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NEET 2024 Paper — NEET Complete Yearwise Papers MCQs with Solutions

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Sample questions with solutions

Q1 — Physics · medium · numerical
A tightly wound 100 turns coil of radius $10\ \text{cm}$ carries a current of $7\ \text{A}$. The magnitude of the magnetic field at the centre of the coil is (take the permeability of free space as $4\pi \times 10^{-7}$ SI units):
A. $4.4\ \text{mT}$  ✓ Correct
B. $44\ \text{T}$
C. $44\ \text{mT}$
D. $4.4\ \text{T}$
Solution: At the centre of a circular coil, $B = \dfrac{\mu_0 N I}{2r}$. $B = \dfrac{4\pi \times 10^{-7} \times 100 \times 7}{2 \times 10 \times 10^{-2}} = 4.4 \times 10^{-3}\ \text{T} = 4.4\ \text{mT}$
Q2 — Physics · medium · theory
Match List-I with List-II. List-I (Material) (A) Diamagnetic (B) Ferromagnetic (C) Paramagnetic (D) Non-magnetic List-II (Susceptibility $\chi$) (I) $\chi = 0$ (II) $0 > \chi \geq -1$ (III) $\chi \gg 1$ (IV) $0 < \chi < \varepsilon$ (a small positive number) Choose the correct answer from the options given below:
A. A-III, B-II, C-I, D-IV
B. A-IV, B-III, C-II, D-I
C. A-II, B-III, C-IV, D-I  ✓ Correct
D. A-II, B-I, C-III, D-IV
Solution: Diamagnetic materials have a small negative susceptibility ($-10^{-5}$ to $-10^{-9}$), so $0 > \chi \geq -1$ — that is (II). Ferromagnetic materials have $\chi \gg 1$ — that is (III). Paramagnetic materials have a small positive susceptibility ($10^{-5}$ to $10^{-3}$) — that is (IV). A non-magnetic material has $\chi = 0$ — that is (I).
Q3 — Physics · medium · numerical
A thermodynamic system is taken through the cycle $abcda$. The work done by the gas along the path $bc$ is:
A. $-90\ \text{J}$
B. $-60\ \text{J}$
C. zero  ✓ Correct
D. $30\ \text{J}$
Solution: On the $P$–$V$ diagram the path $bc$ is vertical, so the volume stays constant along it. $W = \int P\,dV = 0$
Q4 — Physics · medium · theory
An unpolarised light beam strikes a glass surface at Brewster's angle. Then
A. both the reflected and refracted light will be completely polarized.
B. the reflected light will be completely polarized but the refracted light will be partially polarized.  ✓ Correct
C. the reflected light will be partially polarized.
D. the refracted light will be completely polarized.
Solution: At Brewster's angle only the reflected beam is completely polarized (perpendicular to the plane of incidence). The refracted beam is only partially polarized, because it still carries both components.
Q5 — Physics · medium · numerical
In an ideal transformer, the turns ratio is $\dfrac{N_p}{N_s} = \dfrac{1}{2}$. The ratio $V_s : V_p$ is equal to (the symbols carry their usual meaning):
A. $1 : 1$
B. $1 : 4$
C. $1 : 2$
D. $2 : 1$  ✓ Correct
Solution: For an ideal transformer the transformation ratio is $r = \dfrac{N_s}{N_p} = \dfrac{V_s}{V_p}$. Given $\dfrac{N_p}{N_s} = \dfrac{1}{2}$, so $\dfrac{V_s}{V_p} = \dfrac{2}{1}$, i.e. $V_s : V_p = 2 : 1$.
