NEET Full Mock Test: Practice Test 9
Question paper with answer key. Selection Pro
- Zero
- $\frac{\mu_0 I}{\pi d}$
- $\frac{2\mu_0 I}{\pi d}$
- $\frac{\mu_0 I}{2\pi d}$
Answer: (C) $\frac{2\mu_0 I}{\pi d}$
At the midpoint, fields due to both wires point in the same direction. $B_{net} = \frac{\mu_0 I}{2\pi (d/2)} + \frac{\mu_0 I}{2\pi (d/2)} = \frac{\mu_0 I}{\pi d} + \frac{\mu_0 I}{\pi d} = \frac{2\mu_0 I}{\pi d}$.
- $8.0 \times 10^{14}\text{ Hz}$
- $4.0 \times 10^{14}\text{ Hz}$
- $1.2 \times 10^{15}\text{ Hz}$
- $2.5 \times 10^{14}\text{ Hz}$
Answer: (A) $8.0 \times 10^{14}\text{ Hz}$
$\Phi = h \nu_0 \implies \nu_0 = \frac{3.3 \times 1.6 \times 10^{-19}\text{ J}}{6.6 \times 10^{-34}\text{ J\cdot s}} = 8.0 \times 10^{14}\text{ Hz}$.
- $8$
- $16$
- $32$
- $64$
Answer: (C) $32$
For an adiabatic process, $P_1 V_1^{\gamma} = P_2 V_2^{\gamma} \implies \frac{P_1}{P_2} = \left(\frac{V_2}{V_1}\right)^{\gamma} = (8)^{5/3} = (2^3)^{5/3} = 2^5 = 32$.
- $1\text{ cm}$
- $1\text{ mm}$
- $10^{-6}\text{ m}$
- $10^{-10}\text{ m}$
Answer: (C) $10^{-6}\text{ m}$
The width of the depletion region in a semiconductor $p-n$ junction is typically on the order of $10^{-6}\text{ m}$ ($1\ \mu\text{m}$).
- $\frac{m v^2}{r} \cdot 2\pi r$
- Zero
- $\frac{1}{2} m v^2$
- $m v^2$
Answer: (B) Zero
The centripetal force acts perpendicular to the direction of motion at every instant ($ \theta = 90^\circ$). Hence, work done $W = F d \cos(90^\circ) = 0$.
- $\sqrt{\mu_0 \varepsilon_0}$
- $\frac{1}{\sqrt{\mu_0 \varepsilon_0}}$
- $\frac{\mu_0}{\varepsilon_0}$
- $\sqrt{\frac{\mu_0}{\varepsilon_0}}$
Answer: (B) $\frac{1}{\sqrt{\mu_0 \varepsilon_0}}$
Maxwell proved that the velocity of electromagnetic radiation in free space is given by $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$.
- Increase by $2\%$
- Decrease by $2\%$
- Increase by $4\%$
- Decrease by $4\%$
Answer: (B) Decrease by $2\%$
$T = 2\pi \sqrt{\frac{L}{g}} \implies \frac{\Delta T}{T} = -\frac{1}{2}\frac{\Delta g}{g} = -\frac{1}{2}(4\%) = -2\%$. The time period decreases by $2\%$.
- $20$
- $500$
- $105$
- $95$
Answer: (A) $20$
Magnifying power in normal adjustment is $m = \frac{f_o}{f_e} = \frac{100}{5} = 20$.
- $\text{LiOH}$
- $\text{NaOH}$
- $\text{KOH}$
- $\text{CsOH}$
Answer: (D) $\text{CsOH}$
Down group 1, ionic radius increases, reducing $M-OH$ bond energy. Thus, $\text{CsOH}$ dissociates most readily, making it the most basic.
- Trigonal bipyramidal
- Square pyramidal
- Octahedral
- Tetrahedral
Answer: (A) Trigonal bipyramidal
$\text{PCl}_5$ has 5 bond pairs and $sp^3d$ hybridization, yielding a trigonal bipyramidal shape.
- Primary amine
- Secondary amine
- Tertiary amine
- Quaternary ammonium salt
Answer: (B) Secondary amine
Secondary amines react with Hinsberg's reagent (benzenesulfonyl chloride) to form $N,N$-dialkylbenzenesulfonamides, which lack an acidic hydrogen and are insoluble in aqueous $\text{KOH}$.
- $0$
- $+2$
- $+4$
- $-2$
Answer: (A) $0$
Carbon monoxide ($\text{CO}$) is a neutral ligand. $x + 4(0) = 0 \implies x = 0$.
- $\Delta H < 0$ and $\Delta S > 0$
- $\Delta H > 0$ and $\Delta S < 0$
- $\Delta H < 0$ and $\Delta S < 0$
- $\Delta H > 0$ and $\Delta S > 0$
Answer: (A) $\Delta H < 0$ and $\Delta S > 0$
Since $\Delta G = \Delta H - T\Delta S$, when $\Delta H$ is negative and $\Delta S$ is positive, $\Delta G$ is always negative at any temperature.
- Fluorine
- Chlorine
- Bromine
- Iodine
Answer: (A) Fluorine
Fluorine is highly electronegative and lacks $d$-orbitals, forming only one oxoacid, fluoric(I) acid ($\text{HOF}$).
