Top 50 Physics Numericals in NEET That Guarantee 120+ Marks

Top 50 Physics Numericals in NEET That Guarantee 120+ Marks PDF

The following list of 50 Physics numericals can be helpful to aspirants preparing for NEET 2027. The numericals include questions from significant topics of Mechanics, Thermodynamics, Electrostatics, Current Electricity, Magnetism, Optics & Modern Physics.

Students must first try the important NEET questions without referring to their solutions. After checking the answer, students need to revise the formula/concept applied in that question. It will prove more beneficial than just memorising the right choice.

NEET Physics Most Asked Numericals To Score 120+ Marks

Ques: 1

An AC voltage $V = 220\sin(2\times10^3t)$ Volt is applied to a series LCR circuit. Then the current amplitude in this circuit is:

Given: $L = 10,\text{mH}$, $C = 25,\mu\text{F}$, $R = 100,\Omega$

Option 1) $5.5,\text{A}$

Option 2) $11.0,\text{A}$

Option 3) $22.0,\text{A}$

Option 4) $2.2,\text{A}$

Chapter: Alternating Current

Correct Answer: Option 4

Explanation

The angular frequency is:

$\omega = 2\times10^3,\text{rad/s}$

Inductive reactance:

$X_L = \omega L$

$X_L = 2\times10^3 \times 10\times10^{-3}$

$X_L = 20,\Omega$

Capacitive reactance:

$X_C = \frac{1}{\omega C}$

$X_C = \frac{1}{2\times10^3 \times 25\times10^{-6}}$

$X_C = 20,\Omega$

Since $X_L = X_C$, the circuit is at resonance. Therefore, the impedance is:

$Z = \sqrt{R^2 + (X_L-X_C)^2}$

$Z = \sqrt{100^2 + (20-20)^2}$

$Z = 100,\Omega$

The voltage amplitude is $V_0 = 220,\text{V}$. Hence, the current amplitude is:

$I_0 = \frac{V_0}{Z}$

$I_0 = \frac{220}{100}$

$I_0 = 2.2,\text{A}$

Hence, the correct answer is Option 4: $2.2,\text{A}$.

Ques: 2

Bob B of mass $m$ at rest is hanging vertically from the ceiling via a massless string of length $10,\text{m}$. Point mass A of mass $m$ travelling horizontally with speed $10,\text{m/s}$ hits bob B elastically. The bob B rises $h$ meter after the collision. Taking the acceleration due to gravity $g = 10,\text{m/s}^2$ and neglecting the size of the bob, the value of $h$ is:

Option 1) $7,\text{m}$

Option 2) $5,\text{m}$

Option 3) $2.5,\text{m}$

Option 4) $8,\text{m}$

Chapter: Systems of Particles

Correct Answer: Option 2

Explanation

For a perfectly elastic collision between two bodies of equal mass, the velocities are completely exchanged.

Therefore, after the collision, mass A comes to rest and bob B moves with the initial speed of A:

$v_B = 10,\text{m/s}$

As bob B swings upward, its kinetic energy is converted into gravitational potential energy. Using conservation of mechanical energy:

$\frac{1}{2}mv_B^2 = mgh$

Cancelling $m$ from both sides:

$h = \frac{v_B^2}{2g}$

Substituting $v_B = 10,\text{m/s}$ and $g = 10,\text{m/s}^2$:

$h = \frac{(10)^2}{2\times10}$

$h = \frac{100}{20}$

$h = 5,\text{m}$

Hence, the correct answer is Option 2: $5,\text{m}$.

