(b) Grana are stacks of thylakoids found within plastids, specifically chloroplasts, where photosynthesis takes place.
2. Ovules symmetrical and straight with chalaza at the base and micropyle at the tip are called:
Anatropous
Amphitropous
Campylotropous
Orthotropous
(d) Orthotropous ovules are erect and straight, with the micropyle, chalaza, and hilum lying in a straight line.
3. Which hormone is primarily responsible for cell division in plants?
Abscicic acid (ABA)
Indole Acetic acid
Gibberelin
Cytokinin
(d) Cytokinins are plant hormones that promote cell division, or cytokinesis.
4. Br⪕ ⚭ \((K_((5) ) C_((5) ) )\) As \((G_((2) ) )\)
Is the floral formula of the family.
Cruciferae
Solanaceae
Gramineae
Malvaceae
(b) The given floral formula Br⊕⚥ K(5) C(5) A∞ G(2) represents the family Solanaceae (Potato family).
5. Potometer is used to measure:
Transpiration
Respiration
Photosynthesis
Growth of plants
(a) A potometer is a device used to measure the rate of water uptake by a plant, which is directly related to the rate of transpiration.
6. Volume of CO2 absorbed in photosynthesis is almost:
Double of O2 liberated
Equal to O2 liberated
Half of O2 liberated
Negligible to O2 liberated
(b) During photosynthesis, the volume of carbon dioxide absorbed is equal to the volume of oxygen liberated, according to the balanced chemical equation.
7. The organism which has the property of changing sugar into alcohol is:
Saccharomyces
Virus
Mucor
Spirogyra
(a) Saccharomyces (yeast) is a fungus that performs fermentation, converting sugars into alcohol and carbon dioxide.
8. The amount of energy usually released in the aerobic respiration is:
204 kCal
284 kCal
464 kCal
674 kCal
(d) Aerobic respiration typically yields around 674 kCal of energy per mole of glucose.
9. Gossypium hirsutum is the botanical name of
Potato
Sugarcane
Cotton plant
Soyabean
(c) Gossypium hirsutum is the botanical name for upland cotton, the most widely cultivated species of cotton plant.
10. The hypogynous flower is seen in:
China rose
Rose
Plum
Cucumber
(a) In a hypogynous flower, the gynoecium is superior, and other floral parts (sepals, petals, and stamens) are inserted below it, as seen in the China rose.
11. The association of fungus and roots of higher plants are called:
Pneumatophores
Assimilatory roots
Floating roots
Mycorrhizal roots
(d) Mycorrhizae are symbiotic associations between a fungus and the roots of a vascular plant.
12. Leafy stipules are found in:
Chrysanthemum
Lathyrus
Polygonium
Geranium
(b) Lathyrus (sweet pea) is known for having leafy stipules, which are leaf-like appendages at the base of the petiole.
13. In the floral leaves, sepals and petals simply touch each other, but do not overlap, this type of aestivation is called:
Quincuncial
Imbricate
Twisted
Valvate
(d) Valvate aestivation is characterized by floral leaves (sepals or petals) that meet at their edges without overlapping.
14. Interchange of the parts of the chromatids of a pair of chromosome is known as:
Gene mutation
Crossing over
Linkage
Dihybrid cross
(b) Crossing over is the process during meiosis where homologous chromosomes exchange segments, leading to genetic recombination.
15. Aerial mode of animal adaptation is known as:
Cursorial
Fossorial
Scansorial
Volant
(d) Volant adaptation refers to the ability of animals to fly or glide through the air.
16. Diphyllobotrium of phylum platyhelminthes is known as:
Fish tapeworm
Pork tapeworm
Dog 🐶 tapeworm
Dwarf tape worm
(a) Diphyllobothrium latum is a tapeworm commonly known as the broad fish tapeworm.
17. The carrier of Trypanosoma in man is:
Tse-tse fly
Mosquito
Sandfly
Housefly
(a) The Tse-tse fly is the vector for the transmission of Trypanosoma brucei, the parasite that causes sleeping sickness (African trypanosomiasis) in humans.
18. The sponge differs from other animals by the presence of:
Collar cells
Hollow body
Coelenteron cavity
Only one mouth
(a) Sponges are unique in having choanocytes, or collar cells, which are specialized flagellated cells used for filter-feeding.
19. Liver-rot is caused by:
Fasciola
Schistosoma
Taenia
Echinococcus
(a) Liver fluke disease, or liver-rot, is caused by the parasitic trematode Fasciola hepatica.
