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Sindh Board • XI

PHYSICS • CLASS XI

Understand the idea. Derive the result. Apply it. Test yourself.

A structured first-year Physics path for Sindh Board and BIEK students, aligned with the current Class XI sequence.

An initiative of Merit 'n' Merit Educational Network ↗

STUDY PATH

Model Paper 2026 + Units 1-14

STBB Class XI Physics (English medium, 2025-26) • BIEK Physics Paper-I Model Paper 2026

MODELModel Paper 2026 1Physics and Measurements2Kinematics3Dynamics4Rotational and Circular Motion5Work, Energy and Power6Fluid Statics7Fluid Dynamics8Electric Fields9Capacitors10D.C. Circuits11Oscillations12Acoustics13Physical Optics14Communication

OFFICIAL BIEK MATERIAL

Model Paper 2026

BOARD OF INTERMEDIATE EDUCATION, KARACHI • INTERMEDIATE EXAMINATION • PHYSICS PAPER - I • According to New Book

Official Model PaperThis transcription is kept separate from PrepMode-generated practice. Mathematical typography has been normalized where the PDF extraction is visibly damaged.
SECTION A - MULTIPLE CHOICE QUESTIONS 17 marks

Time: 20 Minutes. Select the correct answer for each from the given options.

  1. A ball is thrown upward with a velocity of \(100\,\mathrm{m\,s^{-1}}\). The time it takes to reach the ground is \((g=10\,\mathrm{m\,s^{-2}})\):

    1. 5 second
    2. 10 second
    3. 20 second
    4. 40 second

  2. If the dot product of two non-zero vectors vanishes, then vectors are:

    1. perpendicular
    2. parallel
    3. in opposite directions
    4. at an angle of 45°

  3. This force is also called a self-adjusting force:

    1. Friction
    2. Tension
    3. Weight
    4. Thrust

  4. A wooden block of volume \(0.05\,\mathrm{m^3}\) is floating on the surface of water. The buoyant force acting on the block is \((\text{Density of water}=1000\,\mathrm{kg\,m^{-3}})\):

    1. 500 N
    2. 50 N
    3. 5000 N
    4. 5 N

  5. The SI-unit of conductance is:

    1. Ohms \((\Omega)\)
    2. siemens \((S)\)
    3. ampere \((A)\)
    4. volt \((V)\)

  6. Coulomb’s force in a dielectric medium is less than that in vacuum, due to:

    1. Ionization
    2. Electric polarization
    3. Magnetization
    4. Superposition

  7. The displacement-time graph for simple harmonic motion is:

    1. straight line
    2. circle
    3. ellipse
    4. sine curve

  8. The Doppler Effect is used in this medical imaging technique:

    1. Ultrasound
    2. X-rays
    3. Magnetic resonance image
    4. Computed Tomography Scan

  9. The superposition of signal wave on a carrier wave is called:

    1. Modulation
    2. Diffraction
    3. Polarization
    4. Refraction

  10. The unit of electric flux is:

    1. \(\mathrm{V/m}\)
    2. \(\mathrm{N/m}\)
    3. \(\mathrm{N\,m^2/C}\)
    4. \(\mathrm{N/C}\)

  11. The resistance of a superconductor below a critical temperature is:

    1. Infinite
    2. Zero
    3. Finite
    4. Unchanged

  12. In SHM, kinetic energy is maximum at:

    1. extreme position
    2. mean position
    3. between mean and extreme positions
    4. mean and extreme positions

  13. The dimensions of angular momentum are given by:

    1. \(ML^{-1}T^{-2}\)
    2. \(ML^2T^{-1}\)
    3. \(M^0L^2T^{-1}\)
    4. \(MLT^{-3}\)

  14. A body weighs \(10\,\mathrm N\) out of water and \(7\,\mathrm N\) when submerged in water. The buoyant force on the body will be:

    1. 3 N
    2. 5 N
    3. 7 N
    4. 10 N

  15. A mass spring-oscillator has a time period \(T\), if the mass is doubled, the time period will become:

    1. \(T\)
    2. \(2T\)
    3. \(\sqrt2T\)
    4. \(T/\sqrt2\)

  16. This electromagnetic wave has the shortest wavelength:

    1. Radio wave
    2. Ultraviolet wave
    3. Microwaves
    4. Infrared wave

  17. The maximum number of beats that can be heard by human is:

    1. 3
    2. 5
    3. 7
    4. 9
SECTION B - SHORT-ANSWER QUESTIONS 36 marks

Answer any Nine part questions from this section. All part questions carry equal marks. Draw diagrams where necessary. Use of scientific calculator is allowed.

  1. What is an ideal banking angle for a turn of 1.20 km radius on a highway with a 105 km/h speed limit?
  2. Prove the following equations are dimensionally correct: \(F=mv^2/r\) and \(2aS=v_f^2-v_i^2\).
  3. Two tug boats are towing a ship, each exerting a force of 6000 N and the angle between two forces is \(60^\circ\). Calculate the resultant force on the ship.
  4. A 50g bullet is fired into a 10kg block suspended by a cord. The center of gravity of block rises by 10cm. What is the speed of bullet?
  5. A car starts from rest and moves with a constant acceleration. During the 5th second of its motion it covers a distance of 36m. Calculate acceleration of the car.
  6. A 70kg man runs up a hill through a height of 3m in 2s. Calculate the work done and power output?
    OR
    A 20m long wire has a cross sectional area of 1mm² and a resistance of \(5\Omega\). Calculate the conductance of the material of wire.
  7. A particle having charge \(2\times10^{-19}\,\mathrm C\), is held in an electric field between two parallel metal plates 4cm apart, is acted upon by a force of \(10^{-4}\,\mathrm N\). What is the intensity of the electric field?
    OR
    The period of oscillation of an object of an ideal mass-spring system is 0.50s and the amplitude is 5cm. What is the speed of object at equilibrium position?
  8. If the speed of sound in air at \(27^\circ\mathrm C\) is \(345\,\mathrm{m/s}\), find the speed at \(127^\circ\mathrm C\).
    OR
    A source of sound and a listener are moving toward each other with velocities 0.5 times and 0.2 times the speed of sound respectively. If the frequency of emitted sound is 2000Hz, calculate the change in the frequency heard by the listener.
  9. Derive an equation for balanced Wheatstone bridge.
    OR
    Define Electric flux, write its SI unit. Under what conditions the electric flux through a surface will be: Maximum; Minimum.
  10. Prove that the gravitational field of earth is conservative.
    OR
    What is interference of light? Write three conditions of interference of light.
SECTION C - DETAILED-ANSWER QUESTIONS 32 marks

Answer any Four questions from this section. All questions carry equal marks. Draw diagrams where necessary.

Question 3

Two vectors \(A_1\) and \(A_2\) are making angles \(\theta_1\) and \(\theta_2\) with x-axis respectively. Derive the formulae for the magnitude and direction of the resultant vector.

Question 4

Define Simple Harmonic Motion. A particle is moving in a circle with a constant speed; prove that its projection executes simple harmonic motion along the diameter of the circle.

Question 5

Define the capacitance of a capacitor. Derive mathematical relations for the capacitance of parallel plate capacitor when: (i) Air is present between the plates (ii) A dielectric slab is present between the plates.
OR
Two smooth, rigid and not-rotating spheres of masses \(m_1\) and \(m_2\), moving with initial velocities \(U_1\) and \(U_2\) respectively, collide elastically in one dimension. Derive the expression for the velocity of any one of the sphere after collision.

Question 6

State Bernoulli’s Theorem and derive its equation.
OR
What are Newton’s Rings? Explain the process of their formation. Derive the expressions for the radii of nth bright and dark rings.

Question 7

What are Stationary Waves? If stationary waves are set up in a stretched string, derive expressions for the frequencies when string is vibrating in: (i) One loop (ii) Two loops (iii) Three loops (iv) n loops.
OR
What is an Electric Dipole? Derive the expression for electric field intensity at a point at perpendicular distance ‘y’ from the center of the dipole.

CLASS XI PHYSICS

Units

The learning sequence stays consistent: Concept → Derivation → Worked Example → MCQs.

Physics describes matter, energy, motion and interactions through measurable quantities and mathematical relationships.

RememberEvery measured quantity should be written with a numerical value and an appropriate unit.
\[Q=n\times u\]

Important ideas include SI base and derived units, dimensions, significant figures, uncertainty and dimensional analysis.

\[[F]=MLT^{-2},\qquad [E]=ML^2T^{-2}\]

Coverage check

  • SI base and derived quantities; scientific notation and prefixes.
  • Accuracy, precision, least count, uncertainty and significant figures.
  • Dimensions and dimensional homogeneity.
  • Scalars and vectors, vector addition, components and resultant.
\[R_x=\sum A_x,\qquad R_y=\sum A_y,\qquad R=\sqrt{R_x^2+R_y^2}\]

Complete chapter revision

Use this as the chapter-level revision pass before moving to derivations and numericals.

