pragativeer-v2/neet_sample_questions.md

NEET Sample Questions

Here are some realistic sample questions for a NEET exam.

Q1. The dimension of \frac{1}{\sqrt{\mu_0 \varepsilon_0}} is that of:

Subject: Physics | Topic: Units and Measurements | Difficulty: Medium

• A: Velocity
• B: Time
• C: Capacitance
• D: Distance

Correct Answer: A

Solution: We know that c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}, where c is the speed of light in vacuum. Therefore, the dimension of \frac{1}{\sqrt{\mu_0 \varepsilon_0}} is that of velocity, which is [L T^{-1}].

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Q2. Two projectiles are thrown with the same initial velocity at angles of 30^\circ and 60^\circ with the horizontal. The ratio of their maximum heights is:

Subject: Physics | Topic: Kinematics | Difficulty: Easy

• A: 1:3
• B: 3:1
• C: 1:\sqrt{3}
• D: 1:2

Correct Answer: A

Solution: Maximum height H = \frac{u^2 \sin^2 \theta}{2g}. For \theta_1 = 30^\circ, H_1 = \frac{u^2 (1/2)^2}{2g} = \frac{u^2}{8g}. For \theta_2 = 60^\circ, H_2 = \frac{u^2 (\sqrt{3}/2)^2}{2g} = \frac{3u^2}{8g}. Ratio H_1 : H_2 = 1 : 3.

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Q3. A Carnot engine takes heat from a reservoir at 627^\circ\text{C} and rejects heat to a sink at 27^\circ\text{C}. Its efficiency is:

Subject: Physics | Topic: Thermodynamics | Difficulty: Medium

• A: 2/3
• B: 1/3
• C: 3/4
• D: 1/2

Correct Answer: A

Solution: Temperature of source, T_1 = 627 + 273 = 900\text{ K}. Temperature of sink, T_2 = 27 + 273 = 300\text{ K}. Efficiency \eta = 1 - \frac{T_2}{T_1} = 1 - \frac{300}{900} = 1 - \frac{1}{3} = \frac{2}{3}.

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Q4. A parallel plate capacitor with air between the plates has a capacitance of 8\text{ pF}. What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?

Subject: Physics | Topic: Electrostatics | Difficulty: Hard

• A: 96 pF
• B: 8 pF
• C: 48 pF
• D: 12 pF

Correct Answer: A

Solution: Initial capacitance C_0 = \frac{\varepsilon_0 A}{d} = 8\text{ pF}. New distance d' = d/2 and new dielectric constant K = 6. New capacitance C' = \frac{K \varepsilon_0 A}{d'} = \frac{6 \varepsilon_0 A}{d/2} = 12 \left(\frac{\varepsilon_0 A}{d}\right) = 12 \times 8\text{ pF} = 96\text{ pF}.

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Q5. In a meter bridge experiment, the null point is obtained at 20\text{ cm} from one end of the wire when resistance X is balanced against another resistance Y. If X < Y, then what will be the new position of the null point from the same end, if one decides to balance a resistance of 4X against Y?

Subject: Physics | Topic: Current Electricity | Difficulty: Medium

• A: 50 cm
• B: 40 cm
• C: 80 cm
• D: 20 cm

Correct Answer: A

Solution: Initially, \frac{X}{Y} = \frac{20}{100-20} = \frac{20}{80} = \frac{1}{4}. So Y = 4X. For new resistance 4X against Y, ratio is \frac{4X}{Y} = \frac{4X}{4X} = 1. Let new null point be l, then \frac{l}{100-l} = 1 \implies l = 100 - l \implies 2l = 100 \implies l = 50\text{ cm}.

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Q6. The magnetic field at the centre of a circular current carrying coil of radius R is B_0. If the radius is doubled keeping the current same, the new magnetic field at the centre will be:

Subject: Physics | Topic: Magnetism | Difficulty: Easy

• A: B_0/2
• B: B_0/4
• C: 2B_0
• D: B_0

Correct Answer: A

Solution: Magnetic field at the centre of a circular coil is B = \frac{\mu_0 I}{2R}. Since B \propto \frac{1}{R}, doubling the radius will make the magnetic field half. Thus, new field is B_0/2.

