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NEET 2024 Paper 2 — NEET Complete Yearwise Papers MCQs with Solutions
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Sample questions with solutions
Q1 — Physics · medium · numerical
The magnetic potential energy, when a magnetic bar of magnetic moment $\vec{m}$ is placed perpendicular to the magnetic field $\vec{B}$, is
A. $-\dfrac{mB}{2}$
B. Zero ✓ Correct
C. $-mB$
D. $mB$
Solution: The potential energy stored in an external magnetic field is $U = -\vec{m} \cdot \vec{B} = -mB\cos\theta$.
Here the bar is perpendicular to the field, so $\theta = 90^\circ$ and
$U = -mB\cos 90^\circ = 0$
Q2 — Physics · medium · numerical
A bob is whirled in a horizontal circle by means of a string at an initial speed of 10 rpm. If the tension in the string is quadrupled while keeping the radius constant, the new speed is:
A. 20 rpm ✓ Correct
B. 40 rpm
C. 5 rpm
D. 10 rpm
Solution: In horizontal circular motion, $T = m\omega^2 r$. For constant $m$ and $r$, $T \propto \omega^2$.
Given $T' = 4T$, so $\omega' = 2\omega = 2 \times 10 = 20\ \text{rpm}$.
Q3 — Physics · medium · numerical
A metal cube of side 5 cm is charged with $6\ \mu C$. The surface charge density on the cube is
A. $0.125 \times 10^{-3}\ \text{C m}^{-2}$
B. $0.25 \times 10^{-3}\ \text{C m}^{-2}$
C. $4 \times 10^{-3}\ \text{C m}^{-2}$
D. $0.4 \times 10^{-3}\ \text{C m}^{-2}$ ✓ Correct
Solution: In a metal all the charge resides on the surface, and a cube has six faces.
$S = 6a^2 = 6 \times (5 \times 10^{-2})^2 = 150 \times 10^{-4}\ \text{m}^2$
$\sigma = \dfrac{Q}{S} = \dfrac{6 \times 10^{-6}}{150 \times 10^{-4}} = 0.4 \times 10^{-3}\ \text{C m}^{-2}$
Q4 — Physics · medium · theory
The incorrect relation for a diamagnetic material (all the symbols carry their usual meaning, and $\varepsilon$ is a small positive number) is:
A. $\mu < \mu_0$
B. $0 \leq \mu_r < 1$
C. $-1 \leq \chi < 0$
D. $1 < \mu_r < 1 + \varepsilon$ ✓ Correct
Solution: For a diamagnetic material $0 \leq \mu_r < 1$, and since $\chi = \mu_r - 1$ this gives $-1 \leq \chi < 0$. With $\mu_r < 1$ it also follows that $\mu < \mu_0$.
So (a), (b) and (c) are all correct. The relation $1 < \mu_r < 1 + \varepsilon$ describes a PARAMAGNETIC material, so it is the incorrect one here.
Q5 — Physics · medium · theory
An ideal fluid is flowing in a non-uniform cross-sectional tube $XY$ (as shown in the figure) from end $X$ to end $Y$. If $K_1$ and $K_2$ are the kinetic energy per unit volume of the fluid at $X$ and $Y$ respectively, then the correct option is:
A. $K_1 = K_2$
B. $2K_1 = K_2$
C. $K_1 > K_2$ ✓ Correct
D. $K_1 < K_2$
Solution: By Bernoulli's principle, kinetic energy per unit volume + potential energy per unit volume + pressure is constant:
$\dfrac{1}{2}\rho v^2 + \rho g h + P = \text{constant}$
Applying it at $X$ (on the ground) and $Y$ (at height $h$) with the same pressure:
$P + K_1 + \rho g(0) = P + K_2 + \rho g h \Rightarrow K_1 = K_2 + \rho g h$
So $K_1 > K_2$.
Q6 — Physics · medium · numerical
The escape velocity for earth is $v$. A planet having 9 times the mass of earth and radius 16 times that of earth has the escape velocity of:
A. $\dfrac{v}{3}$
B. $\dfrac{2v}{3}$
C. $\dfrac{3v}{4}$ ✓ Correct
D. $\dfrac{9v}{4}$
Solution: $v_e = \sqrt{\dfrac{2GM}{R}}$, so $v_e \propto \sqrt{\dfrac{M}{R}}$.
