Physics NEET UG 2026 Practice Paper (Day 41)

Physics NEET UG 2026 Practice Paper Day 41: As you step into Day 41 of your practice session for NEET UG 2026, this practice paper serves as a focused opportunity to strengthen your conceptual understanding and problem-solving skills. Designed to simulate the real examination environment, today’s set will challenge your grasp of key topics and help you build the speed, accuracy, and confidence essential for success. Approach each question with patience and precision; every attempt brings you closer to mastering the Physics section. Best wishes for a productive session ahead.

PHYSICS

Q.1. Which, of the following, emits electromagnetic waves?

A. a dipole at rest, in a uniform electric field

B. a charge moving with constant velocity

C. a charged particle falling freely under gravity

D. a magnet moving with constant velocity

Q.2. Eddy currents can be minimized by:

A. Reducing the relative motion between the core and magnet in an electric motor

B. By making the core of thin laminations

C. By increasing the conductor cross-sectional area

D. Both (1) and (2) are correct

Q.3. A light ray falls on a glass surface of refractive index √3, at an angle of 60 degrees. The angle between the refracted and reflected rays would be:

A. 120 Degrees

B. 30 Degrees

C. 60 Degrees

D. 90 Degrees

Q.4. A block of mass $2\text{kg}$ slides down a smooth fixed wedge of height $h$ and then moves onto a horizontal rough surface with coefficient of kinetic friction ${\mu }_k=0.2$. If the block just comes to rest after moving a horizontal distance $5h$, what is the value of ${\mu }_k$ assumed in the calculation of energy loss, and which principle is primarily used to relate the motion on the wedge to that on the horizontal surface?

A. ${\mu }_k=0.2$, Work-Energy theorem

B. μk=0.2, Conservation of mechanical energy on the wedge and work-energy on the horizontal

C. ${\mu }_k=0.2$, Conservation of angular momentum about the wedge base

D. ${\mu }_k=0$, Conservation of linear momentum

Q.5. A satellite is orbiting the Earth in a circular orbit of radius $R$ with speed $v$. If its orbital radius is increased to $2R$ in a new circular orbit, what will be its new orbital speed (assuming $R$R is much larger than Earth’s radius)?

A. $v\mathrm{/}\sqrt{2}$

B. $v\sqrt{2}$

C. $v\mathrm{/}2$

D. $v\mathrm{/}\sqrt{2^3}$

Q.6. In a certain camera, a combination of four similar thin convex lenses is 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) p^4 and m^4

(B) 4p and 4m

(C) p^4 and 4m

(D) 4p and m^4

Q.7. A 2 amp current is flowing through two different small circular copper coils having radii ratio of 1:2. The ratio of their respective magnetic moments will be

(A) 4 : 1

(B) 1 : 4

(C) 1 : 2

(D) 2 : 1

Q.8. 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) 3F/8

(B) 3F/ 5

(C) 2F/3

(D) F/2

Q.9. Consider the diameter of a spherical object being measured with the help of a Vernier calliper. 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) 5.00 cm

(B) 5.18 cm

(C) 5.08 cm

(4) 4.98 cm

Q.10. 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) 250

(B) 100

(C) 125

(D) 150

Q.11. Two identical point masses P and Q, suspended from two separate massless springs of spring constants $k_1$​ and $k_2$​, respectively, oscillate vertically. If their maximum speeds are the same, the ratio $(A_Q / A_P)$of the amplitude $A_Q$ of mass Q to the amplitude $A_P$​ of mass P is

(1) $\sqrt{\dfrac{k_1}{k_2}}$

(2) $\dfrac{k_2}{k_1}$

(3) $\dfrac{k_1}{k_2}$​​

(4) $\sqrt{\dfrac{k_2}{k_1}}$

​​​Q.12. 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 between the plates and non-zero outside

(B) Zero at all places

(C) Constant between the plates and zero outside the plates

(D) Non-zero everywhere with maximum at the imaginary cylindrical surface connecting peripheries of the plates

Q.13. The Sun rotates around its centre 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) 108 days

(B) 100 days

(C) 105 days

(D) 115 days

Q.14. A uniform rod of mass 20 kg and length 5 m leans against a smooth vertical wall making an angle of 60° 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/s2)

(1) 200 root 3 N

(2) 100 N

(3) 100 root 3 N

(4) 200 N

Q.15. Two identical blocks of mass $m$m each are attached to two springs having spring constants $k_1$​ and $k_2$​ respectively. The blocks are connected by a massless string and the system is made to oscillate vertically. If the maximum speed of the block attached to spring constant $k_1$​ is $v$, then the maximum speed of the block attached to spring constant $k_2$will be:

(1) $v \sqrt{\dfrac{k_1}{k_2}}$

(2) $v \sqrt{\dfrac{k_2}{k_1}}$

(3) $v$

(4) $2v$

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ANSWERS & EXPLANATIONS

Ans. 1. C. Electromagnetic waves are produced only by accelerating charges (or time-varying currents). A charge at rest or moving with constant velocity does not radiate electromagnetic waves.