Q6 — Physics · medium · theory
A logic circuit provides the output $Y$ as per the following truth table: A | B | Y 0 | 0 | 1 0 | 1 | 0 1 | 0 | 1 1 | 1 | 0
A. $\bar{B}$  ✓ Correct
B. $B$
C. $A \cdot B + \bar{A}$
D. $A \cdot \bar{B} + \bar{A}$
Solution: Reading the table, $Y$ is 1 exactly when $B$ is 0 and 0 exactly when $B$ is 1, whatever $A$ does. So $Y = \bar{B}$. Checking the other choices against all four rows, none of them reproduces the column $1, 0, 1, 0$.
Q7 — Physics · medium · numerical
In a vernier calipers, $(N + 1)$ divisions of the vernier scale coincide with $N$ divisions of the main scale. If 1 MSD represents $0.1\ \text{mm}$, the vernier constant (in cm) is:
A. $100N$
B. $10(N + 1)$
C. $\dfrac{1}{10N}$
D. $\dfrac{1}{100(N + 1)}$  ✓ Correct
Solution: $(N+1)\ \text{VSD} = N\ \text{MSD}$, so $1\ \text{VSD} = \dfrac{N}{N+1}\ \text{MSD}$. Vernier constant $= 1\ \text{MSD} - 1\ \text{VSD} = \left(1 - \dfrac{N}{N+1}\right)\text{MSD} = \dfrac{1}{N+1}\ \text{MSD}$. With $1\ \text{MSD} = 0.1\ \text{mm} = \dfrac{0.1}{10}\ \text{cm}$, this is $\dfrac{1}{100(N+1)}\ \text{cm}$.
Q8 — Physics · medium · numerical
The maximum elongation of a steel wire of $1\ \text{m}$ length, if the elastic limit of steel and its Young's modulus are respectively $8 \times 10^8\ \text{N m}^{-2}$ and $2 \times 10^{11}\ \text{N m}^{-2}$, is:
A. $40\ \text{mm}$
B. $8\ \text{mm}$
C. $4\ \text{mm}$  ✓ Correct
D. $0.4\ \text{mm}$
Solution: $(\text{Stress})_{max} = \text{Young's modulus} \times (\text{Strain})_{max}$ $(\text{Strain})_{max} = \dfrac{8 \times 10^8}{2 \times 10^{11}} = 4 \times 10^{-3}$ For a $1\ \text{m}$ wire, $(\Delta \ell)_{max} = 4 \times 10^{-3}\ \text{m} = 4\ \text{mm}$.
Q9 — Physics · medium · numerical
A horizontal force of $10\ \text{N}$ is applied to a block $A$ as shown in the figure. The masses of blocks $A$ and $B$ are $2\ \text{kg}$ and $3\ \text{kg}$ respectively. The blocks slide over a frictionless surface. The force exerted by block $A$ on block $B$ is:
A. $6\ \text{N}$  ✓ Correct
B. $10\ \text{N}$
C. zero
D. $4\ \text{N}$
Solution: The two blocks move together, so the common acceleration is $a = \dfrac{F_{net}}{M_{total}} = \dfrac{10}{2 + 3} = 2\ \text{m s}^{-2}$ From the free-body diagram of $B$, the only horizontal force on it is the push from $A$: $F_{BA} = m_B a = 3 \times 2 = 6\ \text{N}$
Q10 — Physics · medium · theory
If the monochromatic source in Young's double slit experiment is replaced by white light, then
A. there will be a central bright white fringe surrounded by a few coloured fringes.  ✓ Correct
B. all bright fringes will be of equal width.
C. interference pattern will disappear.
D. there will be a central dark fringe surrounded by a few coloured fringes.
Solution: At the centre the path difference is zero for every wavelength, so all colours arrive in phase and the central fringe is white. Away from the centre each colour has its own fringe width, so the maxima of different colours fall at different places and we see a few coloured fringes on either side before the pattern washes out.
Q11 — Physics · medium · theory
The graph which shows the variation of $\dfrac{1}{\lambda^2}$ with the kinetic energy $E$ is (where $\lambda$ is the de Broglie wavelength of a free particle):
A.
B.  ✓ Correct
C.
D.