Measurement and SI

Physical quantities are expressed as a numerical value multiplied by a unit. Distinguish base quantities from derived quantities and use SI prefixes correctly. Convert units before substitution, especially cm to m, g to kg and powers of ten.

Accuracy and precision

Accuracy describes closeness to the accepted value; precision describes repeatability. Least count limits instrument resolution. Absolute, fractional and percentage uncertainty should be carried consistently through measurements.

Significant figures

Record only justified digits. In multiplication/division, the result normally follows the least number of significant figures; in addition/subtraction, it follows the least number of decimal places.

Dimensions

Write a derived quantity in powers of M, L and T. Dimensional homogeneity is a necessary condition for a physical equation, but dimensional correctness alone does not prove an equation physically complete.

Vectors

A vector has magnitude and direction. Resolve vectors into rectangular components, add components algebraically, then reconstruct the resultant magnitude and direction.

Exam focus

Be ready to check equations dimensionally, convert units, interpret significant figures, and determine a resultant using components or the parallelogram law.

Dimensional checking: an equation is dimensionally consistent when both sides have the same dimensions.

For centripetal force,

\[F=\frac{mv^2}{r}\]

Dimensions of the right side are

\[\frac{M(LT^{-1})^2}{L}=MLT^{-2}\]

which are the dimensions of force. Hence

\[\boxed{\left[\frac{mv^2}{r}\right]=[F]}\]

Resultant of two vectors

Resolve each vector into rectangular components. If the two vectors make angles \(\theta_1\) and \(\theta_2\) with the x-axis,

\[R_x=A_1\cos\theta_1+A_2\cos\theta_2,\quad R_y=A_1\sin\theta_1+A_2\sin\theta_2\]
\[\boxed{R=\sqrt{R_x^2+R_y^2}},\qquad \boxed{\tan\phi=\frac{R_y}{R_x}}\]

Derivation bank

Revise the starting relation, the key transformation and the final result for each.

1

Dimensions of acceleration and force

\[a=\frac{\Delta v}{\Delta t}\Rightarrow[a]=LT^{-2},\qquad F=ma\Rightarrow[F]=MLT^{-2}\]
2

Resultant from rectangular components

\[R_x=\sum A_i\cos\theta_i,\quad R_y=\sum A_i\sin\theta_i,\quad R=\sqrt{R_x^2+R_y^2},\quad \tan\phi=\frac{R_y}{R_x}\]
3

Percentage uncertainty

\[\text{percentage uncertainty}=\frac{\Delta Q}{Q}\times100\%\]

Given \(v=20.0\,\mathrm{m\,s^{-1}}\), \(t=4.0\,\mathrm s\)

Required Distance assuming constant speed.

Formula \(s=vt\)

Substitution \(s=(20.0)(4.0)\)

Calculation \(s=80\,\mathrm m\)

Final Answer \(\boxed{80\,\mathrm m}\)

Second worked example

Given \(A=3.0\,\mathrm N\) east, \(B=4.0\,\mathrm N\) north.

Required Resultant.

Formula \(R=\sqrt{A^2+B^2}\)

Calculation \(R=5.0\,\mathrm N\), \(\theta=\tan^{-1}(4/3)=53.1^\circ\) north of east.

Final Answer \(\boxed{5.0\,\mathrm N\text{ at }53.1^\circ}\)

Additional revision examples

These PrepMode examples reinforce the same chapter relationships and are not labelled as official BIEK questions.

Revision Example 1: Percentage uncertainty

Given \(\;Q=50.0\pm0.5\,\mathrm{cm}\;\)

Required Determine the stated quantity.

Formula / Substitution / Calculation \(\frac{0.5}{50.0}\times100=1.0\%\)

Final Answer \(\boxed{1.0\%}\)

Revision Example 2: Vector components

Given \(\;A=10\,\mathrm N,\;\theta=30^\circ\;\)

Required Determine the stated quantity.

Formula / Substitution / Calculation \(A_x=A\cos\theta,\;A_y=A\sin\theta\Rightarrow A_x=8.66\,\mathrm N,\;A_y=5.00\,\mathrm N\)

Final Answer \(\boxed{A_x=8.66\,\mathrm N,\;A_y=5.00\,\mathrm N}\)

PREPMODE PRACTICE • NOT OFFICIAL BIEK QUESTIONS

1. The SI base unit of length is:

Answer: metre
The metre is the SI base unit of length.

2. The dimensions of velocity are:

Answer: \(LT^{-1}\)
Velocity is displacement divided by time.

3. Dimensional analysis can be used to check:

Answer: dimensional consistency
Both sides of a valid physical equation must have matching dimensions.

4. A quantity with dimensions \(ML^2T^{-1}\) is:

Answer: angular momentum
For \(L=rp\), dimensions are \(ML^2T^{-1}\).

5. Which is a derived SI unit?

Answer: newton
Newton is kg m s^-2.

6. Precision refers mainly to:

Answer: repeatability of readings
Precision concerns spread/reproducibility.

7. The dimensional formula of energy is:

Answer: \(ML^2T^{-2}\)
Energy is force times distance.

8. Two perpendicular vectors 3 and 4 units have resultant magnitude:

Answer: 5
Use Pythagoras.

9. A scalar quantity is:

Answer: mass
Mass has magnitude only.

10. Dimensional analysis cannot determine:

Answer: dimensionless numerical constants
It cannot recover pure numbers such as 1/2 or 2π.

Kinematics describes motion without considering its causes. The central quantities are displacement, velocity and acceleration.

\[v=u+at\]
\[s=ut+\frac12at^2\]
\[v^2=u^2+2as\]

Graphs are also important: the slope of a displacement-time graph gives velocity, while the slope of a velocity-time graph gives acceleration.

Coverage check

  • Position, displacement, speed, velocity and acceleration.
  • Uniformly accelerated motion and motion graphs.
  • Free fall and vertical motion under gravity.
  • Projectile motion as independent horizontal and vertical motions.
\[T=\frac{2u\sin\theta}{g},\quad H=\frac{u^2\sin^2\theta}{2g},\quad R=\frac{u^2\sin2\theta}{g}\]

Complete chapter revision

Use this as the chapter-level revision pass before moving to derivations and numericals.

Motion variables

Position locates an object; displacement is the vector change in position. Average velocity is displacement per time, while instantaneous velocity is the slope of the position-time curve.

Acceleration

Acceleration is the rate of change of velocity. Constant acceleration allows use of the standard equations; non-uniform motion must be treated through graphs or calculus.

Graphs

Slope of x-t gives velocity; slope of v-t gives acceleration; area under a v-t graph gives displacement. A horizontal v-t line represents constant velocity.

Projectile motion

Resolve initial velocity into horizontal and vertical components. Horizontal acceleration is zero while vertical acceleration is g downward, if air resistance is neglected.

Relative motion

Relative velocity is the velocity of one object as observed from another and is found by vector subtraction.

Exam focus

Know the assumptions behind constant-acceleration equations and distinguish distance/speed from displacement/velocity.

Equation without time: start with \(v=u+at\), so

\[t=\frac{v-u}{a}\]

Using average velocity for constant acceleration, \(s=\frac{u+v}{2}t\). Substitute for \(t\):

\[s=\frac{u+v}{2}\frac{v-u}{a}=\frac{v^2-u^2}{2a}\]
\[\boxed{v^2=u^2+2as}\]

Projectile range

Horizontal speed remains \(u\cos\theta\). The time of flight is \(T=2u\sin\theta/g\). Therefore

\[R=(u\cos\theta)T=\frac{u^2(2\sin\theta\cos\theta)}{g}\]
\[\boxed{R=\frac{u^2\sin2\theta}{g}}\]

Derivation bank

Revise the starting relation, the key transformation and the final result for each.

1

Second equation of motion

\[s=\bar vt=\frac{u+v}{2}t,\quad v=u+at\Rightarrow \boxed{s=ut+\frac12at^2}\]
2

Third equation of motion

\[s=\frac{u+v}{2}\frac{v-u}{a}\Rightarrow \boxed{v^2=u^2+2as}\]
3

Projectile range

\[T=\frac{2u\sin\theta}{g},\quad R=(u\cos\theta)T\Rightarrow\boxed{R=\frac{u^2\sin2\theta}{g}}\]

Given A car starts from rest, \(u=0\), and covers \(36\,\mathrm m\) during the 5th second.

