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NEET Physics Assessment

Practice with Discipline, Revise with Purpose

Dear Students,

This NEET Physics assessment paper is based on Units and Dimensions, Kinematics, and Motion in Two Dimensions / Motion in a Plane. These chapters form the foundation of Class 11 Physics, NEET Physics, and IIT-JEE Physics. They test dimensional analysis, error calculation, equations of motion, projectile motion, vector resolution, relative velocity, circular motion basics, graph understanding, and calculation accuracy.

This paper has been prepared and solved by Kumar Sir, an experienced Physics Tutor in Altamount Road – Mumbai. Kumar Sir is one of the best Physics teachers for serious NEET aspirants, and this paper has been designed in a systematic, conceptual, and exam-focused manner. Students should attempt this paper sincerely, patiently, and with full concentration.

If students are searching for Physics Tutor, NEET Physics Tutor, or Physics Tutor in Altamount Road – Mumbai and they are unable to solve these questions properly, they should contact Kumar Sir for one-to-one online Physics classes.

This paper should be attempted only after revising the important formulas of Units and Dimensions, Kinematics, and Motion in Two Dimensions / Motion in a Plane. First revise the formula bank, then solve the complete question paper under timed conditions. Do not open the solution immediately. First think, calculate, choose your answer, and then compare it with the official solution. Every mistake should be treated as a learning point.

Formula Bank

Important Formula Revision for NEET Physics: Units, Kinematics and Motion in Two Dimensions

Before starting this paper, revise the important formulas of Units and Dimensions, Kinematics, and Motion in Two Dimensions. NEET Physics often tests whether a student can select the correct formula, apply it correctly, interpret motion graphs, resolve vectors properly, and avoid calculation mistakes. Many students remember formulas but still lose marks because they do not know where and how to apply them. This formula bank is added to help students quickly revise the major concepts before attempting the paper.

Units and Dimensions

Velocity

[v] = LT^-1

Used to check dimensional consistency in basic motion equations.

Acceleration

[a] = LT^-2

Used in equations of motion and dimensional analysis.

Force

[F] = MLT^-2

Used in numerical problems based on Newton's laws.

Work / Energy

[W] = ML²T^-2

Used when comparing work, energy, torque, and related quantities.

Power

[P] = ML²T^-3

Used in rate-of-work and dimensional formula questions.

Momentum

[p] = MLT^-1

Used in impulse, collision, and kinetic energy relation questions.

Impulse

[J] = MLT^-1

Used because impulse equals change in momentum.

Pressure

[P] = ML^-1T^-2

Used in force per unit area applications.

Density

[ρ] = ML^-3

Used in mass-volume dimensional analysis.

Homogeneity

Dimensions of LHS = Dimensions of RHS

Used to test whether an equation can be physically meaningful.

Product / Quotient Error

% error = sum of individual percentage errors

Used in maximum percentage error problems.

Power Error

If Z = Aⁿ, % error in Z = n × % error in A

Used when a measured quantity is raised to a power.

Absolute Error

Δa = |measured value - true value|

Used to measure direct deviation from the accepted value.

Relative Error

relative error = Δa / a

Used to compare error with the size of the measured value.

Percentage Error

percentage error = (Δa / a) × 100

Used in NEET numerical error calculations.

Kinematics

Average Velocity

v_avg = Δx / Δt

Used for displacement per unit time.

Average Speed

average speed = total distance / total time

Used when distance, not displacement, is required.

Acceleration

a = Δv / Δt

Used to measure rate of change of velocity.

First Equation

v = u + at

Used in uniformly accelerated motion.

Second Equation

s = ut + 1/2 at²

Used in displacement-time numerical problems.

Third Equation

v² = u² + 2as

Used when time is not directly involved.

nth Second

s_n = u + a(2n - 1)/2

Used for distance covered in a particular second.

Uniform Acceleration Distance

S = ut + 1/2 at²

Used in graph and direct motion questions.

Free Fall

v = u + gt

Used when downward motion is taken positive.

Vertical Height

h = ut - 1/2 gt²

Used in upward vertical motion.

Maximum Height

H = u² / 2g

Used for vertical upward throw.

Time of Ascent

t = u / g

Used to find time to reach the highest point.

Total Vertical Flight

T = 2u / g

Used when body returns to the same level.

Motion in Two Dimensions / Motion in a Plane

Resultant of Two Vectors

R = √(A² + B² + 2ABcosθ)

Used in vector addition problems.

Direction of Resultant

tanα = Bsinθ / (A + Bcosθ)

Used to find angle of resultant vector.