20. In earthworm, septa are absent in segments:
8-12
17-21
1-4
14-26
(c) Septa are absent in the first four segments of the earthworm's body.
21. Parapodia are found in:
Oligochaeta
Polychaeta
Arthropoda
Hirudinea
(b) Parapodia, which are fleshy lateral appendages used for locomotion, are characteristic of the class Polychaeta (bristle worms).
22. Most distinguish feature of eutherian mammals is presence of:
Pinna
Hairs
Diaphragm
Placenta
(d) While other features are present in mammals, the placenta, which provides nourishment to the developing fetus within the uterus, is the defining characteristic of eutherian mammals.
23. When lizards are threatened they can break off their tails and escape. This phenomenon is called:
Fragmentation
Hypertomy
Autotomy
Hypotomy
(c) Autotomy is the ability of an animal to voluntarily shed a part of its body, such as a tail, usually as a self-defense mechanism against predators.
24. Liver of frog consists of:
Single lobe
Two lobes
3 lobes
4 lobes
(c) The liver of a frog is typically composed of three lobes.
25. Response of organisms to stimulus of touch is called:
Phototaxis
Thermotaxis
Hydrotaxis
Thigmotaxis
(d) Thigmotaxis is the behavioral response of an organism to a physical touch stimulus.
26. In the following nuclear reactions what are X and Y respectively:
7N14 + 2He4 = 8O17 +X
13Al27 + 1H2 = 14Si28 +Y
1H1 and 0n1
0n1 and 1H1
2He4 + 0n1
0n1 + 2He4
(a) For the first reaction, conservation of mass number (14+4 = 17+1) and atomic number (7+2 = 8+1) indicates X is 1H1 (proton). For the second reaction, conservation of mass number (27+2 = 28+1) and atomic number (13+1 = 14+0) indicates Y is 0n1 (neutron).
27. For the reacions: N2+3H2⇋2NH3+Heat
Kp=Kc
Kp=Kc RT
Kp=Kc (RT)(−2)
KP=KC (RT)(−1)
(c) The relationship between Kp and Kc is given by Kp = Kc(RT)^Δn, where Δn is the change in the number of moles of gas. In this reaction, Δn = (2) - (1+3) = -2. Therefore, Kp = Kc(RT)−2.
28. Which of the following shows Tautomerism?
(CH3 )2 NH
(CH3 )3 CNO
R3 CNO2
RCH2 NO2
(d) Tautomerism is a phenomenon where a single chemical compound exists in two or more interconvertible structures. Nitro compounds with an α-hydrogen (RCH2NO2) exhibit tautomerism, existing in nitro and aci-nitro forms.
29. Ethyl alcohol and Acetic acid are mixed in equimolecular proportions; equilibrium is attained when (2/3)rd of the acid and alcohol are consumed. The equilibrium constant of the reaction will be:
0.04
0.4
4.0
40
(c) The reaction is CH3COOH + C2H5OH ⇋ CH3COOC2H5 + H2O. Initially, let moles of acid and alcohol be 'a' each, and moles of ester and water be 0. At equilibrium, (2/3)a of acid and alcohol are consumed, so the moles of each reactant remaining are a - (2/3)a = (1/3)a. The moles of ester and water formed are each (2/3)a. The equilibrium constant Kc = [(2/3)a * (2/3)a] / [(1/3)a * (1/3)a] = (4/9) / (1/9) = 4.0.
30. On reaction with Baeyer's Reagent, Ethylene and Acetylene form:
Formic acid and acetic acid
Oxalic acid and ethylene glycol
Ethylene glycol and oxalic acid
Acetic acid and Formic acid
(c) Baeyer's reagent (cold, dilute alkaline KMnO4) oxidizes alkenes to vicinal diols (glycols) and alkynes to α-diketones or carboxylic acids depending on the conditions. Ethylene forms ethylene glycol, and acetylene forms oxalic acid.
31. In the following chain of reactions, which is the product C?
CH3 COOH→(NH3 ) A→△ B→(P2 O5 ) C
Ammonium acetate
CH4
Acetonitrile
CH3 OH
(c) The reaction sequence is: CH3COOH + NH3 → CH3COONH4 (A - Ammonium acetate) →△ CH3CONH2 (B - Acetamide) →P2O5 CH3CN (C - Acetonitrile). P2O5 acts as a dehydrating agent.
32. Which of the following reacts with KCN to form Benzoin?
C6 H5 COOH
C6 H5 Cl
C6 H5 CHO
C6 H5 OH
(c) Benzoin condensation involves the reaction of benzaldehyde (C6H5CHO) with KCN (as a catalyst) to form benzoin (C6H5CH(OH)COC6H5).
