NEET Full Mock Test: Practice Test 4
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- $\frac{A}{2}$
- $\frac{A}{\sqrt{2}}$
- $\frac{A}{4}$
- $\frac{\sqrt{3}A}{2}$
Answer: (B) $\frac{A}{\sqrt{2}}$
Kinetic Energy $KE = \frac{1}{2}m\omega^2(A^2 - x^2)$ and Potential Energy $PE = \frac{1}{2}m\omega^2 x^2$. Equating $KE = PE$ gives $A^2 - x^2 = x^2 \implies 2x^2 = A^2 \implies x = \frac{A}{\sqrt{2}}$.
- $\frac{12.27}{\sqrt{V}}\text{ \AA}$
- $\frac{12.27}{V}\text{ \AA}$
- $\frac{\sqrt{V}}{12.27}\text{ \AA}$
- $12.27 \times \sqrt{V}\text{ \AA}$
Answer: (A) $\frac{12.27}{\sqrt{V}}\text{ \AA}$
De Broglie wavelength for an electron is $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2meV}} = \frac{12.27}{\sqrt{V}}\text{ \AA}$.
- $1\text{ A}$
- $2\text{ A}$
- $3\text{ A}$
- $0.67\text{ A}$
Answer: (C) $3\text{ A}$
Equivalent resistance $R_{eq} = \frac{R_1 R_2}{R_1 + R_2} = \frac{6 \times 3}{6 + 3} = 2\ \Omega$. Total current $I = \frac{V}{R_{eq}} = \frac{6}{2} = 3\text{ A}$.
- $\frac{2}{5}MR^2$
- $\frac{2}{3}MR^2$
- $\frac{1}{2}MR^2$
- $\frac{7}{5}MR^2$
Answer: (A) $\frac{2}{5}MR^2$
The moment of inertia of a solid sphere about an axis passing through its center (diameter) is $I = \frac{2}{5}MR^2$.
- $30^\circ$
- $60^\circ$
- $120^\circ$
- $150^\circ$
Answer: (C) $120^\circ$
Angle of deviation for a plane mirror is given by $\delta = 180^\circ - 2i = 180^\circ - 2(30^\circ) = 120^\circ$.
- $2\text{ m/s}^2$
- $2.8\text{ m/s}^2$
- $1.4\text{ m/s}^2$
- $10\text{ m/s}^2$
Answer: (A) $2\text{ m/s}^2$
Net force $F_{net} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = 10\text{ N}$. Acceleration $a = \frac{F_{net}}{m} = \frac{10}{5} = 2\text{ m/s}^2$.
- $2E$
- $4E$
- $8E$
- $16E$
Answer: (D) $16E$
According to Stefan-Boltzmann Law, $E \propto T^4$. $T_1 = 27+273 = 300\text{ K}$, $T_2 = 327+273 = 600\text{ K}$. $\frac{E_2}{E_1} = \left(\frac{600}{300}\right)^4 = 2^4 = 16 \implies E_2 = 16E$.
- $4.83 \times 10^{14}\text{ Hz}$
- $3.0 \times 10^{15}\text{ Hz}$
- $6.63 \times 10^{14}\text{ Hz}$
- $1.2 \times 10^{15}\text{ Hz}$
Answer: (A) $4.83 \times 10^{14}\text{ Hz}$
Work function $\Phi_0 = h \nu_0 \implies \nu_0 = \frac{\Phi_0}{h} = \frac{2.0 \times 1.6 \times 10^{-19}\text{ J}}{6.63 \times 10^{-34}\text{ J s}} \approx 4.83 \times 10^{14}\text{ Hz}$.
- Oxygen
- Chlorine
- Fluorine
- Nitrogen
Answer: (C) Fluorine
Fluorine is the most electronegative element in the periodic table with a Pauling value of 4.0.
- Tetrahedral, $sp^3$
- Square planar, $sp^3d^2$
- Octahedral, $sp^3d^2$
- Trigonal bipyramidal, $sp^3d$
Answer: (B) Square planar, $sp^3d^2$
$\text{XeF}_4$ has 4 bond pairs and 2 lone pairs on Xe (steric number = 6). Its steric hybridization is $sp^3d^2$, and shape/geometry of atoms is square planar.
- Methanol
- Ethanol
- Propan-1-ol
- 2-Methylpropan-2-ol
Answer: (D) 2-Methylpropan-2-ol
Lucas reagent reacts via $S_N1$ mechanism involving carbocation intermediate. 2-Methylpropan-2-ol is a tertiary alcohol and forms carbocation instantly, giving immediate turbidity.
- $0.01\text{ min}^{-1}$
- $0.1\text{ min}^{-1}$
- $0.001\text{ min}^{-1}$
- $0.693\text{ min}^{-1}$
Answer: (A) $0.01\text{ min}^{-1}$
For a first-order reaction, $k = \frac{0.693}{t_{1/2}} = \frac{0.693}{69.3} = 0.01\text{ min}^{-1}$.
- Ethanal ($\text{CH}_3\text{CHO}$)
- Propanone ($\text{CH}_3\text{COCH}_3$)
- Benzaldehyde ($\text{C}_6\text{H}_5\text{CHO}$)
- Propanal ($\text{CH}_3\text{CH}_2\text{CHO}$)
Answer: (C) Benzaldehyde ($\text{C}_6\text{H}_5\text{CHO}$)
Aldol condensation requires at least one $\alpha$-hydrogen atom. Benzaldehyde lacks $\alpha$-hydrogens and hence undergoes Cannizzaro reaction instead.
