Physics भौतिकशास्त्र

Quick summary

Physics is built on measurement. Every measured value has a number, a unit and some uncertainty. This topic covers SI units, dimensions, dimensional analysis, and the types of errors in measurement and how they combine.

Key points

• A physical quantity = numerical value × unit (e.g. 5 m).
• The SI system has 7 base quantities: length (metre, m), mass (kilogram, kg), time (second, s), electric current (ampere, A), temperature (kelvin, K), amount of substance (mole, mol) and luminous intensity (candela, cd).
• Derived quantities (force, energy, pressure) are made from base quantities.
• Dimensions show how a quantity depends on M, L, T, etc. Force = [M L T⁻²].
• Principle of homogeneity: every term in a correct physical equation has the same dimensions.
• Uses of dimensional analysis: check equations, convert units, derive relations between quantities.
• Limits of dimensional analysis: cannot find dimensionless constants (like 2π or ½), cannot handle sums of several terms, fails for trigonometric, log and exponential relations.
• Systematic errors come from faulty instruments (zero error), wrong method or personal bias. They act in one direction and can be corrected.
• Random errors vary by chance from reading to reading. Taking the mean of many readings reduces them.
• Least count is the smallest value an instrument can measure (vernier calliper 0.01 cm, screw gauge 0.001 cm usually).
• Significant figures are the reliable digits plus the first doubtful digit.

Important terms

• Accuracy – how close a measured value is to the true value.
• Precision – how close repeated measurements are to each other (अचूकता vs. सुस्पष्टता).
• Absolute error – difference between the measured value and the true (mean) value.
• Relative error – absolute error divided by the true value.
• Percentage error – relative error × 100.
• Dimensionless quantity – has no dimensions, e.g. strain, refractive index, angle.

Formulas

• Mean value: a_mean = (a₁ + a₂ + … + aₙ)/n
• Absolute error of i-th reading: Δaᵢ = |a_mean − aᵢ|
• Mean absolute error: Δa_mean = (Σ|Δaᵢ|)/n
• Relative error = Δa_mean / a_mean; percentage error = (Δa_mean / a_mean) × 100
• Sum or difference Z = A ± B: ΔZ = ΔA + ΔB (absolute errors add)
• Product or quotient Z = AB or A/B: ΔZ/Z = ΔA/A + ΔB/B (relative errors add)
• Power Z = Aⁿ: ΔZ/Z = n (ΔA/A)
• Unit conversion: n₂ = n₁ [M₁/M₂]ᵃ [L₁/L₂]ᵇ [T₁/T₂]ᶜ

5 important questions

Q1. State the principle of homogeneity of dimensions.

In a correct physical equation, the dimensions of every term on both sides must be the same. For example, in v = u + at, each term has dimensions [L T⁻¹].

Q2. Distinguish between systematic and random errors.

Systematic errors are in one direction (always more or always less), come from instrument or method, and can be removed by correction. Random errors change irregularly between readings and are reduced by taking the mean of many readings.

Q3. Using dimensions, derive the time period of a simple pendulum.

1. Assume T depends on mass m, length l and g: T = k mᵃ lᵇ gᶜ.
2. Dimensions: [T] = [M]ᵃ [L]ᵇ [L T⁻²]ᶜ = Mᵃ Lᵇ⁺ᶜ T⁻²ᶜ.
3. Compare powers: M: a = 0; T: −2c = 1 so c = −½; L: b + c = 0 so b = ½.
4. So T = k √(l/g). Experiment gives k = 2π, so T = 2π√(l/g).
5. Note: dimensional analysis cannot give the value 2π.

Q4. Convert 1 newton into dyne using dimensions.

1. Force = [M L T⁻²]; a = 1, b = 1, c = −2.
2. n₂ = n₁ (M₁/M₂)¹ (L₁/L₂)¹ (T₁/T₂)⁻².
3. M₁ = 1 kg = 1000 g, L₁ = 1 m = 100 cm, T₁ = T₂ = 1 s.
4. n₂ = 1 × 1000 × 100 × 1 = 10⁵.
5. So 1 N = 10⁵ dyne.

Q5. The length and breadth of a rectangle are (5.0 ± 0.1) cm and (2.0 ± 0.1) cm. Find the area with its error.

1. Area A = l × b = 5.0 × 2.0 = 10.0 cm².
2. Relative error ΔA/A = Δl/l + Δb/b = 0.1/5.0 + 0.1/2.0 = 0.02 + 0.05 = 0.07.
3. ΔA = 0.07 × 10.0 = 0.7 cm².
4. Area = (10.0 ± 0.7) cm²; percentage error = 7%.

Exam tip

In error questions, remember: absolute errors add for sums and differences; relative (percentage) errors add for products, quotients and powers. Always write units and the final answer as value ± error.

Quick summary

Many physical quantities (force, velocity, field) have both size and direction, so physics needs vectors. This topic covers unit vectors, components, the scalar (dot) product, the vector (cross) product, the scalar and vector triple products, and simple uses such as work, torque and area. These tools are used in almost every later topic of B.Sc. physics.

Key points

• A scalar has only magnitude (mass, time, work). A vector has magnitude and direction and adds by the parallelogram (or triangle) law (velocity, force).
• In Cartesian form A = Ax î + Ay ĵ + Az k̂, where î, ĵ, k̂ are unit vectors along x, y, z. Magnitude |A| = √(Ax² + Ay² + Az²).
• Unit vector  = A ÷ |A|; it gives only direction.
• Dot product A·B = AB cos θ = AxBx + AyBy + AzBz. It is a scalar and is commutative. If A·B = 0 (both non-zero), A ⟂ B.
• Cross product A×B = AB sin θ n̂. It is a vector perpendicular to both A and B (right-hand rule). It is not commutative: A×B = −(B×A). If A×B = 0, A ∥ B.
• î·î = 1, î·ĵ = 0; î×ĵ = k̂, ĵ×k̂ = î, k̂×î = ĵ, î×î = 0.
• |A×B| = area of the parallelogram formed by A and B.
• Scalar triple product A·(B×C) = volume of the parallelepiped; it is zero when the three vectors are coplanar.
• Vector triple product A×(B×C) = B(A·C) − C(A·B) ("BAC − CAB" rule).
• Physics uses: work W = F·s, power P = F·v, torque τ = r×F, angular momentum L = r×p, magnetic force F = q(v×B).

