NEET Physics Previous Year Questions (2024–2025)
NEET Physics PYQs 2025: 95 Solved Questions - Online Practice — Practice Questions
1. (Orig Q1) Water tank has one wall at x=L. Liquid of surface tension S, density rho makes angle theta0 (<<1) with x-axis at x=L. y(x)=height of surface. Equation for y(x) is: (take theta(x)=sin theta(x)=tan theta(x)=dy/dx) Figure of water tank with curved surface and angle theta0 at x=L]
- (A) d2y/dx2 = (rho g/S) x
- (B) d2y/dx2 = (rho g/S) y
- (C) d2y/dx2 = sqrt(rho g/S)
- (D) dy/dx = sqrt(rho g/S) x
Answer: (B) d2y/dx2 = (rho g/S) y
Force balance on liquid column: S(d2y/dx2)=rho g y, from d(theta)/dx=d2y/dx2 and Fy=weight of liquid lifted.
2. (Orig Q2) Microscope: objective focal length 2 cm, eyepiece focal length 4 cm, tube length 40 cm, distance of distinct vision 25 cm. Magnification of microscope is:
- (A) 100
- (B) 125
- (C) 150
- (D) 250
Answer: (B) 125
m = (L/fo)(D/fe) = (40/2)(25/4) = 20*6.25 = 125
3. (Orig Q3) Electron (mass 9e-31 kg, charge 1.6e-19 C) moving at c/100 enters magnetic field B=9e-4 T perpendicular to velocity. To keep it undeflected, apply uniform E with B. Then (c=3e8 m/s):
- (A) E perp to B, magnitude 27x10^4 V/m
- (B) E perp to B, magnitude 27x10^2 V/m
- (C) E parallel to B, magnitude 27x10^2 V/m
- (D) E parallel to B, magnitude 27x10^4 V/m
Answer: (B) E perp to B, magnitude 27x10^2 V/m
For zero net force E must be perpendicular to B; E=vB=(3e6)(9e-4)=27x10^2 V/m.
4. (Orig Q4) Two inclined surfaces of equal length L, angle 45 deg, one rough one smooth. A body takes 2x time to slide on rough vs smooth surface. Coefficient of kinetic friction mu_k is close to:
- (A) 0.25
- (B) 0.40
- (C) 0.5
- (D) 0.75
Answer: (D) 0.75
a_s=g sin45; a_r=g sin45-mu g cos45=a_s/4 (since t doubles, a quarters). Solving gives mu_k=0.75.
5. (Orig Q5) KE of cars A and B are 100 J and 225 J. Car A stops after 1000 m, car B after 1500 m. Ratio FA/FB of braking forces is:
- (A) 3/2
- (B) 2/3
- (C) 1/3
- (D) 1/2
Answer: (B) 2/3
F=KE/s. FA=100/1000=0.10 N, FB=225/1500=0.15 N, ratio=2/3.
6. Current passing through the battery in the given Wheatstone-bridge-like circuit is:
- (A) 2.0 A
- (B) 0.5 A
- (C) 2.5 A
- (D) 1.5 A
Answer: (B) 0.5 A
Bridge is balanced so the 6 ohm resistor carries no current and is removed; Resistance 5 Ω and 3 Ω are in series and 2 . 5 Ω and 1 . 5 Ω are in series. Then both the series combinations are in parallel with each other. Thus equivalent resistance is 8/3 Now, the given circuit can be redrawn as R = 8/3 +1/3 +1.5 + 5.5 = 10 Ω total resistance =10 ohm; V=iR gives i=0.5 A.
7. Bob of mass m on string length l given horizontal velocity v0. String goes slack at point P at angle theta from horizontal. Ratio v/v0 at P is: Circular pendulum diagram showing point P at angle theta and velocity v0 at bottom
- (A) (sin theta)^(1/2)
- (B) (1/(2+3 sin theta))^(1/2)
- (C) (cos theta/(2+3 sin theta))^(1/2)
- (D) (sin theta/(2+3 sin theta))^(1/2)
Answer: (D) (sin theta/(2+3 sin theta))^(1/2)
At P: mg sin theta = mv^2/l so v^2=gl sin theta. Energy conservation gives v0^2=3gl sin theta+2gl, so v/v0=sqrt(sin theta/(2+3 sin theta)).
