Vardaan Learning Institute - Science Mock Test Solutions

1. (a) $CaO + H_2O \rightarrow Ca(OH)_2 + Heat$
Calcium oxide reacts with water to form calcium hydroxide (slaked lime), which is used for white-washing.

2. (b) Bubbles of a colourless and odourless gas.
Ethanoic acid reacts with sodium carbonate to release $CO_2$ gas, which is colourless and odourless.

3. (c) Magnesium (Mg)
Magnesium is above hydrogen in the reactivity series, so it can displace hydrogen. Copper, Silver, and Mercury are below hydrogen.

4. (b) Small intestine
Bile juice from the liver is released into the small intestine (duodenum) to emulsify fats.

5. (b) Mitochondria
Aerobic respiration (breakdown of pyruvate using oxygen) occurs in the mitochondria.

6. (c) Cytokinin
Cytokinins promote cell division and are abundant in areas of rapid cell division like fruits and seeds.

7. (c) Anther and ovary
Anther produces pollen (male gametes) and ovary produces ovules (containing female gametes).

8. (a) Round and yellow
In a dihybrid cross, the F1 generation exhibits the dominant traits. Round (R) and Yellow (Y) are dominant over wrinkled (r) and green (y). Genotype: RrYy.

9. (d) Between the pole of the mirror and its principal focus.
A concave mirror forms a virtual, erect, and magnified image only when the object is placed between the Pole (P) and Focus (F).

10. (c) Refraction, dispersion and internal reflection.
Formation of a rainbow involves refraction (entry), dispersion (splitting), internal reflection (inside drop), and refraction (exit). Note: Often referred to as "Total Internal Reflection" in older texts, but "Internal Reflection" is scientifically more accurate for rainbows as not all light is reflected. Option (c) is the most appropriate standard answer.

11. (c) $2A$
Resistance $R \propto \frac{l}{A}$. For the first wire, $R = \rho \frac{l}{A}$. For the second wire with resistance $R$ and length $2l$, let area be $A'$. Then $R = \rho \frac{2l}{A'}$. Equating both: $\rho \frac{l}{A} = \rho \frac{2l}{A'} \Rightarrow A' = 2A$.

12. (b) Parallel straight lines.
Inside a solenoid, the magnetic field is uniform, represented by parallel straight lines.

13. (a) Carnivores
1st level: Producers -> 2nd level: Herbivores -> 3rd level: Small Carnivores.

14. (b) $HCl$
$HCl$ loses hydrogen to form $Cl_2$, so it is oxidized. $MnO_2$ loses oxygen to form $MnCl_2$, so it is reduced.

15. (d) Diamond > Water > Kerosene
Refractive index: Diamond (2.42) > Kerosene (1.44) > Water (1.33). Note: Kerosene is optically denser than water.

16. (c) Banana peel
Biological waste like fruit peels are biodegradable. Plastic, glass, and metals are non-biodegradable.

17. (a) Both A and R are true, and R is the correct explanation of A.

18. (a) Both A and R are true, and R is the correct explanation of A.
The father produces two types of sperms (X and Y). The fusion of these determines the sex.

19. (a) Both A and R are true, and R is the correct explanation of A.
Magnetic field strength is inversely proportional to the distance from the wire.

20. (c) A is true but R is false.
Producers (plants) synthesise complex organic substances (starch) from simple inorganic substances ($CO_2, H_2O$). They do not break them down as their primary definition; Decomposers do that.

21. (a) The powder is Baking Powder. It is a mixture of Baking Soda (Sodium Hydrogen Carbonate, $NaHCO_3$) and a mild edible acid like Tartaric acid.
(b) $2NaHCO_3 \xrightarrow{Heat} Na_2CO_3 + H_2O + CO_2$
(Note: The actual reaction in baking involves the acid: $NaHCO_3 + H^+ \rightarrow Na^+ + CO_2 + H_2O$)

22. (a) Thyroxin hormone regulates carbohydrate, protein, and fat metabolism. It is secreted by the Thyroid gland.
(b) Iodised salt provides iodine, which is essential for the synthesis of thyroxin. Its deficiency causes Goitre.

23. Water and minerals are transported by Xylem tissue. The main driving force during the day is Transpiration Pull, created by the evaporation of water from the leaves (stomata). Root pressure also plays a minor role.

24. (a) Power of a lens is the reciprocal of its focal length in meters ($P = 1/f$). Unit: Dioptre (D).
(b) $P = +1.5 D$. Focal length $f = 1/P = 1/1.5 = 0.67$ meters (or 66.7 cm). Since power is positive, it is a Convex (Converging) lens.

25. Resistance of bulb $R = V^2/P = 220^2/100 = 48400/100 = 484 \, \Omega$.
Power at 110 V: $P' = V'^2/R = 110^2/484 = 12100/484 = 25$ W.
Power consumed = 25 W.

