NEET Full Mock Test: Practice Test 2
Question paper with answer key. Selection Pro
- $-12\text{ m/s}$
- $0\text{ m/s}$
- $6\text{ m/s}$
- $-6\text{ m/s}$
Answer: (A) $-12\text{ m/s}$
Velocity $v = \frac{dx}{dt} = 6t - 12$ and acceleration $a = \frac{dv}{dt} = 6\text{ m/s}^2$ (constant). Since acceleration is constant ($6\text{ m/s}^2$), velocity when $t=0$ or at any reference is calculated using expression $v(t)$. At $t=0$, $v = -12\text{ m/s}$.
- $nRT \ln\left(\frac{V_2}{V_1}\right)$
- $nRT \left(\frac{V_2}{V_1}\right)$
- $nRT \ln\left(\frac{V_1}{V_2}\right)$
- Zero
Answer: (A) $nRT \ln\left(\frac{V_2}{V_1}\right)$
For an isothermal process ($T = \text{constant}$), work done $W = \int P dV = nRT \ln\left(\frac{V_2}{V_1}\right)$.
- Equal
- Double
- Half
- Four times
Answer: (B) Double
Maximum torque $\tau_{\max} = pE$. For dipole moment $2p$, $\tau_{\max} = 2pE$, which is double that of the first.
- Number of turns $N$
- Square of number of turns $N^2$
- $1/N$
- $1/N^2$
Answer: (B) Square of number of turns $N^2$
Self-inductance of solenoid $L = \frac{\mu_0 N^2 A}{l}$, so $L \propto N^2$.
- Unchanged
- Doubled
- Four times
- Halved
Answer: (C) Four times
Fringe width $\beta = \frac{\lambda D}{d}$. New fringe width $\beta' = \frac{\lambda (2D)}{(d/2)} = 4\beta$.
- $1:2$
- $2:1$
- $1:4$
- $1:\sqrt{2}$
Answer: (A) $1:2$
At height $h=R$, $g' = g/4$. Time period $T \propto 1/\sqrt{g}$. Thus $T_2 = 2 T_1$, so $T_1/T_2 = 1/2$.
- $4v$
- $v/4$
- $v/2$
- $2v$
Answer: (B) $v/4$
Speed of sound $v = \sqrt{\frac{\gamma RT}{M}}$. Molecular mass of $O_2$ is $16$ times that of $H_2$. Therefore, $v_{O_2} = \frac{v_{H_2}}{\sqrt{16}} = \frac{v}{4}$.
- Forward biased
- Reverse biased
- Unbiased
- Lightly doped
Answer: (A) Forward biased
In forward bias, depletion layer width decreases and resistance becomes low, so it acts like a closed switch.
- $\text{O}_2$
- $\text{B}_2$
- $\text{N}_2$
- $\text{NO}$
Answer: (C) $\text{N}_2$
$\text{N}_2$ has $14$ electrons with all electrons paired in molecular orbitals, making it diamagnetic.
- $1.0 \times 10^{-5}\text{ M}$
- $1.0 \times 10^{-10}\text{ M}$
- $2.0 \times 10^{-5}\text{ M}$
- $1.0 \times 10^{-2.5}\text{ M}$
Answer: (A) $1.0 \times 10^{-5}\text{ M}$
For a 1:1 salt like $\text{AgCl}$, $K_{sp} = S^2 \implies S = \sqrt{K_{sp}} = \sqrt{10^{-10}} = 10^{-5}\text{ M}$.
- $\text{NH}_3$
- $\text{N}_2\text{H}_4$
- $\text{HNO}_3$
- $\text{NO}_2$
Answer: (C) $\text{HNO}_3$
In $\text{HNO}_3$, oxidation state of N is $+5$, which is its maximum oxidation state.
- $\text{Br}_2/\text{CCl}_4$
- Ammoniacal $\text{AgNO}_3$ solution
- Bayer's reagent
- Cold dilute $\text{KMnO}_4$
Answer: (B) Ammoniacal $\text{AgNO}_3$ solution
Ethyne (terminal alkyne) contains acidic hydrogens and reacts with ammoniacal $\text{AgNO}_3$ (Tollens' reagent) to give a white precipitate, while ethane does not.
- $4.0$
- $3.3$
- $3.5$
- $4.5$
Answer: (B) $3.3$
$[\text{H}^+]_1 = 10^{-3}$, $[\text{H}^+]_2 = 10^{-5}$. Total $[\text{H}^+] = \frac{10^{-3} + 10^{-5}}{2} \approx 5 \times 10^{-4}\text{ M}$. $\text{pH} = -\log(5 \times 10^{-4}) \approx 3.3$.