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Q7. In a series LCR circuit, the voltage across the resistance, capacitance, and inductance is 10\text{ V} each. If the capacitance is short-circuited, the voltage across the inductance will be:

Subject: Physics | Topic: Alternating Current | Difficulty: Medium

• A: 10 V
• B: 10\sqrt{2}\text{ V}
• C: 10/\sqrt{2}\text{ V}
• D: Zero

Correct Answer: C

Solution: Initially, V_R = 10\text{ V}, V_L = 10\text{ V}, V_C = 10\text{ V}. Since V_L = V_C, the circuit is in resonance. The applied voltage V = \sqrt{V_R^2 + (V_L - V_C)^2} = V_R = 10\text{ V}. If capacitance is shorted, it becomes an LR circuit. Impedance Z = \sqrt{R^2 + X_L^2}. Since V_R = V_L initially, R = X_L, so Z = R\sqrt{2}. New current I' = \frac{V}{Z} = \frac{10}{R\sqrt{2}}. New voltage across inductance V_L' = I' X_L = \frac{10}{R\sqrt{2}} \times R = \frac{10}{\sqrt{2}}\text{ V}.

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Q8. In Young's double slit experiment, if the separation between coherent sources is halved and the distance of the screen from the coherent sources is doubled, then the fringe width becomes:

Subject: Physics | Topic: Optics | Difficulty: Medium

• A: Half
• B: Four times
• C: One-fourth
• D: Double

Correct Answer: B

Solution: Fringe width \beta = \frac{\lambda D}{d}. New slit separation d' = d/2. New screen distance D' = 2D. New fringe width \beta' = \frac{\lambda (2D)}{(d/2)} = 4 \frac{\lambda D}{d} = 4\beta.

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Q9. When the light of frequency 2\nu_0 (where \nu_0 is threshold frequency) is incident on a metal plate, the maximum velocity of electrons emitted is v_1. When the frequency of the incident radiation is increased to 5\nu_0, the maximum velocity of electrons emitted from the same plate is v_2. The ratio of v_1 to v_2 is:

Subject: Physics | Topic: Modern Physics | Difficulty: Hard

• A: 1:2
• B: 1:4
• C: 4:1
• D: 2:1

Correct Answer: A

Solution: From Einstein's photoelectric equation: K_{\max} = h\nu - h\nu_0 = \frac{1}{2} m v^2. Case 1: \frac{1}{2} m v_1^2 = h(2\nu_0) - h\nu_0 = h\nu_0. Case 2: \frac{1}{2} m v_2^2 = h(5\nu_0) - h\nu_0 = 4h\nu_0. Dividing the two equations: \frac{v_1^2}{v_2^2} = \frac{1}{4} \implies \frac{v_1}{v_2} = \frac{1}{2}.

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Q10. Which logic gate is represented by the boolean expression Y = \overline{A \cdot B}?

Subject: Physics | Topic: Semiconductors | Difficulty: Easy

• A: NOR
• B: OR
• C: NAND
• D: AND

Correct Answer: C

Solution: The boolean expression Y = \overline{A \cdot B} represents the NAND gate. It gives an output of 0 only when both inputs A and B are 1, and 1 otherwise.

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Q11. The number of water molecules is maximum in:

Subject: Chemistry | Topic: Physical Chemistry | Difficulty: Easy

• A: 18 grams of water
• B: 18 moles of water
• C: 18 molecules of water
• D: 1.8 grams of water

Correct Answer: B

Solution: 18 grams of water = 1 mole = N_A molecules. 18 moles of water = 18 N_A molecules. 1.8 grams of water = 0.1 mole = 0.1 N_A molecules. Thus, 18 moles of water contains the maximum number of molecules.

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Q12. The maximum number of electrons in an atom that can have the quantum numbers n = 4 and m_l = +1 is:

Subject: Chemistry | Topic: Physical Chemistry | Difficulty: Medium

• A: 2
• B: 4
• C: 6
• D: 8

Correct Answer: C

Solution: For n = 4, the possible values of l are 0, 1, 2, 3. The condition m_l = +1 is possible for l = 1, l = 2, and l = 3. Each subshell (l=1, 2, 3) will have one orbital with m_l = +1. Since each orbital can hold 2 electrons, the total number of electrons is 3 \times 2 = 6.

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Q13. The species having pyramidal shape is:

Subject: Chemistry | Topic: Inorganic Chemistry | Difficulty: Medium

• A: \text{SO}_3
• B: \text{BrF}_3
• C: \text{SiO}_3^{2-}
• D: \text{OSF}_2

Correct Answer: D

Solution: In \text{OSF}_2, the central sulfur atom has 4 electron domains (1 double bond to O, 2 single bonds to F, and 1 lone pair). This corresponds to an sp^3 hybridization with a trigonal pyramidal shape. \text{SO}_3 is trigonal planar, \text{BrF}_3 is T-shaped.