Required Acceleration \(a\)

Formula Distance in the \(n\)-th second: \(s_n=u+\frac a2(2n-1)\)

Substitution \(36=0+\frac a2(9)\)

Calculation \(a=8\,\mathrm{m\,s^{-2}}\)

Final Answer \(\boxed{8\,\mathrm{m\,s^{-2}}}\)

Second worked example

Given \(u=20\,\mathrm{m/s},\theta=30^\circ,g=9.8\,\mathrm{m/s^2}\).

Required Maximum height.

Formula \(H=u^2\sin^2\theta/(2g)\)

Calculation \(H=400(0.25)/19.6=5.10\,\mathrm m\).

Final Answer \(\boxed{5.10\,\mathrm m}\)

Additional revision examples

These PrepMode examples reinforce the same chapter relationships and are not labelled as official BIEK questions.

Revision Example 1: Stopping distance

Given \(u=20\,\mathrm{m\,s^{-1}},\;v=0,\;a=-4\,\mathrm{m\,s^{-2}}\)

Required Determine the stated quantity.

Formula / Substitution / Calculation \(v^2=u^2+2as\Rightarrow0=400-8s\Rightarrow s=50\,\mathrm m\)

Final Answer \(\boxed{50\,\mathrm m}\)

Revision Example 2: Projectile maximum height

Given \(u=20\,\mathrm{m\,s^{-1}},\;\theta=30^\circ,\;g=9.8\,\mathrm{m\,s^{-2}}\)

Required Determine the stated quantity.

Formula / Substitution / Calculation \(u_y=10;\;H=u_y^2/(2g)=100/19.6=5.10\,\mathrm m\)

Final Answer \(\boxed{5.10\,\mathrm m}\)

PREPMODE PRACTICE • NOT OFFICIAL BIEK QUESTIONS

1. The slope of a velocity-time graph gives:

Answer: acceleration
Acceleration is the rate of change of velocity.

2. For free fall near Earth, acceleration is approximately:

Answer: \(g\) downward
Ignoring air resistance, free-fall acceleration is directed downward.

3. A body moving with constant velocity has acceleration:

Answer: zero
Velocity is not changing.

4. At the highest point of ideal projectile motion, vertical velocity is:

Answer: zero
The vertical component changes sign through zero at the top.

5. Area under a velocity-time graph gives:

Answer: displacement
Integral of velocity over time is displacement.

6. At the highest point of ideal projectile motion, vertical velocity is:

Answer: zero
The vertical component momentarily vanishes.

7. Horizontal acceleration of an ideal projectile is:

Answer: zero
Air resistance is neglected.

8. Uniform acceleration means acceleration is:

Answer: constant
Uniform acceleration is constant acceleration.

9. If a body returns to its starting point, total displacement is:

Answer: zero
Initial and final positions coincide.

10. The equation \(v=u+at\) applies directly when acceleration is:

Answer: constant
It is a constant-acceleration relation.

Dynamics connects motion with its causes. Newton's laws describe the relation between force, mass and acceleration.

\[\vec F_{\text{net}}=m\vec a\]

Momentum is \(\vec p=m\vec v\), and impulse equals the change in momentum:

\[\vec J=\Delta\vec p=\vec F\,\Delta t\]

Friction is a contact force that opposes relative motion or its tendency.

Coverage check

  • Newton's three laws, inertia and free-body diagrams.
  • Static and kinetic friction; limiting friction and coefficient of friction.
  • Linear momentum, impulse and conservation of momentum.
  • Elastic and inelastic collisions in one dimension.
\[f_s\le \mu_sN,\qquad f_k=\mu_kN\]

Complete chapter revision

Use this as the chapter-level revision pass before moving to derivations and numericals.

Newton's laws

The first law defines inertial behavior, the second connects net force with momentum change, and the third states that interaction forces occur in equal and opposite pairs on different bodies.

Free-body diagrams

Draw only forces acting on the chosen object. Resolve forces along convenient axes before applying \(\sum F=ma\).

Friction

Static friction adjusts up to a limiting value; kinetic friction acts during sliding. Friction opposes relative motion or its tendency, not necessarily the object's velocity in every problem.

Momentum and impulse

Linear momentum is \(p=mv\). Impulse \(J=F\Delta t\) equals \(\Delta p\). In an isolated system total momentum remains constant.

Collisions

Momentum is conserved in isolated collisions. Kinetic energy is additionally conserved only in perfectly elastic collisions.

Exam focus

Separate action-reaction pairs from balanced forces on one object and use momentum conservation for recoil/collision problems.

Conservation of linear momentum: for an isolated two-body system, internal forces occur in equal and opposite pairs. The total external force is zero:

\[\frac{d}{dt}(\vec p_1+\vec p_2)=0\]

Therefore total momentum remains constant:

\[\boxed{m_1\vec u_1+m_2\vec u_2=m_1\vec v_1+m_2\vec v_2}\]

One-dimensional elastic collision

Use conservation of momentum together with conservation of kinetic energy. For masses \(m_1,m_2\), the relative speed of separation equals the relative speed of approach:

\[v_2-v_1=u_1-u_2\]

Combining with momentum conservation gives the standard post-collision velocities. This is the key extra condition that distinguishes an elastic collision.

Derivation bank

Revise the starting relation, the key transformation and the final result for each.

1

Impulse-momentum theorem

\[F=\frac{\Delta p}{\Delta t}\Rightarrow \boxed{F\Delta t=\Delta p}\]
2

Momentum conservation

\[F_{\rm ext}=0\Rightarrow\frac{dP_{\rm total}}{dt}=0\Rightarrow\boxed{P_i=P_f}\]
3

Friction on an incline

\[N=mg\cos\theta,\qquad f_{\max}=\mu_sN=\mu_smg\cos\theta\]

Given \(m=2.0\,\mathrm{kg}\), \(a=3.0\,\mathrm{m\,s^{-2}}\)

Required Net force.

Formula \(F=ma\)

Substitution \(F=(2.0)(3.0)\)

Calculation \(F=6.0\,\mathrm N\)

Final Answer \(\boxed{6.0\,\mathrm N}\)

Second worked example

Given \(m=5\,\mathrm{kg},\mu_k=0.20,g=9.8\,\mathrm{m/s^2}\).

Required Kinetic friction on a horizontal surface.

Formula \(f_k=\mu_kmg\)

Calculation \(f_k=9.8\,\mathrm N\).

Final Answer \(\boxed{9.8\,\mathrm N}\)

Additional revision examples

These PrepMode examples reinforce the same chapter relationships and are not labelled as official BIEK questions.

Revision Example 1: Impulse

Given \(F=200\,\mathrm N,\;\Delta t=0.050\,\mathrm s\)

Required Determine the stated quantity.

Formula / Substitution / Calculation \(J=F\Delta t=200(0.050)=10\,\mathrm{N\,s}\)

Final Answer \(\boxed{10\,\mathrm{N\,s}}\)

Revision Example 2: Friction

Given \(m=5\,\mathrm{kg},\;\mu_k=0.20\)

Required Determine the stated quantity.

Formula / Substitution / Calculation \(N=mg=49\,\mathrm N,\;f_k=\mu_kN=9.8\,\mathrm N\)

Final Answer \(\boxed{9.8\,\mathrm N}\)

PREPMODE PRACTICE • NOT OFFICIAL BIEK QUESTIONS

1. Newton's second law relates net force to:

Answer: rate of change of momentum
For constant mass it becomes \(F=ma\).

2. A self-adjusting force is commonly:

Answer: static friction
Static friction adjusts up to its limiting value.

3. Impulse has the same dimensions as:

Answer: momentum
Impulse equals change in momentum.

4. In an isolated collision, the quantity always conserved is:

Answer: linear momentum
Total linear momentum is conserved when external impulse is negligible.

5. Action and reaction forces act on:

Answer: different bodies
Newton's third-law pair acts on two interacting bodies.

6. Net force zero implies momentum is:

Answer: constant
From F=dp/dt.

7. Static friction is:

Answer: self-adjusting up to a limit
It matches the required opposing force until limiting friction.

8. Momentum SI unit is:

Answer: kg m s\(^{-1}\)
Momentum is mass times velocity.

9. In an isolated collision, conserved quantity is always:

Answer: total momentum
Kinetic energy is not conserved in all collisions.

10. Impulse equals:

Answer: change in momentum
J=Δp.

