# 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 $[L T^{-1}]$. --- ### 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 $H_1 : H_2 = 1 : 3$. --- ### 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 $\eta = 1 - \frac{T_2}{T_1} = 1 - \frac{300}{900} = 1 - \frac{1}{3} = \frac{2}{3}$. --- ### 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 $C' = \frac{K \varepsilon_0 A}{d'} = \frac{6 \varepsilon_0 A}{d/2} = 12 \left(\frac{\varepsilon_0 A}{d}\right) = 12 \times 8\text{ pF} = 96\text{ pF}$. --- ### 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 $\frac{l}{100-l} = 1 \implies l = 100 - l \implies 2l = 100 \implies l = 50\text{ cm}$. --- ### 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 $B_0/2$. --- ### 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 $V_L' = I' X_L = \frac{10}{R\sqrt{2}} \times R = \frac{10}{\sqrt{2}}\text{ V}$. --- ### 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 $\beta' = \frac{\lambda (2D)}{(d/2)} = 4 \frac{\lambda D}{d} = 4\beta$. --- ### 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: $\frac{v_1^2}{v_2^2} = \frac{1}{4} \implies \frac{v_1}{v_2} = \frac{1}{2}$. --- ### 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. --- ### 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. --- ### 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 $3 \times 2 = 6$. --- ### 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. --- ### 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. --- ### 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. --- ### 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. --- ### 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 ($-\text{CH}=\text{CH}_2$). --- ### 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). --- ### 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 $\text{mol}\cdot\text{L}^{-1}\cdot\text{s}^{-1}$. --- ### 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. --- ### 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. --- ### 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. --- ### 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. --- ### 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). --- ### 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. --- ### 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. --- ### 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. --- ### 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. --- ### 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. --- ### 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). ---