Atomic Models – Thomson & Rutherford Model for NEET \| NCERT Class 11 Structure of Atom

 - Dr.Sanjaykumar Pawar   

Chemistry Teacher IQ: free notes, MCQs and mock tests for CBSE, NEET and JEE chemistry.

Atomic Models for NEET: Thomson Plum Pudding & Rutherford Nuclear Model Explained

IntroductionThomsonRutherfordFormulasNumericalsComparisonPracticeFAQs

Introduction

What are atomic models?

An atomic model is a scientific representation of the internal structure of an atom: how protons, neutrons and electrons are arranged inside it. Atoms are too small to observe directly, so scientists proposed models based on the experimental evidence available at the time.

Think of a sealed box. You can't see inside, but by shaking it, weighing it and listening, you can guess what is in it. That is what Thomson, Rutherford and later Bohr did with the atom.

Why is this topic important for NEET?

• Atomic models are the foundation of the "Structure of Atom" chapter.
• NEET often asks conceptual and statement-based questions on Thomson's and Rutherford's models.
• Questions on drawbacks, features and the alpha-particle scattering experiment appear almost every alternate year.
• They are essential before studying Bohr's model and quantum mechanics.

NCERT chapter and class

Detail	Information
Class	11
Chapter	Chapter 2 – Structure of Atom
Section	2.3 – Developments Leading to the Bohr's Model of Atom

Approximate NEET weightage

Sub-topic	Expected questions
Thomson's model	0–1 question
Rutherford's model & alpha scattering	1–2 questions
Total from atomic models	1–2 questions (4–8 marks)

Previous year trends

• 2023 NEET: statement-based question on Rutherford's experiment observations.
• 2022 NEET: match-the-column on features of atomic models.
• 2021 NEET: assertion–reason on stability of Rutherford's atom.
• 2020 NEET: direct question on a drawback of Thomson's model.
• Trend: NEET is shifting toward NCERT line-by-line statement questions and assertion–reason formats.

Learning objectives

1. Describe Thomson's plum pudding model and its key features.
2. Explain Rutherford's nuclear model and the alpha-particle scattering experiment.
3. Identify the drawbacks of both models.
4. Solve numericals on distance of closest approach and impact parameter.
5. Answer NEET-style MCQs and assertion–reason questions confidently.

1. Thomson's Atomic Model (Plum Pudding Model) – 1898

Who was J.J. Thomson?

Sir Joseph John Thomson discovered the electron in 1897 through cathode ray experiments. The next question was: how are electrons arranged inside a neutral atom?

Definition

Thomson's model states that an atom is a uniform sphere of positive charge in which negatively charged electrons are embedded, like plums in a pudding or seeds in a watermelon.

In simple language

Picture a watermelon. The red flesh is the positive charge spread through the atom; the black seeds are the electrons embedded in it. Hence the names "Watermelon model" and "Plum pudding model".

Key features

Feature	Description
Shape of atom	Uniform sphere
Positive charge	Spread uniformly over the entire volume
Electrons	Embedded (stuck) inside the positive sphere
Mass distribution	Uniform over the whole atom
Overall charge	Electrically neutral (positive charge = total negative charge)
Size of atom	About 10−10 m (1 Å)

Analogy: a blueberry muffin. The batter (positive charge) is everywhere and the blueberries (electrons) are stuck in it. There is no separate core or nucleus.

✔ What it could explain

• Overall electrical neutrality of the atom.
• The presence of electrons inside the atom.

✘ Drawbacks

Drawback	Explanation
Could not explain alpha scattering	With uniformly spread positive charge, alpha particles should pass with almost no deflection. Rutherford observed large-angle deflections.
Could not explain atomic spectra	No discrete energy levels or line spectra.
Could not explain stability of electrons	No reason given for why embedded electrons stay put.
No nucleus	Misses the dense central core.

Common misconceptions

Misconception	Reality
"Thomson discovered the proton."	He discovered the electron. The proton was identified later.
"Electrons orbit the nucleus in Thomson's model."	Electrons are embedded, not orbiting.
"Thomson's model has a nucleus."	No. The nucleus concept came from Rutherford.