Ques: 3

A wire of resistance $R$ is cut into 8 equal pieces. From these pieces, two equivalent resistances are made by adding four of these together in parallel. Then these two sets are added in series. The net effective resistance of the combination is

Option 1) $\frac{R}{8}$
Option 2) $\frac{R}{64}$
Option 3) $\frac{R}{32}$
Option 4) $\frac{R}{16}$

Chapter: Current Electricity

Correct Answer: Option 4

Explanation:
The wire is cut into 8 equal pieces, so each piece has resistance
$R_{piece}=\frac{R}{8}$

The resistance of each group is
$\frac{1}{R_{group}}=\frac{1}{R/8}+\frac{1}{R/8}+\frac{1}{R/8}+\frac{1}{R/8}=\frac{32}{R}$

$R_{group}=\frac{R}{32}$

Since the groups are connected in series, the net resistance is
$R_{net}=\frac{R}{32}+\frac{R}{32}=\frac{R}{16}$

Hence, the answer is option (4).

Ques: 4

In a potentiometer circuit, a cell of EMF 1.5 V gives a balance point at 36 cm length of wire. If another cell of EMF 2.5 V replaces the first cell, then at what length of the wire, the balance point occur?

Option 1) 60 cm
Option 2) 21.6 cm
Option 3) 64 cm
Option 4) 62 cm

Chapter: Current Electricity

Correct Answer: Option 1

Explanation:
$\frac{E_1}{E_2}=\frac{l_1}{l_2}$

$\frac{1.5}{2.5}=\frac{36}{l_2}$

$l_2=60cm$

Hence, the answer is option (1).

Ques: 5

An electric dipole with dipole moment $5\times10^{-6}Cm$ is aligned with the direction of a uniform electric field of magnitude $4\times10^5NC^{-1}$. The dipole is then rotated through an angle of $60^\circ$ with respect to the electric field. The change in the potential energy of the dipole is

Option 1) 1.5 J
Option 2) 0.8 J
Option 3) 1.0 J
Option 4) 1.2 J

Chapter: Electrostatic Potential and Capacitance

Correct Answer: Option 3

Explanation:
Given,
$p=5\times10^{-6}Cm$
$E=4\times10^5NC^{-1}$
$\theta_i=0^\circ,\ \theta_f=60^\circ$

$\Delta U=U_f-U_i$

$\Delta U=pE(\cos\theta_i-\cos\theta_f)$

$\Delta U=5\times10^{-6}\times4\times10^5\times(1-\frac{1}{2})=1J$

Hence, the answer is option (3).

Ques: 6

The radius of the Martian orbit around the Sun is about 4 times the radius of the orbit of Mercury. The Martian year is 687 Earth days. Then, which of the following is the length of 1 year on Mercury?

Option 1) 124 earth days
Option 2) 88 earth days
Option 3) 225 earth days
Option 4) 172 earth days

Chapter: Gravitation

Correct Answer: Option 2

Explanation:
From Kepler’s third law, $T^2\propto R^3$

$\frac{T_{Mercury}}{T_{Mars}}=\left(\frac{R_{Mercury}}{R_{Mars}}\right)^{3/2}=\left(\frac{1}{4}\right)^{3/2}=\frac{1}{8}$

$T_{Mercury}=\frac{687}{8}\approx88days$

Hence, the answer is option (2).

Ques: 7

The minimum energy required to launch a satellite of mass $m$ from the surface of the earth of mass $M$ and radius $R$ in a circular orbit at an altitude of $2R$ from the surface of the earth is:

Option 1) $\frac{5GMm}{6R}$
Option 2) $\frac{2GMm}{3R}$
Option 3) $\frac{GMm}{2R}$
Option 4) $\frac{GMm}{R}$

Chapter: Gravitation

Correct Answer: Option 2

Explanation:
Using energy conservation,

$U_i+K_i=U_f+K_f$

$-\frac{GMm}{R}+K_i=-\frac{GMm}{3R}+\frac{1}{2}m\frac{GM}{3R}$

$K_i=\frac{5GMm}{6R}$

Hence, the answer is option (2).

