Physics — NEET Complete Yearwise Papers MCQs with Solutions
Free NEET Complete Yearwise Papers Physics MCQs with step-by-step solutions (45 questions). Part of NEET 2025 Paper. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Physics · medium · numerical
Consider a water tank shown in the figure. It has one wall at $x = L$ and can be taken to be very wide in the $z$ direction. When filled with a liquid of surface tension $S$ and density $\rho$, the liquid surface makes angle $\theta_0\,(\theta_0 \ll 1)$ with the $x$-axis at $x = L$. If $y(x)$ is the height of the surface then the equation for $y(x)$ is: (take $\theta(x) = \sin\theta(x) = \tan\theta(x) = \dfrac{dy}{dx}$, $g$ is the acceleration due to gravity)
A. $\dfrac{d^{2}y}{dx^{2}} = \dfrac{\rho g}{S}x$
B. $\dfrac{d^{2}y}{dx^{2}} = \dfrac{\rho g}{S}y$ ✓ Correct
C. $\dfrac{d^{2}y}{dx^{2}} = \sqrt{\dfrac{\rho g}{S}}$
D. $\dfrac{dy}{dx} = \sqrt{\dfrac{\rho g}{S}}\,x$
Solution: Balancing the weight of a thin element against the difference in surface-tension force gives $S\,\dfrac{d\theta}{dx} = \rho g y$. Since $\theta = \dfrac{dy}{dx}$ for small angles, differentiating gives $\dfrac{d^{2}y}{dx^{2}} = \dfrac{\rho g}{S}y$.
Q2 — Physics · medium · numerical
A microscope has an objective of focal length 2 cm, eyepiece of focal length 4 cm and the tube length of 40 cm. If the distance of distinct vision of eye is 25 cm, the magnification in the microscope is:
A. 100
B. 125 ✓ Correct
C. 150
D. 250
Solution: In normal adjustment $m = \dfrac{L}{f_0} \times \dfrac{D}{f_e} = \dfrac{40}{2} \times \dfrac{25}{4} = 20 \times 6.25 = 125$.
Q3 — Physics · medium · numerical
An electron (mass $9 \times 10^{-31}\ kg$ and charge $1.6 \times 10^{-19}\ C$) moving with speed $c/100$ ($c$ = speed of light) is injected into a magnetic field $\vec{B}$ of magnitude $9 \times 10^{-4}\ T$ perpendicular to its direction of motion. We wish to apply an uniform electric field $\vec{E}$ together with the magnetic field so that the electron does not deflect from its path. Then (speed of light $c = 3 \times 10^{8}\ ms^{-1}$)
A. $\vec{E}$ is perpendicular to $\vec{B}$ and its magnitude is $27 \times 10^{4}\ V\,m^{-1}$
B. $\vec{E}$ is perpendicular to $\vec{B}$ and its magnitude is $27 \times 10^{2}\ V\,m^{-1}$ ✓ Correct
C. $\vec{E}$ is parallel to $\vec{B}$ and its magnitude is $27 \times 10^{2}\ V\,m^{-1}$
D. $\vec{E}$ is parallel to $\vec{B}$ and its magnitude is $27 \times 10^{4}\ V\,m^{-1}$
Solution: For no deflection the electric and magnetic forces must cancel, so $\vec{E}$ must be perpendicular to $\vec{B}$ with $E = vB = \dfrac{3 \times 10^{8}}{100} \times 9 \times 10^{-4} = 27 \times 10^{2}\ V\,m^{-1}$.
Q4 — Physics · medium · numerical
There are two inclined surface of equal length (L) and same angle of inclination $45^{\circ}$ with the horizontal. One of them is rough and the other is perfectly smooth. A given body takes 2 times as much time to slide down on rough surface than on the smooth surface. The coefficient of kinetic friction $(\mu_k)$ between the object and the rough surface is close to:
A. 0.25
B. 0.40
C. 0.5
D. 0.75 ✓ Correct
Solution: With $L = \tfrac{1}{2}at^{2}$ the same distance gives $\dfrac{a_{smooth}}{a_{rough}} = \left(\dfrac{t_{rough}}{t_{smooth}}\right)^{2} = 4$. On a $45^{\circ}$ incline $a_{smooth} = g\sin 45^{\circ}$ and $a_{rough} = g\sin 45^{\circ} - \mu_k g\cos 45^{\circ}$, so $1 - \mu_k = \tfrac{1}{4}$, giving $\mu_k = 0.75$.