$\dfrac{(v_e)_p}{(v_e)_e} = \sqrt{\dfrac{9M_e}{16R_e} \times \dfrac{R_e}{M_e}} = \sqrt{\dfrac{9}{16}} = \dfrac{3}{4}$
So $(v_e)_p = \dfrac{3v}{4}$.
Q7 — Physics · medium · theory
An electron and an alpha particle are accelerated by the same potential difference. Let $\lambda_e$ and $\lambda_\alpha$ denote the de Broglie wavelengths of the electron and the alpha particle, respectively, then:
A. $\lambda_e > \lambda_\alpha$ ✓ Correct
B. $\lambda_e = 4\lambda_\alpha$
C. $\lambda_e = \lambda_\alpha$
D. $\lambda_e < \lambda_\alpha$
Solution: $\lambda = \dfrac{h}{p} = \dfrac{h}{\sqrt{2mqV}}$, so for the same potential difference $\lambda \propto \dfrac{1}{\sqrt{mq}}$.
$\dfrac{\lambda_\alpha}{\lambda_e} = \sqrt{\dfrac{m_e q_e}{m_\alpha q_\alpha}}$
The alpha particle is far heavier and carries twice the charge, so $\lambda_e > \lambda_\alpha$.
Q8 — Physics · medium · numerical
An object moving along the horizontal x-direction with kinetic energy 10 J is displaced through $\vec{x} = (3\hat{i})\ \text{m}$ by the force $\vec{F} = (-2\hat{i} + 3\hat{j})\ \text{N}$. The kinetic energy of the object at the end of the displacement $x$ is:
A. 10 J
B. 16 J
C. 4 J ✓ Correct
D. 6 J
Solution: By the work-energy theorem, $W_{all} = \Delta K.E.$
$K_f - K_i = \vec{F} \cdot \Delta\vec{x} = (-2\hat{i} + 3\hat{j}) \cdot (3\hat{i}) = -6\ \text{J}$
$K_f - 10 = -6 \Rightarrow K_f = 4\ \text{J}$
Q9 — Physics · medium · numerical
An object falls from a height of 10 m above the ground. After striking the ground it loses 50% of its kinetic energy. The height up to which the object can rebound from the ground is:
A. 7.5 m
B. 10 m
C. 2.5 m
D. 5 m ✓ Correct
Solution: Kinetic energy just before striking the ground: $K_1 = mgh_1 = mg(10)$.
After losing 50%: $K_2 = \dfrac{K_1}{2}$.
That kinetic energy carries it back up to $h_2$:
$\dfrac{mg(10)}{2} = mgh_2 \Rightarrow h_2 = 5\ \text{m}$
Q10 — Physics · medium · theory
In the circuit shown below, the inductance $L$ is connected to an ac source. The current flowing in the circuit is $I = I_0\sin\omega t$. The voltage drop $(V_L)$ across $L$ is
A. $\omega L\,I_0\sin\omega t$
B. $\dfrac{I_0}{\omega L}\sin\omega t$
C. $\dfrac{I_0}{\omega L}\cos\omega t$
D. $\omega L I_0\cos\omega t$ ✓ Correct
Solution: Across a pure inductor the voltage LEADS the current by $\dfrac{\pi}{2}$.
With $I = I_0\sin\omega t$,
$V_L = V_0\sin\left(\omega t + \dfrac{\pi}{2}\right) = I_0 X_L\cos\omega t = \omega L I_0\cos\omega t$
Q11 — Physics · medium · numerical
A 12 pF capacitor is connected to a 50 V battery. The electrostatic energy stored in the capacitor, in nJ, is
A. 15 ✓ Correct
B. 7.5
C. 0.3
D. 150
Solution: $U = \dfrac{1}{2}CV^2 = \dfrac{1}{2} \times 12 \times 10^{-12} \times (50)^2$
$U = 6 \times 25 \times 10^{-10} = 15 \times 10^{-9}\ \text{J} = 15\ \text{nJ}$
Q12 — Physics · medium · numerical
A uniform wire of diameter $d$ carries a current of 100 mA when the mean drift velocity of electrons in the wire is $v$. For a wire of diameter $\dfrac{d}{2}$ of the same material to carry a current of 200 mA, the mean drift velocity of electrons in the wire is
A. $4v$
B. $8v$ ✓ Correct
C. $v$
D. $2v$
Solution: $i = nAv_de = n\left(\dfrac{\pi D^2}{4}\right)v_de$, so $i \propto D^2 v_d$.