2. Rutherford's Atomic Model (Nuclear Model) – 1911

The alpha-particle scattering experiment (Geiger–Marsden)

Rutherford, with students Geiger and Marsden, performed a landmark experiment in 1909.

Setup: a narrow beam of alpha particles (He²⁺ nuclei) from a radioactive source such as polonium is aimed at a very thin gold foil (about 100 nm, roughly 1000 atoms thick). A circular zinc sulfide (ZnS) fluorescent screen around the foil detects scattered particles, and a microscope is used to observe the flashes (scintillations).

Observations

Observation	Share	Meaning
Most alpha particles passed straight through	~99%	Most of the atom is empty space
Some were deflected by small angles	~1%	A concentrated positive charge repels the alpha particles
Very few (about 1 in 8000) bounced back at angles > 90° (some 180°)	~0.01%	Positive charge and mass sit in an extremely tiny, dense region

Rutherford compared the result to firing a 15-inch shell at tissue paper and having it come back and hit you.

Definition

Rutherford's nuclear model states that the entire positive charge and almost all the mass of an atom is concentrated in an extremely small, dense central region called the nucleus, and electrons revolve around it in circular orbits, held by electrostatic attraction.

Key features

Feature	Description
Nucleus	Tiny, dense, positively charged core at the centre
Size of nucleus	~10−15 m (1 fm), about 1/100,000 of the atom
Size of atom	~10−10 m (1 Å)
Electrons	Revolve in circular orbits (like planets around the Sun)
Force	Electrostatic (Coulombic) attraction between nucleus and electrons
Empty space	Most of the atom
Mass	Almost all in the nucleus (protons + neutrons)

Analogy: if a cricket stadium is the atom, the nucleus is a marble at the centre of the pitch and the electrons are tiny gnats around the outermost stands. Everything between is empty air.

✔ What it could explain

• The results of the alpha scattering experiment.
• A dense, positively charged nucleus.
• The vast empty space in the atom, and why most alpha particles pass undeflected.

✘ Drawbacks (most tested in NEET)

Drawback	Explanation
1. Stability of the atom (major flaw)	By Maxwell's electromagnetic theory, a charge moving in a circle must continuously radiate energy. The electron would spiral into the nucleus in about 10−8 s. Atoms are stable, so the model fails.
2. No electron arrangement	Says nothing about how many orbits or what energies.
3. Cannot explain atomic spectra	A spiralling electron would give a continuous spectrum, but atoms show discrete line spectra.
4. No energy levels	No specific energy values for orbits.

Common misconceptions

Misconception	Reality
"Rutherford discovered the neutron."	Chadwick discovered the neutron in 1932. Rutherford discovered the nucleus.
"Electrons are stationary in Rutherford's model."	They revolve around the nucleus.
"Rutherford's model explains line spectra."	It does not. Bohr's model does.
"The nucleus contains electrons."	The nucleus contains protons and neutrons only.

Formula Handbook

1. Distance of closest approach (d₀)

$$d_0=\frac{1}{4\pi\varepsilon_0}\cdot\frac{2Ze^2}{K}$$

• Variables: Z = atomic number of target, e = 1.6 × 10⁻¹⁹ C, K = kinetic energy of alpha particle. Unit: m. Dimensions: [M⁰L¹T⁰].
• Conditions: head-on collision (impact parameter = 0); alpha particle momentarily stops.
• Assumptions: stationary nucleus; only Coulomb repulsion; no nuclear forces. Valid for low-energy alpha particles.
• Shortcut: d₀ ∝ Z/K. If KE doubles, d₀ halves.
• Common mistake: forgetting that the alpha particle has charge 2e, not e.

2. Coulomb's law

$$F=\frac{1}{4\pi\varepsilon_0}\cdot\frac{q_1q_2}{r^2}$$

• Unit: N. Dimensions: [M¹L¹T⁻²]. Point charges, stationary.
• Shortcut: 1/(4πε₀) = 9 × 10⁹ N·m²/C².
• Common mistake: using cm instead of m.