Q5 — Physics · medium · numerical
The kinetic energies of two similar cars A and B are 100 J and 225 J respectively. On applying breaks, car A stops after 1000 m and car B stops after 1500 m. If $F_A$ and $F_B$ are the forces applied by the breaks on cars A and B, respectively, then the ratio $F_A/F_B$ is:
A. $\dfrac{3}{2}$
B. $\dfrac{2}{3}$ ✓ Correct
C. $\dfrac{1}{3}$
D. $\dfrac{1}{2}$
Solution: The braking force does work equal to the kinetic energy, so $F = KE/d$. Then $F_A = 100/1000 = 0.1\ N$ and $F_B = 225/1500 = 0.15\ N$, giving $F_A/F_B = 2/3$.
Q6 — Physics · medium · numerical
The current passing through the battery in the given circuit, is:
A. 2.0 A
B. 0.5 A ✓ Correct
C. 2.5 A
D. 1.5 A
Solution: Reducing the network gives an equivalent resistance of $10\ \Omega$ across the 5 V cell, so the battery current is $i = 5/10 = 0.5\ A$.
Q7 — Physics · medium · numerical
A bob of heavy mass $m$ is suspended by a light string of length $l$. The bob is given a horizontal velocity $v_0$ as shown in figure. If the string gets slack at some point P making an angle $\theta$ from the horizontal, the ratio of the speed $v$ of the bob at point P to its initial speed $v_0$ is:
A. $(\sin\theta)^{\frac{1}{2}}$
B. $\left(\dfrac{1}{2 + 3\sin\theta}\right)^{\frac{1}{2}}$
C. $\left(\dfrac{\cos\theta}{2 + 3\sin\theta}\right)^{\frac{1}{2}}$
D. $\left(\dfrac{\sin\theta}{2 + 3\sin\theta}\right)^{\frac{1}{2}}$ ✓ Correct
Solution: The string goes slack when the tension vanishes, so $mg\sin\theta = \dfrac{mv_P^{2}}{l}$, i.e. $v_P^{2} = gl\sin\theta$. Energy conservation gives $\tfrac{1}{2}v_0^{2} = gl(1 + \sin\theta) + \tfrac{1}{2}gl\sin\theta$, so $\dfrac{v_P}{v_0} = \left(\dfrac{\sin\theta}{2 + 3\sin\theta}\right)^{1/2}$.
Q8 — Physics · medium · theory
The output (Y) of the given logic implementation is similar to the output of an/a ____ gate.
A. AND
B. NAND
C. OR
D. NOR ✓ Correct
Solution: Working the combination through gives $Y = \overline{A + B}$, which is the NOR operation.
Q9 — Physics · medium · numerical
The electric field in a plane electromagnetic wave is given by $E_z = 60\cos(5x + 1.5 \times 10^{9}t)\ V/m$. Then expression for the corresponding magnetic field is (here subscripts denote the direction of the field):
A. $B_y = 2 \times 10^{-7}\cos(5x + 1.5 \times 10^{9}t)\ T$ ✓ Correct
B. $B_x = 2 \times 10^{-7}\cos(5x + 1.5 \times 10^{9}t)\ T$
C. $B_z = 60 \cdot \cos(5x + 1.5 \times 10^{9}t)\ T$
D. $B_y = 60\sin(5x + 1.5 \times 10^{9}t)\ T$
Solution: The magnetic field is in phase with the electric field, perpendicular to both $\vec{E}$ and the direction of travel, with $B_0 = E_0/c = 60/(3 \times 10^{8}) = 2 \times 10^{-7}\ T$. With $\vec{E}$ along $z$ and propagation along $-x$, $\vec{B}$ lies along $y$.
Q10 — Physics · medium · numerical
A ball of mass 0.5 kg is dropped from a height of 40 m. The ball hits the ground and rises to a height of 10 m. The impulse imparted to the ball during its collision with the ground is (Take $g = 9.8\ m/s^{2}$).
