete breakdown, end products are Ethanol/\(CO_2\) or Lactic acid, releases less energy (2 ATP).
Organism: Yeast.

11. Blood Flow: Deoxygenated blood from body -> Vena Cava -> Right Atrium -> Right Ventricle -> Pulmonary Artery -> Lungs (Oxygenation) -> Pulmonary Vein -> Left Atrium -> Left Ventricle -> Aorta -> Body.
Double Circulation: Blood passes through the heart twice in one complete cycle (once through the right side to lungs, and once through the left side to body). This prevents mixing of oxygenated and deoxygenated blood, ensuring efficient oxygen supply.

12. How are fats digested in our bodies? Where does this process take place? Mention the role of bile salts. (3 Marks)

Process: Digestion of fats takes place primarily in the small intestine.
Role of Bile Salts: Fats are present as large globules. Bile salts (from liver) break them down into smaller globules (Emulsification), increasing the surface area for enzyme action.
Enzyme Action: The enzyme Lipase (from pancreatic juice) breaks down emulsified fats into Fatty Acids and Glycerol.

13. (a) Diagram Reference: Please refer to NCERT Class 10 Science, Chapter 6, Figure 6.1 (a) Structure of Neuron. Label Dendrite (branching tips), Cell Body (center with nucleus), Axon (long tail).
(b) Synapse Transmission: When an electrical impulse reaches the axonal end, it releases chemicals called neurotransmitters. These chemicals diffuse across the gap (synapse) and start a similar electrical impulse in the dendrite of the next neuron.

14. Urine Formation:
1. Filtration: Blood enters the glomerulus under pressure. Nitrogenous wastes (urea), glucose, water, and salts are filtered into the Bowman's capsule.
2. Reabsorption: As the filtrate moves through the tubular part of the nephron, useful substances like glucose, amino acids, salts, and water are selectively reabsorbed into blood capillaries.
3. Secretion: Excess water and wastes like urea remain in the tubule, forming urine, which collects in the collecting duct.

15. (a) Diagram Reference: Refer to NCERT Class 10 Science, Chapter 7, Figure 7.7 Longitudinal section of flower.
(b) Pollination vs Fertilization: Pollination is the transfer of pollen from anther to stigma. Fertilization is the fusion of the male gamete with the female gamete (egg) inside the ovule.
(c) Post-Fertilization: The Zygote divides to form an Embryo. The Ovule develops a hard coat and becomes the Seed. The Ovary grows rapidly and ripens to form the Fruit. Petals/sepals fall off.
OR
(a) Organs: Testes, Vas deferens, Seminal vesicles, Prostate gland, Urethra, Penis.
(b) Functions: (i) Testis: Produces sperms and Testosterone. (ii) Vas deferens: Transports sperms. (iii) Prostate: Adds fluid to make sperm transport easier and provide nutrition.
(c) Location: Sperm formation requires a temperature 2-3°C lower than body temperature. The scrotum outside the body provides this optimal temperature.

16. (i) F2 Ratio: 3 Tall : 1 Short.
(ii) F1 Conclusion: Only the dominant trait (Tallness) is expressed in the F1 generation, even though the plant has genes for both.
(iii) (a) Short Plants: 25% of 100 = 25 plants.
(iii) (b) Law of Dominance: In a heterozygote (Tt), one trait will conceal the presence of another trait for the same characteristic. The trait that is expressed (Tall) is dominant, and the one hidden (Short) is recessive.

17. (b) \(4Na + O_2 \rightarrow 2Na_2O\)
This is a combination reaction where two elements (Sodium and Oxygen) combine to form a single compound.

18. (d) 10
Red litmus turning blue indicates a Basic solution. pH > 7 is basic. 10 is the only basic pH option.

19. (c) Magnesium (Mg)
Na, K, Ca react with cold water. Magnesium reacts only with hot water to form Magnesium hydroxide/oxide and hydrogen.

20. (c) Ketone
The suffix '-one' indicates a Ketone group (\(>C=O\)).

21. (i) Iron with Steam: \(3Fe(s) + 4H_2O(g) \rightarrow Fe_3O_4(s) + 4H_2(g)\)
(ii) Calcium with Water: \(Ca(s) + 2H_2O(l) \rightarrow Ca(OH)_2(aq) + H_2(g)\)

22. Distilled water is a pure covalent compound and does not contain free ions to conduct electricity. Rainwater dissolves gases like \(CO_2\) and \(SO_2\) from the atmosphere, forming acids (Carbonic acid) which dissociate into ions (\(H^+, CO_3^{2-}\)), making it conductive.