Important terms

• Position vector – vector from the origin to a point, r = x î + y ĵ + z k̂.
• Direction cosines – cos α = Ax/A, cos β = Ay/A, cos γ = Az/A; cos²α + cos²β + cos²γ = 1.
• Null vector – vector of zero magnitude.
• Coplanar vectors – vectors lying in one plane.
• Projection of A on B – A·B̂ = (A·B)/|B|.

Formulas

• A·B = AB cos θ = AxBx + AyBy + AzBz
• cos θ = (A·B) ÷ (AB)
• A×B = determinant with rows (î ĵ k̂), (Ax Ay Az), (Bx By Bz) = (AyBz − AzBy) î − (AxBz − AzBx) ĵ + (AxBy − AyBx) k̂
• Area of triangle with sides A and B = ½|A×B|
• A·(B×C) = determinant of rows (Ax Ay Az), (Bx By Bz), (Cx Cy Cz)
• A×(B×C) = B(A·C) − C(A·B)

5 important questions

Q1. Distinguish between scalar product and vector product.

Q2. Find the angle between A = 2î + 2ĵ − k̂ and B = 6î − 3ĵ + 2k̂.

A·B = 12 − 6 − 2 = 4. |A| = √(4+4+1) = 3, |B| = √(36+9+4) = 7.
cos θ = 4 ÷ 21 = 0.190, so θ ≈ 79°.

Q3. Find A×B for A = î + 2ĵ + 3k̂ and B = 2î − ĵ + k̂, and the area of the parallelogram they form.

• x part: (2)(1) − (3)(−1) = 2 + 3 = 5
• y part: −[(1)(1) − (3)(2)] = −(1 − 6) = 5
• z part: (1)(−1) − (2)(2) = −1 − 4 = −5
• A×B = 5î + 5ĵ − 5k̂; area = |A×B| = √75 = 5√3 ≈ 8.66 units².
• Check: (A×B)·A = 5 + 10 − 15 = 0, so the result is ⟂ to A, as it must be.

Q4. Prove the vector triple product rule A×(B×C) = B(A·C) − C(A·B) (outline).

1. B×C is ⟂ to the plane of B and C, so A×(B×C) lies in that plane: A×(B×C) = mB + nC.
2. Dotting with A gives 0 = m(A·B) + n(A·C), so m = λ(A·C), n = −λ(A·B).
3. Check with unit vectors, e.g. A = B = î, C = ĵ: left side = î×k̂ = −ĵ; right side = î(0) − ĵ(1) = −ĵ, so λ = 1.
4. Hence A×(B×C) = B(A·C) − C(A·B).

Q5. Show that A = î − 2ĵ + 3k̂, B = 2î + ĵ − k̂ and C = 3î − ĵ + 2k̂ are coplanar.

A·(B×C) = 1[(1)(2) − (−1)(−1)] − (−2)[(2)(2) − (−1)(3)] + 3[(2)(−1) − (1)(3)]
= 1(2 − 1) + 2(4 + 3) + 3(−2 − 3) = 1 + 14 − 15 = 0.
The scalar triple product is zero, so the volume is zero and the vectors are coplanar.

Exam tip

Always write the cross product as a 3×3 determinant and remember the minus sign on the ĵ term. Check your answer: A×B must give zero when dotted with A or B.

Quick summary

Kinematics describes motion without asking what causes it. This topic covers displacement, velocity and acceleration in one dimension, the equations of uniformly accelerated motion (derived by calculus), motion under gravity, and projectile motion in two dimensions with its time of flight, maximum height, range and path.

Key points

• Displacement is the change in position (a vector); distance is the total path length (a scalar).
• Instantaneous velocity v = dx/dt; instantaneous acceleration a = dv/dt = d²x/dt² = v dv/dx.
• Area under a v–t graph = displacement; slope of a v–t graph = acceleration; slope of an x–t graph = velocity.
• For constant acceleration, integrating a = dv/dt and v = dx/dt gives the three equations of motion.
• Freely falling bodies have a = g ≈ 9.8 m s⁻² downward (air resistance neglected).
• A projectile has constant horizontal velocity (no horizontal force) and uniformly accelerated vertical motion (a = −g). The two motions are independent.
• The path of a projectile is a parabola.
• Range is maximum at θ = 45°; angles θ and (90° − θ) give the same range.
• At the highest point the vertical velocity is zero but the horizontal velocity u cos θ remains.

Important terms

• Uniform motion – equal displacements in equal intervals of time.
• Relative velocity – v_AB = v_A − v_B, velocity of A as seen by B.
• Time of flight – total time a projectile stays in air.
• Horizontal range – horizontal distance covered during the time of flight.
• Angle of projection – angle between the initial velocity and the horizontal.

Formulas

• v = u + at; s = ut + ½at²; v² = u² + 2as (u = initial velocity, a = constant acceleration, s = displacement, m)
• Distance in the nth second: s_n = u + (a/2)(2n − 1)
• Time of flight T = 2u sin θ / g
• Maximum height H = u² sin²θ / 2g
• Range R = u² sin 2θ / g; R_max = u²/g at θ = 45°
• Path: y = x tan θ − g x² / (2u² cos²θ)
• Projectile (speed u, angle θ, g in m s⁻²):
• Horizontal projection from height h: T = √(2h/g), R = u√(2h/g)

5 important questions

Q1. Derive v = u + at and s = ut + ½at² using calculus.