A. 21 NS ✓ Correct
B. 7 NS
C. 0
D. 84 NS
Solution: Striking speed $= \sqrt{2 \times 9.8 \times 40} = 28\ m/s$ and rebound speed $= \sqrt{2 \times 9.8 \times 10} = 14\ m/s$. Impulse $= m(v_1 + v_2) = 0.5 \times 42 = 21\ N\,s$.
Q11 — Physics · medium · numerical
AB is a part of an electrical circuit (see figure). The potential difference "$V_A - V_B$", at the instant when current $i = 2\ A$ and is increasing at a rate of 1 amp/second is:
A. 5 volt
B. 6 volt
C. 9 volt
D. 10 volt ✓ Correct
Solution: Walking from A to B: the inductor drops $L\dfrac{di}{dt} = 1 \times 1 = 1\ V$, the cell adds 5 V and the resistor drops $iR = 2 \times 2 = 4\ V$. So $V_A - V_B = 1 + 5 + 4 = 10\ V$.
Q12 — Physics · medium · numerical
A 2 amp current is flowing through two different small circular copper coils having radii ratio $1 : 2$. The ratio of their respective magnetic moments will be
A. $1 : 4$ ✓ Correct
B. $1 : 2$
C. $2 : 1$
D. $4 : 1$
Solution: Magnetic moment $M = iA = i\pi r^{2} \propto r^{2}$ for the same current, so the ratio is $1^{2} : 2^{2} = 1 : 4$.
Q13 — Physics · medium · theory
In a certain camera, a combination of four similar thin convex lenses are arranged axially in contact. Then the power of the combination and the total magnification in comparison to the power $(p)$ and magnification $(m)$ for each lens will be, respectively
A. $4p$ and $4m$
B. $p^{4}$ and $4m$
C. $4p$ and $m^{4}$ ✓ Correct
D. $p^{4}$ and $m^{4}$
Solution: Powers of lenses in contact add, giving $4p$, while magnifications multiply, giving $m^{4}$.
Q14 — Physics · medium · numerical
An oxygen cylinder of volume 30 liter has 18.20 moles of oxygen. After some oxygen is withdrawn from the cylinder, its gauge pressure drops to 11 atmospheric pressure at temperature $27^{\circ}C$. The mass of the oxygen withdrawn from the cylinder is nearly equal to: [Given, $R = \dfrac{100}{12}\ J\ mol^{-1}\ K^{-1}$, and molecular mass of $O_2 = 32$, 1 atm pressure $= 1.01 \times 10^{5}\ N/m^{2}$]
A. 0.125 kg
B. 0.144 kg
C. 0.116 kg ✓ Correct
D. 0.156 kg
Solution: A gauge pressure of 11 atm means an absolute pressure of 12 atm. Then $n_f = \dfrac{PV}{RT} = \dfrac{12 \times 1.01 \times 10^{5} \times 30 \times 10^{-3}}{(100/12) \times 300} = 14.54$ mol, so $18.2 - 14.54 = 3.66$ mol were removed — a mass of $3.66 \times 32\ g \approx 0.116\ kg$.
Q15 — Physics · medium · numerical
In some appropriate units, time $(t)$ and position $(x)$ relation of a moving particle is given by $t = x^{2} + x$. The acceleration of the particle is
A. $-\dfrac{2}{(x+2)^{3}}$
B. $-\dfrac{2}{(2x+1)^{3}}$ ✓ Correct
C. $+\dfrac{2}{(x+1)^{3}}$
D. $+\dfrac{2}{2x+1}$
Solution: Differentiating, $\dfrac{dt}{dx} = 2x + 1$, so $v = \dfrac{1}{2x+1}$. Then $a = v\dfrac{dv}{dx} = \dfrac{1}{2x+1} \times \dfrac{-2}{(2x+1)^{2}} = -\dfrac{2}{(2x+1)^{3}}$.
Q16 — Physics · medium · numerical
To an ac power supply of 220 V at 50 Hz, a resistor of $20\ \Omega$, a capacitor of reactance $25\ \Omega$ and an inductor of reactance $45\ \Omega$ are connected is series. The corresponding current in the circuit and the phase angle between the current and the voltage is, respectively-
A. 7.8 A and $30^{\circ}$
B. 7.8 A and $45^{\circ}$ ✓ Correct
C. 15.6 and $30^{\circ}$
D. 15.6 and $45^{\circ}$
Solution: Impedance $Z = \sqrt{R^{2} + (X_L - X_C)^{2}} = \sqrt{20^{2} + 20^{2}} = 20\sqrt{2} \approx 28.3\ \Omega$, so $i = 220/28.3 \approx 7.8\ A$. Since $\tan\phi = (X_L - X_C)/R = 1$, the phase angle is $45^{\circ}$.