Solution: $\lambda = \dfrac{h}{p} = \dfrac{h}{\sqrt{2mE}}$, so $\lambda^2 = \dfrac{h^2}{2mE}$ and therefore $\dfrac{1}{\lambda^2} = \dfrac{2mE}{h^2}$. So $\dfrac{1}{\lambda^2} \propto E$ — a straight line through the origin.
Q12 — Physics · medium · numerical
In the following circuit, the equivalent capacitance between terminal $A$ and terminal $B$ is:
A. $0.5\ \mu F$
B. $4\ \mu F$
C. $2\ \mu F$  ✓ Correct
D. $1\ \mu F$
Solution: The four $2\ \mu F$ capacitors form a balanced Wheatstone bridge, so no charge sits on the bridging capacitor in the middle and it can be removed. Each arm is then two $2\ \mu F$ capacitors in series $= 1\ \mu F$, and the two arms are in parallel: $C_{AB} = 1 + 1 = 2\ \mu F$
Q13 — Physics · medium · theory
In the diagram, a strong bar magnet is moving towards solenoid-2 from solenoid-1. The direction of the induced current in solenoid-1 and that in solenoid-2, respectively, are through the directions:
A. $AB$ and $CD$
B. $BA$ and $DC$
C. $AB$ and $DC$  ✓ Correct
D. $BA$ and $CD$
Solution: By Lenz's law each solenoid opposes the change it experiences. Solenoid-1 is losing flux as the magnet moves away, so its induced current keeps the flux going — it flows through $AB$. Solenoid-2 is gaining flux as the magnet approaches, so its induced current opposes it by presenting a like pole to the magnet — it flows through $DC$.
Q14 — Physics · medium · theory
Consider the following statements $A$ and $B$ and identify the correct answer: A. For a solar cell, the I–V characteristic lies in the IV quadrant of the given graph. B. In a reverse biased pn junction diode, the current measured in $\mu A$ is due to majority charge carriers.
A. Both A and B are correct.
B. Both $A$ and $B$ are incorrect.
C. $A$ is correct but $B$ is incorrect.  ✓ Correct
D. A is incorrect but B is correct.
Solution: A is correct: a solar cell supplies current to the load rather than drawing it, so its I–V curve is drawn in the fourth quadrant. B is incorrect: in reverse bias the majority carriers are swept away from the junction, and the small $\mu A$ reverse current is a drift current carried by the MINORITY carriers of both the n-side and the p-side.
Q15 — Physics · medium · numerical
A light ray enters through a right angled prism at point $P$ with an angle of incidence $30^\circ$ as shown in the figure. It travels through the prism parallel to its base $BC$ and emerges along the face $AC$. The refractive index of the prism is:
A. $\dfrac{\sqrt{3}}{4}$
B. $\dfrac{\sqrt{3}}{2}$
C. $\dfrac{\sqrt{5}}{4}$
D. $\dfrac{\sqrt{5}}{2}$  ✓ Correct
Solution: At face $AB$: $1 \cdot \sin 30^\circ = \mu \sin r$ …(i) The ray emerges along $AC$, i.e. grazing, so at face $AC$: $\mu \sin(90^\circ - r) = 1 \Rightarrow \mu \cos r = 1 \Rightarrow \cos r = \dfrac{1}{\mu}$ …(ii) Putting (ii) into (i): $\dfrac{1}{2} = \mu\sqrt{1 - \dfrac{1}{\mu^2}} \Rightarrow \dfrac{1}{4} = \mu^2 - 1 \Rightarrow \mu = \dfrac{\sqrt{5}}{2}$
Q16 — Physics · medium · theory
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: The potential $(V)$ at any axial point, at $2\ \text{m}$ distance $(r)$ from the centre of a dipole of dipole moment vector $\vec{P}$ of magnitude $4 \times 10^{-6}\ \text{C m}$, is $\pm 9 \times 10^3\ \text{V}$. (Take $\dfrac{1}{4\pi\epsilon_0} = 9 \times 10^9$ SI units) Reason R: $V = \pm \dfrac{2P}{4\pi\epsilon_0 r^2}$, where $r$ is the distance of any axial point situated at $2\ \text{m}$ from the centre of the dipole. In the light of the above statements, choose the correct answer from the options given below:
A. A is true but $R$ is false.  ✓ Correct
B. $A$ is false but $R$ is true.
C. Both $A$ and $R$ are true and $R$ is the correct explanation of $A$.
D. Both $A$ and $R$ are true and $R$ is NOT the correct explanation of A.
Solution: The potential of a dipole is $V = \dfrac{Kp\cos\theta}{r^2}$, and on the axis $\theta = 0^\circ$ or $180^\circ$, so $V = \pm\dfrac{Kp}{r^2}$. $V = \pm\dfrac{9 \times 10^9 \times 4 \times 10^{-6}}{2^2} = \pm 9 \times 10^3\ \text{V}$ — so the Assertion is true. The Reason quotes the formula with an extra factor of 2 in the numerator, so R is false.
Q17 — Physics · medium · numerical
The moment of inertia of a thin rod about an axis passing through its mid point and perpendicular to the rod is $2400\ \text{g cm}^2$. The length of the $400\ \text{g}$ rod is nearly:
A. $20.7\ \text{cm}$
B. $72.0\ \text{cm}$
C. $8.5\ \text{cm}$  ✓ Correct
D. $17.5\ \text{cm}$
Solution: For a rod about its centre, $I = \dfrac{ML^2}{12}$. $\dfrac{400 \times L^2}{12} = 2400 \Rightarrow L^2 = \dfrac{2400 \times 12}{400} = 72 \Rightarrow L = \sqrt{72} \approx 8.5\ \text{cm}$
Q18 — Physics · medium · numerical
The terminal voltage of the battery, whose emf is $10\ \text{V}$ and internal resistance $1\ \Omega$, when connected through an external resistance of $4\ \Omega$ as shown in the figure, is:
A. $8\ \text{V}$  ✓ Correct
B. $10\ \text{V}$
C. $4\ \text{V}$
D. $6\ \text{V}$
Solution: Current in the loop: $10 = i(4 + 1) \Rightarrow i = 2\ \text{A}$. Terminal voltage $= \varepsilon - i r = 10 - 2 \times 1 = 8\ \text{V}$.
Q19 — Physics · medium · numerical
Match the List-I with List-II. List-I (Spectral lines of hydrogen for transitions from) (A) $n_2 = 3$ to $n_1 = 2$ (B) $n_2 = 4$ to $n_1 = 2$ (C) $n_2 = 5$ to $n_1 = 2$ (D) $n_2 = 6$ to $n_1 = 2$ List-II (Wavelengths in nm) (I) 410.2 (II) 434.1 (III) 656.3 (IV) 486.1 Choose the correct answer from the options given below:
A. A-IV, B-III, C-I, D-II
B. A-I, B-II, C-III, D-IV
C. A-II, B-I, C-IV, D-III
D. A-III, B-IV, C-II, D-I  ✓ Correct
Solution: For hydrogen, $\lambda = 91.2\left(\dfrac{n_1^2 n_2^2}{n_2^2 - n_1^2}\right)\text{nm}$. $3 \to 2$: $\lambda = 656.3\ \text{nm}$ (III) $4 \to 2$: $\lambda = 486.1\ \text{nm}$ (IV) $5 \to 2$: $\lambda = 434.1\ \text{nm}$ (II) $6 \to 2$: $\lambda = 410.2\ \text{nm}$ (I)
Q20 — Physics · medium · theory
If $c$ is the velocity of light in free space, the correct statements about a photon among the following are: A. The energy of a photon is $E = h\nu$. B. The velocity of a photon is $c$. C. The momentum of a photon is $p = \dfrac{h\nu}{c}$. D. In a photon-electron collision, both total energy and total momentum are conserved. E. Photon possesses positive charge. Choose the correct answer from the options given below:
A. A, C and D only
B. A, B, D and E only
C. A and B only
D. A, B, C and D only  ✓ Correct
Solution: A, B, C and D are all standard properties of a photon: $E = h\nu$, it travels at $c$, and since $E = pc$ its momentum is $p = \dfrac{h\nu}{c}$. In a photon-electron collision (the Compton effect) both energy and momentum are conserved. E is wrong — a photon is electrically neutral.