Vector Resolution

A_x = Acosθ, A_y = Asinθ

Used in projectile and force resolution questions.

Relative Velocity

v_AB = v_A - v_B

Used in one-dimensional and two-dimensional relative motion.

Projectile Horizontal Component

u_x = ucosθ

Used because horizontal velocity remains constant.

Projectile Vertical Component

u_y = usinθ

Used in vertical motion of projectile.

Projectile Time of Flight

T = 2usinθ / g

Used when projectile returns to the same level.

Projectile Maximum Height

H = u²sin²θ / 2g

Used to calculate highest point of projectile.

Projectile Range

R = u²sin2θ / g

Used for horizontal range on level ground.

Trajectory

y = xtanθ - gx²/(2u²cos²θ)

Used in equation-of-path questions.

Maximum Range

R_max = u² / g

Used when launch angle is 45 degrees.

Circular Speed

v = rω

Used in uniform circular motion basics.

Angular Velocity

ω = 2π/T = 2πf

Used to connect frequency and time period.

Centripetal Acceleration

a_c = v²/r = rω²

Used in circular motion numerical problems.

Centripetal Force

F_c = mv²/r = mrω²

Used in NEET conceptual applications of circular motion.

Time Period

T = 2π / ω

Used to convert angular velocity into time period.

Study Discipline

Build Your Physics Foundation for NEET

Dear students, Units and Dimensions, Kinematics, and Motion in Two Dimensions are not small chapters. They create the base for the entire mechanics portion of Physics. If these chapters are weak, students often struggle later in Laws of Motion, Work-Energy, Rotation, Gravitation, SHM, Waves, and even advanced NEET numerical questions.

This paper should be solved like a real exam. Sit with a timer, attempt every question honestly, and do not open the solution before trying properly. If you are living in Altamount Road – Mumbai and searching for a Physics Tutor for NEET, IB, ICSE, IIT-JEE, CBSE, IGCSE, AP Physics or any serious Physics preparation, contact Kumar Sir for one-to-one online Physics guidance. Kumar Sir helps students understand concepts deeply, solve difficult numericals, and build confidence for competitive exams.

Preparation Mindset

Why Strong Physics Preparation Is Now More Important Than Ever

NEET Physics is becoming more conceptual and competitive. Students need conceptual clarity, calculation accuracy, speed, and the confidence to solve unfamiliar problems. Memorising formulas is not enough; a serious student must understand when to apply a formula, where it fits in the situation, and how to use it without losing marks in calculation or interpretation.

Future Aspirants

Important Message for NEET 2027, 2028, 2029, 2030 and Future Aspirants

Future NEET aspirants must prepare seriously for online-style or changing exam patterns, where question variation and concept application may become more important. Practise papers under timed conditions, revise formulas regularly, analyse every mistake, and strengthen weak chapters before they become a larger problem in advanced mechanics and numerical Physics.

One-to-One Guidance

Why Study Physics with Kumar Sir?

Kumar Sir provides personalised one-to-one online Physics classes for students who want serious conceptual improvement and exam-oriented preparation. He clears each and every concept, explains difficult topics in simple language, and helps students prepare for NEET, CBSE, JEE, IB, ICSE, IGCSE, AP Physics and other exams. His teaching style focuses on conceptual clarity, numerical practice, doubt-solving, graph understanding, and step-by-step problem solving. If you are struggling in Units and Dimensions, Kinematics, Motion in Two Dimensions or any Physics topic, Kumar Sir can guide you patiently and systematically until your basics become strong.

Contact Kumar Sir

Premium Physics Support for Serious Students

If you are searching for a Physics Tutor in Altamount Road – Mumbai for NEET, IB, ICSE, IIT-JEE, CBSE, IGCSE, AP Physics or any advanced Physics preparation, contact Kumar Sir. Kumar Sir explains Units and Dimensions, Kinematics, Motion in Two Dimensions and other Physics topics in a very clear, step-by-step, and exam-oriented way.