Q17 — Physics · medium · numerical
The Sun rotates around its center once in 27 days. What will be the period of revolution if the Sun were to expand to twice its present radius without any external influence? Assume the Sun to be a sphere of uniform density.
A. 100 days
B. 105 days
C. 115 days
D. 108 days ✓ Correct
Solution: Angular momentum is conserved, and $I \propto R^{2}$, so doubling the radius makes $I$ four times larger and $\omega$ four times smaller. The period therefore becomes $4 \times 27 = 108$ days.
Q18 — Physics · medium · numerical
A model for quantized motion of an electron in a uniform magnetic field B states that the flux passing through the orbit of the electron is $n(h/e)$ where $n$ is an integer, $h$ is Planck's constant and $e$ is the magnitude of electron's charge. According to the model, the magnetic moment of an electron in its lowest energy state will be ($m$ is the mass of the electron)
A. $\dfrac{he}{\pi m}$
B. $\dfrac{he}{2\pi m}$ ✓ Correct
C. $\dfrac{heB}{\pi m}$
D. $\dfrac{heB}{2\pi m}$
Solution: With $M = iA = \dfrac{ev}{2\pi r}\pi r^{2}$, the quantization $B\pi r^{2} = h/e$ (for $n = 1$) gives $r^{2} = \dfrac{h}{\pi eB}$, and the circular-motion condition gives $v = \dfrac{eBr}{m}$. Substituting yields $M = \dfrac{he}{2\pi m}$.
Q19 — Physics · medium · numerical
Three identical heat conducting rods are connected in series as shown in the figure. The rods on the sides have thermal conductivity 2K while that in the middle has thermal conductivity K. The left end of the combination is maintained at temperature 3T and the right end at T. The rods are thermally insulated from outside. In steady state, temperature at the left junction is $T_1$ and that at the right junction is $T_2$. The ratio $T_1/T_2$ is
A. $\dfrac{3}{2}$
B. $\dfrac{4}{3}$
C. $\dfrac{5}{3}$ ✓ Correct
D. $\dfrac{5}{4}$
Solution: In the steady state the same heat current flows through all three rods: $2K(3T - T_1) = K(T_1 - T_2) = 2K(T_2 - T)$. Solving the two equations gives $\dfrac{T_1}{T_2} = \dfrac{5}{3}$.
Q20 — Physics · medium · numerical
The plates of a parallel plate capacitor are separated by $d$. Two slabs of different dielectric constant $K_1$ and $K_2$ with thickness $\dfrac{3}{8}d$ and $\dfrac{d}{2}$, respectively are inserted in the capacitor. Due to this, the capacitance becomes two times larger than when there is nothing between the plates. (If $K_1 = 1.25K_2$, the value of $K_1$ is:
A. 2.66 ✓ Correct
B. 2.33
C. 1.60
D. 1.33
Solution: The empty gap left is $d - \tfrac{3d}{8} - \tfrac{d}{2} = \tfrac{d}{8}$. Doubling the capacitance requires $\tfrac{1}{8} + \dfrac{3}{8K_1} + \dfrac{1}{2K_2} = \tfrac{1}{2}$. Putting $K_2 = K_1/1.25$ gives $\dfrac{1}{K_1} = \tfrac{3}{8}$, so $K_1 \approx 2.66$.
Q21 — Physics · medium · numerical
Two cities X and Y are connected by a regular bus service with a bus leaving in either direction every T min. A girl is driving scooty with a speed of 60 km/h in the direction X to Y notices that a bus goes past her every 30 minutes in the direction of her motion, and every 10 minutes in the opposite direction. Choose the correct option for the period T of the bus service and the speed (assumed constant) of the buses.