3. Kinetic energy of alpha particle

$$K=\tfrac12mv^2=qV$$

• m = mass, v = velocity, q = 2e, V = accelerating potential. Units: J or eV.
• Shortcut: 1 eV = 1.6 × 10⁻¹⁹ J. Mistake: confusing MeV with eV (1 MeV = 10⁶ eV).

4. Impact parameter (b)

$$b=\frac{Ze^2\cot(\theta/2)}{4\pi\varepsilon_0\,K}$$

• θ = scattering angle. Single scattering, Coulomb interaction only.
• Shortcut: b ∝ cot(θ/2). For θ = 180°, b = 0 (head-on). For θ = 0°, b = ∞.
• Mistake: using θ instead of θ/2 inside cot.

Derivation: distance of closest approach

In a head-on collision the alpha particle slows down due to repulsion. At d₀ all its kinetic energy has become electrostatic potential energy.

1. Initial KE: $K=\tfrac12 m_\alpha v^2$
2. At closest approach KE = 0 and $U=\dfrac{1}{4\pi\varepsilon_0}\dfrac{(2e)(Ze)}{d_0}$
3. Conservation of energy, K = U: $K=\dfrac{1}{4\pi\varepsilon_0}\dfrac{2Ze^2}{d_0}$
4. Solving: $$\boxed{d_0=\frac{1}{4\pi\varepsilon_0}\cdot\frac{2Ze^2}{K}}$$

NEET tip: the derivation is not asked directly, but the formula and d₀ ∝ Z/K are frequently tested.

Graphs and data interpretation

Graph 1: number of scattered particles (N) vs scattering angle (θ). θ from 0° to 180° on the x-axis, N on a logarithmic y-axis. The curve starts very high near 0° and falls steeply toward zero at large angles.

Feature	Interpretation
Shape	Steeply decreasing, $N\propto \dfrac{1}{\sin^4(\theta/2)}$
θ ≈ 0°	Maximum particles (most go straight)
θ ≈ 180°	Almost zero (very few bounce back)
NEET question	"N at angle θ is proportional to?" → $1/\sin^4(\theta/2)$

Graph 2: d₀ vs K. A rectangular hyperbola (d₀ ∝ 1/K) with no finite intercepts. If KE is doubled, d₀ becomes half.

Solved Numericals

Numerical 1: distance of closest approach

Q. A 5 MeV alpha particle is fired at a gold nucleus (Z = 79). Find d₀.

K = 5 × 10⁶ × 1.6 × 10⁻¹⁹ = 8 × 10⁻¹³ J

$$d_0=\frac{9\times10^9\times2\times79\times(1.6\times10^{-19})^2}{8\times10^{-13}}=\frac{3.64\times10^{-26}}{8\times10^{-13}}=4.55\times10^{-14}\ \text{m}$$

Answer: d₀ ≈ 4.55 × 10⁻¹⁴ m ≈ 45.5 fm.

Shortcut: d₀ (fm) ≈ 2.88 Z / K(MeV) = 2.88 × 79 / 5 ≈ 45.5 fm. Mistake: not converting MeV to J, or using e instead of 2e.

Numerical 2: effect of changing KE

Q. KE rises from 2 MeV to 8 MeV. By what factor does d₀ change?

$$\frac{d_{0_2}}{d_{0_1}}=\frac{K_1}{K_2}=\frac{2}{8}=\frac14$$

Answer: d₀ becomes 1/4 of the original. Mistake: writing 4 instead of 1/4 (mixing direct and inverse proportion).

Numerical 3: effect of changing target

Q. Same KE, copper (Z = 29) and silver (Z = 47) foils. Find the ratio of d₀.

d₀ ∝ Z at constant K, so $d_{0,Cu}:d_{0,Ag}=29:47\approx0.617$.

Mistake: inverting the ratio.