23. Roasting: Heating sulphide ores in excess air. \(2ZnS + 3O_2 \rightarrow 2ZnO + 2SO_2\).
Calcination: Heating carbonate ores in limited air. \(ZnCO_3 \rightarrow ZnO + CO_2\).

24. (a) Both A and R are true and R is the correct explanation of A.
Carbon forms covalent bonds by sharing electrons because gaining 4 electrons (C4-) or losing 4 electrons (C4+) requires too much energy or makes the nucleus unstable.

25. (a) Product: Ethanoic Acid (\(CH_3COOH\)). Reaction: \(CH_3CH_2OH + 2[O] \xrightarrow{Alk. KMnO_4} CH_3COOH + H_2O\).
(b) Cleansing Action: Soaps form micelles. The hydrophobic tail attaches to dirt/oil, and the hydrophilic head attaches to water. Agitation pulls the dirt out into the water.
Hard Water: Hard water contains Calcium and Magnesium ions. Soap reacts with them to form insoluble precipitate called Scum (\(RCOO\)_2Ca), wasting soap and not cleaning.

26. (a) Oxidised: \(HCl\) (Loses H to form \(Cl_2\)). Reduced: \(MnO_2\) (Loses O to form \(MnCl_2\)).
(b) Redox Reaction: A reaction where oxidation and reduction occur simultaneously. Example: \(CuO + H_2 \rightarrow Cu + H_2O\).

27. (i) Mild Base: Magnesium Hydroxide (Milk of Magnesia) or Sodium Hydrogen Carbonate.
(ii) Tooth Decay: Below pH 5.5, the acid produced by bacteria corrodes the calcium phosphate (enamel) of teeth.
(iii) (a) Bee sting injects Methanoic Acid. Remedy: Apply a mild base like Baking Soda.
OR
(iii) (b) The soil is Acidic. Quick lime (CaO) is basic and is used to neutralize the excess acid in the soil.

28. (a) Amphoteric Oxides: Metal oxides that react with both acids and bases to produce salt and water. Examples: \(Al_2O_3\) (Aluminium Oxide) and \(ZnO\) (Zinc Oxide).
(b) (i) \(Al_2O_3 + 6HCl \rightarrow 2AlCl_3 + 3H_2O\)
(ii) \(Al_2O_3 + 2NaOH \rightarrow 2NaAlO_2 + H_2O\) (Sodium Aluminate).
(c) Sodium is highly reactive and reacts vigorously with oxygen and moisture in the air (catching fire). Kerosene cuts off contact with air/moisture.
OR
(a) Structures: (i) \(CH_3COOH\) (ii) \(H-S-H\) (Electron dots sharing).
(b) Homologous Series: A series of compounds with same functional group. Characteristics: Same general formula, gradual change in physical properties, similar chemical properties.
(c) Alkene Member 2: General formula \(C_nH_{2n}\). 1st is Ethene (n=2). 2nd is Propene (n=3) -> \(C_3H_6\).

29. Since X reacts with Zn to give Hydrogen, X is an Acid. Since it reacts with \(CH_3COOH\) (Ethanoic Acid) to give C (likely Ester), X must be an Alcohol. Let's re-evaluate. Alcohol + Acid = Ester. Alcohol + Zn = ? Usually Acid + Metal = H2. Wait, ethanol reacts with sodium to give H2, not Zn. Let's assume X is an Acid (Ethanoic Acid? No, X reacts with Ethanoic acid).
Correction: If X reacts with Carbonate/Base, it's acid. If X reacts with Acid, it could be Base/Alcohol.
Reaction (iii) implies Esterification. So one is Alcohol, one is Acid. X + Ethanoic Acid -> Ester. So X is Ethanol (\(C_2H_5OH\)).
Wait, Ethanol + Zn does not typically give H2. Ethanol + Na gives H2. Maybe X is a metal? No.
Let's assume X is Sodium Hydroxide (NaOH) (Base).
(i) \(2NaOH + Zn \rightarrow Na_2ZnO_2 + H_2\) (Sodium Zincate + Hydrogen). Fits.
(ii) \(NaOH + HCl \rightarrow NaCl + H_2O\). Fits.
(iii) \(NaOH + CH_3COOH \rightarrow CH_3COONa + H_2O\). Fits.
Answers:
X = Sodium Hydroxide (\(NaOH\))
A = Sodium Zincate (\(Na_2ZnO_2\))
B = Sodium Chloride (\(NaCl\))
C = Sodium Ethanoate (\(CH_3COONa\))

30. (d) Between the pole of the mirror and its principal focus
Only in this position does a concave mirror form a Virtual, Erect, and Magnified image.