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Q14. For the reversible reaction \text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightleftharpoons 2\text{NH}_3(\text{g}) + \text{Heat}, the equilibrium shifts in forward direction:

Subject: Chemistry | Topic: Physical Chemistry | Difficulty: Medium

• A: By increasing the concentration of \text{NH}_3(\text{g})
• B: By decreasing the pressure
• C: By decreasing the concentrations of \text{N}_2(\text{g}) and \text{H}_2(\text{g})
• D: By increasing pressure and decreasing temperature

Correct Answer: D

Solution: According to Le Chatelier's principle, an exothermic reaction (\Delta H < 0) is favored by low temperatures. A reaction involving a decrease in the number of gaseous moles (4 \rightarrow 2) is favored by high pressures. Therefore, increasing pressure and decreasing temperature will shift the equilibrium in the forward direction.

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Q15. Given E^\circ_{\text{Cr}^{3+}/\text{Cr}} = -0.74\text{ V}; E^\circ_{\text{MnO}_4^-/\text{Mn}^{2+}} = 1.51\text{ V}; E^\circ_{\text{Cr}_2\text{O}_7^{2-}/\text{Cr}^{3+}} = 1.33\text{ V}; E^\circ_{\text{Cl}/\text{Cl}^-} = 1.36\text{ V}.

Based on the data given above, strongest oxidizing agent will be: Subject: Chemistry | Topic: Physical Chemistry | Difficulty: Hard

• A: \text{Cl}
• B: \text{Cr}^{3+}
• C: \text{Mn}^{2+}
• D: \text{MnO}_4^-

Correct Answer: D

Solution: Higher the standard reduction potential (E^\circ), stronger is the oxidizing power. The E^\circ value for \text{MnO}_4^-/\text{Mn}^{2+} is the highest (+1.51\text{ V}). Therefore, \text{MnO}_4^- is the strongest oxidizing agent among the given species.

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Q16. Which of the following carbocations is most stable?

Subject: Chemistry | Topic: Organic Chemistry | Difficulty: Medium

• A: Allyl carbocation
• B: Primary alkyl carbocation
• C: Tert-butyl carbocation
• D: Triphenylmethyl carbocation

Correct Answer: D

Solution: Stability of carbocations is determined by delocalization of charge. The triphenylmethyl carbocation, (\text{C}_6\text{H}_5)_3\text{C}^+, has 3 phenyl rings providing extensive resonance stabilization, making it exceptionally stable compared to the others.

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Q17. Ozonolysis of an organic compound gives formaldehyde as one of the products. This confirms the presence of:

Subject: Chemistry | Topic: Organic Chemistry | Difficulty: Hard

• A: Two isomeric double bonds
• B: A vinyl group
• C: An isopropyl group
• D: An acetylenic triple bond

Correct Answer: B

Solution: Ozonolysis cleaves a carbon-carbon double bond and adds oxygen. Formaldehyde (\text{HCHO}) is produced if the original double bond was at the end of a chain, specifically involving a terminal =\text{CH}_2 group, which is characteristic of a vinyl group (-\text{CH}=\text{CH}_2).

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Q18. The IUPAC name of [\text{Pt}(\text{NH}_3)_2\text{Cl}(\text{NO}_2)] is:

Subject: Chemistry | Topic: Inorganic Chemistry | Difficulty: Medium

• A: Platinum diaminechloronitrite
• B: Chloronitrito-N-ammineplatinum(II)
• C: Diamminechloridonitrito-N-platinum(II)
• D: Diamminechloronitrito-O-platinate(II)

Correct Answer: C

Solution: Ligands are named in alphabetical order: ammine, chlorido, nitrito-N (since it coordinates through nitrogen). The oxidation state of Pt is x + 0 + (-1) + (-1) = 0 \implies x = +2. Thus, the correct name is Diamminechloridonitrito-N-platinum(II).

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Q19. The unit of rate constant for a zero order reaction is:

Subject: Chemistry | Topic: Physical Chemistry | Difficulty: Easy

• A: \text{s}^{-1}
• B: \text{mol}\cdot\text{L}^{-1}\cdot\text{s}^{-1}
• C: \text{L}\cdot\text{mol}^{-1}\cdot\text{s}^{-1}
• D: \text{L}^2\cdot\text{mol}^{-2}\cdot\text{s}^{-1}

Correct Answer: B

Solution: For a zero-order reaction, Rate = k[A]^0 = k. The unit of rate is concentration per unit time (\text{mol}\cdot\text{L}^{-1}\cdot\text{s}^{-1}). Thus, the unit of the rate constant k is also \text{mol}\cdot\text{L}^{-1}\cdot\text{s}^{-1}.

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Q20. Which of the following hormones is produced under the condition of stress which stimulates glycogenolysis in the liver of human beings?

Subject: Chemistry | Topic: Organic Chemistry | Difficulty: Medium

• A: Thyroxin
• B: Insulin
• C: Adrenaline
• D: Estradiol

Correct Answer: C

Solution: Adrenaline (epinephrine) is the 'fight or flight' hormone released during stress. It promotes the breakdown of glycogen into glucose (glycogenolysis) in the liver to increase blood glucose levels for energy.