Question Index

Question 1

Resistors in Series — Percentage Error

The value of two resistors are (5.0 ± 0.2) Ω) and (10.0 ± 0.1) Ω). The maximum percentage error in the equivalent resistance when they are connected in series will be:

Correct Answer: Option 3

Official Solution

For series combination,

R = R₁ + R₂

R = 5.0 + 10.0 = 15.0 Ω

Maximum absolute error:

Δ R = 0.2 + 0.1 = 0.3 Ω

Percentage error:

Δ R/R × 100 = 0.3/15.0 × 100 = 2%

Question 2

Error in Kinetic Energy

If maximum percentage error in measurement of momentum is (1.5%), and there is no error in mass, then maximum percentage error in the measurement of kinetic energy is:

Correct Answer: Option 1

Official Solution

Kinetic energy in terms of momentum is:

K = p^2/2m

Since mass has no error,

Δ K/K × 100 = 2 (Δ p/p × 100)

= 2 × 1.5% = 3%

Question 3

Significant Figures in Division

If (62.24 cm) is divided by (2.0 s), the result up to appropriate significant figures is:

Correct Answer: Option 4

Official Solution

62.24/2.0 = 31.12 cm/s

In multiplication or division, the result should have the same number of significant figures as the least precise quantity.

Here, (2.0) has 2 significant figures. Therefore,

31.12 ≈ 31 cm/s

This can be written as:

3.1 × 10 cm/s

Question 4

Most Accurate Measurement

Which of the following measurement is most accurate for measurement of length of a metre stick?

Correct Answer: Option 4

Official Solution

A metre stick has true length close to (1 m). The measurement closest to (1 m) is:

1.001 m

It also has the highest precision among the given values.

Question 5

Area with Correct Significant Figures

If length of side of a square is (1.2 m), then area of square with correct number of significant figures is:

Correct Answer: Option 2

Official Solution

A = a²

A = (1.2)² = 1.44 m²

The given side (1.2) has 2 significant figures, so area must also be written up to 2 significant figures.

1.44 ≈ 1.4

Question 6

Work and Torque — Assertion Reason

Assertion (A):Work and torque have same dimensions, but the two are not physically same.Reason (R):Dimensional correctness of a relation does not ensure its physical correctness.

Correct Answer: Option 1

Official Solution

Work:

W = F × s

[W] = [MLT^-2] [L] = [ML²T^-2]

Torque:

τ = F × r

[τ] = [MLT^-2] [L] = [ML²T^-2]

So work and torque have same dimensions.

But work is scalar, while torque is vector. Therefore, they are physically different.

The reason is also true because dimensional correctness alone cannot prove physical correctness.

Question 7

Rounding Off to Significant Figures

Choose the correct statement.

Correct Answer: Option 1

Official Solution

For (17.684), first three significant figures are (1,7,6). The next digit is (8), so (6) is increased to (7).

17.684 ≈ 17.7

So statement (1) is correct.

Question 8

Significant Figures in Subtraction

Taking into account significant figures, what is value of:

9.99 m - 0.009 m

Correct Answer: Option 1

Official Solution

9.99 - 0.009 = 9.981

For addition or subtraction, final answer is written according to the least number of decimal places.

Here (9.99) has 2 decimal places. Therefore,

9.981 ≈ 9.98

Question 9

Dimension of Planck’s Constant

Planck’s constant (h) has the same dimensions as:

Correct Answer: Option 4

Official Solution

From Planck’s equation:

E = hν

h = E/ν

So,

[h] = [ML²T^-2]/[T^-1] = [ML²T^-1]

Angular momentum also has dimension:

[L] = [mvr] = [M][LT^-1][L] = [ML²T^-1]

So (h) has same dimensions as energy/frequency and angular momentum.

Question 10

Mean Value of Measurements

A person measures the length of a rod as (10 cm, 11 cm, 10 cm, 10 cm, 9 cm). The true value of length of rod is:

Correct Answer: Option 1

Official Solution

The best estimate of the true value is the mean value.

Mean = (10+11+10+10+9)/5

= 50/5 = 10 cm

Question 11

Rounding Off Numbers

The numbers (3.865) and (3.875) on rounding off to 3 significant figures will give:

Correct Answer: Option 2

Official Solution

Using the standard rounding rule for 5:

For (3.865), the digit before 5 is (6), which is even. So it remains unchanged.

3.865 ≈ 3.86

For (3.875), the digit before 5 is (7), which is odd. So it is increased.

3.875 ≈ 3.88

Question 12

Maximum Percentage Error

A physical quantity

P = √(ABC²)/D³

is determined by measuring (A, B, C) and (D) separately with percentage errors of (2%, 2%, 4%) and (1%) respectively. Maximum percentage error in the value of (P) will be:

Correct Answer: Option 2

Official Solution

P = √(ABC²)/D³

P = A^1/2B^1/2C¹D^-3

Maximum percentage error:

Δ P/P× 100 = 1/2(2%) + 1/2(2%) + 1(4%) + 3(1%)

= 1% + 1% + 4% + 3%

= 9%

Question 13

Dimension of (pr/q)

The acceleration (a) of a particle at time (t) is given by:

a = pt + q/(t²+r)

where (p, q) and (r) are constants. The dimension of

pr/q

is:

Correct Answer: Option 1

Official Solution

Since (t²) and (r) are added,

[r] = [t²] = [T²]

Now,

a = pt

[p][T] = [LT^-2]

[p] = [LT^-3]

Also,

q/(t²+r)

has dimension of acceleration.