31. (b) Accommodation
Accommodation is the ability of the eye lens to adjust its focal length to see objects at varying distances.

32. (b) \(IR^2\)
Power \(P = VI = I^2R = V^2/R\). \(IR^2\) is incorrect dimensionally.

33. (b) Parallel straight lines
This indicates a uniform magnetic field inside the solenoid.

34. (a) Defects: -5.5D (Negative) -> Myopia (Near-sightedness). +1.5D (Positive) -> Hypermetropia (Far-sightedness) or Presbyopia.
(b) (i) Distant: \(f = 1/P = 1/(-5.5) = -0.18\) m or -18.18 cm.
(ii) Near: \(f = 1/P = 1/(+1.5) = +0.66\) m or +66.67 cm.

35. Total Power \(P = P1 + P2 = 100W + 60W = 160W\).
Voltage \(V = 220V\).
Current \(I = P/V = 160/220 = 16/22 = 0.727\) A.
Current = 0.73 A (approx).

36. (a) Right-Hand Thumb Rule: If you hold a current-carrying straight conductor in your right hand such that the thumb points towards the direction of current, then your fingers will wrap around the conductor in the direction of the field lines of the magnetic field.
(b) Diagram Reference: Refer to NCERT Class 10 Science, Chapter 12, Figure 12.6. It shows concentric circles around the wire. Arrows on circles indicate field direction (Clockwise if current is Down, Anti-clockwise if Up).

37. (a) Principal Focus (Concave): A point on the principal axis where rays of light parallel to the principal axis meet (converge) after reflection.
(b) Data: \(u = -10\) cm, \(f = +15\) cm (Convex mirror f is positive).
Formula: \(1/v + 1/u = 1/f \Rightarrow 1/v = 1/15 - 1/(-10) = 1/15 + 1/10 = (2+3)/30 = 5/30 = 1/6\).
\(v = +6\) cm.
Nature: Since v is positive, image is behind the mirror, Virtual and Erect. Position: 6 cm behind the mirror.

38. (a) Ohm's Law: Potential difference (V) across ends of a wire is directly proportional to current (I), provided temperature is constant. \(V=IR\). Diagram: Battery, Switch, Ammeter (series), Voltmeter (parallel to R).
(b) Series Derivation: \(V = V_1 + V_2 + V_3\). Since \(V=IR\), \(IR_{eq} = IR_1 + IR_2 + IR_3\). Divide by I: \(R_{eq} = R_1 + R_2 + R_3\).
(c) Square: Total R=20. 4 sides, so each side = 20/4 = 5 Ohms. Diagonal connects two corners. Path 1 (2 sides) = 5+5=10. Path 2 (2 sides) = 5+5=10. These two 10 Ohm paths are in Parallel. \(1/R_{eq} = 1/10 + 1/10 = 2/10 = 1/5\). \(R_{eq} = 5 \Omega\).
OR
(a) Heating Effect: Dissipation of electrical energy as heat in a resistor. \(W = Q \times V = (It) \times (IR) = I^2Rt\).
(b) Alloys: Higher resistivity than constituent metals and do not oxidize (burn) easily at high temperatures.
(c) \(P = 840W, V = 220V\).
Current \(I = P/V = 840/220 = 3.82\) A.
Resistance \(R = V/I = 220/3.82 = 57.6 \Omega\).

39. (i) Refraction.
(ii) Snell's Law: The ratio of sine of angle of incidence to the sine of angle of refraction is a constant for a given pair of media. \(\frac{\sin i}{\sin r} = n\).
(iii) (a) Diagram Reference: Refer to NCERT Class 10 Science, Chapter 9, Figure 9.10. Shows ray bending towards normal entering glass, away from normal exiting glass. Emergent ray parallel to incident ray.
(iii) (b) \(n = c/v\).
\(1.5 = (3 \times 10^8) / v\).
\(v = (3 \times 10^8) / 1.5 = 2 \times 10^8\) m/s.
Speed in glass = \(2 \times 10^8\) m/s.