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Q21. Crossing over occurs during which stage of meiosis?

Subject: Biology | Topic: Cell Structure and Function | Difficulty: Medium

• A: Leptotene
• B: Pachytene
• C: Diplotene
• D: Zygotene

Correct Answer: B

Solution: Crossing over, the exchange of genetic material between non-sister chromatids of homologous chromosomes, occurs during the pachytene stage of prophase I of meiosis.

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Q22. A cross between two tall plants resulted in offspring having few dwarf plants. What would be the genotypes of both the parents?

Subject: Biology | Topic: Genetics and Evolution | Difficulty: Easy

• A: TT and Tt
• B: Tt and Tt
• C: TT and TT
• D: Tt and tt

Correct Answer: B

Solution: Since both parents are tall but they produce some dwarf offspring, both parents must carry the recessive allele for dwarfness. Thus, their genotype must be heterozygous (Tt). A cross of Tt x Tt yields a 3:1 ratio of tall to dwarf.

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Q23. Which of the following RNAs should be most abundant in an animal cell?

Subject: Biology | Topic: Genetics and Evolution | Difficulty: Medium

• A: tRNA
• B: mRNA
• C: miRNA
• D: rRNA

Correct Answer: D

Solution: Ribosomal RNA (rRNA) constitutes about 80% of the total RNA in an actively growing cell. It forms the structural and catalytic core of ribosomes.

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Q24. Which of the following terms describes human dentition?

Subject: Biology | Topic: Human Physiology | Difficulty: Hard

• A: Thecodont, Diphyodont, Homodont
• B: Thecodont, Diphyodont, Heterodont
• C: Pleurodont, Monophyodont, Homodont
• D: Pleurodont, Diphyodont, Heterodont

Correct Answer: B

Solution: Human dentition is Thecodont (teeth embedded in jaw sockets), Diphyodont (two sets of teeth: deciduous and permanent), and Heterodont (different types of teeth: incisors, canines, premolars, molars).

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Q25. Which hormone is responsible for the ovulation in human females?

Subject: Biology | Topic: Reproduction | Difficulty: Medium

• A: FSH (Follicle Stimulating Hormone)
• B: Progesterone
• C: LH (Luteinizing Hormone)
• D: Estrogen

Correct Answer: C

Solution: A surge in LH (Luteinizing Hormone) mid-cycle triggers the rupture of the Graafian follicle and the subsequent release of the ovum, a process known as ovulation.

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Q26. The first stable product of CO2 fixation in C3 plants is:

Subject: Biology | Topic: Plant Physiology | Difficulty: Medium

• A: Phosphoglyceraldehyde (PGAL)
• B: Malic acid
• C: Oxaloacetic acid (OAA)
• D: 3-Phosphoglyceric acid (3-PGA)

Correct Answer: D

Solution: In the Calvin cycle of C3 plants, CO2 combines with RuBP to form a 6-carbon unstable intermediate that immediately splits into two molecules of 3-phosphoglyceric acid (3-PGA), the first stable product.

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Q27. The pyramid of energy is always:

Subject: Biology | Topic: Ecology | Difficulty: Easy

• A: Upright
• B: Inverted
• C: Spindle shaped
• D: Variable

Correct Answer: A

Solution: The pyramid of energy is always upright because energy is lost as heat at each trophic level (10% law), so energy always decreases as it flows from producers to higher trophic levels.

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Q28. The cutting of DNA at specific locations became possible with the discovery of:

Subject: Biology | Topic: Biotechnology | Difficulty: Medium

• A: Selectable markers
• B: Ligases
• C: Restriction enzymes
• D: Probes

Correct Answer: C

Solution: Restriction endonucleases (or restriction enzymes) cut DNA at specific recognition sequences, making recombinant DNA technology possible.

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Q29. Receptor sites for neurotransmitters are present on:

Subject: Biology | Topic: Human Physiology | Difficulty: Hard

• A: Pre-synaptic membrane
• B: Tips of axons
• C: Post-synaptic membrane
• D: Membrane of synaptic vesicles

Correct Answer: C

Solution: Neurotransmitters are released from synaptic vesicles in the pre-synaptic neuron and diffuse across the synaptic cleft to bind with specific receptor sites located on the post-synaptic membrane.

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Q30. Which of the following are found in extreme saline conditions?

Subject: Biology | Topic: Diversity in Living World | Difficulty: Medium

• A: Eubacteria
• B: Cyanobacteria
• C: Mycobacteria
• D: Archaebacteria

Correct Answer: D

Solution: Archaebacteria are known to live in some of the most harsh habitats. Specifically, halophiles live in extreme saline conditions (e.g., Dead Sea).

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