1. a = dv/dt, so dv = a dt. Integrate from (t = 0, v = u) to (t, v): v − u = at, so v = u + at.
2. v = dx/dt = u + at, so dx = (u + at) dt. Integrate from 0 to t: s = ut + ½at².
3. Using a = v dv/dx: ∫v dv = ∫a dx gives ½(v² − u²) = as, so v² = u² + 2as.

Q2. A body starts from rest with acceleration 2 m s⁻². Find the distance covered in the 5th second.

s_n = u + (a/2)(2n − 1) = 0 + 1 × (10 − 1) = 9 m.
Check: s(5) − s(4) = ½·2·25 − ½·2·16 = 25 − 16 = 9 m.

Q3. Show that the path of a projectile is a parabola and derive its range.

1. Horizontal: x = (u cos θ) t, so t = x / (u cos θ).
2. Vertical: y = (u sin θ) t − ½ g t².
3. Substituting t: y = x tan θ − g x² / (2u² cos²θ). This has the form y = bx − cx², which is a parabola.
4. y = 0 again when t = T = 2u sin θ / g. Then R = u cos θ × T = 2u² sin θ cos θ / g = u² sin 2θ / g.
5. R is largest when sin 2θ = 1, i.e. θ = 45°, giving R_max = u²/g.

Q4. A ball is thrown at 20 m s⁻¹ at 30° to the horizontal. Find T, H and R (g = 10 m s⁻²).

• u sin θ = 20 × 0.5 = 10 m s⁻¹; u cos θ = 20 × 0.866 = 17.32 m s⁻¹
• T = 2 × 10 / 10 = 2 s
• H = (10)² / (2 × 10) = 5 m
• R = 17.32 × 2 = 34.6 m (check: u² sin 60° / g = 400 × 0.866 / 10 = 34.6 m)

Q5. Prove that for the same speed, angles θ and (90° − θ) give equal range. Which one gives more height?

R₁ = u² sin 2θ / g. For (90° − θ): R₂ = u² sin(180° − 2θ) / g = u² sin 2θ / g = R₁.
Heights: H₁ = u² sin²θ / 2g and H₂ = u² cos²θ / 2g. The larger angle gives the greater height and a longer time of flight. Also H₁ × H₂ = R²/16 (a useful result).

Exam tip

Always split the projectile's velocity into u cos θ (horizontal, constant) and u sin θ (vertical, changes with g). Write the units of every answer.

Quick summary

Newton's three laws explain how forces change motion. From them come the great conservation laws: conservation of linear momentum, of energy and of angular momentum. This topic also introduces inertial and non-inertial frames and the work–energy theorem.

Key points

• First law (law of inertia): a body stays at rest or in uniform motion in a straight line unless an external force acts on it.
• Second law: the rate of change of momentum equals the applied force, F = dp/dt. For constant mass, F = ma.
• Third law: to every action there is an equal and opposite reaction; the two forces act on different bodies.
• An inertial frame is one in which Newton's first law holds (not accelerating). A frame that accelerates is non-inertial; in it we must add pseudo forces (e.g. centrifugal force).
• Conservation of linear momentum: if the net external force on a system is zero, total momentum stays constant. Examples: recoil of a gun, rocket motion.
• Work–energy theorem: work done by the net force equals the change in kinetic energy.
• A force is conservative if work done by it does not depend on the path (gravity, spring force, electrostatic force). Friction is non-conservative.
• For a conservative force, F = −dU/dx (force is the negative gradient of potential energy).
• Conservation of mechanical energy: with only conservative forces, K + U = constant.
• Elastic collision: momentum and kinetic energy both conserved. Inelastic: only momentum conserved.
• Conservation of angular momentum: if net external torque is zero, L stays constant (a skater pulling in arms spins faster).

Important terms

• Momentum (संवेग) – p = mv, a vector, unit kg m s⁻¹.
• Impulse – force × time = change in momentum, J = FΔt = Δp.
• Inertia – the tendency of a body to resist change in its state.
• Pseudo force – an imaginary force −ma added in an accelerated frame.
• Coefficient of restitution (e) – relative speed of separation ÷ relative speed of approach (e = 1 elastic, e = 0 perfectly inelastic).

Formulas

• F = dp/dt = ma (N)
• p = mv (kg m s⁻¹); J = FΔt = Δp (N s)
• Work W = F·s = Fs cos θ (J)
• K = ½mv²; gravitational U = mgh (near Earth); spring U = ½kx²
• Work–energy theorem: W_net = K_f − K_i
• Rocket thrust: F = u (dm/dt); velocity v = v₀ + u ln(m₀/m) (no gravity)
• 1-D elastic collision (m₂ at rest): v₁ = (m₁ − m₂)u₁/(m₁ + m₂), v₂ = 2m₁u₁/(m₁ + m₂)
• Angular momentum L = r × p = Iω; torque τ = dL/dt

5 important questions

Q1. What is an inertial frame of reference?

A frame of reference in which Newton's first law holds, i.e. a body with no net force stays at rest or moves with constant velocity. Any frame moving with constant velocity relative to an inertial frame is also inertial.

Q2. Show that impulse equals change in momentum.

From F = dp/dt, F dt = dp. Integrating from time t₁ to t₂: ∫F dt = p₂ − p₁. The left side is impulse J. So J = Δp.

Q3. State and prove the law of conservation of linear momentum for two colliding bodies.