Q21 — Physics · medium · numerical
$^{290}_{82}X \xrightarrow{\ \alpha\ } Y \xrightarrow{\ e^+\ } Z \xrightarrow{\ \beta^-\ } P \xrightarrow{\ e^-\ } Q$ In the nuclear emission stated above, the mass number and atomic number of the product $Q$ respectively, are:
A. 288, 82
B. 286, 81  ✓ Correct
C. 280, 81
D. 286, 80
Solution: $\alpha$ emission: mass number falls by 4 and atomic number by 2 $\Rightarrow\ ^{286}_{80}Y$. $e^+$ (positron) emission: mass number unchanged, atomic number falls by 1 $\Rightarrow\ ^{286}_{79}Z$. $\beta^-$ emission: mass number unchanged, atomic number rises by 1 $\Rightarrow\ ^{286}_{80}P$. The last $e^-$ emission again raises the atomic number by 1 $\Rightarrow\ ^{286}_{81}Q$.
Q22 — Physics · medium · numerical
At any instant of time $t$, the displacement of a particle is given by $2t - 1$ (SI unit) under the influence of a force of $5\ \text{N}$. The value of the instantaneous power is (in SI unit):
A. 7
B. 6
C. 10  ✓ Correct
D. 5
Solution: $s = 2t - 1 \Rightarrow v = \dfrac{ds}{dt} = 2\ \text{m s}^{-1}$ (constant). Instantaneous power $P = \vec{F} \cdot \vec{v} = 5 \times 2 = 10\ \text{W}$.
Q23 — Physics · medium · theory
The output $(Y)$ of the given logic gate is similar to the output of an/a:
A. OR gate
B. AND gate  ✓ Correct
C. NAND gate
D. NOR gate
Solution: Each input first passes through a gate acting as an inverter, giving $\bar{A}$ and $\bar{B}$. These feed a NOR gate: $Y = \overline{\bar{A} + \bar{B}} = \bar{\bar{A}} \cdot \bar{\bar{B}} = A \cdot B$ which is an AND gate.
Q24 — Physics · medium · numerical
The mass of a planet is $\dfrac{1}{10}$ th that of the earth and its diameter is half that of the earth. The acceleration due to gravity on that planet is:
A. $4.9\ \text{m s}^{-2}$
B. $3.92\ \text{m s}^{-2}$  ✓ Correct
C. $19.6\ \text{m s}^{-2}$
D. $9.8\ \text{m s}^{-2}$
Solution: On earth, $g = \dfrac{GM}{R^2} = 9.8\ \text{m s}^{-2}$. For the planet, $M' = \dfrac{M}{10}$ and $R' = \dfrac{R}{2}$, so $g' = \dfrac{GM/10}{(R/2)^2} = \dfrac{4}{10}\cdot\dfrac{GM}{R^2} = 0.4 \times 9.8 = 3.92\ \text{m s}^{-2}$
Q25 — Physics · medium · theory
Given below are two statements: Statement I: Atoms are electrically neutral as they contain equal number of positive and negative charges. Statement II: Atoms of each element are stable and emit their characteristic spectrum. In the light of the above statements, choose the most appropriate answer from the options given below:
A. Statement I is correct but Statement II is incorrect.  ✓ Correct
B. Statement I is incorrect but Statement II is correct.
C. Both Statement I and Statement II are correct.
D. Both Statement I and Statement II are incorrect.
Solution: Statement I is correct — an atom has equal numbers of protons and electrons, so it is neutral. Statement II is not correct as a blanket claim: atoms of MOST elements are stable and emit a characteristic spectrum, but not those of every element (radioactive elements are unstable).