33. Chlorine reacts with Benzaldehyde to give
Benzal chloride
Benzoyl chloride
Chlorobenzene
Benzyl chloride
(b) Chlorine reacts with benzaldehyde in the presence of sunlight or a Lewis acid catalyst to undergo side-chain chlorination forming benzoyl chloride (C6H5COCl).
34. In cannizaro's reaction:
Acid is coverted into amine
Aldehyde is converted into alcohol
Alcohol is converted into aldehyde
Primary amine is converted into cyanide
(b) Cannizzaro reaction is a disproportionation reaction of aldehydes lacking α-hydrogen atoms in the presence of a strong base, resulting in the formation of a primary alcohol and a carboxylate salt.
35. The rate of diffusion of a gaa 'A' is 5 times that of a gas 'B', what will be the ratio of the densities?
1/25
1/5
25
5
(a) According to Graham's law of diffusion, the rate of diffusion of a gas is inversely proportional to the square root of its density. So, (Rate of A / Rate of B) = √(Density of B / Density of A). Given that the rate of diffusion of A is 5 times that of B, we have 5 = √(Density of B / Density of A). Squaring both sides, we get 25 = Density of B / Density of A. Therefore, the ratio of their densities (Density of A / Density of B) is 1/25.
36. The solubility product of a sparingly soluble salt AB2, is 1.08 x 10-13 at 25 °C. Its molar solubility is:
3 x 10-4 M
3 x 10-5 M
3 x 10-6 M
3 x 10-7 M
(a) For a salt AB2, the solubility equilibrium is AB2 ⇋ A2+ + 2B−. If the molar solubility is 's', then [A2+] = s and [B−] = 2s. The solubility product Ksp = [A2+][B−]2 = s(2s)2 = 4s3. Given Ksp = 1.08 x 10−13, we have 4s3 = 1.08 x 10−13. So, s3 = (1.08 / 4) x 10−13 = 0.27 x 10−13 = 27 x 10−15. Taking the cube root, s = 3 x 10−5 M. There seems to be a calculation error in my thought process. Let me re-calculate. s3 = 0.27 x 10-13 = 270 x 10-16. s = (270)1/3 x 10-16/3 which is not a clean answer. Let's check the options. If s = 3 x 10-4, then Ksp = 4 * (3 x 10-4)3 = 4 * 27 * 10-12 = 108 x 10-12 = 1.08 x 10-10. This doesn't match. If s = 3 x 10-5, then Ksp = 4 * (3 x 10-5)3 = 4 * 27 * 10-15 = 108 x 10-15 = 1.08 x 10-13. This matches the given Ksp.
37. Sodium is commercially obtained by the electrolysis of its fused chloride. The process is called:
Castner process
Downs process
Kellner process
Solvay process
(b) Sodium is commercially extracted by the electrolysis of molten sodium chloride in the Downs process.
38. What type of hybridization is involved in [Fe^(+2) (CN)6 ](4−)?
Sp
Sp2
Sp3d2
Sp3
(d) In [Fe+2(CN)6]4−, Fe+2 has a d6 configuration. In the presence of strong field ligands like CN−, inner orbital hybridization (d2sp3) occurs, leading to an octahedral complex.
39. The distance between an object and a diverging lens is 'p' times the focal length of the lens. The lateral magnification 'm' produced by the lens will be:
P
P-1
1+P
(1+P)-1
(d) For a diverging lens, the focal length (f) is negative. Given object distance (u) = -pf (since p times focal length and object is real). Using the lens formula: 1/v - 1/u = 1/f, we get 1/v - 1/(-pf) = 1/(-f) => 1/v + 1/(pf) = -1/f => 1/v = -1/f - 1/(pf) = (-p - 1)/(pf) => v = -pf/(p+1). Lateral magnification m = v/u = [-pf/(p+1)] / [-pf] = 1/(p+1) = (1+P)-1.
40. Curie is the temperature at which:
A diamagnet becomes a paramagnet
Magnetism is completely destroyed in a paramagnet
A ferromagnet becomes a diamagnet
A ferromagnet becomes a paramagnet
(d) The Curie temperature is the specific temperature above which a ferromagnetic material loses its spontaneous magnetization and becomes paramagnetic.