Circular motion requires a centripetal acceleration directed toward the center.

\[a_c=\frac{v^2}{r}=\omega^2r\]
\[F_c=\frac{mv^2}{r}\]

For rotational motion, torque is the turning effect of force:

\[\tau=rF\sin\theta\]

Angular momentum for a rigid body is \(L=I\omega\).

Coverage check

  • Angular displacement, angular velocity and angular acceleration.
  • Centripetal acceleration and force; banking of roads.
  • Torque, moment of inertia and angular momentum.
  • Gravitation, gravitational field/potential and conservative nature of gravity.
\[L=I\omega,\qquad \tau=I\alpha,\qquad U=-\frac{GMm}{r}\]

Complete chapter revision

Use this as the chapter-level revision pass before moving to derivations and numericals.

Angular variables

Angular displacement \(\theta\), angular velocity \(\omega\) and angular acceleration \(\alpha\) correspond to linear displacement, velocity and acceleration.

Circular motion

Velocity is tangent to the path while centripetal acceleration points toward the center. The centripetal force is the net inward force, not an additional new force.

Torque

Torque is \(\tau=rF\sin\theta\). Rotational equilibrium requires zero net torque, and translational equilibrium requires zero net force.

Moment of inertia

Moment of inertia measures rotational inertia and depends on both mass and its distribution about the axis.

Angular momentum

For a rigid body \(L=I\omega\). If external torque is zero, angular momentum is conserved.

Banking

On an ideally banked curve, the horizontal component of the normal reaction provides the required centripetal force.

Banking of a road without friction: resolve the normal reaction \(N\).

\[N\cos\theta=mg\]
\[N\sin\theta=\frac{mv^2}{r}\]

Divide the equations:

\[\tan\theta=\frac{v^2}{rg}\]
\[\boxed{\theta=\tan^{-1}\!\left(\frac{v^2}{rg}\right)}\]

Conservative gravitational field

The work done by gravity between radii \(r_1\) and \(r_2\) is

\[W=\int_{r_1}^{r_2}-\frac{GMm}{r^2}\,dr=GMm\left(\frac1{r_2}-\frac1{r_1}\right)\]

It depends only on the end points, not the path. Hence gravity is conservative.

Derivation bank

Revise the starting relation, the key transformation and the final result for each.

1

Centripetal acceleration

\[\Delta v\approx v\Delta\theta,\quad \Delta\theta=\frac{v\Delta t}{r}\Rightarrow\boxed{a_c=\frac{v^2}{r}=\omega^2r}\]
2

Torque

\[\tau=rF_\perp=rF\sin\theta\]
3

Ideal banking

\[N\cos\theta=mg,\;N\sin\theta=\frac{mv^2}{r}\Rightarrow\boxed{\tan\theta=\frac{v^2}{rg}}\]

Given \(r=1.20\,\mathrm{km}=1200\,\mathrm m\), \(v=105\,\mathrm{km\,h^{-1}}=29.17\,\mathrm{m\,s^{-1}}\)

Required Ideal banking angle.

Formula \(\tan\theta=v^2/(rg)\)

Substitution \(\tan\theta=(29.17)^2/[1200(9.8)]\)

Calculation \(\theta\approx4.14^\circ\)

Final Answer \(\boxed{\theta\approx4.1^\circ}\)

Second worked example

Given \(m=0.50\,\mathrm{kg},v=4.0\,\mathrm{m/s},r=2.0\,\mathrm m\).

Required Centripetal force.

Formula \(F=mv^2/r\)

Calculation \(F=0.5(16)/2=4.0\,\mathrm N\).

Final Answer \(\boxed{4.0\,\mathrm N}\)

Additional revision examples

These PrepMode examples reinforce the same chapter relationships and are not labelled as official BIEK questions.

Revision Example 1: Centripetal force

Given \(m=0.50\,\mathrm{kg},\;v=6\,\mathrm{m\,s^{-1}},\;r=2.0\,\mathrm m\)

Required Determine the stated quantity.

Formula / Substitution / Calculation \(F_c=mv^2/r=0.5(36)/2=9.0\,\mathrm N\)

Final Answer \(\boxed{9.0\,\mathrm N}\)

Revision Example 2: Torque

Given \(r=0.25\,\mathrm m,\;F=40\,\mathrm N,\;\theta=90^\circ\)

Required Determine the stated quantity.

Formula / Substitution / Calculation \(\tau=rF\sin\theta=0.25(40)=10\,\mathrm{N\,m}\)

Final Answer \(\boxed{10\,\mathrm{N\,m}}\)

PREPMODE PRACTICE • NOT OFFICIAL BIEK QUESTIONS

1. Centripetal acceleration is directed:

Answer: toward the center
It changes the direction of velocity.

2. Angular momentum has dimensions:

Answer: \(ML^2T^{-1}\)
For a particle, \(L=rp\).

3. For ideal banking without friction, \(\tan\theta\) equals:

Answer: \(v^2/rg\)
This follows from resolving the normal reaction.

4. For uniform circular motion, centripetal force does work equal to:

Answer: zero
The force is perpendicular to instantaneous displacement.

5. Centripetal force does work in uniform circular motion equal to:

Answer: zero
Force is perpendicular to instantaneous displacement.

6. SI unit of torque is:

Answer: N m
Torque is force times perpendicular distance.

7. If speed doubles at same radius, centripetal force becomes:

Answer: four times
Fc∝v².

8. Angular momentum is conserved when external torque is:

Answer: zero
dL/dt=τ_ext.

9. Linear speed and angular speed satisfy:

Answer: \(v=\omega r\)
Tangential speed equals angular speed times radius.

10. On an ideally banked road, inward force is provided by a component of:

Answer: normal reaction
The horizontal normal component is centripetal.

Work is the transfer of energy by a force acting through a displacement.

\[W=Fs\cos\theta\]
\[K=\frac12mv^2\]
\[U_g=mgh\]

Power measures the rate of doing work:

\[P=\frac{W}{t}=Fv\cos\theta\]

Coverage check

  • Work by constant and variable forces.
  • Kinetic and potential energy; work-energy theorem.
  • Conservative versus non-conservative forces and mechanical-energy conservation.
  • Power and efficiency.
\[E_{mech}=K+U,\qquad \eta=\frac{P_{out}}{P_{in}}\times100\%\]

Complete chapter revision

Use this as the chapter-level revision pass before moving to derivations and numericals.

Work

Work is the scalar product \(W=Fs\cos\theta\). It is positive when force has a component along displacement, negative when opposite, and zero when perpendicular.

Energy

Kinetic energy is \(K=\tfrac12mv^2\). Near Earth, gravitational potential energy change is \(\Delta U=mgh\). Elastic energy is stored in a stretched/compressed spring.

Work-energy theorem

Net work done on a particle equals its change in kinetic energy.

Conservation

For conservative forces, mechanical energy \(K+U\) remains constant when no non-conservative work is done.

Power

Average power is \(W/t\); instantaneous mechanical power can be written \(P=Fv\cos\theta\).

Efficiency

Efficiency is useful output divided by input, usually expressed as a percentage and cannot exceed 100% for a real machine.

Work-energy theorem: for one-dimensional motion with constant net force,

\[W=Fs=mas\]

Using \(v^2-u^2=2as\),

\[as=\frac{v^2-u^2}{2}\]

Hence

\[\boxed{W=\frac12mv^2-\frac12mu^2=\Delta K}\]

Power for a moving body

Starting from \(P=dW/dt\) and \(dW=\vec F\cdot d\vec s\),

\[P=\vec F\cdot\frac{d\vec s}{dt}=\boxed{\vec F\cdot\vec v}\]

Derivation bank

Revise the starting relation, the key transformation and the final result for each.

1

Work-energy theorem

\[W=Fs=mas,\quad v^2-u^2=2as\Rightarrow\boxed{W=\frac12mv^2-\frac12mu^2}\]
2

Gravitational potential-energy change

\[W_{\rm against\,g}=mgh\Rightarrow\boxed{\Delta U=mgh}\]
3

Power at constant velocity

\[P=\frac{dW}{dt}=\vec F\cdot\frac{d\vec s}{dt}\Rightarrow\boxed{P=Fv\cos\theta}\]

Given \(m=70\,\mathrm{kg}\), \(h=3\,\mathrm m\), \(t=2\,\mathrm s\)

Required Work and power.

Formula \(W=mgh\), \(P=W/t\)

Substitution \(W=70(9.8)(3)=2058\,\mathrm J\)

Calculation \(P=2058/2=1029\,\mathrm W\)

Final Answer \(\boxed{W\approx2.06\,\mathrm{kJ},\;P\approx1.03\,\mathrm{kW}}\)

Second worked example

Given \(F=50\,\mathrm N,v=3\,\mathrm{m/s}\), same direction.