Q26 — Physics · medium · theory
A wheel of a bullock cart is rolling on a level road as shown in the figure below. If its linear speed is $v$ in the direction shown, which one of the following options is correct ($P$ and $Q$ are the highest and lowest points on the wheel respectively)?
A. Both the points $P$ and $Q$ move with equal speed.
B. Point $P$ has zero speed.
C. Point $P$ moves slower than point $Q$.
D. Point $P$ moves faster than point $Q$.  ✓ Correct
Solution: In rolling without slipping the point of contact $Q$ is instantaneously at rest, so its speed is zero. The topmost point $P$ moves at twice the speed of the centre of mass, i.e. $2v$. So $P$ moves faster than $Q$.
Q27 — Physics · medium · theory
A particle moving with uniform speed in a circular path maintains:
A. constant velocity but varying acceleration.
B. varying velocity and varying acceleration.  ✓ Correct
C. constant velocity.
D. constant acceleration.
Solution: The speed is constant but the direction of motion changes continuously, so the velocity vector varies. The centripetal acceleration has constant magnitude but its direction (always towards the centre) also keeps changing, so the acceleration vector varies too.
Q28 — Physics · medium · numerical
A thin flat circular disc of radius $4.5\ \text{cm}$ is placed gently over the surface of water. If the surface tension of water is $0.07\ \text{N m}^{-1}$, then the excess force required to take it away from the surface is:
A. $1.98\ \text{mN}$
B. $99\ \text{N}$
C. $19.8\ \text{mN}$  ✓ Correct
D. $198\ \text{N}$
Solution: For a flat circular disc lying on the surface, $F_{excess} = 2\pi r T$. $F = 2 \times \dfrac{22}{7} \times 4.5 \times 10^{-2} \times 7 \times 10^{-2} = 198 \times 10^{-4}\ \text{N} = 19.8\ \text{mN}$
Q29 — Physics · medium · numerical
In a uniform magnetic field of $0.049\ \text{T}$, a magnetic needle performs 20 complete oscillations in 5 seconds as shown. The moment of inertia of the needle is $9.8 \times 10^{-6}\ \text{kg m}^2$. If the magnitude of the magnetic moment of the needle is $x \times 10^{-5}\ \text{A m}^2$, then the value of $x$ is:
A. $50\pi^2$
B. $1280\pi^2$  ✓ Correct
C. $5\pi^2$
D. $128\pi^2$
Solution: Time period $T = \dfrac{5}{20} = \dfrac{1}{4}\ \text{s}$, and $T = 2\pi\sqrt{\dfrac{I}{mB}}$, so $m = \dfrac{4\pi^2 I}{T^2 B}$. $m = \dfrac{4\pi^2 \times 9.8 \times 10^{-6}}{\left(\frac{1}{4}\right)^2 \times 4.9 \times 10^{-2}} = 1280\pi^2 \times 10^{-5}\ \text{A m}^2$ So $x = 1280\pi^2$.
Q30 — Physics · medium · numerical
Two bodies $A$ and $B$ of same mass undergo a completely inelastic one dimensional collision. The body $A$ moves with velocity $v_1$ while body $B$ is at rest before collision. The velocity of the system after collision is $v_2$. The ratio $v_1 : v_2$ is:
A. $4 : 1$
B. $1 : 4$
C. $1 : 2$
D. $2 : 1$  ✓ Correct
Solution: The two stick together, so by conservation of momentum $m v_1 + 0 = 2m v_2 \Rightarrow \dfrac{v_1}{v_2} = \dfrac{2}{1}$ So $v_1 : v_2 = 2 : 1$.