26. The energy captured by autotrophs does not revert to the sun. The energy that passes to herbivores does not come back to autotrophs. As it moves progressively through trophic levels, it is no longer available to the previous level. Hence, flow is unidirectional.

27. (a) $Mg (2,8,2) \rightarrow Mg^{2+} + 2e^-$
$2 [Cl (2,8,7) + e^- \rightarrow Cl^- (2,8,8)]$
Magnesium transfers its 2 valence electrons, one to each Chlorine atom. Product: $MgCl_2$.
(b) In solid state, ions are held in fixed positions by strong electrostatic forces and cannot move. In molten state (or aqueous solution), these forces are overcome by heat, allowing ions to move freely and conduct electricity.

28. (a) Modern Periodic Law: Properties of elements are a periodic function of their atomic numbers.
(b) Electronic configurations:
A (1): 1
B (8): 2, 6
C (11): 2, 8, 1
D (19): 2, 8, 8, 1
Odd element: B (Oxygen). Reason: B has 6 valence electrons (non-metal), while A, C, and D have 1 valence electron (alkali metals/hydrogen group).

29. (a) (i) Fertilization takes place in the Fallopian Tube (Oviduct).
(ii) Implantation occurs in the Uterus.
(b) If the egg is not fertilized, the thick and spongy lining of the uterus (endometrium), which was prepared to receive the embryo, is not needed. It breaks down and comes out through the vagina as blood and mucus. This is called menstruation.

30. (a) All flowers in F1 will be Blue (since Blue B is dominant over White b). Genotype: Bb.
(b) In F2, the ratio is 3 Blue : 1 White. So, percentage of white flowers = $\frac{1}{4} \times 100 = 25\%$.
(c) Genotypic ratio in F2: 1 BB : 2 Bb : 1 bb. Ratio of BB : Bb is 1 : 2.

31. (a) Dispersion: The splitting of white light into its component seven colours is called dispersion. The sequence of colours is VIBGYOR (Violet, Indigo, Blue, Green, Yellow, Orange, Red). [Diagram should show a prism splitting white beam].
(b) Stars act as point sources of light. Atmospheric refraction causes the path of light to vary, causing the apparent position to fluctuate and intensity to flicker (twinkling). Planets are closer and act as extended sources (collection of point sources); the variations cancel out, so they don't twinkle.

32. (a) The difference is in pattern: Inside the solenoid, field lines are parallel straight lines (indicating uniform field). Outside, they form curved loops similar to a bar magnet (emerging from North, entering South).
(b) Clock Face Rule: If current flows clockwise at a face, it is South Pole. If anticlockwise, it is North Pole.
(c) Strength can be increased by: (i) Increasing current, (ii) Increasing number of turns, or (iii) Inserting a soft iron core.

33. Food Chain: Grass $\rightarrow$ Insects $\rightarrow$ Frog $\rightarrow$ Snake $\rightarrow$ Hawk.
(a) Hawk will have the highest concentration. Phenomenon: Biomagnification.
(b) 10% Law:
Grass: 1% of 10,000 J (Solar) = 100 J (Available to grass/producers usually capture 1% of sun energy, but if 10,000J is 'available to grass' as biomass energy, then start there).
Assuming 10,000 J is the energy contained in Grass:
Grass (10,000 J) $\rightarrow$ Insects (1,000 J) $\rightarrow$ Frog (100 J) $\rightarrow$ Snake (10 J).
(If 10,000 J was solar energy hitting the plant, Grass captures 1% = 100J, Insects 10J, Frog 1J, Snake 0.1J). *Standard interpretation is usually energy at producer level.*

34. (a) Homologous Series: A series of compounds with the same functional group and similar chemical properties in which successive members differ by a $-CH_2$ unit or 14u mass.
Characteristics: (i) Same general formula, (ii) Similar chemical properties, (iii) Gradual change in physical properties.
(b) Electron dot structure of Ethene ($C_2H_4$): Two carbon atoms share two pairs of electrons (double bond), and each carbon shares one pair with two hydrogen atoms.
(c) Soaps vs Detergents: Soaps form insoluble precipitate (scum) with Calcium/Magnesium ions in hard water and don't clean well. Detergents do not form scum and clean effectively in hard water.

OR
(a) (i) $2C_2H_5OH + 2Na \rightarrow 2C_2H_5ONa \text{ (Sodium Ethoxide)} + H_2$
(ii) $C_2H_5OH \xrightarrow{Hot \, Conc. \, H_2SO_4} C_2H_4 \text{ (Ethene)} + H_2O$
(b) Cleansing action: Soap molecules have a hydrophilic ionic head and a hydrophobic hydrocarbon tail. The tails attach to oil/dirt (forming the core), and heads face outward into water. This formation is called a Micelle. Agitation pulls the oil out into the water as an emulsion, cleaning the cloth.