Required Power.

Formula \(P=Fv\)

Calculation \(P=150\,\mathrm W\).

Final Answer \(\boxed{150\,\mathrm W}\)

Additional revision examples

These PrepMode examples reinforce the same chapter relationships and are not labelled as official BIEK questions.

Revision Example 1: Kinetic energy

Given \(m=4.0\,\mathrm{kg},\;v=5.0\,\mathrm{m\,s^{-1}}\)

Required Determine the stated quantity.

Formula / Substitution / Calculation \(K=\tfrac12mv^2=0.5(4)(25)=50\,\mathrm J\)

Final Answer \(\boxed{50\,\mathrm J}\)

Revision Example 2: Efficiency

Given \(E_{\rm in}=500\,\mathrm J,\;E_{\rm useful}=375\,\mathrm J\)

Required Determine the stated quantity.

Formula / Substitution / Calculation \(\eta=(375/500)\times100=75\%\)

Final Answer \(\boxed{75\%}\)

PREPMODE PRACTICE • NOT OFFICIAL BIEK QUESTIONS

1. The SI unit of power is:

Answer: watt
One watt is one joule per second.

2. If force is perpendicular to displacement, work done is:

Answer: zero
\(W=Fs\cos90^\circ=0\).

3. The work-energy theorem states net work equals change in:

Answer: kinetic energy
\(W_{net}=\Delta K\).

4. Mechanical energy is conserved when only ___ forces do work:

Answer: conservative
Conservative-force work changes potential and kinetic energy without dissipating total mechanical energy.

5. Work is negative when force component is:

Answer: opposite displacement
Cosθ is negative for obtuse angle.

6. Potential energy is associated with:

Answer: configuration/position
Potential energy belongs to interactions/configuration.

7. A conservative force has path-independent:

Answer: work between two points
Conservative work depends only on endpoints.

8. One watt equals:

Answer: one joule per second
W=J/s.

9. Mechanical energy is conserved when only:

Answer: conservative forces do work
Non-conservative work changes mechanical energy.

10. Efficiency of a real machine is generally:

Answer: less than 100%
Losses prevent perfect conversion.

Fluid statics studies fluids at rest. Pressure is normal force per unit area.

\[P=\frac{F}{A}\]

In a liquid of density \(\rho\), pressure increases with depth:

\[P=P_0+\rho gh\]

Archimedes' principle states that the buoyant force equals the weight of displaced fluid:

\[F_B=\rho_{fluid}gV_{displaced}\]

Coverage check

  • Density, pressure and pressure variation with depth.
  • Pascal's principle and hydraulic machines.
  • Archimedes' principle, buoyancy and flotation.
  • Apparent weight and relative density.
\[F_B=\rho_f gV_{disp}\]

Complete chapter revision

Use this as the chapter-level revision pass before moving to derivations and numericals.

Density and pressure

Density is mass per volume. Pressure is normal force per area and acts in all directions at a point in a static fluid.

Hydrostatic pressure

Pressure increases with depth by \(\rho gh\). At the same depth in a connected static liquid, pressure is the same.

Pascal's principle

A pressure change applied to an enclosed fluid is transmitted throughout the fluid, enabling hydraulic multiplication of force.

Archimedes' principle

An immersed body experiences an upward force equal to the weight of displaced fluid.

Floating

For a floating body in equilibrium, buoyant force equals its weight. The submerged fraction depends on the ratio of object density to fluid density.

Exam focus

Use consistent volumes and densities, and distinguish apparent weight from true weight.

Hydrostatic pressure: consider a liquid column of height \(h\) and area \(A\).

Its mass is \(m=\rho Ah\), so its weight is \(\rho Ahg\). Therefore

\[\Delta P=\frac{\rho Ahg}{A}=\rho gh\]
\[\boxed{P=P_0+\rho gh}\]

Hydraulic principle

Pressure applied to an enclosed fluid is transmitted undiminished:

\[\frac{F_1}{A_1}=\frac{F_2}{A_2}\]
\[\boxed{F_2=F_1\frac{A_2}{A_1}}\]

Derivation bank

Revise the starting relation, the key transformation and the final result for each.

1

Hydrostatic pressure

\[m=\rho Ah,\quad W=\rho Ahg,\quad \Delta P=\frac{W}{A}\Rightarrow\boxed{\Delta P=\rho gh}\]
2

Hydraulic press

\[P_1=P_2\Rightarrow\frac{F_1}{A_1}=\frac{F_2}{A_2}\Rightarrow\boxed{F_2=F_1\frac{A_2}{A_1}}\]
3

Floating condition

\[F_B=W\Rightarrow \rho_f gV_{\rm sub}=\rho_o gV_o\Rightarrow\boxed{\frac{V_{\rm sub}}{V_o}=\frac{\rho_o}{\rho_f}}\]

Given A body weighs \(10\,\mathrm N\) in air and \(7\,\mathrm N\) when submerged.

Required Buoyant force.

Formula \(F_B=W_{air}-W_{submerged}\)

Substitution \(F_B=10-7\)

Final Answer \(\boxed{F_B=3\,\mathrm N}\)

Second worked example

Given Water depth \(h=5.0\,\mathrm m\).

Required Gauge pressure.

Formula \(P=\rho gh\)

Calculation \(P=1000(9.8)(5)=4.90\times10^4\,\mathrm{Pa}\).

Final Answer \(\boxed{49.0\,\mathrm{kPa}}\)

Additional revision examples

These PrepMode examples reinforce the same chapter relationships and are not labelled as official BIEK questions.

Revision Example 1: Pressure at depth

Given \(\rho=1000\,\mathrm{kg\,m^{-3}},\;h=5.0\,\mathrm m\)

Required Determine the stated quantity.

Formula / Substitution / Calculation \(\Delta P=\rho gh=1000(9.8)(5)=4.90\times10^4\,\mathrm{Pa}\)

Final Answer \(\boxed{4.90\times10^4\,\mathrm{Pa}}\)

Revision Example 2: Hydraulic press

Given \(A_1=5\,\mathrm{cm^2},\;A_2=100\,\mathrm{cm^2},\;F_1=50\,\mathrm N\)

Required Determine the stated quantity.

Formula / Substitution / Calculation \(F_2=F_1A_2/A_1=50(100/5)=1000\,\mathrm N\)

Final Answer \(\boxed{1000\,\mathrm N}\)

PREPMODE PRACTICE • NOT OFFICIAL BIEK QUESTIONS

1. Buoyant force equals the weight of:

Answer: displaced fluid
This is Archimedes' principle.

2. Pressure in a static liquid increases with:

Answer: depth
\(P=P_0+\rho gh\).

3. Pressure has SI unit:

Answer: pascal
One pascal is one newton per square metre.

4. A floating body displaces fluid whose weight is:

Answer: equal to the body's weight
For equilibrium, buoyant force equals weight.

5. Pressure at a point in static fluid acts:

Answer: in all directions
Static fluid pressure is isotropic.

6. Buoyant force acts generally:

Answer: upward
It results from larger pressure at greater depth.

7. A floating body displaces fluid whose weight is:

Answer: equal to body weight
Equilibrium requires FB=W.

8. Hydraulic machines use:

Answer: Pascal's principle
Pressure is transmitted through enclosed fluid.

9. Hydrostatic pressure difference depends on:

Answer: vertical depth
ΔP=ρgh.

10. Relative density is:

Answer: dimensionless
It is a ratio of densities.

Fluid dynamics studies fluids in motion. For steady incompressible flow, mass conservation gives the continuity equation.

\[A_1v_1=A_2v_2\]

Bernoulli's equation expresses mechanical-energy conservation along a streamline for ideal flow:

\[P+\frac12\rho v^2+\rho gh=\text{constant}\]

Viscosity produces resistance to motion through a fluid.

Coverage check

  • Steady flow, streamlines and equation of continuity.
  • Bernoulli's theorem and applications.
  • Viscosity, viscous drag and terminal speed.
  • Laminar versus turbulent flow and Reynolds-number idea.
\[F_d=6\pi\eta rv\]

Complete chapter revision

Use this as the chapter-level revision pass before moving to derivations and numericals.

Steady flow

In steady flow, conditions at a fixed point do not change with time. Streamlines indicate the local direction of flow.

Continuity

For an incompressible fluid, \(Av\) is constant; a smaller cross-section therefore corresponds to greater speed.

Bernoulli

Along a streamline for ideal steady flow, pressure energy, kinetic energy and gravitational potential energy per unit volume trade with one another.