$\dfrac{100}{200} = \dfrac{d^2}{\left(\frac{d}{2}\right)^2} \times \dfrac{v}{v'}$
$\dfrac{1}{2} = 4 \times \dfrac{v}{v'} \Rightarrow v' = 8v$
Q13 — Physics · medium · numerical
In an electrical circuit, the voltage is measured as $V = (200 \pm 4)$ volt and the current is measured as $I = (20 \pm 0.2)\ \text{A}$. The value of the resistance is:
A. $(10 \pm 4.2)\ \Omega$
B. $(10 \pm 0.3)\ \Omega$ ✓ Correct
C. $(10 \pm 0.1)\ \Omega$
D. $(10 \pm 0.8)\ \Omega$
Solution: $R = \dfrac{V}{I} = \dfrac{200}{20} = 10\ \Omega$
$\dfrac{\Delta R}{R} = \dfrac{\Delta V}{V} + \dfrac{\Delta I}{I} = \dfrac{4}{200} + \dfrac{0.2}{20} = \dfrac{6}{200}$
$\Delta R = \dfrac{6}{200} \times 10 = 0.3\ \Omega$, so $R = (10 \pm 0.3)\ \Omega$.
Q14 — Physics · medium · numerical
A step up transformer is connected to an ac mains supply of 220 V to operate at 11000 V, 88 watt. The current in the secondary circuit, ignoring the power loss in the transformer, is
A. 8 mA ✓ Correct
B. 4 mA
C. 0.4 A
D. 4 A
Solution: In the secondary circuit, $P = Vi$.
$88 = 11000 \times i \Rightarrow i = \dfrac{88}{11 \times 10^3} = 8 \times 10^{-3}\ \text{A} = 8\ \text{mA}$
Q15 — Physics · medium · numerical
A particle is moving along the $x$-axis with its position $(x)$ varying with time $(t)$ as $x = \alpha t^4 + \beta t^2 + \gamma t + \delta$. The ratio of its initial velocity to its initial acceleration, respectively, is:
A. $2\alpha : \delta$
B. $\gamma : 2\delta$
C. $4\alpha : \beta$
D. $\gamma : 2\beta$ ✓ Correct
Solution: $v = \dfrac{dx}{dt} = 4\alpha t^3 + 2\beta t + \gamma$, so the initial velocity is $v(0) = \gamma$.
$a = \dfrac{dv}{dt} = 12\alpha t^2 + 2\beta$, so the initial acceleration is $a(0) = 2\beta$.
$\dfrac{v(0)}{a(0)} = \dfrac{\gamma}{2\beta}$
Q16 — Physics · medium · numerical
The radius of gyration of a solid sphere of mass 5 kg about $XY$ is 5 m as shown in the figure. The radius of the sphere is $\dfrac{5x}{\sqrt{7}}$ m, then the value of $x$ is:
A. 5
B. $\sqrt{2}$
C. $\sqrt{3}$
D. $\sqrt{5}$ ✓ Correct
Solution: The axis $XY$ is tangential to the sphere, so by the parallel axis theorem
$I_{XY} = I_{CM} + MR^2 = \dfrac{2}{5}MR^2 + MR^2 = \dfrac{7}{5}MR^2$
Also $I_{XY} = MK^2 = M(5)^2$, so $\dfrac{7}{5}R^2 = 25 \Rightarrow R = \sqrt{\dfrac{5}{7}} \times 5 = \dfrac{5\sqrt{5}}{\sqrt{7}}$
Comparing with $\dfrac{5x}{\sqrt{7}}$ gives $x = \sqrt{5}$.
Q17 — Physics · medium · theory
The I–V characteristics shown above are exhibited by a
A. Light emitting diode
B. Zener diode
C. Photodiode
D. Solar cell ✓ Correct
Solution: The curve lies in the region of NEGATIVE current with positive voltage — the fourth quadrant. A device that delivers power rather than absorbing it behaves this way.
That is the I–V characteristic of a solar cell.