A. 9 min, 40 km/h
B. 25 min, 100 km/h
C. 10 min, 90 km/h
D. 15 min, 120 km/h ✓ Correct
Solution: If the buses run at $v$ with spacing $vT$, then $\dfrac{vT}{v-60} = 30$ and $\dfrac{vT}{v+60} = 10$. Dividing gives $v + 60 = 3(v - 60)$, so $v = 120$ km/h, and then $T = 15$ min.
Q22 — Physics · medium · numerical
A uniform rod of mass 20 kg and length 5 m leans against a smooth vertical wall making an angle of $60^{\circ}$ with it. The other end rests on a rough horizontal floor. The friction force that the floor exerts on the rod is (take $g = 10\ m/s^{2}$)
A. 100 N
B. $100\sqrt{3}$ N ✓ Correct
C. 200 N
D. $200\sqrt{3}$ N
Solution: The rod makes $30^{\circ}$ with the floor. Taking torques about the foot of the rod, $N_{wall} \times L\sin 30^{\circ} = mg \times \tfrac{L}{2}\cos 30^{\circ}$, giving $N_{wall} = 100\sqrt{3}\ N$. Since the wall is smooth, the floor friction must balance this: $f = 100\sqrt{3}\ N$.
Q23 — Physics · medium · theory
In an oscillating spring mass system, a spring is connected to a box filled with sand. As the box oscillates, sand leaks slowly out of the box vertically so that the average frequency $\omega(t)$ and average amplitude $A(t)$ of the system change with time $t$. Which one of the following options schematically depicts these changes correctly?
A.
B. ✓ Correct
C.
D.
Solution: As sand leaks out the mass falls, so $\omega = \sqrt{k/m}$ rises. The escaping sand carries energy away, so the total energy $E = \tfrac{1}{2}kA^{2}$ falls and the amplitude decreases with it. The correct graph shows $\omega$ rising and $A$ falling.
Q24 — Physics · medium · numerical
A balloon is made of a material of surface tension S and its inflation outlet (from where gas is filled in it) has small area A. It is filled with a gas of density $\rho$ and takes a spherical shape of radius R. When the gas is allowed to flow freely out of it, its radius changes from R to 0 (zero) in time T. If the speed $v(r)$ of gas coming out of the balloon depends on $r$ as $r^{a}$ and $T \propto S^{\alpha}A^{\beta}\rho^{\gamma}R^{\delta}$ then
A. $a = \dfrac{1}{2}, \alpha = \dfrac{1}{2}, \beta = -1, \gamma = +1, \delta = \dfrac{3}{2}$
B. $a = -\dfrac{1}{2}, \alpha = -\dfrac{1}{2}, \beta = -1, \gamma = -\dfrac{1}{2}, \delta = \dfrac{5}{2}$
C. $a = -\dfrac{1}{2}, \alpha = -\dfrac{1}{2}, \beta = -1, \gamma = \dfrac{1}{2}, \delta = \dfrac{7}{2}$ ✓ Correct
D. $a = \dfrac{1}{2}, \alpha = \dfrac{1}{2}, \beta = -\dfrac{1}{2}, \gamma = \dfrac{1}{2}, \delta = \dfrac{7}{2}$
Solution: Dimensional analysis on $T \propto S^{\alpha}A^{\beta}\rho^{\gamma}R^{\delta}$ with $[S] = MT^{-2}$, $[A] = L^{2}$, $[\rho] = ML^{-3}$: matching $T^{1}$ gives $\alpha = -\tfrac{1}{2}$, matching $M^{0}$ gives $\gamma = +\tfrac{1}{2}$, and matching $L^{0}$ then gives $\beta = -1$, $\delta = \tfrac{7}{2}$; the outflow speed varies as $r^{-1/2}$.
Q25 — Physics · medium · numerical
Consider the diameter of a spherical object being measured with the help of a Vernier callipers. Suppose its 10 Vernier Scale Divisions (V.S.D.) are equal to its 9 Main Scale Divisions (M.S.D.). The least division in the M.S. is 0.1 cm and the zero of V.S. is at $x = 0.1$ cm when the jaws of Vernier callipers are closed. If the main scale reading for the diameter is $M = 5$ cm and the number of coinciding vernier division is 8, the measured diameter after zero error correction, is
A. 5.18 cm
B. 5.08 cm
C. 4.98 cm ✓ Correct
D. 5.00 cm
Solution: Least count $= 1\ MSD - 1\ VSD = 0.01$ cm. The reading is $5 + 8 \times 0.01 = 5.08$ cm, and the zero error of $+0.1$ cm must be subtracted: $5.08 - 0.1 = 4.98$ cm.