Viscosity

Viscosity represents internal resistance to flow. Real fluids dissipate mechanical energy.

Terminal speed

A falling body in a viscous fluid reaches terminal speed when its net force becomes zero after drag and buoyancy balance weight.

Applications

Venturi meters, atomizers and aerodynamic effects are applications of pressure-speed relationships; note the ideal-flow assumptions.

Bernoulli equation: consider a fluid element moving between two points. Pressure work changes its kinetic and gravitational potential energy.

\[P_1V-P_2V=\frac12\rho V(v_2^2-v_1^2)+\rho Vg(h_2-h_1)\]

Divide by \(V\) and rearrange:

\[\boxed{P+\frac12\rho v^2+\rho gh=\text{constant}}\]

Terminal speed of a small sphere

At terminal speed, effective weight equals Stokes drag. For sphere density \(\rho_s\) in fluid density \(\rho_f\):

\[\frac43\pi r^3(\rho_s-\rho_f)g=6\pi\eta rv_t\]
\[\boxed{v_t=\frac{2r^2(\rho_s-\rho_f)g}{9\eta}}\]

Derivation bank

Revise the starting relation, the key transformation and the final result for each.

1

Continuity equation

\[\dot m=\rho Av=\text{constant}\Rightarrow\boxed{A_1v_1=A_2v_2}\quad(\rho=\text{constant})\]
2

Bernoulli equation

\[P_1+\frac12\rho v_1^2+\rho gh_1=P_2+\frac12\rho v_2^2+\rho gh_2\]
3

Stokes terminal speed

\[mg-\rho_fVg-6\pi\eta rv_t=0\Rightarrow\boxed{v_t=\frac{2r^2(\rho_s-\rho_f)g}{9\eta}}\]

Given \(A_1=4.0\,\mathrm{cm^2}\), \(v_1=2.0\,\mathrm{m\,s^{-1}}\), \(A_2=2.0\,\mathrm{cm^2}\)

Required \(v_2\)

Formula \(A_1v_1=A_2v_2\)

Substitution \(4(2)=2v_2\)

Final Answer \(\boxed{v_2=4.0\,\mathrm{m\,s^{-1}}}\)

Second worked example

Given Pipe radius halves.

Required Speed change for incompressible steady flow.

Formula \(A_1v_1=A_2v_2\), and \(A\propto r^2\).

Calculation \(A_2=A_1/4\Rightarrow v_2=4v_1\).

Final Answer \(\boxed{v_2=4v_1}\)

Additional revision examples

These PrepMode examples reinforce the same chapter relationships and are not labelled as official BIEK questions.

Revision Example 1: Continuity

Given \(A_1=6\,\mathrm{cm^2},\;v_1=2\,\mathrm{m\,s^{-1}},\;A_2=3\,\mathrm{cm^2}\)

Required Determine the stated quantity.

Formula / Substitution / Calculation \(v_2=A_1v_1/A_2=4\,\mathrm{m\,s^{-1}}\)

Final Answer \(\boxed{4\,\mathrm{m\,s^{-1}}}\)

Revision Example 2: Bernoulli, same height

Given \(\rho=1000,\;v_1=2,\;v_2=5\,\mathrm{m\,s^{-1}},\;P_1=2.0\times10^5\,\mathrm{Pa}\)

Required Determine the stated quantity.

Formula / Substitution / Calculation \(P_2=P_1+\tfrac12\rho(v_1^2-v_2^2)=1.895\times10^5\,\mathrm{Pa}\)

Final Answer \(\boxed{1.90\times10^5\,\mathrm{Pa}}\)

PREPMODE PRACTICE • NOT OFFICIAL BIEK QUESTIONS

1. The continuity equation represents conservation of:

Answer: mass
For incompressible steady flow, \(Av\) is constant.

2. According to Bernoulli's effect, higher fluid speed is associated with:

Answer: lower static pressure
Along the same streamline at equal height, pressure falls as speed rises.

3. A Venturi meter operates mainly using:

Answer: Bernoulli's principle
Pressure differences are related to flow speed.

4. For steady incompressible flow through a narrowing pipe, speed:

Answer: increases
Continuity requires \(Av=constant\).

5. For incompressible flow through a narrower pipe, speed:

Answer: increases
A v is constant.

6. Bernoulli equation is based on conservation of:

Answer: mechanical energy
It balances pressure, kinetic and potential terms.

7. Viscosity represents:

Answer: internal fluid friction
It resists relative motion between fluid layers.

8. Terminal velocity occurs when net force is:

Answer: zero
Acceleration becomes zero.

9. A Venturi meter measures:

Answer: flow rate
It uses pressure differences in varying pipe area.

10. Ideal Bernoulli flow assumes negligible:

Answer: viscosity
Bernoulli is derived for non-viscous flow.

An electric field is a region where a charge experiences electric force.

\[F=\frac{1}{4\pi\varepsilon_0}\frac{|q_1q_2|}{r^2}\]
\[\vec E=\frac{\vec F}{q_0}\]

For a point charge,

\[E=\frac{1}{4\pi\varepsilon_0}\frac{|q|}{r^2}\]

Electric flux through a flat surface in a uniform field is \(\Phi_E=EA\cos\theta\).

Coverage check

  • Coulomb's law and superposition.
  • Electric-field intensity and field lines.
  • Electric flux and Gauss-law idea.
  • Electric potential, potential difference and potential energy.
  • Electric dipole and dipole field.
\[V=\frac{1}{4\pi\varepsilon_0}\frac{q}{r},\qquad U=qV\]

Complete chapter revision

Use this as the chapter-level revision pass before moving to derivations and numericals.

Coulomb law

The force between point charges varies directly with the product of charges and inversely with the square of separation.

Electric field

Field strength is force per unit positive test charge. Direction is the direction a positive test charge would accelerate.

Superposition

Fields from several charges add vectorially. Field lines begin on positive charge and end on negative charge; they never cross.

Flux and Gauss law

Electric flux measures field passing through a surface. For a closed surface, Gauss's law links net flux to enclosed charge.

Electric dipole

A dipole consists of equal opposite charges separated by a small distance; its dipole moment is directed from negative to positive charge.

Potential idea

Electric potential is scalar energy per unit charge. Equipotential surfaces are everywhere perpendicular to electric-field lines.

Field due to a point charge: place a small positive test charge \(q_0\) at distance \(r\) from source charge \(q\).

\[F=\frac{1}{4\pi\varepsilon_0}\frac{qq_0}{r^2}\]

Since \(E=F/q_0\),

\[\boxed{E=\frac{1}{4\pi\varepsilon_0}\frac{q}{r^2}}\]

Electric field on the perpendicular bisector of a dipole

For charges \(\pm q\) separated by \(2a\), transverse components cancel and axial components add. At distance \(y\) from the center,

\[E=\frac{1}{4\pi\varepsilon_0}\frac{2qa}{(y^2+a^2)^{3/2}}\]

For \(y\gg a\), with \(p=2qa\),

\[\boxed{E\approx\frac{1}{4\pi\varepsilon_0}\frac{p}{y^3}}\]

Derivation bank

Revise the starting relation, the key transformation and the final result for each.

1

Point-charge field

\[F=k\frac{qq_0}{r^2},\quad E=\frac{F}{q_0}\Rightarrow\boxed{E=k\frac{q}{r^2}}\]
2

Electric flux

\[\Phi_E=\vec E\cdot\vec A=\boxed{EA\cos\theta}\]
3

Dipole field on equatorial line, far field

\[E_{\rm eq}\approx\boxed{\frac{1}{4\pi\varepsilon_0}\frac{p}{r^3}}\quad\text{directed opposite to }\vec p\]

Given Force \(F=1.0\times10^{-4}\,\mathrm N\), charge \(q=2.0\times10^{-19}\,\mathrm C\)

Required Electric-field intensity.

Formula \(E=F/q\)

Substitution \(E=(1.0\times10^{-4})/(2.0\times10^{-19})\)

Final Answer \(\boxed{E=5.0\times10^{14}\,\mathrm{N\,C^{-1}}}\)

Second worked example

Given \(q=2.0\,\mu C,r=0.30\,\mathrm m\).

Required Potential in vacuum.

Formula \(V=kq/r\)

Calculation \(V=(8.99\times10^9)(2\times10^{-6})/0.30=5.99\times10^4\,\mathrm V\).

Final Answer \(\boxed{6.0\times10^4\,\mathrm V}\)

Additional revision examples

These PrepMode examples reinforce the same chapter relationships and are not labelled as official BIEK questions.