[q]/[T²] = [LT^-2]

[q] = [L]

Now,

[pr/q] [LT^-3][T²]/[L]

= [T^-1]

= [M⁰L⁰T^-1]

Question 14

Pitch of Screw Gauge

A screw gauge has least count of (0.01 mm) and there are 50 divisions in its circular scale. The pitch of the screw gauge is:

Correct Answer: Option 2

Official Solution

Least count = Pitch/Number of circular scale divisions

0.01 = Pitch/50

Pitch = 0.01 × 50 = 0.5 mm

Question 15

Meaningful Dimensional Expression

If (A, B) and (C) are physical quantities such that (A) and (B) have same dimensions while (C) has different dimensions, then choose the meaningful expression.

Correct Answer: Option 4

Official Solution

Only quantities with same dimensions can be added or subtracted.

Since (A) and (B) have same dimensions, the expression:

2A - 3B

is meaningful.

Dividing it by (C) is also mathematically allowed.

So,

2A-3B/C

is meaningful.

Question 16

Matching Dimensions

Match the physical quantities in List-I with their dimensions in List-II.

List-I:a. Velocity b. Acceleration c. Force d. Pressure

List-II:(i) ([M¹L¹T^-2]) (ii) ([M⁰L¹T^-2]) (iii) ([M¹L^-1T^-2]) (iv) ([M⁰L¹T^-1])

Correct Answer: Option 1

Official Solution

Velocity:

[v] = [LT^-1] = [M⁰L¹T^-1]

So,

a arrow (iv)

Acceleration:

[a] = [LT^-2] = [M⁰L¹T^-2]

So,

b arrow (ii)

Force:

[F] = [MLT^-2]

So,

c arrow (i)

Pressure:

P = F/A

[P] = [MLT^-2]/[L²] = [ML^-1T^-2]

So,

d arrow (iii)

Question 17

Dimensions of Current

If (Q, R) and (C) represent charge, resistance and capacitance respectively, then dimensions of current are same as dimensions of:

Correct Answer: Option 4

Official Solution

We know:

RC = time constant

So,

[RC] = [T]

Current is:

I = Q/t

Therefore,

I = Q/RC

Question 18

Screw Gauge Reading

The reading of screw gauge shown in the figure is, where least count is (0.01 mm).

$Screw gauge diagram shows main scale reading (6.5\ \text{mm}), and circular scale division approximately (1) coincides with the reference line after 50.$

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Correct Answer: Option 1

Official Solution

Main scale reading:

MSR = 6.5 mm

Circular scale reading:

CSR = 1

Least count:

LC = 0.01 mm

Total reading:

= MSR + CSR × LC

= 6.5 + 1 × 0.01

= 6.51 mm

Question 19

Zero Error in Vernier Calipers

In an experiment to measure diameter of a wire, a vernier calliper is used in which 9 main scale divisions of the calliper coincide with 10 vernier scale divisions. One main scale division is (2 mm). When nothing is in between the jaws, it is found that zero of vernier lies slightly to the left of zero of main scale and 5th division of Vernier scale coincides with some division of main scale. The zero error has a magnitude of:

Correct Answer: Option 1

Official Solution

Given:

1 MSD = 2 mm

Also,

10 VSD = 9 MSD

10 VSD = 9 × 2 = 18 mm

1 VSD = 1.8 mm

Least count:

LC = 1 MSD - 1 VSD

LC = 2 - 1.8 = 0.2 mm

Since zero of vernier lies to the left of zero of main scale, the error is negative. Magnitude of zero error:

= (10 - 5) × LC

= 5 × 0.2 = 1 mm

Question 20

Vernier Calipers Reading

The main scale of a vernier callipers is calibrated in mm and 18 divisions of main scale are equal in length to 20 divisions of vernier scale. In measuring the diameter of a cylinder by this instrument, the main scale reads 25 divisions and 5th division of vernier scale coincides with a main scale division. The reading of diameter is:

Correct Answer: Option 2

Official Solution

Given:

1 MSD = 1 mm

20 VSD = 18 MSD = 18 mm

1 VSD = 18/20 = 0.9 mm

Least count:

LC = 1 - 0.9 = 0.1 mm

Main scale reading:

MSR = 25 mm

Vernier reading:

VR = 5 × 0.1 = 0.5 mm

Total reading:

= 25 + 0.5 = 25.5 mm

Question 21

Least Count of Vernier Calipers

Consider a Vernier callipers in which each (1 cm) on main scale is divided into 8 equal divisions. If 5 divisions of the Vernier scale coincide with 4 divisions on the main scale, then least count is:

Correct Answer: Option 4

Official Solution

Each (1 cm) is divided into 8 equal divisions.

1 MSD = 1/8 cm

= 10/8 mm

= 1.25 mm

Given:

5 VSD = 4 MSD

1 VSD = 4/5 MSD

1 VSD = 4/5 × 1.25 = 1.0 mm

Least count:

LC = 1 MSD - 1 VSD

LC = 1.25 - 1.0 = 0.25 mm

0.25 mm = 1/4 mm

Question 22

Vertical Motion — Average Speed

A particle is projected vertically upward with velocity (v) and it returns to the point of projection. For the complete journey:

Correct Answer: Option 3

Official Solution

Maximum height:

H = v^2/2g

Total distance:

= 2H = v²/g

Total time of flight:

T = 2v/g

Average speed:

= Total distance/Total time

= v²/g/2v/g

= v/2

Question 23

Height Passed Twice in Vertical Motion

A body thrown vertically up with initial velocity (52 m/s) from the ground passes a point twice at height (h) above ground in an interval of (10 s). The height (h) is:

g = 10 m/s²

Correct Answer: Option 2

Official Solution

For vertical motion:

h = ut - 1/2gt²

h = 52t - 5t²

5t² - 52t + h = 0

Let the two times be (t₁) and (t₂).

Sum of roots:

t₁ + t₂ = 52/5 = 10.4

Given:

t₂ - t₁ = 10

Solving,

t₂ = 10.2 s, t₁ = 0.2 s

Now,

h = 52(0.2) - 5(0.2)²

h = 10.4 - 0.2 = 10.2 m

Question 24

Distance in 5th Half Second

A particle starts moving with acceleration (2 m/s²). Distance travelled by it in 5th half second is:

Correct Answer: Option 2

Official Solution

5th half second means time interval:

2 s to 2.5 s

Since particle starts from rest,

s = 1/2at²

Distance in 5th half second:

s = 1/2a(2.5)² - 1/2a(2)²

= 1/2(2)[(2.5)² - (2)²]

= 1(6.25 - 4)

= 2.25 m

Question 25

Two Balls Meeting

From the top of a building of height (100 m), a ball is dropped. At the same time another ball is projected vertically upwards from ground with a velocity (50 m/s) along the same line. The height from ground where the two balls will meet each other is:

g = 10 m/s²

Correct Answer: Option 3

Official Solution

Let upward direction be positive and ground be origin.

For dropped ball:

y₁ = 100 - 1/2gt²

y₁ = 100 - 5t²

For upward projected ball:

y₂ = 50t - 5t²

They meet when:

y₁ = y₂

100 - 5t² = 50t - 5t²

100 = 50t

t = 2 s

Height from ground:

y = 50(2) - 5(2)²

y = 100 - 20 = 80 m

Question 26

Velocity-Time Graph

A car is moving on a straight road according to the given velocity-time (v-t) graph. Then:

$Velocity increases uniformly from (0) to (4\ \text{m/s}) in (0) to (2\ \text{s}), remains constant at (4\ \text{m/s}) from (2) to (6\ \text{s}), then decreases uniformly to zero from (6) to (8\ \text{s}).$

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Correct Answer: Option 4

Official Solution

Displacement is area under (v-t) graph.

First triangle:

A₁ = 1/2 × 2 × 4 = 4 m

Rectangle:

A₂ = 4 × 4 = 16 m

Second triangle:

A₃ = 1/2 × 2 × 4 = 4 m

Total displacement:

S = 4 + 16 + 4 = 24 m

Question 27

Acceleration from (v²-t) Graph

A graph between square of speed (v²) with time (t) for a particle moving on straight line is shown in the figure. The magnitude of acceleration of the particle at (t = 1 s) is:

$A straight line graph of (v^2) vs (t) decreases from (v^2 = 8) at (t=0) to (v^2 = 0) at (t=2\ \text{s}). At (t=1\ \text{s}), (v^2=4).$