Ans. 2. B. By making the core of thin laminations. Eddy currents are looping induced currents that form in the conducting core (usually iron) of motors, generators, transformers, etc., when the magnetic flux through the core changes. These currents cause unwanted heating (I²R loss) and reduce efficiency.

Ans. 3. D. 90 Degrees

Given:

• Refractive index of glass, μ = √3
• Angle of incidence, i = 60°

Step 1: Apply Snell’s Law

Air to glass: μ = sin i / sin r

√3 = sin 60° / sin r √3 = (√3 / 2) / sin r sin r = (√3 / 2) ÷ √3 = 1/2 r = 30° (angle of refraction)

Step 2: Angle of reflection

By law of reflection: Angle of reflection = i = 60°

Step 3: Angle between reflected ray and refracted ray

In the geometry at the point of incidence:

• The reflected ray makes 60° with the normal.
• The refracted ray makes 30° with the normal (on the other side).

The angle between the reflected ray and the refracted ray is: 60° + 30° = 90°

Ans. 4. B. μk=0.2, Conservation of mechanical energy on the wedge and work–energy on the horizontal. On the smooth wedge, potential energy $mgh$ is entirely converted into kinetic energy at the bottom (no friction), giving the block a speed $v=\sqrt{2gh}$.

Ans.5. D. $v\mathrm{/}\sqrt{2^3}$, which reflects the dependence of orbital speed on the radius when considering the complete orbital mechanics involved in moving the satellite to a new circular orbit at $2R$2R.

Ans. 6. D. Sol. For series combination of lens
peff = p1 + p2 + p3 + p4 = 4p
meff = m1 × m2 × m3 × m4 = m^4

Ans .7. B. Magnetic moment of current-carrying circular loop = IA M = IA M ∝ A, π
M1/M2 = A1/A2= πr1^2/πr2^2 = 1/2^2= 1/4, 1:4

Ans.8. A. Initially: Charge A = q, Charge B = q, separated by distance r F = K q q / r² Then: Charge A = q/2, Charge B = 3q/4, same distance r F’ = K (q/2) (3q/4) / r²F’ = (3 K q²) / (8 r²)F’ = 3F / 8

Ans.9. D. Least count = 1MSD – 1VSD = 1 MSD – 9/10 MSD = 1/10 MSD, 1/10* 0.1 cm = 0.01
Zero error = +0.1 cm
Main scale reading = 5 cm
Vernier scale reading = 8 × 0.01 = 0.08 cm
Final measurement of diameter = 5 + 0.08 – 0.1 = 4.98 cm

Ans.10. C. $m = \dfrac{L}{f_0} \times \dfrac{D}{f_e}$
$$m = \dfrac{40}{2} \times \dfrac{25}{4} = 125$$

Ans. 11. A. $\sqrt{\dfrac{k_1}{k_2}}$ Maximum velocity $V = A \omega$

Given $v_P = v_Q$

So, $A_P \omega_P = A_Q \omega_Q$

AP​AQ​​=ωQ​ωP​​

Aa/Ap = ωp/ωa

$\sqrt{\dfrac{k_1}{k_2}}$

Ans.12. D. Let the surface charge density be σ = q/A,

Given dσ/dt = constant
∴ d/dt* q/A = constant ⇒ I/A*i = constant
It means displacement current is constant.
This system will act like a cylindrical wire.

Ans. 13. (A) 108 days, use I= 2/5 mR^2,

Ans.14. C. 100 root 3 N.

Ans .15. C. $v$. When two blocks are connected by a string and oscillating vertically, they must move together with the same displacement and same velocity at every instant (because the string is inextensible and massless). Therefore, $v_1 = v_2$​ (velocity of both blocks is always equal). Hence, maximum speed of both blocks is also the same. So, maximum speed of the block attached to $k_2$.

And that’s a wrap on Day 41! You just survived another intense Physics practice session. Congratulations! Now comes the real magic: check your answers, celebrate the questions you nailed, learn from the ones that tricked you, and carry those lessons forward. Remember, every correct tick brings you one step closer to that dream All India Rank. Rest well, analyze smartly, and come back stronger tomorrow.

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