Revision Example 1: Point-charge field

Given \(q=2.0\,\mu\mathrm C,\;r=0.30\,\mathrm m\)

Required Determine the stated quantity.

Formula / Substitution / Calculation \(E=kq/r^2=8.99\times10^9(2\times10^{-6})/0.09=2.00\times10^5\,\mathrm{N\,C^{-1}}\)

Final Answer \(\boxed{2.00\times10^5\,\mathrm{N\,C^{-1}}}\)

Revision Example 2: Electric flux

Given \(E=500\,\mathrm{N\,C^{-1}},\;A=0.020\,\mathrm{m^2},\;\theta=60^\circ\)

Required Determine the stated quantity.

Formula / Substitution / Calculation \(\Phi=EA\cos\theta=500(0.020)(0.5)=5.0\,\mathrm{N\,m^2\,C^{-1}}\)

Final Answer \(\boxed{5.0\,\mathrm{N\,m^2\,C^{-1}}}\)

PREPMODE PRACTICE • NOT OFFICIAL BIEK QUESTIONS

1. Electric-field strength is force per unit:

Answer: positive test charge
\(E=F/q_0\).

2. Coulomb force in a dielectric is reduced mainly because of:

Answer: electric polarization
Polarization weakens the effective field.

3. The SI unit of electric flux can be written as:

Answer: \(\mathrm{N\,m^2\,C^{-1}}\)
\(\Phi_E=EA\), so units are \(\mathrm{N\,m^2\,C^{-1}}\).

4. Electric potential is a:

Answer: scalar
Potential adds algebraically rather than vectorially.

5. Electric-field direction is defined using a:

Answer: positive test charge
By convention E points in the force direction on positive test charge.

6. Field of a point charge varies as:

Answer: \(1/r^2\)
Coulomb field follows inverse-square law.

7. Electric flux is maximum for surface normal:

Answer: parallel to E
Φ=EA cosθ.

8. Net electric flux through a closed surface depends on:

Answer: enclosed charge
Gauss law.

9. Electric potential is a:

Answer: scalar
Potential adds algebraically.

10. Dipole moment direction is from:

Answer: negative to positive charge
This is the conventional electric dipole direction.

A capacitor stores separated electric charge and electrostatic energy.

\[C=\frac{Q}{V}\]

For parallel plates in air or vacuum,

\[C=\frac{\varepsilon_0A}{d}\]

With a dielectric of relative permittivity \(\kappa\),

\[C=\frac{\kappa\varepsilon_0A}{d}\]

Stored energy is \(U=\frac12CV^2\).

Coverage check

  • Capacitance and factors affecting parallel-plate capacitance.
  • Dielectrics and relative permittivity.
  • Series and parallel combinations.
  • Energy stored and energy density of an electric field.
\[\frac1{C_s}=\sum\frac1{C_i},\qquad C_p=\sum C_i\]
\[U=\frac12CV^2=\frac{Q^2}{2C}=\frac12QV\]

Complete chapter revision

Use this as the chapter-level revision pass before moving to derivations and numericals.

Capacitance

Capacitance \(C=Q/V\) measures charge storage per potential difference and depends on geometry and dielectric, not directly on Q or V for an ideal linear capacitor.

Parallel plates

For large plates, \(C=\varepsilon A/d\); capacitance increases with area and dielectric permittivity and decreases with separation.

Dielectrics

A dielectric polarizes in an electric field, reducing the effective internal field and increasing capacitance.

Combinations

Parallel capacitors share voltage and add directly; series capacitors carry equal magnitude charge and combine through reciprocal addition.

Energy

Stored electrostatic energy can be written \(U=\tfrac12CV^2=\tfrac12QV=Q^2/(2C)\).

Exam focus

Identify whether voltage or charge remains fixed when a dielectric or geometry change is introduced.

Parallel-plate capacitance: for large parallel plates, \(E=\sigma/\varepsilon_0\) and \(V=Ed\).

\[V=\frac{\sigma d}{\varepsilon_0}=\frac{Qd}{\varepsilon_0A}\]

Therefore

\[\boxed{C=\frac{Q}{V}=\frac{\varepsilon_0A}{d}}\]

If the space is filled with dielectric, replace \(\varepsilon_0\) with \(\kappa\varepsilon_0\).

Energy stored in a capacitor

During charging, instantaneous potential is \(V=q/C\). The incremental work is \(dW=Vdq\):

\[U=\int_0^Q\frac{q}{C}dq=\boxed{\frac{Q^2}{2C}}\]

Using \(Q=CV\) gives \(U=\tfrac12CV^2=\tfrac12QV\).

Derivation bank

Revise the starting relation, the key transformation and the final result for each.

1

Parallel-plate capacitance

\[E=\frac{\sigma}{\varepsilon_0},\;V=Ed=\frac{Qd}{\varepsilon_0A}\Rightarrow\boxed{C=\frac{\varepsilon_0A}{d}}\]
2

Series combination

\[V=\sum V_i=Q\sum\frac1{C_i}\Rightarrow\boxed{\frac1{C_{\rm eq}}=\sum\frac1{C_i}}\]
3

Stored energy

\[dW=V\,dq=\frac{q}{C}dq\Rightarrow U=\int_0^Q\frac{q}{C}dq=\boxed{\frac{Q^2}{2C}=\frac12CV^2}\]

Given \(C=10\,\mu\mathrm F\), \(V=12\,\mathrm V\)

Required Stored energy.

Formula \(U=\frac12CV^2\)

Substitution \(U=\frac12(10\times10^{-6})(12)^2\)

Calculation \(U=7.2\times10^{-4}\,\mathrm J\)

Final Answer \(\boxed{0.72\,\mathrm{mJ}}\)

Second worked example

Given \(C_1=3\,\mu F,C_2=6\,\mu F\) in series.

Required Equivalent capacitance.

Formula \(1/C=1/C_1+1/C_2\)

Calculation \(C=2\,\mu F\).

Final Answer \(\boxed{2\,\mu F}\)

Additional revision examples

These PrepMode examples reinforce the same chapter relationships and are not labelled as official BIEK questions.

Revision Example 1: Parallel-plate capacitance

Given \(A=0.020\,\mathrm{m^2},\;d=1.0\,\mathrm{mm}\)

Required Determine the stated quantity.

Formula / Substitution / Calculation \(C=\varepsilon_0A/d=8.85\times10^{-12}(0.020)/10^{-3}=1.77\times10^{-10}\,\mathrm F\)

Final Answer \(\boxed{177\,\mathrm{pF}}\)

Revision Example 2: Two capacitors in series

Given \(C_1=6\,\mu\mathrm F,\;C_2=3\,\mu\mathrm F\)

Required Determine the stated quantity.

Formula / Substitution / Calculation \(1/C=1/6+1/3=1/2\;(\mu\mathrm F)^{-1}\)

Final Answer \(\boxed{C_{\rm eq}=2\,\mu\mathrm F}\)

PREPMODE PRACTICE • NOT OFFICIAL BIEK QUESTIONS

1. Capacitance is defined as:

Answer: \(Q/V\)
Capacitance is charge stored per unit potential difference.

2. Increasing plate area of a parallel-plate capacitor causes capacitance to:

Answer: increase
\(C=\varepsilon A/d\).

3. A dielectric between capacitor plates generally:

Answer: increases capacitance
The dielectric reduces the effective field for a given free charge.

4. Capacitors in parallel have the same:

Answer: potential difference
Parallel branches share the same terminal voltage.

5. Capacitance SI unit is:

Answer: farad
C=Q/V.

6. For parallel capacitors, equivalent capacitance is:

Answer: sum of capacitances
They share voltage.

7. For series capacitors, each carries the same:

Answer: charge magnitude
Charge delivered through the series path is the same.

8. Inserting a dielectric generally:

Answer: increases capacitance
C becomes κ times for full filling.

9. Energy stored at fixed C varies as:

Answer: \(V^2\)
U=1/2 CV².

10. Doubling plate separation at fixed geometry otherwise makes capacitance:

Answer: half
C∝1/d.

Direct-current circuits involve charge flow that maintains one direction. Ohm's law for an ohmic conductor is

\[V=IR\]

Resistance depends on material and geometry:

\[R=\rho\frac{L}{A}\]

Conductance is \(G=1/R\). Kirchhoff's laws express conservation of charge and energy in networks.

Coverage check

  • Current, current density and microscopic drift picture.
  • Resistance, resistivity, conductance and temperature dependence.
  • Electrical power and Joule heating.
  • EMF, terminal potential and internal resistance.
  • Kirchhoff's junction/loop laws and Wheatstone bridge.
\[P=VI=I^2R=\frac{V^2}{R}\]

Complete chapter revision

Use this as the chapter-level revision pass before moving to derivations and numericals.