Q26 — Physics · medium · theory
A parallel plate capacitor made of circular plates is being charged such that the surface charge density on its plates is increasing at a constant rate with time. The magnetic field arising due to displacement current is:
A. zero at all places
B. constant between the plates and zero outside the plates
C. non-zero everywhere with maximum at the imaginary cylindrical surface connecting peripheries of the plates ✓ Correct
D. zero between the plates and non-zero outside
Solution: The steadily growing field between the plates is a constant displacement current $I_d = A\,\dfrac{d\sigma}{dt}$. Its magnetic field grows with distance from the axis and peaks at the plate edge — the imaginary cylinder joining the plate peripheries — then falls off outside.
Q27 — Physics · medium · numerical
An unpolarized light beam travelling in air is incident on a medium of refractive index 1.73 at Brewster's angle. Then-
A. reflected light is completely polarized and the angle of reflection is close to $60^{\circ}$ ✓ Correct
B. reflected light is partially polarized and the angle of reflection is close to $30^{\circ}$
C. both reflected and transmitted light are perfectly polarized with angles of reflection and refraction close to $60^{\circ}$ and $30^{\circ}$, respectively
D. transmitted light is completely polarized with angle of refraction close to $30^{\circ}$
Solution: At Brewster's angle $\tan\theta_B = \mu = 1.73 = \sqrt{3}$, so $\theta_B = 60^{\circ}$. The reflected beam is completely polarized (the transmitted beam only partially), and the angle of reflection equals the angle of incidence, $60^{\circ}$.
Q28 — Physics · medium · numerical
Two identical charged conducting spheres A and B have their centres separated by a certain distance. Charge on each sphere is q and the force of repulsion between them is F. A third identical uncharged conducting sphere is brought in contact with sphere A first and then with B and finally removed from both. New force of repulsion between spheres A and B (Radii of A and B are negligible compared to the distance of separation so that for calculating force between them they can be considered as point charges) is best given as:
A. $\dfrac{3F}{5}$
B. $\dfrac{2F}{3}$
C. $\dfrac{F}{2}$
D. $\dfrac{3F}{8}$ ✓ Correct
Solution: Touching A leaves $q/2$ on each; touching B then shares $(q/2 + q)/2 = 3q/4$. The new force is $\propto (q/2)(3q/4) = \tfrac{3}{8}q^{2}$, i.e. $\dfrac{3F}{8}$.
Q29 — Physics · medium · numerical
A container has two chambers of volumes $V_1 = 2$ litres and $V_2 = 3$ liters separated by a partition made of a thermal insulator. The chambers contain $n_1 = 5$ and $n_2 = 4$ moles of ideal gas at pressures $p_1 = 1$ atm and $p_2 = 2$ atm, respectively. When the partition is removed, the mixture attains an equilibrium pressure of:
A. 1.3 atm
B. 1.6 atm ✓ Correct
C. 1.4 atm
D. 1.8 atm
Solution: With the total energy conserved, $P = \dfrac{P_1V_1 + P_2V_2}{V_1 + V_2} = \dfrac{1 \times 2 + 2 \times 3}{5} = 1.6$ atm.
Q30 — Physics · medium · numerical
A particle of mass m is moving around the origin with a constant force F pulling it towards the origin. If Bohr model is used to describe its motion, the radius r of the n$^{th}$ orbit and the particle's speed v in the orbit depend on n as
A. $r \propto n^{1/3};\ v \propto n^{1/3}$
B. $r \propto n^{1/3};\ v \propto n^{2/3}$
C. $r \propto n^{2/3};\ v \propto n^{1/3}$ ✓ Correct
D. $r \propto n^{4/3};\ v \propto n^{-1/3}$
Solution: A constant force gives $\dfrac{mv^{2}}{r} = F$, so $v^{2} \propto r$. Combining with the Bohr condition $mvr = n\hbar$ gives $r^{3/2} \propto n$, i.e. $r \propto n^{2/3}$ and $v \propto n^{1/3}$.