Numerical 4: charge on the nucleus

Q. For KE = 7.7 MeV, d₀ = 3 × 10⁻¹⁴ m. Estimate Z.

K = 7.7 × 1.6 × 10⁻¹³ = 1.232 × 10⁻¹² J

$$Z=\frac{d_0K}{2e^2\times9\times10^9}=\frac{3.696\times10^{-26}}{4.608\times10^{-28}}\approx80$$

Answer: Z ≈ 80 (close to gold, Z = 79; the difference is rounding). Shortcut: Z ≈ d₀(fm) × K(MeV) / 2.88 = 30 × 7.7 / 2.88 ≈ 80.

Numerical 5: nuclear vs atomic volume

Q. Nucleus radius 10⁻¹⁵ m, atom radius 10⁻¹⁰ m. What fraction of the atomic volume does the nucleus occupy?

$$\left(\frac{10^{-15}}{10^{-10}}\right)^3=(10^{-5})^3=10^{-15}$$

Answer: 10⁻¹⁵. Mistake: squaring instead of cubing the radius ratio.

Comparison Tables

Thomson vs Rutherford

Feature	Thomson	Rutherford
Year	1898	1911
Also known as	Plum pudding / Watermelon	Nuclear / Planetary
Positive charge	Spread uniformly	Concentrated in a tiny nucleus
Electrons	Embedded in positive sphere	Revolve around nucleus
Nucleus	✘ No	✔ Yes
Empty space	✘ No	✔ Most of atom
Mass distribution	Uniform	Concentrated in nucleus
Explains neutrality	✔ Yes	✔ Yes
Explains alpha scattering	✘ No	✔ Yes
Explains stability	✘ No	✘ No
Explains line spectra	✘ No	✘ No
Force on electrons	Not specified	Electrostatic (Coulombic)

Size comparison

Entity	Size	Ratio
Atom	~10⁻¹⁰ m (1 Å)	1
Nucleus	~10⁻¹⁵ m (1 fm)	1/100,000
Electron	< 10⁻¹⁸ m (point-like)	Negligible

Common mistakes

#	Mistake	Correction
1	Using MeV directly without converting to J	1 MeV = 1.6 × 10⁻¹³ J
2	Misplacing powers of 10 in e²	e² = 2.56 × 10⁻³⁸, not 10⁻¹⁹
3	Using q = e for alpha particle	He²⁺ → charge = 2e
4	Treating alpha–nucleus force as attractive	Both positive → repulsive
5	Confusing 10⁻¹⁴ m with 10⁻¹⁵ m	Nucleus ≈ 10⁻¹⁵ m; d₀ ≈ 10⁻¹⁴ m
6	Saying Rutherford's model explains stability	It does not, its biggest drawback
7	Giving Thomson orbiting electrons	Thomson → embedded; Rutherford → orbiting
8	Using radius ratio for volume	Volume ratio = (radius ratio)³

Quick Revision

Constant	Value
Charge of electron	1.6 × 10⁻¹⁹ C
Mass of alpha particle	6.64 × 10⁻²⁷ kg
1/(4πε₀)	9 × 10⁹ N·m²/C²
1 eV / 1 keV / 1 MeV	1.6 × 10⁻¹⁹ J / 1.6 × 10⁻¹⁶ J / 1.6 × 10⁻¹³ J
1 Å / 1 nm / 1 fm	10⁻¹⁰ m / 10⁻⁹ m / 10⁻¹⁵ m

• Key graphs: N vs θ is steeply decreasing (∝ 1/sin⁴(θ/2)); d₀ vs K is a rectangular hyperbola; d₀ vs Z is a straight line through the origin.
• Shortcuts: d₀ doubles → KE halves. Z doubles → d₀ doubles. Head-on → θ = 180°, b = 0. Grazing → θ ≈ 0°, b maximum. Nucleus : atom = 10⁻⁵.