41. A projectile's time of flight 'T' is related to horizontal range by equation gT2=2R. The angle of projection in degree is:
45 °
90 °
30 °
60 °
(a) We know that Time of flight T = (2u sinθ)/g and Horizontal Range R = (u2 sin2θ)/g = (2u2 sinθ cosθ)/g. Given gT2 = 2R => g * (2u sinθ/g)2 = 2 * (2u2 sinθ cosθ)/g => g * (4u2 sin2θ)/g2 = (4u2 sinθ cosθ)/g => (4u2 sin2θ)/g = (4u2 sinθ cosθ)/g => sin2θ = sinθ cosθ => sinθ = cosθ (since sinθ ≠ 0 for a projectile) => tanθ = 1 => θ = 45°.
42. A stone is dropped from the top of a tower of height 'h'. It reaches the ground in 't' secs. The position of the stone after t/3 secs will be …….from the ground.
3/4 h
h9
8/9 h
1/4 h
(c) Using the equation of motion, s = ut + (1/2)at2. Here, u = 0, a = g. So, h = 0 * t + (1/2)gt2 => h = (1/2)gt2. The distance covered in t/3 secs from the top is s' = 0 * (t/3) + (1/2)g(t/3)2 = (1/2)g(t2/9) = (1/9) * (1/2)gt2 = h/9. Therefore, the position of the stone from the ground after t/3 secs will be h - s' = h - h/9 = 8h/9.
43. When two bodies are rubbed against each other they are electrified. Which one of the following is correct?
A glass rod becomes negatively charged if rubbed with a metal.
A glass rod becomes positively charged if rubbed with a flannel.
A glass rod becomes positively charged if rubbed with ebonite.
A glass rod becomes negatively charged if rubbed with amber.
(b) Triboelectric series shows that glass becomes positively charged when rubbed with flannel.
44. Two stars radiate maximum energy at 320 nm and 400 nm respectively. The ratio of their Kelvin temperate is:
5:4
4:5
2:5
4:25
(a) According to Wien's displacement law, λmaxT = constant. So, λ1T1 = λ2T2. Given λ1 = 320 nm, λ2 = 400 nm. Therefore, 320 * T1 = 400 * T2 => T1/T2 = 400/320 = 40/32 = 5/4. The ratio of their Kelvin temperatures is 5:4.
45. A cell of emf 'X' is connected across a resistor 'R'. The potential difference across the wire is measured as 'Y'. The internal resistance of the cell should be:
(X−Y) R/Y
(X−Y) R/X
(X−Y)R
X−Y/R
(a) Let the internal resistance of the cell be 'r'. The current flowing through the circuit is I = X / (R + r). The potential difference across the external resistor R is Y = IR = [X / (R + r)] * R => Y(R + r) = XR => YR + Yr = XR => Yr = XR - YR => r = (XR - YR) / Y = (X - Y)R / Y.
46. Ball 'A' of mass 0.1 kg moving with a velocity of 6 ms-1 collides with ball 'B' of mass 0.2 kg at rest. If ball 'A' rebounds with a velocity of 2 ms-1 in opposite direction after collision, what would be the velocity of 'B'?
2 ms-1
4 ms-1
6 ms-1
8 ms-1
(b) According to the law of conservation of momentum, total momentum before collision = total momentum after collision. (mAuA + mBuB) = (mAvA + mBvB). Given mA = 0.1 kg, uA = 6 ms-1, mB = 0.2 kg, uB = 0, vA = -2 ms-1 (rebounds in opposite direction). So, (0.1 * 6 + 0.2 * 0) = (0.1 * -2 + 0.2 * vB) => 0.6 = -0.2 + 0.2vB => 0.8 = 0.2vB => vB = 0.8 / 0.2 = 4 ms-1.
47. A man can swim in still water with a velocity of 4 ms-1. He crosses a river directly perpendicular to the river when he heads 30 ° with the normal to the river bank. The velocity of the river is:
2 ms-1
2 √3 ms-1
4 √3 ms-1
4/√3 ms-1
(a) Let vm be the velocity of the man in still water (4 ms-1), vr be the velocity of the river, and vmr be the resultant velocity of the man with respect to the ground. The man heads at an angle of 30° with the normal. For the man to cross directly perpendicular, the horizontal component of his velocity should cancel out the river velocity. So, vr = vm sin30° = 4 * (1/2) = 2 ms-1.