1. Bodies A and B collide; A exerts F_AB on B, B exerts F_BA on A.
2. By third law, F_BA = −F_AB.
3. Change in momentum of A: Δp_A = F_BA Δt; of B: Δp_B = F_AB Δt.
4. Adding: Δp_A + Δp_B = (F_BA + F_AB)Δt = 0.
5. So total momentum before = total momentum after: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂.

Q4. A 4 kg gun fires a 20 g bullet with speed 400 m/s. Find the recoil velocity of the gun.

1. Initial momentum = 0 (both at rest).
2. Final: m_b v_b + m_g v_g = 0.
3. 0.02 × 400 + 4 × v_g = 0 → 8 + 4v_g = 0.
4. v_g = −2 m/s. The gun recoils backward at 2 m/s.

Q5. A ball of mass 2 kg moving at 6 m/s hits a ball of 1 kg at rest in a 1-D elastic collision. Find their velocities after collision.

1. v₁ = (m₁ − m₂)u₁/(m₁ + m₂) = (2 − 1) × 6/3 = 2 m/s.
2. v₂ = 2m₁u₁/(m₁ + m₂) = 2 × 2 × 6/3 = 8 m/s.
3. Check momentum: before 2 × 6 = 12; after 2 × 2 + 1 × 8 = 12 ✓.
4. Check KE: before ½ × 2 × 36 = 36 J; after ½ × 2 × 4 + ½ × 1 × 64 = 4 + 32 = 36 J ✓.

Exam tip

Always state the condition with each conservation law (zero external force, zero external torque, only conservative forces). Checking momentum and KE at the end of a collision numerical earns marks and catches mistakes.

Quick summary

Friction is the force that opposes relative motion between two surfaces in contact. Circular motion needs a force directed towards the centre (centripetal force). This topic covers the laws of friction, the angle of friction and angle of repose, motion on an inclined plane, uniform circular motion, banking of roads and motion in a vertical circle.

Key points

• Static friction adjusts itself up to a maximum value fs(max) = μs N. Kinetic friction fk = μk N, and usually μk is less than μs.
• Laws of limiting friction: friction is independent of the area of contact, depends on the nature of the surfaces, and is proportional to the normal reaction N.
• Rolling friction is much smaller than sliding friction; that is why wheels and ball bearings are used.
• Angle of friction λ: tan λ = μ. Angle of repose α: the largest incline angle at which a body stays at rest; tan α = μs. So α = λ.
• In uniform circular motion speed is constant but velocity changes direction, so there is an acceleration v²/r towards the centre.
• Centripetal force is not a new kind of force: tension, friction, gravity or a normal reaction supplies it.
• Centrifugal force is a pseudo force seen only in the rotating (non-inertial) frame.
• On a banked road, the horizontal component of the normal reaction supplies the centripetal force, so vehicles need less friction.
• In a vertical circle, speed is not constant; the minimum speed at the top is √(gr) and at the bottom √(5gr).

Important terms

• Coefficient of friction (μ) – ratio of limiting friction to normal reaction.
• Angular velocity (ω) – rate of change of angular position, ω = dθ/dt = v/r (rad s⁻¹).
• Banking angle (θ) – angle by which the outer edge of a curved road is raised.
• Conical pendulum – a bob moving in a horizontal circle while the string traces a cone.

Formulas

• f = μN; tan λ = μ; tan α = μs
• Body sliding down a rough incline: a = g(sin θ − μk cos θ)
• Centripetal acceleration a = v²/r = ω²r; force F = mv²/r
• Banking without friction: tan θ = v²/(rg)
• Banking with friction μ: v_max = √[rg(μ + tan θ)/(1 − μ tan θ)]
• Level curve: v_max = √(μrg)
• Vertical circle (string): v_top(min) = √(gr), v_bottom(min) = √(5gr); T_bottom − T_top = 6mg
• Conical pendulum: tan θ = v²/(rg), period T = 2π√(L cos θ/g)

5 important questions

Q1. Define angle of repose and show that it equals the angle of friction.

At the angle of repose α the body is just about to slide. Along the incline: mg sin α = fs(max) = μs N. Perpendicular: N = mg cos α. Dividing, tan α = μs. Since tan λ = μs too, α = λ.

Q2. Derive the expression for centripetal acceleration.

Let a particle move with speed v on a circle of radius r. In a short time Δt it turns through Δθ, and the velocity vector turns by the same angle, so |Δv| = vΔθ. Then a = |Δv|/Δt = v(Δθ/Δt) = vω = v²/r, directed towards the centre.

Q3. Derive the angle of banking for a road (no friction) and solve a problem.

Forces: weight mg down, normal reaction N perpendicular to the road.

• Vertical: N cos θ = mg
• Horizontal: N sin θ = mv²/r
• Dividing: tan θ = v²/(rg)

Example: r = 100 m, v = 72 km h⁻¹ = 20 m s⁻¹, g = 10 m s⁻². tan θ = 400/1000 = 0.4, so θ ≈ 21.8°.

Q4. A block slides down a 30° incline with μk = 0.2. Find its acceleration (g = 9.8 m s⁻²).

a = g(sin θ − μk cos θ) = 9.8(0.5 − 0.2 × 0.866) = 9.8(0.5 − 0.1732) = 9.8 × 0.3268 ≈ 3.2 m s⁻².

Q5. Discuss motion in a vertical circle and find the minimum speed at the top and the bottom.

1. At the top: T + mg = mv_top²/r. The string just stays tight when T = 0, so v_top(min) = √(gr).
2. Energy conservation between bottom and top (height 2r): ½mv_b² = ½mv_top² + mg(2r).
3. With v_top² = gr: v_b² = gr + 4gr = 5gr, so v_bottom(min) = √(5gr).
4. Tension difference: T_b − T_t = 6mg.

Example: r = 0.9 m, g = 10 m s⁻²: v_top = 3 m s⁻¹, v_bottom = √45 ≈ 6.7 m s⁻¹.

