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12 real questions from the NEET 2026 paper, in the interface the 2027 exam moves to: one question at a time, a palette, a countdown, +4 / −1. No login, nothing to install.

NEET Mini Mock · 2026 PYQs

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Question 1 · Physics · NEET 2026 · +4 / −1
A cylindrical cork of uniform density floats in a liquid of density $\rho_1$ . If the cork is depressed slightly and released, it oscillates harmonically with time period $T$ . If the same cork floats in another liquid of density $\rho_2$ , then the similar oscillation has time period $2T$ . The value of $\rho_2/\rho_1$ is:
Answer and explanation
Answer: D · Oscillations
Depressing the cork by $x$ creates an extra buoyant force $A\rho g x$ directed back towards equilibrium, so the effective spring constant is $k = A\rho g$ and $T = 2\pi\sqrt{\frac{m}{A\rho g}}$ . With $m$ and $A$ unchanged, $T \propto \rho^{-1/2}$ . Doubling the period therefore requires the density to fall by a factor of four: $\frac{\rho_2}{\rho_1} = \frac{1}{4}$ . NCERT Class 11, Oscillations (SHM examples).
Question 2 · Physics · NEET 2026 · +4 / −1
The angular speed of a flywheel is increased from 600 rpm to 1200 rpm in 10 s. The number of revolutions completed by the flywheel during this time is:
Answer and explanation
Answer: C · System of Particles and Rotational Motion
With uniform angular acceleration the number of revolutions equals the average angular speed times the time: $\frac{600+1200}{2} = 900$ rpm over $\frac{10}{60}$ min = $\frac{1}{6}$ min, giving $900 \times \frac{1}{6} = 150$ revolutions, option (3). Equivalently, using $\omega_2^2 = \omega_1^2 + 2\alpha\theta$ gives $\theta = 300\pi$ rad, so $\theta/2\pi = 150$ .
Question 3 · Physics · NEET 2020 · +4 / −1
A series LCR circuit is connected to an ac voltage source. When L is removed from the circuit, the phase difference between current and voltage is $\pi$ /3. If instead C is removed from the circuit, the phase difference is again $\pi$ /3 between current and voltage. The power factor of the circuit is
Answer and explanation
Answer: B · Alternating Current
For the RC circuit, tan( $\pi$ /3) = Xc/R; for the RL circuit, tan( $\pi$ /3) = XL/R. Hence XL = Xc in the complete series circuit, so it is at resonance and its power factor is cos 0 = 1. Chapter: Alternating Current.
Question 4 · Physics · NEET 2020 · +4 / −1
A spherical conductor of radius 10 cm has a charge of 3.2 × 10⁻⁷ C distributed uniformly. What is the magnitude of electric field at a point 15 cm from the centre of the sphere? (1/(4πε₀) = 9 × 10⁹ N m² C⁻²)
Answer and explanation
Answer: A · Electric Charges and Fields
The electric field outside a charged spherical conductor is given by $E = \frac{1}{4\pi \epsilon_0} \frac{Q}{r^2}$ . Substituting the values, $E = 9 * 10^9 * \frac{3.2 * 10^{-7}}{(0.15)^2} = 1.28 * 10^5\ N/C$ . Chapter: Electric Charges and Fields.
Question 5 · Chemistry · NEET 2026 · +4 / −1
Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R. Assertion (A): The first ionization enthalpy of O is lower than that of N and F. Reason (R): The loss of an electron from O leads to stable half-filled p orbital In light of the above statements, choose the most appropriate answer from the options given below:
Answer and explanation
Answer: A · Classification of Elements and Periodicity in Properties
Oxygen is $[\mathrm{He}]2\mathrm{s}^2 2\mathrm{p}^4$ ; the fourth 2p electron must pair up in an already-occupied orbital and suffers extra repulsion, so it is comparatively easy to remove. Losing it gives $\mathrm{O}^{+}$ with the $2\mathrm{p}^3$ half-filled configuration, which is extra stable through exchange energy. That is precisely why oxygen's first ionization enthalpy dips below nitrogen's despite nitrogen having a smaller nuclear charge, and stays below fluorine's. R therefore explains A. NCERT Class 11, Classification of Elements and Periodicity in Properties.
Question 6 · Chemistry · NEET 2026 · +4 / −1
The number of hydrogen atoms present in 5.4 g of urea is: (Given: Molar mass of urea = $60\ \text{g mol}^{-1}$ ; $N_A = 6.022 \times 10^{23}\ \text{particles mol}^{-1}$ )
Answer and explanation
Answer: D · Some Basic Concepts of Chemistry
Moles of urea = 5.4/60 = 0.09 mol. Urea, $\mathrm{NH}_{2}\mathrm{CONH}_{2}$ , contains 4 hydrogen atoms per molecule. So number of H atoms = 0.09 * 4 * 6.022 * $10^{23}$ $\approx$ 2.168 * $10^{23}$ . The distractors arise from using the wrong number of H atoms per molecule or a wrong power of ten.