35. (a) Diagram should show: Left Ventricle (thick wall, bottom left), Aorta (arising from LV), Vena Cava (entering Right Atrium).
(b) Double Circulation: Blood passes through the heart twice in one complete cycle (Pulmonary and Systemic circulation). It is necessary to prevent mixing of oxygenated and deoxygenated blood, ensuring highly efficient oxygen supply for warm-blooded animals (energy needs).
(c) Functions of Lymph: (i) Carries digested and absorbed fat from the intestine. (ii) Drains excess fluid from extracellular space back into the blood. (iii) Protects against infection (contains lymphocytes).

OR
(a) Diagram should show: Kidneys (bean-shaped), Ureters (tubes), Urinary Bladder (sac).
(b) Urine Formation: 1. Filtration: Blood is filtered in the Glomerulus (Bowman's capsule) under pressure. Nitrogenous waste, glucose, water, salts pass out. 2. Reabsorption: Useful substances (glucose, amino acids, salts, water) are selectively reabsorbed in the tubular part of the nephron. 3. Secretion: Some ions ($K^+, H^+$) are secreted into the urine to maintain ionic balance. The final fluid is urine.

36. (a) Ohm's Law: The potential difference (V) across the ends of a metallic wire is directly proportional to the current (I) flowing through it, provided temperature remains constant. $V = IR$.
Diagram: Battery, Switch, Ammeter (series), Voltmeter (parallel across resistor), Rheostat.
(b) Parallel combination ($R_p$): $1/R_p = 1/5 + 1/10 = 3/10 \Rightarrow R_p = 10/3 \, \Omega$.
Total Resistance ($R_{eq}$): $R_p + 5 \Omega = 10/3 + 5 = 25/3 \, \Omega$.
Total Current ($I$): $I = V/R_{eq} = 6 / (25/3) = 18/25 = 0.72$ A.

OR
(a) Heating Effect: When current flows through a resistor, electrical energy is dissipated as heat.
Derivation: Work done $W = Q \times V$. Since $Q = I \times t$ and $V = IR$, $W = (It) \times (IR) = I^2Rt$. Heat $H = W = I^2Rt$.
(b) Data: $R = 20 \Omega$, $I = 5 A$, $t = 30 s$.
$H = I^2Rt = 5^2 \times 20 \times 30 = 25 \times 600 = 15,000$ Joules (or 15 kJ).

37. (i) Mercury (Hg).
(ii) $Al_2O_3 + 2NaOH \rightarrow 2NaAlO_2 \text{ (Sodium Aluminate)} + H_2O$.
(iii) (a) Calcination: Heating carbonate ores in limited supply of air to convert them into oxides. E.g., $ZnCO_3 \xrightarrow{Heat} ZnO + CO_2$.
OR
(iii) (b) Electrolytic Refining: - Anode: Impure Copper block. - Cathode: Thin strip of pure Copper. - Electrolyte: Acidified Copper Sulphate solution. - Process: On passing current, pure Cu from anode dissolves into electrolyte and deposits on cathode. Impurities settle as anode mud.

38. (i) Garden peas have short life cycles, visible contrasting traits, and are easy to cross-pollinate.
(ii) Genotype: The genetic constitution of an organism (e.g., TT, Tt, or tt).
(iii) (a) In F2 generation, ratio is 3 Tall : 1 Short.
OR
(iii) (b) Law of Independent Assortment: When two pairs of traits are combined in a hybrid, segregation of one pair of characters is independent of the other pair of characters. E.g., In a cross between Round-Yellow (RRYY) and Wrinkled-Green (rryy), the inheritance of seed shape is independent of seed colour, leading to a 9:3:3:1 ratio in F2.

39. (i) The image is Virtual, Erect, and Magnified.
(ii) $P = +5 D$. Focal length $f = 1/P = 1/5 = 0.2$ meters or 20 cm.
(iii) (a) Ray diagram: Object at $2F_1$ forms a real, inverted image of the same size at $2F_2$.
OR
(iii) (b) Data: $h = 5$ cm, $u = -25$ cm, $f = +10$ cm.
Lens Formula: $1/v - 1/u = 1/f \Rightarrow 1/v = 1/10 + 1/(-25) = 1/10 - 1/25 = (5-2)/50 = 3/50$.
$v = 50/3 = +16.67$ cm.
Magnification $m = v/u = (50/3) / -25 = -2/3$.
Image height $h' = m \times h = (-2/3) \times 5 = -3.33$ cm.
Result: Image is formed 16.67 cm on the other side, real, inverted, and 3.33 cm tall.