- $A > B > C$
- $C > B > A$
- $B > C > A$
- $A > C > B$
Answer: (B) $C > B > A$
More negative reduction potential corresponds to a stronger reducing agent. Hence order is $C > B > A$.
- Propanone and Ethanal
- Ethanal and Methanal
- Propanone and Methanal
- Butanone and Ethanal
Answer: (A) Propanone and Ethanal
Cleaving the double bond in $(\text{CH}_3)_2\text{C}=\text{CH}-\text{CH}_3$ via ozonolysis gives propanone ($(\text{CH}_3)_2\text{C}=\text{O}$) and ethanal ($\text{CH}_3\text{CHO}$).
- Chloroplast, Peroxisome, Mitochondrion
- Chloroplast, Mitochondrion, Peroxisome
- Peroxisome, Chloroplast, Mitochondrion
- Mitochondrion, Peroxisome, Chloroplast
Answer: (A) Chloroplast, Peroxisome, Mitochondrion
Photorespiration occurs in three cell organelles in sequence: Chloroplast $\rightarrow$ Peroxisome $\rightarrow$ Mitochondrion.
- One codon codes for more than one amino acid
- More than one codon can code for a single amino acid
- Codons overlap during translation
- It varies between prokaryotes and eukaryotes
Answer: (B) More than one codon can code for a single amino acid
Degeneracy means that most amino acids are specified by more than one codon (e.g., 6 different codons code for Leucine).
- Xylem vessels
- Sieve tube elements
- Sclereids
- Tracheids
Answer: (B) Sieve tube elements
Sieve tubes along with companion cells in phloem conduct organic food from leaves to sink tissues.
- Rhizobium
- Azotobacter
- Frankia
- Anabaena azollae
Answer: (B) Azotobacter
*Azotobacter* and *Beijerinckia* are free-living aerobic nitrogen-fixing bacteria in soil.
- Fabaceae
- Solanaceae
- Liliaceae
- Brassicaceae
Answer: (B) Solanaceae
Solanaceae exhibits actinomorphic symmetry, fused calyx (5), fused corolla (5), 5 epipetalous stamens, and bicarpellary syncarpous superior ovary.
- Are linear double-stranded RNA
- Replicate independently of genomic DNA
- Integrate into host chromosome permanently
- Lack origin of replication
Answer: (B) Replicate independently of genomic DNA
Plasmids are extra-chromosomal, autonomous, self-replicating circular double-stranded DNA molecules.
- Leptotene
- Zygotene
- Pachytene
- Diplotene
Answer: (B) Zygotene
During zygotene stage of Prophase I, homologous chromosomes pair up (synapsis) mediated by synaptonemal complex.
- Chlorophyll a
- Chlorophyll b
- Carotenoids
- Xanthophylls
Answer: (A) Chlorophyll a
Chlorophyll a acts as the chief reaction-center pigment in light reaction of photosynthesis.
- Cerebellum
- Thalamus
- Hypothalamus
- Medulla oblongata
Answer: (C) Hypothalamus
The hypothalamus contains centers that control body temperature, urge for eating (hunger), and drinking (thirst).
- Primary follicle
- Graafian follicle
- Secondary oocyte
- Corpus albicans
Answer: (B) Graafian follicle
Following ovulation, the ruptured Graafian follicle transforms under LH influence into the endocrine structure called corpus luteum.
- IgA
- IgM
- IgG
- IgE
Answer: (C) IgG
IgG is the most abundant immunoglobulin (~80%) and the only class capable of crossing the placental barrier.
- $125\text{ mL/min}$ ($180\text{ L/day}$)
- $500\text{ mL/min}$
- $50\text{ mL/min}$
- $12.5\text{ L/day}$
Answer: (A) $125\text{ mL/min}$ ($180\text{ L/day}$)
Normal GFR in a healthy individual is around $125\text{ mL/minute}$, amounting to $180\text{ liters}$ per day.
- Atrioventricular Node (AVN)
- Sinoatrial Node (SAN)
- Bundle of His
- Purkinje fibers
Answer: (B) Sinoatrial Node (SAN)
SAN generates maximum action potentials ($70-75/\text{min}$) and sets the rhythm of heart contractions, acting as the pacemaker.
- Oxytocin
- Anti-Diuretic Hormone (ADH / Vasopressin)
- Aldosterone
- Atrial Natriuretic Factor (ANF)
Answer: (B) Anti-Diuretic Hormone (ADH / Vasopressin)
ADH facilitates water reabsorption from distal parts of the tubule and collecting duct to prevent diuresis.
- Sutures in skull
- Joint between adjacent vertebrae
- Shoulder joint (ball and socket)
- Knee joint (hinge)
Answer: (B) Joint between adjacent vertebrae
Joints between adjacent vertebrae in the vertebral column are cartilaginous joints permitting limited movement.
- $\text{CO}_2 + \text{H}_2\text{O} \rightleftharpoons \text{H}_2\text{CO}_3$
- $\text{O}_2 + \text{Hb} \rightleftharpoons \text{HbO}_2$
- Glucose $\rightarrow$ Pyruvate
- Pepsinogen $\rightarrow$ Pepsin
Answer: (A) $\text{CO}_2 + \text{H}_2\text{O} \rightleftharpoons \text{H}_2\text{CO}_3$
Carbonic anhydrase speeds up the hydration of carbon dioxide into carbonic acid ($\text{H}_2\text{CO}_3$) for transport as bicarbonate ions.