- cis-$[\text{Co}(\text{en})_2\text{Cl}_2]^+$
- trans-$[\text{Co}(\text{en})_2\text{Cl}_2]^+$
- $[\text{Co}(\text{NH}_3)_4\text{Cl}_2]^+$
- $[\text{Ni}(\text{CN})_4]^{2-}$
Answer: (A) cis-$[\text{Co}(\text{en})_2\text{Cl}_2]^+$
cis-$[\text{Co}(\text{en})_2\text{Cl}_2]^+$ lacks a plane of symmetry and is non-superimposable on its mirror image, showing optical activity. The trans-isomer has a plane of symmetry.
- $\text{H}_2\text{CO}_3$
- $\text{CO}_3^{2-}$
- $\text{OH}^-$
- $\text{CO}_2$
Answer: (B) $\text{CO}_3^{2-}$
Conjugate base is formed when an acid donates a proton ($H^+$). $\text{HCO}_3^- - \text{H}^+ = \text{CO}_3^{2-}$.
- $(\text{CH}_3)_2\text{NH} > \text{CH}_3\text{NH}_2 > (\text{CH}_3)_3\text{N} > \text{NH}_3$
- $(\text{CH}_3)_3\text{N} > (\text{CH}_3)_2\text{NH} > \text{CH}_3\text{NH}_2 > \text{NH}_3$
- $\text{CH}_3\text{NH}_2 > (\text{CH}_3)_2\text{NH} > (\text{CH}_3)_3\text{N} > \text{NH}_3$
- $(\text{CH}_3)_2\text{NH} > (\text{CH}_3)_3\text{N} > \text{CH}_3\text{NH}_2 > \text{NH}_3$
Answer: (A) $(\text{CH}_3)_2\text{NH} > \text{CH}_3\text{NH}_2 > (\text{CH}_3)_3\text{N} > \text{NH}_3$
In aqueous medium, due to inductive, steric, and solvation factors, the basicity order of methyl-substituted amines is $2^\circ > 1^\circ > 3^\circ > \text{NH}_3$ ($2130$).
- Fertilization of egg cell twice by two sperms
- Syngamy and Triple fusion
- Fusion of sperm with synergid and antipodal cell
- Fusion of two polar nuclei with one male gamete
Answer: (B) Syngamy and Triple fusion
Double fertilization comprises syngamy (fusion of one male gamete with egg) and triple fusion (fusion of second male gamete with two polar nuclei/diploid secondary nucleus).
- Prophase I
- Metaphase I
- Anaphase I
- Metaphase II
Answer: (B) Metaphase I
During Metaphase I of meiosis, homologous chromosome pairs (bivalents) align on the equatorial plate.
- Rhizobium
- Azotobacter
- Anabaena
- Clostridium
Answer: (B) Azotobacter
Azotobacter and Beijerinckia are free-living aerobic nitrogen-fixing bacteria in soil.
- Transpiration pull
- Root pressure
- Osmotic pressure
- Suction pressure
Answer: (B) Root pressure
Positive root pressure created during early morning/night pushes water out through hydathodes (guttation).
- Cistron / Gene
- Codon
- Intron
- Exon
Answer: (A) Cistron / Gene
A gene (or cistron) is defined as a segment of DNA coding for a polypeptide or RNA molecule.
- Oxaloacetic acid (OAA)
- 3-Phosphoglyceric acid (PGA)
- Phosphoenolpyruvate (PEP)
- Glucose
Answer: (B) 3-Phosphoglyceric acid (PGA)
In Calvin cycle (C3 pathway), carboxylation of RuBP yields two molecules of 3-Phosphoglyceric acid (3-PGA).
- Mutualism
- Commensalism
- Parasitism
- Amensalism
Answer: (B) Commensalism
Commensalism is an interaction where one species benefits (orchid gets support) and the other is neither harmed nor benefited (mango tree).
- Ethylene
- Gibberellin
- Cytokinin
- Auxin
Answer: (A) Ethylene
Ethylene is a gaseous plant hormone that promotes fruit ripening and enhances respiration rate (climacteric rise).
- Estrogen
- Progesterone
- LH
- FSH
Answer: (B) Progesterone
Ruptured Graafian follicle transforms into corpus luteum, which secretes large amounts of progesterone essential for maintaining endometrium.
- Neuron
- Nephron
- Alveolus
- Glomerulus
Answer: (B) Nephron
Nephron is the structural and functional filtration unit of the kidney.
- Tracheal cartilages
- Bronchial lining
- Alveolar walls
- Diaphragm muscles
Answer: (C) Alveolar walls
Cigarette smoking leads to destruction of alveolar walls in emphysema, reducing respiratory surface area.
- Monocytes
- Lymphocytes
- Neutrophils
- Eosinophils
Answer: (B) Lymphocytes
Lymphocytes are specialized agranulocytic white blood cells responsible for adaptive (acquired) immunity.
- Melatonin
- Thymosin
- Oxytocin
- Cortisol
Answer: (A) Melatonin
Melatonin, secreted by the pineal gland, regulates sleep-wake cycles and diurnal body rhythms.
- Crop
- Gizzard (Proventriculus)
- Hepatic caeca
- Mesenteron
Answer: (B) Gizzard (Proventriculus)
The gizzard has an outer thick circular muscle layer and inner chitinous teeth used for grinding food particles.
- Tidal Volume
- Expiratory Reserve Volume
- Residual Volume
- Vital Capacity
Answer: (C) Residual Volume
Residual Volume (RV) is the volume of air left in lungs after forceful expiration (~1100 mL to 1200 mL).
- Tubectomy
- Vasectomy
- CuT insertion
- Oral contraceptive pills
Answer: (B) Vasectomy
Vasectomy is a surgical sterilization procedure in males where a small part of vas deferens is cut and tied to prevent sperm transport.