8. Output Y of given logic gate implementation (two NOR/OR/NAND combination with inputs A,B,A) is similar to output of a/an ____ gate.
- (A) AND
- (B) NAND
- (C) OR
- (D) NOR
Answer: (D) NOR
Y = (A + B)' × (A·BY)' = (A'·B')(A' + (BY)') = A'·B'·A' + B'·A'·(BY)' = A'·B' = (A + B)'. The correct option is D. NOR
9. Electric field of a plane EM wave: Ez=60 cos(5x+1.5x10^9 t) V/m. Expression for corresponding magnetic field is:
- (A) By=2x10^-7 cos(5x+1.5x10^9 t) T
- (B) Bx=2x10^-7 cos(5x+1.5x10^9 t) T
- (C) Bz=60 cos(5x+1.5x10^9 t) T
- (D) By=60 sin(5x+1.5x10^9 t) T
Answer: (A) By=2x10^-7 cos(5x+1.5x10^9 t) T
B0=E0/c=60/3e8=2e-7 T; propagation along -x, so B is along y-axis: By=2x10^-7 cos(5x+1.5x10^9 t) T.
10. Ball of mass 0.5 kg dropped from 40 m, rebounds to 10 m. Impulse imparted during collision with ground is: (g=9.8 m/s2)
- (A) 21 Ns
- (B) 7 Ns
- (D) 84 Ns
Answer: (A) 21 Ns
vi=sqrt(2g*40)=28 m/s down, vf=sqrt(2g*10)=14 m/s up; J=m(vf+vi)=0.5(14+28)=21 Ns.
11. AB is part of a circuit with 1H inductor, 5V battery, 2 ohm resistor. Potential difference VA-VB at i=2A increasing at 1 A/s is:
- (A) 5 volt
- (B) 6 volt
- (C) 9 volt
- (D) 10 volt
Answer: (D) 10 volt
VAB=L(di/dt)+5V+IR = 1(1)+5+(2)(2)=10 V.
12. 2 amp current flows through two small circular copper coils with radii ratio 1:2. Ratio of their magnetic moments is:
- (A) 1:4
- (B) 1:2
- (C) 2:1
- (D) 4:1
Answer: (A) 1:4
Magnetic moment m=NIA is proportional to r^2, so m1/m2=(r1/r2)^2=(1/2)^2=1/4.
13. Camera has four identical thin convex lenses in axial contact. Power and total magnification of combination vs power p and magnification m of each lens are, respectively:
- (A) 4p and 4m
- (B) p^4 and 4m
- (C) 4p and m^4
- (D) p^4 and m^4
Answer: (C) 4p and m^4
Powers of thin lenses in contact add: P=4p. Magnifications multiply: M=m^4.
14. Oxygen cylinder of 30 L has 18.20 mol O2. After withdrawal, gauge pressure drops to 11 atm at 27C. Mass of oxygen withdrawn is nearly (R=100/12 J/mol.K, M(O2)=32, 1 atm=1.01x10^5 N/m2):
- (A) 0.125 kg
- (B) 0.144 kg
- (C) 0.116 kg
- (D) 0.156 kg
Answer: (C) 0.116 kg
n_final = PV/RT = 12(1.01e5)(0.03)/((100/12)(300)) approx 14.55 mol; withdrawn = 18.20-14.55=3.65 mol; mass=3.65*32g approx 0.116 kg.
15. Time t and position x of a particle: t = x^2 + x. Acceleration of the particle is:
- (A) -2/(x+2)^3
- (B) -2/(2x+1)^3
- (C) +2/(x+1)^3
- (D) +2/(2x+1)
Answer: (B) -2/(2x+1)^3
v=dx/dt=1/(2x+1); a=dv/dx * v = -2/(2x+1)^3.