Practice Section

10 MCQs

1. Thomson's atomic model is also known as: A) Nuclear model B) Planetary model C) Plum pudding model D) Quantum model
2. Most alpha particles passed through the gold foil undeflected. This indicates: A) The nucleus is very large B) Most of the atom is empty space C) Alpha particles are very heavy D) Gold foil is not dense
3. The main drawback of Rutherford's model is: A) It cannot explain neutrality B) It cannot explain the nucleus C) It cannot explain the stability of the atom D) It cannot explain electrons
4. In Thomson's model the positive charge is: A) Concentrated at the centre B) Spread uniformly C) Located in electron orbits D) Absent
5. The distance of closest approach is inversely proportional to: A) Atomic number Z B) Kinetic energy of the alpha particle C) Charge of the alpha particle D) Mass of the nucleus
6. If the KE of an alpha particle is doubled, d₀ will: A) Double B) Become four times C) Become half D) Remain the same
7. Size of the nucleus compared with the atom is approximately: A) 1/10 B) 1/100 C) 1/10,000 D) 1/100,000
8. Rutherford's model could NOT explain: A) Alpha scattering B) Existence of nucleus C) Discrete line spectra D) Empty space in the atom
9. The electrostatic force between nucleus and orbiting electron is: A) Gravitational B) Magnetic C) Attractive (Coulombic) D) Repulsive (Coulombic)
10. Alpha particles bounced back at 180° had an impact parameter of approximately: A) Maximum B) Equal to atomic radius C) Zero D) Equal to nuclear radius

Show MCQ answers

Q	Ans	Explanation
1	C	Plum pudding / watermelon model.
2	B	Most going straight means mostly empty space.
3	C	By Maxwell's theory, orbiting electrons radiate energy and spiral into the nucleus.
4	B	Uniformly distributed positive charge.
5	B	d₀ ∝ 1/K.
6	C	Double KE → half d₀.
7	D	10⁻¹⁵ / 10⁻¹⁰ = 10⁻⁵ = 1/100,000.
8	C	Line spectra were explained by Bohr's model.
9	C	Positive nucleus + negative electron → attraction.
10	C	180° scattering = head-on = impact parameter ≈ 0.

3 Assertion–Reason questions

1. A: Rutherford's model could not explain the stability of the atom. R: By classical electromagnetic theory, an accelerating charged particle radiates energy continuously.
2. A: In Thomson's model the atom is electrically neutral. R: The total positive charge equals the total negative charge of the embedded electrons.
3. A: Most alpha particles passed through the gold foil without deflection. R: The nucleus of the atom is positively charged.

Show AR answers

• AR1: Both true, and R explains A. The orbiting electron accelerates, so it radiates energy and would spiral into the nucleus.
• AR2: Both true, and R explains A. Thomson's uniform positive charge balances the electrons' total negative charge.
• AR3: Both true, but R does not explain A. Non-deflection is due to the atom being mostly empty space; the nucleus's positive charge explains the deflection.

3 Numericals for practice

1. A 10 MeV alpha particle approaches a silver nucleus (Z = 47). Find d₀.
2. d₀ for an alpha particle on copper (Z = 29) is 10 fm. What is it for gold (Z = 79) at the same KE?
3. A hydrogen atom has radius 0.53 Å and its nucleus 1.2 fm. Find the ratio of atomic volume to nuclear volume.

Show answers

1) ≈ 13.5 fm   2) ≈ 27.2 fm   3) ≈ 10¹⁴

FAQs

What is Thomson's model of the atom?

Proposed in 1898, it describes the atom as a uniform sphere of positive charge with electrons embedded in it. It explains neutrality but not alpha scattering or atomic stability.

What is Rutherford's model of the atom?

Proposed in 1911, it places all the positive charge and most of the mass in a tiny central nucleus, with electrons revolving around it in circular orbits. It is also called the nuclear or planetary model.

What was the main observation of the alpha scattering experiment?

Most alpha particles passed straight through the gold foil, some were deflected at small angles, and about 1 in 8000 bounced back at angles greater than 90°. This led to the discovery of the nucleus.

Why did Rutherford's model fail?

It could not explain atomic stability: by Maxwell's theory an orbiting electron would radiate energy and spiral into the nucleus in about 10⁻⁸ s. It also could not explain discrete line spectra.

What is the distance of closest approach?

The minimum distance an alpha particle reaches from the nucleus in a head-on collision, where all its KE has become electrostatic potential energy: $d_0=\dfrac{1}{4\pi\varepsilon_0}\dfrac{2Ze^2}{K}$.

Which model is more accurate?