Q18 — Physics · medium · numerical
The magnetic moment and moment of inertia of a magnetic needle as shown are respectively $1.0 \times 10^{-2}\ \text{A m}^2$ and $\dfrac{10^{-6}}{\pi^2}\ \text{kg m}^2$. If it completes 10 oscillations in 10 s, the magnitude of the magnetic field is
A. 0.4 T
B. 4 T
C. 0.4 mT ✓ Correct
D. 4 mT
Solution: Time period of oscillation of a magnet in a magnetic field: $T = 2\pi\sqrt{\dfrac{I}{MB}}$.
$T = \dfrac{t}{n} = \dfrac{10}{10} = 1\ \text{s}$
$1 = 2\pi\sqrt{\dfrac{10^{-6}}{\pi^2 \times 1.0 \times 10^{-2} \times B}} \Rightarrow \dfrac{1}{4} = \dfrac{10^{-4}}{B}$
$B = 4 \times 10^{-4}\ \text{T} = 0.4\ \text{mT}$
Q19 — Physics · medium · theory
The capacitance of a capacitor with charge $q$ and a potential difference $V$ depends on
A. both q and V
B. the geometry of the capacitor ✓ Correct
C. q only
D. V only
Solution: For a parallel plate capacitor $C = \dfrac{A\epsilon_0}{d}$ — the capacitance is fixed by the plate area, their separation and the medium between them.
It is independent of the charge $q$ and the potential $V$; changing one simply changes the other so that $\dfrac{q}{V}$ stays equal to $C$.
Q20 — Physics · medium · theory
Given below are two statements:
Statement I: Image formation needs regular reflection and / or refraction.
Statement II: The variety in colour of objects we see around us is due to the constituent colours of the light incident on them.
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
B. Statement I is incorrect but Statement II is correct
C. Both Statement I and Statement II are correct ✓ Correct
D. Both Statement I and Statement II are incorrect
Solution: Both are correct. Regular reflection is necessary for image formation — which is why we see our image in a mirror but not in a wall.
And the variety of colours we see is due to the constituent colours present in the white light falling on the objects, some of which are absorbed and some reflected.
Q21 — Physics · medium · numerical
A uniform metal wire of length $l$ has $10\ \Omega$ resistance. Now this wire is stretched to a length $2l$ and then bent to form a perfect circle. The equivalent resistance across any arbitrary diameter of that circle is:
A. $10\ \Omega$ ✓ Correct
B. $5\ \Omega$
C. $40\ \Omega$
D. $20\ \Omega$
Solution: Stretching to $n$ times the length raises the resistance to $n^2$ times (volume is conserved):
$R_1 = 2^2 \times 10 = 40\ \Omega$
Bent into a circle, any diameter splits the ring into two halves of $20\ \Omega$ each, in parallel:
$R_{AB} = \dfrac{20 \times 20}{20 + 20} = 10\ \Omega$
Q22 — Physics · medium · theory
The spectral series which corresponds to the electronic transition from the levels $n_2 = 5, 6, \ldots$ to the level $n_1 = 4$ is
A. Pfund series
B. Brackett series ✓ Correct
C. Lyman series
D. Balmer series
Solution: The hydrogen series are named by the level the electron falls TO: $n_1 = 1$ Lyman, $n_1 = 2$ Balmer, $n_1 = 3$ Paschen, $n_1 = 4$ Brackett, $n_1 = 5$ Pfund.
Transitions ending at $n_1 = 4$ are therefore the Brackett series.
Q23 — Physics · medium · theory
Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: Houses made of concrete roofs overlaid with foam keep the room hotter during summer.
Reason R: The layer of foam insulation prohibits heat transfer, as it contains air pockets.
In the light of the above statements, choose the correct answer from the options given below.
A. A is true but R is false.
B. A is false but R is true. ✓ Correct
C. Both A and R are true and R is the correct explanation of A.
D. Both A and R are true but R is NOT the correct explanation of A.
Solution: The Assertion is false — a foam layer creates insulation, so a roof overlaid with foam keeps the room COOLER during summer, not hotter.
The Reason is true: foam contains trapped air pockets, and air is a poor conductor, so the layer prohibits heat transfer.