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Correct Answer: Option 1

Official Solution

Slope of (v²-t) graph:

d(v²)/dt = 0-8/2-0 = -4

But,

d(v²)/dt = 2vdv/dt = 2va

At (t=1 s),

v² = 4

v = 2 m/s

So,

-4 = 2(2)a

a = -1 m/s²

Magnitude:

|a| = 1 m/s²

Question 28

Velocity from Displacement Equation

The displacement (x) of a particle moving along x-axis at time (t) is given by:

x² = 2t² + 6t

The velocity at any time (t) is:

Correct Answer: Option 4

Official Solution

x² = 2t² + 6t

Differentiate both sides with respect to (t):

2xdx/dt = 4t + 6

Since,

v = dx/dt

2xv = 4t + 6

v = (4t+6)/2x

v = (2t+3)/x

Question 29

Maximum Height Above Point Q

A ball is thrown vertically upward. On its way up, it passes point (P) with a speed (2V) and point (Q), (2 m) higher than (P), with a speed (V). The maximum height reached by the ball above point (Q) is:

Correct Answer: Option 4

Official Solution

Between (P) and (Q):

v² = u² - 2gs

At (P),

u = 2V

At (Q),

v = V

Distance:

s = 2 m

So,

V² = (2V)² - 2g(2)

V² = 4V² - 4g

3V² = 4g

V² = 4g/3

Height above (Q):

h = V^2/2g

h = 4g/3/2g

h = 2/3 m

Question 30

Average Velocity under Uniform Acceleration

A body starts from rest and moving with uniform acceleration (a), for time (t). The average velocity over time (t) is:

Correct Answer: Option 2

Official Solution

Initial velocity:

u = 0

Final velocity:

v = u + at = at

Average velocity for uniform acceleration:

v_avg = (u+v)/2

v_avg = (0+at)/2

v_avg = at/2

Question 31

Vertical Motion with Air Drag

A body is thrown vertically upward from the ground with speed (15 m/s) and it reaches at the highest position at time (t = 1 s). Assuming constant air drag, time taken by it to reach the ground from the highest point is nearly:

g = 10 m/s²

Correct Answer: Option 4

Official Solution

During upward motion, final velocity at top is zero.

v = u - a_upt

0 = 15 - a_up(1)

a_up = 15 m/s²

Since during upward motion,

a_up = g + a_d

15 = 10 + a_d

a_d = 5 m/s²

Height reached:

H = (u+v)/2t

H = (15+0)/2(1)

H = 7.5 m

During downward motion, air drag acts upward, so net downward acceleration:

a_down = g - a_d

a_down = 10 - 5 = 5 m/s²

From top:

H = 1/2a_downt²

7.5 = 1/2(5)t²

7.5 = 2.5t²

t² = 3

t = √(3) ≈ 1.7 s

Question 32

Particle at Rest from (x-t) Graph

The figure shows position (x) versus time (t) graph of a particle moving along x-axis. During time interval (t=0) to (t=4 s), how many times the particle comes to rest?

$The (x-t) graph is a wave-like curve. Between (t=0) and (t=4\ \text{s}), it has two maxima and two minima.$

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Correct Answer: Option 1

Official Solution

A particle comes to rest when velocity is zero.

v = dx/dt

On an (x-t) graph, velocity is the slope of the curve. So, particle is at rest where slope is zero, that is, at maxima and minima.

From (t=0) to (t=4 s), the graph has:

2 maxima + 2 minima = 4

Therefore, particle comes to rest 4 times.

Question 33

Acceleration versus Displacement Graph

If velocity of the particle is:

v = kx^3/2

then the acceleration versus displacement graph is:

Correct Answer: Option 1

Official Solution

Acceleration can be written as:

a = vdv/dx

Given:

v = kx^3/2

Differentiate:

dv/dx = 3/2kx^1/2

So,

a = kx^3/2 · 3/2kx^1/2

a = 3/2k²x²

Thus,

a ∝ x²

This represents a parabola.

Question 34

Relative Velocity from (x-t) Graph

Two objects are moving in a straight line. Their position (x) versus time (t) graphs are as in figure. The relative velocity of (A) with respect to (B) is:

sin 37° = 3/5

$Line (A) makes (37^\circ) with positive time axis. Line (B) is descending and makes (53^\circ) with the vertical (x)-axis.$

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Correct Answer: Option 3

Official Solution

Slope of (x-t) graph gives velocity.