Exam tip

Always draw a free-body diagram first. In circular motion, find which real force points to the centre; do not add a "centripetal force" as an extra arrow.

Quick summary

Rotational motion is the turning of a rigid body about an axis. Moment of inertia plays the role that mass plays in straight-line motion. This topic covers angular quantities, torque, the theorems of parallel and perpendicular axes, and the moment of inertia of standard bodies.

Key points

• A rigid body keeps the distance between its particles fixed.
• Angular displacement θ, angular velocity ω = dθ/dt, angular acceleration α = dω/dt.
• Linear and angular quantities are linked: v = rω, a_t = rα.
• Moment of inertia (जडत्व आघूर्ण) I = Σ mr² depends on the mass and how far it is spread from the axis.
• Radius of gyration K: I = MK², the distance at which the whole mass could be placed to give the same I.
• Torque τ = r × F causes angular acceleration: τ = Iα.
• Parallel axis theorem: I = I_cm + Md², where d is the distance between the axis and a parallel axis through the centre of mass.
• Perpendicular axis theorem (only for flat laminas): I_z = I_x + I_y, where x and y lie in the plane and z is perpendicular to it.
• Rotational KE = ½Iω². Angular momentum L = Iω.
• A body rolling without slipping has both translational and rotational KE: KE = ½Mv²(1 + K²/R²).
• A flywheel is a heavy wheel with large I used to keep speed steady in engines.

Important terms

• Axis of rotation – the fixed line about which the body turns.
• Torque (moment of force) – turning effect of a force, unit N m.
• Couple – two equal, opposite, parallel forces not on the same line.
• Radius of gyration – K = √(I/M).

Formulas

• τ = Iα (N m); L = Iω (kg m² s⁻¹); KE_rot = ½Iω² (J)
• Rolling down an incline of height h: v = √(2gh/(1 + K²/R²))
• Flywheel: mgh = ½mv² + ½Iω² + n₁W_f (friction work for n₁ turns)

5 important questions

Q1. State the parallel axis theorem.

The moment of inertia of a body about any axis equals its moment of inertia about a parallel axis through its centre of mass plus Md², where M is its mass and d the distance between the two axes: I = I_cm + Md².

Q2. Define radius of gyration. What is it for a disc about its central perpendicular axis?

Radius of gyration K is the distance from the axis at which the whole mass, if concentrated, gives the same moment of inertia: I = MK². For a disc, I = ½MR², so K = R/√2.

Q3. Derive the moment of inertia of a thin uniform rod about an axis through its centre perpendicular to its length.

1. Rod: mass M, length L; mass per unit length λ = M/L.
2. Take element dx at distance x from centre; mass dm = λ dx.
3. dI = x² dm = λx² dx.
4. I = ∫ from −L/2 to L/2 of λx² dx = λ [x³/3] = λ (L³/24 + L³/24) = λL³/12.
5. Substitute λ = M/L: I = ML²/12.
6. About one end (parallel axis, d = L/2): I = ML²/12 + ML²/4 = ML²/3.

Q4. Find the moment of inertia of a disc about its diameter using the perpendicular axis theorem.

1. About the central perpendicular axis, I_z = ½MR².
2. By symmetry, I about any two perpendicular diameters are equal: I_x = I_y = I_d.
3. Perpendicular axis theorem: I_z = I_x + I_y = 2I_d.
4. I_d = ½ × ½MR² = ¼MR².

Q5. A solid sphere of mass 2 kg and radius 0.1 m rotates about its diameter at 10 rad/s. Find its moment of inertia, rotational KE and angular momentum.

1. I = (2/5)MR² = 0.4 × 2 × 0.01 = 0.008 kg m².
2. KE = ½Iω² = ½ × 0.008 × 100 = 0.4 J.
3. L = Iω = 0.008 × 10 = 0.08 kg m² s⁻¹.

Exam tip

Learn the I-table with axes; most numericals simply pick one formula and add Md² or use the perpendicular axis theorem. Draw a neat figure showing the axis — it earns a mark.

Quick summary

Gravitation is the force of attraction between any two masses. This topic covers Newton's law of gravitation, gravitational field and potential, variation of g with height and depth, Kepler's three laws of planetary motion, orbital velocity, escape velocity and geostationary satellites.

Key points

• Newton's law: F = G m₁m₂/r², along the line joining the masses; always attractive. G = 6.67 × 10⁻¹¹ N m² kg⁻².
• Gravitational field E = F/m = GM/r² (N kg⁻¹); it points towards the mass.
• Gravitational potential V = −GM/r (J kg⁻¹); it is zero at infinity and negative everywhere else. E = −dV/dr.
• Potential energy of two masses U = −Gm₁m₂/r.
• At the Earth's surface g = GM/R² ≈ 9.8 m s⁻².
• g decreases with height: g_h = g R²/(R + h)² ≈ g(1 − 2h/R) for small h.
• g decreases with depth: g_d = g(1 − d/R); g = 0 at the centre.
• Kepler's laws: (1) orbits are ellipses with the Sun at one focus; (2) the line from Sun to planet sweeps equal areas in equal times; (3) T² ∝ a³.
• The second law is a consequence of conservation of angular momentum, because gravity is a central force (torque = 0).
• Escape velocity from Earth ≈ 11.2 km s⁻¹; it does not depend on the mass of the body or the direction of throw.
• A geostationary satellite has period 24 h, moves west to east above the equator, at about 36,000 km height.

Important terms

• Central force – a force always directed along the line joining the body to a fixed point.
• Areal velocity – area swept per unit time by the radius vector, dA/dt = L/2m.
• Orbital velocity – speed needed to move in a circular orbit.
• Binding energy – energy needed to free a satellite from the Earth's pull, = GMm/2r for a circular orbit.