Q24 — Physics · medium · numerical
A particle executing simple harmonic motion with amplitude $A$ has the same potential and kinetic energies at the displacement
A. $2\sqrt{A}$
B. $\dfrac{A}{2}$
C. $\dfrac{A}{\sqrt{2}}$ ✓ Correct
D. $A\sqrt{2}$
Solution: Potential energy $= \dfrac{1}{2}kx^2$ and kinetic energy $= \dfrac{1}{2}kA^2 - \dfrac{1}{2}kx^2$.
Setting them equal:
$\dfrac{1}{2}kx^2 = \dfrac{1}{2}kA^2 - \dfrac{1}{2}kx^2 \Rightarrow 2x^2 = A^2$
$x = \dfrac{A}{\sqrt{2}}$
Q25 — Physics · medium · numerical
Two slits in Young's double slit experiment are 1.5 mm apart and the screen is placed at a distance of 1 m from the slits. If the wavelength of light used is $600 \times 10^{-9}$ m then the fringe separation is
A. $4 \times 10^{-5}\ \text{m}$
B. $9 \times 10^{-8}\ \text{m}$
C. $4 \times 10^{-7}\ \text{m}$
D. $4 \times 10^{-4}\ \text{m}$ ✓ Correct
Solution: Fringe separation $\beta = \dfrac{\lambda D}{d}$.
$\beta = \dfrac{600 \times 10^{-9} \times 1}{1.5 \times 10^{-3}} = \dfrac{6 \times 10^{-7}}{1.5 \times 10^{-3}} = 4 \times 10^{-4}\ \text{m}$
Q26 — Physics · medium · theory
Water is used as a coolant in a nuclear reactor because of its
A. high thermal expansion coefficient
B. high specific heat capacity ✓ Correct
C. low density
D. low boiling point
Solution: Water has an unusually high specific heat capacity, so it can absorb a great deal of heat for only a small rise in its own temperature. That is exactly what a coolant must do.
Q27 — Physics · medium · numerical
The pitch of an error free screw gauge is 1 mm and there are 100 divisions on the circular scale. While measuring the diameter of a thick wire, the pitch scale reads 1 mm and the $63^{rd}$ division on the circular scale coincides with the reference line. The diameter of the wire is:
A. 1.63 cm
B. 0.163 cm ✓ Correct
C. 0.163 m
D. 1.63 m
Solution: Least count $= \dfrac{\text{Pitch}}{\text{No. of divisions on circular scale}} = \dfrac{1}{100} = 0.01\ \text{mm}$
Reading $= \text{MSR} + \text{CSR} \times \text{L.C.} = 1 + 63 \times 0.01 = 1.63\ \text{mm} = 0.163\ \text{cm}$
Q28 — Physics · medium · numerical
Let us consider two solenoids $A$ and $B$, made from the same magnetic material of relative permeability $\mu_r$ and equal area of cross-section. The length of $A$ is twice that of $B$ and the number of turns per unit length in $A$ is half that of $B$. The ratio of self inductances of the two solenoids, $L_A : L_B$, is
A. $1 : 2$ ✓ Correct
B. $2 : 1$
C. $8 : 1$
D. $1 : 8$
Solution: $L = \mu_0\mu_r n^2 A l$, so $L \propto n^2 l$ here.
$\dfrac{L_A}{L_B} = \dfrac{n_A^2}{n_B^2} \times \dfrac{l_A}{l_B} = \left(\dfrac{1}{2}\right)^2 \times 2 = \dfrac{1}{2}$
So $L_A : L_B = 1 : 2$.
Q29 — Physics · medium · theory
When the output of an OR gate is applied as input to a NOT gate, then the combination acts as a
A. NAND gate
B. NOR gate ✓ Correct
C. AND gate
D. OR gate
Solution: An OR gate followed by a NOT gate gives $Y = \overline{A + B}$ — that is precisely the definition of a NOR gate.
Q30 — Physics · medium · theory
Interference pattern can be observed due to superposition of the following waves:
A. $y = a\sin\omega t$
B. $y = a\sin 2\omega t$
C. $y = a\sin(\omega t - \phi)$
D. $y = a\sin 3\omega t$
Choose the correct answer from the options given below.
A. B and C
B. B and D
C. A and C ✓ Correct
D. A and B
Solution: For an interference pattern the sources must be COHERENT — same frequency and a constant phase difference.
Only A and C share the same $\omega$ and differ by the fixed phase $\phi$. B and D have different frequencies ($2\omega$ and $3\omega$), so they cannot interfere steadily.