- Gluconic acid
- Saccharic acid
- Sorbitol
- n-Hexane
Answer: (B) Saccharic acid
Strong oxidizing agent conc. $\text{HNO}_3$ oxidizes both the aldehyde group and primary alcohol group of glucose to dicarboxylic acid (Saccharic acid).
- $\text{Sc}^{3+}$
- $\text{Ti}^{4+}$
- $\text{Cu}^+$
- $\text{Fe}^{2+}$
Answer: (D) $\text{Fe}^{2+}$
$\text{Fe}^{2+}$ has $3d^6$ configuration with unpaired d-electrons permitting d-d transition, making its solution colored. $\text{Sc}^{3+}$ ($3d^0$), $\text{Ti}^{4+}$ ($3d^0$), and $\text{Cu}^+$ ($3d^{10}$) are colorless.
- Benzoic acid
- Salicylaldehyde
- Salicylic acid
- Benzaldehyde
Answer: (B) Salicylaldehyde
Phenol reacts with $\text{CHCl}_3$ and aqueous $\text{NaOH}$ (Reimer-Tiemann reaction) to yield Salicylaldehyde.
- Shoot apex
- Stem
- Leaves
- Roots
Answer: (C) Leaves
Leaves perceive the light stimulus for photoperiodism, which induces the synthesis of florigen (flowering signal).
- Ribosome
- Lysosome
- Golgi apparatus
- Mitochondria
Answer: (B) Lysosome
Lysosomes contain hydrolytic enzymes and can digest damaged cellular components, earning them the name suicide bags.
- Primary consumers
- Producers
- Secondary consumers
- Decomposers
Answer: (B) Producers
According to Lindeman's 10% law, producers capture and store maximum energy, which decreases at subsequent trophic levels.
- Conjoint, collateral, and open
- Conjoint, collateral, and closed
- Radial and open
- Concentric and closed
Answer: (A) Conjoint, collateral, and open
Dicot stem vascular bundles are conjoint, collateral, and open due to presence of cambium between xylem and phloem.
- National Parks
- Botanical Gardens
- Biosphere Reserves
- Sacred Groves
Answer: (B) Botanical Gardens
Botanical gardens and seed banks are ex-situ conservation approaches, whereas National Parks and Biosphere Reserves are in-situ.
- $\text{NADPH}$
- Water ($\text{H}_2\text{O}$)
- Plastocyanin
- Ferredoxin
Answer: (B) Water ($\text{H}_2\text{O}$)
Photolysis of water provides replacement electrons to the reaction center $P_{680}$ of PS II.
- UAA
- UAG
- AUG
- UGA
Answer: (C) AUG
AUG codes for Methionine and serves as the universal start codon for protein translation.
- Cytokinin
- Abscisic Acid (ABA)
- Gibberellic Acid
- Auxin
Answer: (B) Abscisic Acid (ABA)
Abscisic acid (ABA) acts as a stress hormone and triggers rapid closure of stomata to prevent water loss.
- Cerebellum
- Hypothalamus
- Medulla oblongata
- Thalamus
Answer: (B) Hypothalamus
The hypothalamus contains thermoregulatory, satiety, and thirst centers in the brain.
- FSH
- LH / ICSH
- ACTH
- Prolactin
Answer: (A) FSH
FSH acts on Sertoli cells to stimulate factors needed for spermatogenesis.
- Squamous epithelium
- Ciliated epithelium
- Cuboidal epithelium
- Columnar non-ciliated
Answer: (B) Ciliated epithelium
Ciliated columnar epithelium lines fallopian tubes and bronchioles to move particles/cells in a specific direction.
- Thrombokinase
- Thrombin
- Prothrombin
- Renin
Answer: (B) Thrombin
Thrombin converts soluble fibrinogen into insoluble fibrin threads during blood coagulation.
- Bone marrow
- Thymus
- Spleen
- Lymph nodes
Answer: (B) Thymus
T-lymphocytes originate in bone marrow but migrate to the Thymus for maturation.
- AV node
- SA node
- Purkinje fibers
- Bundle of His
Answer: (B) SA node
Sino-atrial node (SA node) acts as natural pacemaker of heart by generating action potentials spontaneously.
- CuT
- Lippes loop
- Progestasert / LNG-20
- Multiload 375
Answer: (C) Progestasert / LNG-20
Progestasert and LNG-20 are hormone-releasing intrauterine devices (IUDs).
- Inspiratory Reserve Volume
- Tidal Volume
- Residual Volume
- Vital Capacity
Answer: (B) Tidal Volume
Tidal Volume (TV) is volume of air inspired or expired during normal respiration (~500 mL).