Formulas

• F = G m₁m₂/r²; g = GM/R²
• V = −GM/r; U = −GMm/r
• Orbital velocity v₀ = √(GM/r) = √(gR²/r); near the surface v₀ = √(gR) ≈ 7.9 km s⁻¹
• Period T = 2πr/v₀ = 2π√(r³/GM)
• Escape velocity v_e = √(2GM/R) = √(2gR) = √2 × v₀(surface)
• Total energy of a satellite E = −GMm/2r (KE = GMm/2r, PE = −GMm/r)
• Areal velocity dA/dt = L/(2m) = constant

5 important questions

Q1. State Kepler's three laws of planetary motion.

1. Law of orbits: every planet moves in an elliptical orbit with the Sun at one focus.
2. Law of areas: the radius vector sweeps equal areas in equal times (areal velocity is constant).
3. Law of periods: the square of the period is proportional to the cube of the semi-major axis, T² ∝ a³.

Q2. Show that Kepler's second law follows from conservation of angular momentum.

In time dt the radius vector r sweeps a small triangle of area dA = ½|r × v dt|. So dA/dt = ½|r × v| = L/(2m). Gravity is a central force, so torque τ = r × F = 0, and L stays constant. Hence dA/dt is constant: equal areas in equal times.

Q3. Derive the expression for escape velocity and calculate it for Earth.

A body of mass m escapes if its kinetic energy can carry it to infinity, where its PE becomes zero:
½mv_e² − GMm/R = 0, so v_e = √(2GM/R) = √(2gR).
With g = 9.8 m s⁻² and R = 6.4 × 10⁶ m: v_e = √(2 × 9.8 × 6.4 × 10⁶) = √(1.254 × 10⁸) ≈ 1.12 × 10⁴ m s⁻¹ = 11.2 km s⁻¹.

Q4. Derive Kepler's third law for a circular orbit and find the height of a geostationary satellite.

Gravity gives the centripetal force: GMm/r² = mv²/r, so v² = GM/r. With v = 2πr/T: 4π²r²/T² = GM/r, so T² = (4π²/GM) r³, i.e. T² ∝ r³.
For T = 86,400 s and GM = gR² = 9.8 × (6.4 × 10⁶)² ≈ 4.01 × 10¹⁴ m³ s⁻²:
r³ = GM T²/4π² ≈ 4.01 × 10¹⁴ × 7.46 × 10⁹ / 39.48 ≈ 7.59 × 10²² m³, so r ≈ 4.23 × 10⁷ m.
Height h = r − R ≈ 4.23 × 10⁷ − 0.64 × 10⁷ ≈ 3.6 × 10⁷ m (about 36,000 km).

Q5. Derive the variation of g with depth.

At depth d, only the inner sphere of radius (R − d) attracts the body (the outer shell gives zero net force). Taking uniform density ρ, mass inside = (4/3)π(R − d)³ρ.
g_d = G(4/3)π(R − d)ρ, while g = G(4/3)πRρ. Dividing: g_d = g(1 − d/R). At the centre (d = R), g = 0.

Exam tip

Remember v_e = √2 × v₀ (near the surface): 11.2 ≈ 1.414 × 7.9. Potential and potential energy are always negative in gravitation; do not drop the minus sign.

Quick summary

Elasticity is the property of a body to regain its original shape and size when the deforming force is removed. This topic covers stress, strain, Hooke's law, the three elastic moduli, Poisson's ratio, their relations, and the bending of beams and torsion of a wire.

Key points

• Stress = restoring force per unit area (N m⁻², Pa). Strain = change in dimension ÷ original dimension (no unit).
• Hooke's law: within the elastic limit, stress ∝ strain. Stress/strain = modulus of elasticity.
• Three moduli: Young's modulus Y (length change), bulk modulus K (volume change), modulus of rigidity η (shape change, shear).
• Poisson's ratio σ = lateral strain ÷ longitudinal strain. Theoretical range −1 < σ < 0.5; for real materials it lies between 0 and 0.5.
• On a stress–strain curve: proportional limit, elastic limit, yield point, ultimate stress, breaking point.
• Work done in stretching a wire is stored as elastic potential energy.
• A beam fixed at one end and loaded at the other is a cantilever. Bending moment depends on Y and the geometrical moment of inertia I_g.
• For a rectangular cross-section, I_g = bd³/12; for a circular one, I_g = πr⁴/4.
• Torsion: twisting a wire fixed at one end produces a restoring couple proportional to the twist angle.
• Ductile materials (copper) show large plastic region; brittle materials (glass) break soon after the elastic limit.

Important terms

• Elastic limit (स्थितिस्थापकता मर्यादा) – maximum stress up to which the body fully recovers.
• Plasticity – the property of not regaining the original shape.
• Elastic fatigue – loss of elastic strength after repeated stress cycles.
• Neutral axis – the line in a bent beam whose length does not change.
• Torsional rigidity (C) – couple per unit twist of a wire.

Formulas

• Y = (F/A)/(Δl/l) = Fl/(AΔl)
• K = −p/(ΔV/V) = −pV/ΔV
• η = (F/A)/θ, where θ is shear angle
• Poisson's ratio σ = (Δd/d)/(Δl/l)
• Relations: Y = 3K(1 − 2σ); Y = 2η(1 + σ); 9/Y = 1/K + 3/η; σ = (3K − 2η)/(6K + 2η)
• Energy per unit volume = ½ × stress × strain; total work W = ½ F Δl
• Bending moment = YI_g/R (R = radius of curvature of neutral axis)
• Cantilever depression at free end: δ = WL³/(3YI_g)
• Twisting couple per unit twist: C = πηr⁴/(2l); couple = Cθ
• Torsional pendulum: T = 2π√(I/C)

5 important questions

Q1. State Hooke's law and define Young's modulus.