Rutherford's, since it identifies the nucleus and explains alpha scattering. Both have limitations that were later resolved by Bohr's model and quantum mechanics.

How many NEET questions come from atomic models?

Typically 1–2 questions (4–8 marks), mostly conceptual and statement-based from NCERT.

Key Takeaways

• Thomson = plum pudding = uniform positive sphere + embedded electrons = no nucleus.
• Rutherford = nuclear model = tiny dense nucleus + orbiting electrons = mostly empty space.
• Alpha scattering proved the existence of the nucleus. Both models fail on stability and line spectra, so Bohr's model was needed.
• Read NCERT Section 2.3 line by line, memorise the drawbacks, and practise the d₀ ∝ Z/K shortcut.
• In assertion–reason questions, check whether R actually explains A or is just independently true.

Revision checklist

• Describe Thomson's model, its features and drawbacks
• Describe the alpha scattering experiment and Rutherford's model
• List the main drawbacks of Rutherford's model
• Calculate d₀ and use the d₀ ∝ Z/K shortcut
• Differentiate the two models
• Solve assertion–reason questions on atomic stability
• Read NCERT Section 2.3 at least twice

Final NEET tip: atomic models are a high-scoring, low-effort topic. Master the features and drawbacks of both models, practise 5–10 numericals on distance of closest approach, and move on to Bohr's model next.

Chemistry Teacher

Class XI – Physics (Practice Question Bank)

Time: 3 hoursMaximum marks: 70

1. All questions are compulsory.
2. Section headings show the marks per question.
3. Use g = 10 m/s² wherever needed.

Section A – Multiple choice questions 1 mark each

1. The SI unit of momentum is:

   1. kg m s⁻¹
   2. kg m s⁻²
   3. kg m² s⁻¹
   4. N m
   Answer: (a) kg m s⁻¹

2. A ball thrown vertically upward is at its highest point. Its acceleration at that instant is:

   1. zero
   2. 10 m/s² downward
   3. 10 m/s² upward
   4. infinite
   Answer: (b) 10 m/s² downward. Only the velocity is zero at the top; gravity still acts.

3. The dimensional formula of work is:

   1. [MLT⁻²]
   2. [ML²T⁻²]
   3. [ML²T⁻¹]
   4. [ML⁻¹T⁻²]
   Answer: (b) [ML²T⁻²]

4. The work done by the centripetal force in uniform circular motion is:

   1. positive
   2. negative
   3. zero
   4. infinite
   Answer: (c) zero, because the force is always perpendicular to displacement.

5. The escape velocity from the surface of the Earth is approximately:

   1. 7.9 km/s
   2. 9.8 km/s
   3. 11.2 km/s
   4. 15.0 km/s
   Answer: (c) 11.2 km/s

Section B – Fill in the blanks 1 mark each

1. The dimensional formula of force is ____________.

   Answer: [MLT⁻²]

2. The slope of a velocity–time graph gives ____________.

   Answer: acceleration

3. The value of acceleration due to gravity at the centre of the Earth is ____________.

   Answer: zero

4. The SI unit of pressure is ____________.

   Answer: pascal (N m⁻²)

Section C – Very short answer questions 1 mark each

1. Define one light year and give its value in metres.

   Answer: The distance travelled by light in vacuum in one year; 1 ly ≈ 9.46 × 10¹⁵ m.

2. What is the SI unit of angular velocity?

   Answer: radian per second (rad s⁻¹).

3. Which physical quantity in rotational motion plays the role of mass in linear motion?

   Answer: Moment of inertia.

4. Name the physical quantity whose unit is N m but which is not work.

   Answer: Torque.

Section D – Assertion and reason 1 mark each

Options: (a) Both A and R are true and R is the correct explanation of A. (b) Both A and R are true but R is not the correct explanation of A. (c) A is true but R is false. (d) A is false but R is true.

1. Assertion (A): A body can have zero average velocity but non-zero average speed.
   Reason (R): Average speed is total path length divided by time, while average velocity is displacement divided by time.

   Answer: (a). In a round trip the displacement is zero but the path length is not.