Question 7 · Chemistry · NEET 2026 · +4 / −1
Identify the incorrect statement from the following:
Answer and explanation
Answer: D · p-Block Elements (Groups 15-18)
Both nitrogen and oxygen are period-2 elements lacking accessible d-orbitals, so nitrogen cannot form a d $\pi$ -p $\pi$ bond with oxygen, making statement D incorrect. Nitrogen does form p $\pi$ -p $\pi$ multiple bonds (as in N2), the heavier elements P and As have vacant d-orbitals allowing d $\pi$ -d $\pi$ overlap with transition metals, and P, As, Sb show catenation, so those statements are correct.
Question 8 · Chemistry · NEET 2020 · +4 / −1
Find out the solubility of Ni(OH)₂ in 0.1 M NaOH. Given that the ionic product of Ni(OH)₂ is 2 × 10⁻¹⁵.
Answer and explanation
Answer: D · Equilibrium
The solubility of $\mathrm{Ni}(\mathrm{OH})_2$ in 0.1 M NaOH can be calculated using the solubility product $K_{sp} = 2 * 10^{-15}$ . Let s be the solubility of $\mathrm{Ni}(\mathrm{OH})_2$ . The concentration of $\mathrm{OH}^-$ from NaOH is 0.1 M, and from $\mathrm{Ni}(\mathrm{OH})_2$ is 2s. Thus, $K_{sp} = s * (0.1 + 2s)^2 $\approx$ s * (0.1)^2 = 2 * 10^{-15} ⇒ s = 2 * 10^{-13}\ M$. Chapter: Equilibrium.
Question 9 · Biology · NEET 2026 · +4 / −1
Which of the following plant produces non-albuminous seeds?
Answer and explanation
Answer: D · Sexual Reproduction in Flowering Plants
Non-albuminous (exalbuminous) seeds consume the entire endosperm during embryo development, so the mature seed stores food in fleshy cotyledons instead. Pea is the standard example, along with groundnut and bean. Wheat, maize, barley, castor and coconut all retain endosperm at maturity and are albuminous. NCERT Class 12, Sexual Reproduction in Flowering Plants.
Question 10 · Biology · NEET 2026 · +4 / −1
The sixth mutant codon of beta globin gene causing polymerization of Haemoglobin and change in RBC shape is ______.
Answer and explanation
Answer: A · Principles of Inheritance and Variation
In sickle cell anaemia the sixth codon of the $\beta$ -globin gene changes from GAG to GUG, so glutamic acid is replaced by valine. This substitution makes haemoglobin polymerize at low oxygen tension and distorts the RBC into a sickle shape, so the mutant codon is GUG.
Question 11 · Biology · NEET 2020 · +4 / −1
The process responsible for facilitating loss of water in liquid form from the tip of grass blades at night and in early morning is
Answer and explanation
Answer: A · Transport in Plants
The liquid-water loss is guttation. Root pressure supplies the positive force that pushes xylem sap out through hydathodes at night or early morning, so root pressure is the facilitating process. Chapter: Transport in Plants.
Question 12 · Biology · NEET 2020 · +4 / −1
The transverse section of a plant shows following anatomical features : Statement I: Large number of scattered vascular bundles surrounded by bundle sheath Statement II: Large conspicuous parenchymatous ground tissue Statement III: Vascular bundles conjoint and closed Statement IV: Phloem parenchyma absent Identify the category of plant and its part :
Answer and explanation
Answer: D · Anatomy of Flowering Plants
The described features-scattered vascular bundles, large parenchymatous ground tissue, conjoint and closed vascular bundles, and absence of phloem parenchyma-are characteristic of a monocotyledonous stem. Chapter: Anatomy of Flowering Plants.

Why practise on this screen

NEET is computer based from 2027. The knowledge transfers from paper practice; the habits do not: clicking instead of shading, a palette instead of page-flicking, a countdown you can always see. What actually changes in the CBT switch covers the details, and reading on a screen is the skill under all of it.

Questions

Is this NEET mock test free?

Yes. No login, no signup, no card. The full BrainIt app is also free until the 2027 exam.

Are these real NEET questions?

Yes. All 12 are from the NEET 2026 NTA paper, graded with the official marking scheme: +4 for a correct answer, -1 for a wrong one, 0 for a skip.

Is NEET 2027 a computer-based test?

Computer-based testing was announced for 2027 by the Education Minister on 15 May 2026. The final details come from the 2027 NTA bulletin. This page offers CBT-style practice with a palette and countdown.

How many questions and how much time?

12 questions in 12 minutes: one minute per question, the same pace as the real exam's 180 questions in 180 minutes.

A free NEET mock test,on the exam screen.

Why practise on this screen

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A free NEET mock test,
on the exam screen.