For object (A):

v_A = tan 37°

Since,

sin 37° = 3/5

cos 37° = 4/5

tan 37° = 3/4

So,

v_A = 3/4 m/s

For object (B), the line is descending, so velocity is negative. Since it makes (53°) with vertical, it makes (37°) with horizontal.

v_B = -tan 37° = -3/4 m/s

Relative velocity of (A) with respect to (B):

v_AB = v_A - v_B

v_AB = 3/4 - (-3/4)

v_AB = 6/4 = 3/2 m/s

Question 35

Acceleration-Time Graph

The acceleration-time graph of a particle moving along straight line is given. At what time velocity of particle becomes equal to its initial velocity?

$Acceleration decreases linearly from (+10\ \text{m/s}^2) at (t=0) to (0) at (t=4\ \text{s}), then to (-10\ \text{m/s}^2) at (t=8\ \text{s}). After that it starts increasing but remains negative.$

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Correct Answer: Option 2

Official Solution

Change in velocity is area under acceleration-time graph.

Velocity becomes equal to initial velocity when net area under (a-t) graph is zero.

Positive area from (0) to (4 s):

A_+ = 1/2 × 4 × 10 = 20

Negative area from (4) to (8 s):

A_- = 1/2 × 4 × 10 = 20

So net change in velocity at (t=8 s):

Δ v = 20 - 20 = 0

Therefore velocity becomes equal to initial velocity at:

t = 8 s

Question 36

Average Speed for Equal Distances

A man covers one third of a distance with (60 km/h), next one-third distance with (20 km/h) and last one-third distance with (10 km/h). His average speed of the entire journey is:

Correct Answer: Option 4

Official Solution

For equal distances, average speed is harmonic mean:

v_avg = 3/(1/60+1/20+1/10)

= 3/((1+3+6)/60)

= 3/10/60

= 3 × 6

= 18 km/h

Question 37

Increasing Velocity from Acceleration-Time Graphs

A particle is moving along positive x-axis with some initial velocity. The acceleration-time graphs are shown. In which case velocity of particle will increase for entire time between (t₁) to (t₂)?

Three acceleration-time graphs I, II and III are shown. In all three graphs, acceleration remains positive between (t_1) and (t_2).

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Correct Answer: Option 4

Official Solution

Velocity increases when acceleration is positive.

a > 0 ⇒ v increases

In all three graphs, acceleration is above the time axis between (t₁) and (t₂). Therefore, velocity increases in all three cases.

Question 38

Retardation from Velocity-Time Equation

A particle starts moving rectilinearly at time (t=0) such that its velocity (v) changes with time (t) according to equation:

v = t² - t

where (t) is in seconds and (v) is in (m/s). The time interval for which the particle retards is:

Correct Answer: Option 2

Official Solution

Given:

v = t² - t

v = t(t-1)

Acceleration:

a = dv/dt = 2t - 1

Retardation occurs when velocity and acceleration have opposite signs.

For:

0 < t < 1/2

v < 0, a < 0

Same sign, so speed increases.

For:

1/2 < t < 1

v < 0, a > 0

Opposite signs, so particle retards.

For:

t > 1

v > 0, a > 0

Same sign, so speed increases.

Question 39

Bullet and Accelerating Car

A policeman fires a bullet on a thief’s car at rest. As the bullet is fired, the car starts accelerating by acceleration (20 m/s²). If the speed of bullet is (720 km/h) and the initial distance of thief’s car from policeman is (2 km), then after what time will the bullet hit the thief’s car?

Correct Answer: Option 4

Official Solution

Bullet speed:

720 km/h = 720 × 5/18

= 200 m/s

Let policeman be at origin.

Position of bullet:

x_b = 200t

Initial distance of car:

2 km = 2000 m

Car starts from rest with acceleration (20 m/s²).

Position of car:

x_c = 2000 + 1/2(20)t²

x_c = 2000 + 10t²

For collision:

x_b = x_c

200t = 2000 + 10t²

10t² - 200t + 2000 = 0

t² - 20t + 200 = 0

Discriminant:

D = (-20)² - 4(1)(200)

D = 400 - 800 = -400

Since discriminant is negative, no real time exists.

Therefore, bullet will not hit the car.

Question 40

Velocity-Time Graph Statements

Given are two statements:

Statement I:Area under velocity-time graph gives the distance travelled.Statement II:For one-dimensional motion, a constant speed in an interval must have non-zero acceleration in that interval.

Choose the most appropriate answer.

Correct Answer: Option 2

Official Solution

Area under velocity-time graph gives displacement, not necessarily distance.