Within the elastic limit, stress is directly proportional to strain. Young's modulus is the ratio of longitudinal (tensile) stress to longitudinal strain: Y = Fl/(AΔl). Its SI unit is N m⁻² (Pa).

Q2. What is Poisson's ratio? Why can it not exceed 0.5?

Poisson's ratio is lateral strain divided by longitudinal strain. From Y = 3K(1 − 2σ), if σ > 0.5, K would be negative, which is not physical (volume would increase under compression). So σ < 0.5.

Q3. Derive the expression for the work done in stretching a wire.

1. Wire: length l, area A, modulus Y. At extension x, restoring force F = YAx/l.
2. Work for small extra extension dx: dW = F dx = (YA/l) x dx.
3. Total work for extension Δl: W = ∫₀^Δl (YA/l) x dx = YA(Δl)²/(2l).
4. Since F = YAΔl/l, W = ½ F Δl.
5. Per unit volume (divide by Al): W/V = ½ stress × strain.

Q4. A steel wire of length 2 m and cross-section 1 mm² is stretched by a load of 10 kg. If Y = 2 × 10¹¹ N m⁻², find the extension (g = 9.8 m s⁻²).

1. F = 10 × 9.8 = 98 N; A = 1 mm² = 1 × 10⁻⁶ m²; l = 2 m.
2. Δl = Fl/(AY) = (98 × 2)/(10⁻⁶ × 2 × 10¹¹).
3. Denominator = 2 × 10⁵.
4. Δl = 196/(2 × 10⁵) = 9.8 × 10⁻⁴ m ≈ 0.98 mm.

Q5. For a material, Y = 2 × 10¹¹ Pa and η = 0.8 × 10¹¹ Pa. Find Poisson's ratio and bulk modulus.

1. Y = 2η(1 + σ) → 1 + σ = Y/(2η) = 2/(1.6) = 1.25 → σ = 0.25.
2. Y = 3K(1 − 2σ) → K = Y/(3(1 − 0.5)) = 2 × 10¹¹/1.5.
3. K ≈ 1.33 × 10¹¹ Pa.

Exam tip

Learn the four relations among Y, K, η and σ together — numericals often give two of them and ask the other two. Remember mm² = 10⁻⁶ m² when converting.

Quick summary

Fluids (liquids and gases) can flow and take the shape of their container. This topic covers pressure in a fluid, Pascal's law and the hydraulic press, Archimedes' principle, types of flow, the equation of continuity, Bernoulli's theorem and its applications such as the venturimeter, Torricelli's theorem and the lift on an aircraft wing.

Key points

• Pressure P = F/A (N m⁻² = Pa). In a liquid at rest it acts equally in all directions at a point.
• Pressure at depth h: P = P₀ + ρgh. It depends only on depth, not on the shape of the vessel.
• Pascal's law: pressure applied to an enclosed fluid is transmitted equally to every part of it. Used in hydraulic lifts and brakes.
• Archimedes' principle: a body in a fluid feels an upward buoyant force equal to the weight of fluid displaced.
• Streamline flow: each particle follows the path of the one before it; velocity at a point does not change with time. Turbulent flow: irregular, with eddies.
• Equation of continuity (from conservation of mass): A₁v₁ = A₂v₂ for an incompressible fluid. Narrow pipe → faster flow.
• Bernoulli's theorem (from conservation of energy): for ideal streamline flow, P + ½ρv² + ρgh = constant along a streamline.
• Where speed is high, pressure is low. This explains the venturimeter, atomiser, lift of wings and roofs blown off in storms.
• Bernoulli's theorem assumes the fluid is non-viscous, incompressible and in steady streamline flow.

Important terms

• Pressure head P/ρg, velocity head v²/2g, gravitational head h – the three terms of Bernoulli's equation in units of length.
• Venturimeter – a tube with a narrow throat used to measure flow rate.
• Pitot tube – device to measure the speed of fluid flow.
• Efflux – liquid flowing out through an opening.

Formulas

• P = P₀ + ρgh (ρ = density in kg m⁻³, h = depth in m)
• Hydraulic press: F₁/A₁ = F₂/A₂
• Buoyant force F_B = ρ_fluid V g
• Continuity: A₁v₁ = A₂v₂ = volume flow rate Q (m³ s⁻¹)
• Bernoulli: P + ½ρv² + ρgh = constant
• Torricelli: speed of efflux v = √(2gh)
• Venturimeter: v₁ = A₂ √[2(P₁ − P₂) / ρ(A₁² − A₂²)], Q = A₁v₁

5 important questions

Q1. State Pascal's law and explain the hydraulic press.

Pressure applied to an enclosed liquid is transmitted undiminished to every point. In a hydraulic press a small force F₁ on a small piston A₁ creates pressure F₁/A₁, which acts on the large piston A₂, giving F₂ = F₁(A₂/A₁). A small force lifts a heavy load.

Q2. Water flows through a pipe of diameter 4 cm at 2 m s⁻¹. Find its speed where the diameter is 2 cm.

A₁v₁ = A₂v₂ and A ∝ d², so v₂ = v₁(d₁/d₂)² = 2 × (4/2)² = 8 m s⁻¹.

Q3. State and prove Bernoulli's theorem.

1. Take a tube of flow. At section 1: area A₁, pressure P₁, speed v₁, height h₁; at section 2: A₂, P₂, v₂, h₂.
2. In time Δt, volume ΔV = A₁v₁Δt = A₂v₂Δt (continuity) flows through each section; mass Δm = ρΔV.
3. Net work done by pressure = P₁ΔV − P₂ΔV.
4. This work equals the gain in KE and PE: (P₁ − P₂)ΔV = ½ρΔV(v₂² − v₁²) + ρΔVg(h₂ − h₁).
5. Rearranging: P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂, i.e. P + ½ρv² + ρgh = constant.