Ques: 8

A body of mass 60 g experiences a gravitational force 3.0 N when placed at a particular point. The magnitude of the gravitational field intensity at that point is:

Option 1) 50 N/kg
Option 2) 20 N/kg
Option 3) 180 N/kg
Option 4) 0.05 N/kg

Chapter: Gravitation

Correct Answer: Option 1

Explanation:
Gravitational field intensity

$g=\frac{F}{m}=\frac{3}{60\times10^{-3}}=50Nkg^{-1}$

Hence, the answer is option (1).

Ques: 9

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

Option 1) $p^4$ and $m^4$
Option 2) $4p$ and $4m$
Option 3) $p^4$ and $4m$
Option 4) $4p$ and $m^4$

Chapter: Ray Optics and Optical Instruments

Correct Answer: Option 4

Explanation:
The total power of lenses in contact

$P_{total}=p+p+p+p=4p$

The total magnification

$m_{total}=m\times m\times m\times m=m^4$

Hence, the answer is option (4).

Ques: 10

The electron concentration in an n-type semiconductor is the same as the hole concentration in a p-type semiconductor. An external electric field is applied across each of them. Compare the currents in them.

Option 1) Current in n-type = current in p-type
Option 2) Current in p-type > current in n-type
Option 3) Current in n-type < current in p-type
Option 4) No current will flow in p-type, current will only flow in n-type

Chapter: Semiconductor Electronics

Correct Answer: Option 1

Explanation:
In both n-type and p-type semiconductors, electric current is due to drift motion of charge carriers when an electric field is applied.

Current in n-type = current in p-type

Hence, the answer is option (1).

NEET Physics 2027: Why Numericals Are the Key to High Scores

NEET Physics questions are largely numerical-based. Almost 70–75% of the questions involve direct or application-based calculations. If students master frequently asked numericals, scoring 120+ marks out of 180 becomes very achievable.

Key reasons why Physics numericals are important:

• Most questions follow repeated patterns

• Direct formula-based problems save time

• NCERT and previous year trends are followed closely

• Accuracy improves with repeated practice

High-Weightage Chapters Covered in NEET Physics Numericals

The Top 50 numericals are selected mainly from chapters that contribute maximum marks in NEET Physics based on the Re NEET 2026 paper analysis:

Why Practising Top 50 NEET Physics Numericals Is Enough for 120+ Marks

Out of 720 marks in NEET, physics constitutes 180 marks and 45 questions, with each question carrying four marks. This means that in order to score 120 marks in physics, the candidate will have to answer 30 questions correctly, assuming there are no wrong answers attempted. In real life, this will require the candidate to attempt more than 30 questions, considering the negative marking.

The selected numerical problems can assist students in developing:

• Better formula memory skills
• Calculation speed
• Ability to determine which formula applies
• Concept application in MCQ form
• Less calculation mistakes
• Time management abilities
• Recognition of repeated NEET question trends
• Confidence in challenging Physics topics

But these 50 questions must not substitute thorough preparation of the entire syllabus or NEET previous year papers.

Smart Strategy to Score 120+ Marks in NEET 2027 Physics

Follow this simple plan:

• Revise all formulas daily

• Practice 5–10 numericals every day

• Focus more on Class 12 Physics

• Avoid lengthy calculations during the exam

• Attempt Modern Physics first

• Do not guess blindly, this will avoid negative marking

• Consistency is more important than solving hundreds of questions.

NEET 2027 Physics Preparation Tips for Average Students

• Stick to NCERT textbooks

• Use one good question source only

• Revise mistakes regularly

• Practice numericals with a timer

• Focus on accuracy over speed initially

• Physics is not about memorisation - it is about understanding and application.

NEET Physics: Top 50 Numericals Selection Method

The Top 50 Physics numericals were selected after:

• Detailed trend analysis of NEET papers

• Identifying high-frequency formulas to solve NEET physics questions.

• Selecting numericals with maximum scoring potential

• Including assertion-reason and application-based questions

• These questions reflect the actual difficulty level of NEET.