48. Two metal bars of thermal conductivity k1 and k2 but of same length and same cross sectional area are joined at the ends thermally as series combination. The equivalent thermal conductance is:
(k1 K2)/(k1+k2 )
k1+k2
√((k12+K22)/2)
(2k1 k2)/(k1+k2 )
(d) For a series combination of thermal conductors, the equivalent thermal resistance R is given by R = R1 + R2. Thermal resistance R = L/(kA), where L is length, k is thermal conductivity, and A is cross-sectional area. So, L/(keqA) = L/(k1A) + L/(k2A) => 1/keq = 1/k1 + 1/k2 = (k1 + k2) / (k1k2) => keq = (k1k2) / (k1 + k2). Thermal conductance is the reciprocal of thermal resistance, so conductance = kA/L. Equivalent thermal conductance = keqA/L = [(k1k2) / (k1 + k2)] * A/L. However, the question asks for equivalent thermal conductance, which might be interpreted as the effective k for the same length and area. If we consider the formula for equivalent thermal conductivity, then the answer would relate to that. Let's revisit the concept of thermal conductance. Thermal conductance C = kA/L. For series combination, the heat flow is the same, and the total temperature difference is the sum of individual temperature differences. ΔT = ΔT1 + ΔT2 => Q/Ceq = Q/C1 + Q/C2 => 1/Ceq = 1/C1 + 1/C2 => Ceq = (C1C2) / (C1 + C2) = [(k1A/L) * (k2A/L)] / [(k1A/L) + (k2A/L)] = (k1k2A2/L2) / [(A/L)(k1 + k2)] = (k1k2A) / [L(k1 + k2)]. If we consider conductance as kA/L, then the equivalent would have the same A and L. So, the equivalent k would be (k1k2) / (k1 + k2). If the question implies equivalent k such that Q = keq A (T1 - T2) / (2L) for total length 2L, then we need to adjust. Let's stick to the equivalent resistance approach. R = L/k. Req = L/k1A + L/k2A = L(k1 + k2) / (k1k2A). Conductance is inverse of resistance. Geq = 1/Req = (k1k2A) / [L(k1 + k2)]. The options seem to relate to equivalent thermal conductivity if the length was the same. In that case, keq = (2k1k2) / (k1 + k2) for the same total length. Let's assume the question meant equivalent thermal conductivity for the same total length. In series, Req = R1 + R2 => L/Akeq = L/Ak1 + L/Ak2 => 1/keq = 1/k1 + 1/k2 = (k1 + k2) / (k1k2) => keq = (k1k2) / (k1 + k2). The question might be about conductance, not conductivity. Let's consider thermal conductance as the rate of heat flow per unit temperature difference. H = KAΔT/L. For series, H is same. ΔT = ΔT1 + ΔT2. ΔT = HL/KeqA = HL/K1A + HL/K2A => 1/Keq = 1/K1 + 1/K2 => Keq = (K1K2)/(K1+K2). If conductance is KA/L, then the equivalent conductance would satisfy 1/Ceq = 1/C1 + 1/C2. Ceq = (C1C2)/(C1+C2) = (k1A/L * k2A/L) / (k1A/L + k2A/L) = (k1k2A) / (L(k1+k2)). Option (d) looks like the formula for equivalent thermal conductivity when two bars of the same length are in series. If the total length becomes 2L, then Req = L/k1A + L/k2A = L(k1+k2)/k1k2A. Then keq for length L would be k1k2/(k1+k2). If the total length is considered 2L, then keq = (2k1k2)/(k1+k2). Let's assume the question implies the equivalent conductivity for the same total length resulting in the same heat flow for the same overall temperature difference. In that case, if lengths are l, total length is 2l. Req = l/k1A + l/k2A = l(k1+k2)/k1k2A. Req = 2l/keqA. So, 2l/keqA = l(k1+k2)/k1k2A. keq = 2k1k2/(k1+k2). This matches option (d).
49. A car 🚙 is travelling with a velocity equal to one-tenth of the velocity of sound. When it is approaching towards a siren 🚨 of frequency 1000 Hz. The velocity of the car is:
17 ms-1
34 ms-1
51 ms-1
68 ms-1
(b) The velocity of sound in air is approximately 340 ms-1. The velocity of the car is one-tenth of this, so vcar = (1/10) * 340 = 34 ms-1.
50. The fringe width interference of monochromatic light produced by double slit experiment is β. The wavelength of light is λ. Then the ratio of the slit separation to the distance the slits and the screen is:
Βλ
λ/β
1/λβ
β2/λ
(b) Fringe width β = λD/d, where λ is the wavelength, D is the distance between the slits and the screen, and d is the slit separation. We need to find the ratio of slit separation (d) to the distance between the slits and the screen (D), which is d/D. From the formula, d/D = λ/β.