2. Assertion (A): Passengers lurch forward when a moving bus stops suddenly.
   Reason (R): A forward force acts on the passengers when the bus brakes.

   Answer: (c). It happens due to inertia of motion; no forward force acts.

3. Assertion (A): The kinetic energy of a body is never negative.
   Reason (R): Kinetic energy = ½mv², and both m and v² are non-negative.

   Answer: (a)

Section E – Statement based questions 1 mark each

Options: (a) Both statements are correct. (b) Only Statement I is correct. (c) Only Statement II is correct. (d) Both statements are incorrect.

1. Statement I: Displacement can be zero even when distance travelled is not zero.
   Statement II: Distance is always greater than or equal to the magnitude of displacement.

   Answer: (a)

2. Statement I: Static friction is self-adjusting up to a limiting value.
   Statement II: Kinetic friction is greater than limiting friction.

   Answer: (b). Kinetic friction is slightly less than limiting friction.

3. Statement I: Work done by the gravitational force depends on the path taken.
   Statement II: Gravitational force is a conservative force.

   Answer: (c). For a conservative force, work depends only on the end points.

Section F – Match the columns 2 marks

Match each quantity in Column I with its dimensional formula in Column II.

Column I (Quantity)	Column II (Dimensional formula)
A. Momentum	p. [ML2T−2]
B. Energy	q. [ML−1T−2]
C. Power	r. [MLT−1]
D. Pressure	s. [ML2T−3]

Answer: A → r, B → p, C → s, D → q

Section G – Short answer questions 3 marks each

1. A constant force of 10 N acts on a body of mass 2 kg, initially at rest, for 5 s. Find the final velocity and the kinetic energy gained.

   Answer: a = F/m = 10/2 = 5 m/s². v = u + at = 0 + 5 × 5 = 25 m/s. KE = ½mv² = ½ × 2 × 25² = 625 J.

2. Define centripetal force. Write its expression and give one example.

   Answer: The net force directed towards the centre that keeps a body in circular motion. F = mv²/r = mω²r. Example: tension in the string when a stone is whirled in a circle.

3. State Kepler's three laws of planetary motion.

   Answer: (1) Law of orbits: planets move in elliptical orbits with the Sun at one focus. (2) Law of areas: the line joining a planet and the Sun sweeps equal areas in equal times. (3) Law of periods: the square of the orbital period is proportional to the cube of the semi-major axis (T² ∝ a³).

Section H – Long answer questions 5 marks each

1. Derive the three equations of uniformly accelerated motion using calculus.

   Answer:
   (i) a = dv/dt, so dv = a dt. Integrating from u to v and 0 to t: v − u = at, so v = u + at.
   (ii) v = dx/dt = u + at. Integrating from 0 to t: x = ut + ½at², so s = ut + ½at².
   (iii) a = dv/dt = v dv/dx, so v dv = a dx. Integrating: ½(v² − u²) = as, so v² = u² + 2as.

2. Derive an expression for the escape velocity of a body from the Earth's surface and find its value (g = 9.8 m/s², R = 6.4 × 10⁶ m).

   Answer: Escape velocity is the minimum speed at which the total energy becomes zero, so the body just reaches infinity.
   ½mv_e² − GMm/R = 0, so v_e = √(2GM/R).
   Since GM = gR², v_e = √(2gR).
   v_e = √(2 × 9.8 × 6.4 × 10⁶) ≈ 1.12 × 10⁴ m/s = 11.2 km/s.

Section I – Case study 4 marks (1 + 1 + 1 + 1)

A ball is thrown vertically upward from the ground with a speed of 20 m/s. Ignore air resistance and take g = 10 m/s².

1. Find the maximum height reached by the ball.

   Answer: H = u²/2g = 400/20 = 20 m.

2. How long does it take to reach the highest point?

   Answer: t = u/g = 20/10 = 2 s.

3. What is the total time of flight?

   Answer: T = 2u/g = 4 s.

4. Find the velocity of the ball at t = 3 s.

   Answer: v = u − gt = 20 − 30 = −10 m/s, i.e. 10 m/s downward.