Distance is obtained from area under speed-time graph or from the magnitude-wise area of velocity-time graph.

So Statement I is incorrect.

Statement II says constant speed must have non-zero acceleration. This is wrong. In one-dimensional motion, if speed is constant and direction is also unchanged, acceleration is zero.

So Statement II is also incorrect.

Question 41

Correct Statements in Motion

Consider the following statements and choose the correct option.

Statement I:If a particle has zero velocity at any instant, then its acceleration must be zero at that instant.Statement II:For uniform motion, velocity-time graph is a straight line parallel to the time axis.Statement III:The magnitude of displacement is always greater than or equal to the distance traversed by an object.Statement IV:For motion with uniform acceleration, position (x)-time (t) graph is a parabola.

Correct Answer: Option 1

Official Solution

Statement I is false. At the highest point of vertical motion, velocity is zero but acceleration is (g).

Statement II is true. In uniform motion, velocity is constant, so (v-t) graph is parallel to time axis.

Statement III is false. Distance is always greater than or equal to magnitude of displacement.

Distance ≥ |Displacement|

Statement IV is true. For uniform acceleration:

x = ut + 1/2at²

This is a quadratic equation in (t), so (x-t) graph is a parabola.

Question 42

Match the Columns

Column-I contains certain physical quantities and Column-II contains various parameters. Match Column-I with Column-II and choose the correct option.

Column-I:a. Change in velocity b. Acceleration c. Displacement d. Instantaneous velocity

Column-II:(i) Slope of velocity vs time curve (ii) Area under velocity vs time curve (iii) Area under acceleration vs time curve (iv) Slope of position vs time curve

Correct Answer: Option 2

Official Solution

Change in velocity:

Δ v = Area under a-t graph

So,

a arrow (iii)

Acceleration:

a = dv/dt

So acceleration is slope of velocity-time graph.

b arrow (i)

Displacement:

s = Area under v-t graph

c arrow (ii)

Instantaneous velocity:

v = dx/dt

So velocity is slope of position-time graph.

d arrow (iv)

Question 43

Maximum Velocity with Acceleration and Deceleration

A car accelerates from rest at a constant rate (a₁) for some time, after which it decelerates at a constant rate (a₂) and comes to rest. If total time elapsed is (t), then maximum velocity acquired by the car is equal to:

Correct Answer: Option 3

Official Solution

Let maximum velocity be (V).

During acceleration:

V = a₁t₁

t₁ = V/a₁

During deceleration:

V = a₂t₂

t₂ = V/a₂

Total time:

t = t₁ + t₂

t = V/a₁ + V/a₂

t = V(1/a₁+1/a₂)

t = V((a₁+a₂)/a₁a₂)

Therefore,

V = ta₁a₂/(a₁+a₂)

Question 44

Velocity-Time Curves I and II

Figure shows velocity-time curves I and II. Which of the following statement(s) are correct?

Both curves are straight lines on a velocity-time graph. Curve I starts with a positive initial velocity and has smaller slope. Curve II starts from origin and has larger slope. The two curves intersect.

**Statements:**
(a) Both the curves are representing uniform acceleration motion.
(b) Acceleration of II is more than acceleration of I.
(c) Displacement covered by particle I is more than II.

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Correct Answer: Option 4

Official Solution

In a velocity-time graph, slope represents acceleration.

Both curves are straight lines, so both represent uniform acceleration motion.

Therefore, statement (a) is correct.

Curve II has greater slope than curve I.

Therefore, acceleration of II is more than acceleration of I.

Statement (b) is correct.

Displacement is area under velocity-time graph. From the given graph, the area under curve I is greater than the area under curve II.

Therefore, statement (c) is also correct.

Question 45

Bolt Falling in Accelerating Elevator

An elevator of cabin height (1.2 m) starts ascending with acceleration (2 m/s²). One second after the start, a loose bolt starts falling from the ceiling. The time after which bolt will hit the floor of elevator is:

g = 10 m/s²

Correct Answer: Option 1

Official Solution

At the instant bolt starts falling, both bolt and elevator have same upward velocity. So relative initial velocity between bolt and elevator is zero.

In elevator frame, since elevator is accelerating upward with acceleration (2 m/s²), effective downward acceleration of bolt is:

g_eff = g + a

g_eff = 10 + 2 = 12 m/s²

Cabin height:

h = 1.2 m

Using relative motion:

h = 1/2g_efft²

1.2 = 1/2(12)t²

1.2 = 6t²

t² = 0.2

t² = 1/5

t = 1/√(5) s

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