Q4. Derive Torricelli's theorem and solve: water stands 5 m above a small hole. Find the speed of efflux (g = 9.8 m s⁻²).

Apply Bernoulli between the top surface (P₀, v ≈ 0, height h) and the hole (P₀, v, height 0): P₀ + ρgh = P₀ + ½ρv². So v = √(2gh), the same as free fall from height h.
v = √(2 × 9.8 × 5) = √98 ≈ 9.9 m s⁻¹.

Q5. Explain the working of a venturimeter.

A venturimeter has a wide inlet (area A₁) and a narrow throat (A₂). By continuity, speed increases at the throat; by Bernoulli (same height), pressure falls there. A manometer measures P₁ − P₂. Combining A₁v₁ = A₂v₂ with P₁ − P₂ = ½ρ(v₂² − v₁²) gives v₁ = A₂√[2(P₁ − P₂)/ρ(A₁² − A₂²)], and the flow rate Q = A₁v₁. It is used to measure flow in water mains and fuel lines.

Exam tip

Continuity comes from conservation of mass and Bernoulli from conservation of energy — examiners like this one-line link. Always convert diameters to areas (A ∝ d²) before using A₁v₁ = A₂v₂.

Quick summary

Surface tension explains why a liquid surface behaves like a stretched membrane — drops are round and water rises in thin tubes. Viscosity is the internal friction of a fluid that opposes relative motion between its layers. This topic covers molecular theory, capillary rise, excess pressure, Newton's law of viscosity, Stokes' law and Poiseuille's equation.

Key points

• Cohesive forces act between like molecules; adhesive forces between unlike molecules.
• A molecule at the surface feels a net inward pull, so the surface tends to contract to minimum area.
• Surface tension T = force per unit length acting along the surface, perpendicular to a line drawn on it. Unit N m⁻¹.
• Surface energy = work done per unit area to increase the surface; numerically equal to T.
• Angle of contact θ: less than 90° for water–glass (liquid wets glass, rises in capillary); more than 90° for mercury–glass (falls).
• Surface tension decreases as temperature rises; detergents lower it.
• Viscosity (श्यानता): the property by which a fluid opposes relative motion of its layers.
• Streamline flow happens at low speed; turbulent flow at high speed. The change is decided by Reynolds number.
• Stokes' law gives the drag on a small sphere moving slowly in a viscous fluid.
• Terminal velocity: the constant speed reached when weight = buoyancy + viscous drag.
• Viscosity of liquids decreases with temperature; of gases it increases.

Important terms

• Capillarity – rise or fall of liquid in a narrow tube.
• Coefficient of viscosity (η) – tangential force per unit area per unit velocity gradient. SI unit N s m⁻² (Pa s); CGS unit poise (1 Pa s = 10 poise).
• Velocity gradient – dv/dx, change of velocity per unit distance between layers.
• Critical velocity – the speed above which flow becomes turbulent.

Formulas

• T = F/l (N m⁻¹); surface energy W = T × ΔA
• Excess pressure in a liquid drop: P = 2T/r; in a soap bubble (two surfaces): P = 4T/r
• Capillary rise (Jurin's law): h = 2T cos θ/(rρg)
• Newton's law of viscosity: F = −ηA (dv/dx)
• Stokes' law: F = 6πηrv
• Terminal velocity: v = 2r²(ρ − σ)g/(9η) (ρ = density of sphere, σ = density of fluid)
• Poiseuille's equation: V/t = πPr⁴/(8ηl)
• Reynolds number: Re = ρvD/η (low Re → streamline)

5 important questions

Q1. Why does mercury fall in a glass capillary while water rises?

For water–glass, adhesion is stronger than cohesion; θ < 90°, cos θ is positive, so h is positive (rise). For mercury, cohesion is stronger; θ > 90°, cos θ is negative, so h is negative (fall).

Q2. State Stokes' law and give the meaning of the terms.

A small sphere of radius r moving with small velocity v through a fluid of viscosity η feels a viscous drag F = 6πηrv. It holds for streamline flow only.

Q3. Derive the excess pressure inside a liquid drop.

1. Drop of radius r, surface tension T, excess pressure P inside.
2. Let radius increase by dr due to P.
3. Work done by pressure = P × 4πr² dr.
4. Increase in surface area = 8πr dr; increase in surface energy = T × 8πr dr.
5. Equate: P × 4πr² dr = 8πrT dr → P = 2T/r.
6. For a soap bubble with two surfaces, P = 4T/r.

Q4. Water rises 3 cm in a capillary of radius 0.5 mm. Find the surface tension of water (θ = 0°, ρ = 1000 kg m⁻³, g = 9.8 m s⁻²).

1. T = hrρg/(2 cos θ).
2. h = 0.03 m, r = 0.5 × 10⁻³ m.
3. T = 0.03 × 0.5 × 10⁻³ × 1000 × 9.8/2.
4. Numerator = 0.03 × 0.5 × 9.8 = 0.147; T = 0.147/2 = 0.0735 N m⁻¹.

Q5. Derive the terminal velocity of a sphere falling in a viscous liquid.

1. Forces: weight W = (4/3)πr³ρg down; buoyancy B = (4/3)πr³σg up; drag F = 6πηrv up.
2. At terminal velocity, net force = 0: W = B + F.
3. (4/3)πr³(ρ − σ)g = 6πηrv.
4. v = 2r²(ρ − σ)g/(9η).
5. So v ∝ r²; bigger drops fall faster.

Exam tip

Remember: drop = 2T/r, bubble = 4T/r (two surfaces). In viscosity numericals convert poise to Pa s (divide by 10) and mm to m before substituting.