Current and emf

Current is rate of charge flow. Emf is energy supplied per unit charge by a source; terminal voltage may differ from emf when internal resistance is present.

Resistance

Ohm's law describes ohmic behavior at constant physical conditions. \(R=\rho L/A\) separates geometry from material resistivity.

Power

Electrical power is \(P=VI=I^2R=V^2/R\) for a resistor.

Networks

Series elements carry the same current; parallel branches share the same potential difference.

Kirchhoff laws

Junction law follows charge conservation; loop law follows energy conservation.

Bridge circuits

A balanced Wheatstone bridge has zero galvanometer current and satisfies the arm-ratio relation.

Balanced Wheatstone bridge: at balance, no current flows through the galvanometer, so the two junctions have equal potential.

For arms \(P,Q,R,S\), potential-drop ratios give

\[\frac{P}{Q}=\frac{R}{S}\]
\[\boxed{PS=QR}\]

Terminal potential of a cell

For cell emf \(\mathcal E\), internal resistance \(r\) and current \(I\), the internal drop is \(Ir\). Therefore while supplying current,

\[\boxed{V=\mathcal E-Ir}\]

Derivation bank

Revise the starting relation, the key transformation and the final result for each.

1

Resistivity relation

\[R\propto L,\;R\propto\frac1A\Rightarrow\boxed{R=\rho\frac{L}{A}}\]
2

Electrical power

\[P=VI,\quad V=IR\Rightarrow\boxed{P=I^2R=\frac{V^2}{R}}\]
3

Balanced Wheatstone bridge

\[\frac{P}{Q}=\frac{R}{S}\Rightarrow\boxed{PS=QR}\]

Given \(R=5\,\Omega\)

Required Conductance.

Formula \(G=1/R\)

Substitution \(G=1/5\)

Final Answer \(\boxed{G=0.20\,\mathrm S}\)

Second worked example

Given \(V=12\,\mathrm V,R=6\,\Omega\).

Required Current and power.

Formula \(I=V/R, P=VI\)

Calculation \(I=2\,\mathrm A,P=24\,\mathrm W\).

Final Answer \(\boxed{2\,\mathrm A;24\,\mathrm W}\)

Additional revision examples

These PrepMode examples reinforce the same chapter relationships and are not labelled as official BIEK questions.

Revision Example 1: Electrical power

Given \(V=12\,\mathrm V,\;R=6\,\Omega\)

Required Determine the stated quantity.

Formula / Substitution / Calculation \(P=V^2/R=144/6=24\,\mathrm W\)

Final Answer \(\boxed{24\,\mathrm W}\)

Revision Example 2: Resistivity

Given \(R=4\,\Omega,\;A=2\times10^{-6}\,\mathrm{m^2},\;L=10\,\mathrm m\)

Required Determine the stated quantity.

Formula / Substitution / Calculation \(\rho=RA/L=4(2\times10^{-6})/10=8\times10^{-7}\,\Omega\mathrm m\)

Final Answer \(\boxed{8.0\times10^{-7}\,\Omega\mathrm m}\)

PREPMODE PRACTICE • NOT OFFICIAL BIEK QUESTIONS

1. The SI unit of conductance is:

Answer: siemens
Conductance is measured in siemens.

2. For an ohmic conductor at constant temperature:

Answer: \(V\propto I\)
This is Ohm's law.

3. At a balanced Wheatstone bridge, galvanometer current is:

Answer: zero
The galvanometer junctions are at equal potential.

4. Kirchhoff's junction rule follows from conservation of:

Answer: charge
Current into a junction equals current out.

5. Current is rate of flow of:

Answer: charge
I=dQ/dt.

6. Resistivity depends primarily on:

Answer: material and temperature
ρ is an intrinsic material property for given conditions.

7. Series resistors have the same:

Answer: current
Only one path exists.

8. Parallel branches have the same:

Answer: potential difference
They connect to common nodes.

9. Kirchhoff junction law expresses conservation of:

Answer: charge
Sum of currents into a node equals sum out.

10. A balanced Wheatstone bridge has galvanometer current:

Answer: zero
Bridge junctions are equipotential.

Simple harmonic motion (SHM) is periodic motion in which acceleration is proportional to displacement from equilibrium and directed toward equilibrium.

\[a=-\omega^2x\]
\[x=A\cos(\omega t+\phi)\]

For a mass-spring system,

\[T=2\pi\sqrt{\frac{m}{k}}\]

Coverage check

  • Conditions for SHM and sinusoidal displacement.
  • Velocity, acceleration and energy in SHM.
  • Mass-spring and simple-pendulum oscillators.
  • Free, damped and forced oscillations; resonance.
\[v=\omega\sqrt{A^2-x^2},\qquad E=\frac12kA^2\]

Complete chapter revision

Use this as the chapter-level revision pass before moving to derivations and numericals.

SHM condition

In simple harmonic motion the restoring acceleration is proportional to displacement and opposite in direction: \(a=-\omega^2x\).

Phase relations

Displacement, velocity and acceleration are sinusoidal but differ in phase. Speed is maximum at equilibrium and zero at extremes.

Energy

Total mechanical energy is constant in ideal SHM; kinetic and potential energy continually interchange.

Mass-spring

For an ideal spring, \(T=2\pi\sqrt{m/k}\). A stiffer spring oscillates faster; a larger mass oscillates more slowly.

Pendulum

For small angular amplitude, a simple pendulum has \(T=2\pi\sqrt{L/g}\), independent of bob mass.

Damping and resonance

Damping removes energy; forced oscillations can show resonance when driving frequency is near natural frequency.

Projection of uniform circular motion: a particle moving in a circle of radius \(A\) has x-coordinate

\[x=A\cos\omega t\]

Differentiating twice,

\[a_x=\frac{d^2x}{dt^2}=-\omega^2A\cos\omega t\]

Since \(x=A\cos\omega t\),

\[\boxed{a_x=-\omega^2x}\]

Thus the projection executes SHM.

Simple pendulum period

For small angles, \(\sin\theta\approx\theta\). Tangential restoring acceleration is

\[a_t=-g\sin\theta\approx-g\theta=-\frac{g}{L}s\]

Thus \(\omega^2=g/L\), so

\[\boxed{T=2\pi\sqrt{\frac{L}{g}}}\]

Derivation bank

Revise the starting relation, the key transformation and the final result for each.

1

Mass-spring period

\[F=-kx=ma\Rightarrow a=-\frac{k}{m}x=-\omega^2x,\quad\boxed{T=2\pi\sqrt{\frac{m}{k}}}\]
2

SHM speed-displacement relation

\[x=A\cos\omega t,\;v=-A\omega\sin\omega t\Rightarrow\boxed{v^2=\omega^2(A^2-x^2)}\]
3

Simple pendulum, small angle

\[\tau\approx-mgL\theta,\;I=mL^2\Rightarrow\ddot\theta=-\frac{g}{L}\theta,\quad\boxed{T=2\pi\sqrt{\frac{L}{g}}}\]

Given \(T=0.50\,\mathrm s\), \(A=5.0\,\mathrm{cm}=0.050\,\mathrm m\)

Required Speed at equilibrium.

Formula \(v_{max}=\omega A=(2\pi/T)A\)

Substitution \(v_{max}=(2\pi/0.50)(0.050)\)

Calculation \(v_{max}\approx0.628\,\mathrm{m\,s^{-1}}\)

Final Answer \(\boxed{0.63\,\mathrm{m\,s^{-1}}}\)

Second worked example

Given \(m=0.25\,\mathrm{kg},k=100\,\mathrm{N/m}\).

Required Period.

Formula \(T=2\pi\sqrt{m/k}\)

Calculation \(T=0.314\,\mathrm s\).

Final Answer \(\boxed{0.314\,\mathrm s}\)

Additional revision examples

These PrepMode examples reinforce the same chapter relationships and are not labelled as official BIEK questions.

Revision Example 1: Spring period

Given \(m=0.50\,\mathrm{kg},\;k=200\,\mathrm{N\,m^{-1}}\)

Required Determine the stated quantity.

Formula / Substitution / Calculation \(T=2\pi\sqrt{m/k}=2\pi\sqrt{0.0025}=0.314\,\mathrm s\)

Final Answer \(\boxed{0.314\,\mathrm s}\)

Revision Example 2: SHM acceleration

Given \(A=0.10\,\mathrm m,\;\omega=5\,\mathrm{rad\,s^{-1}},\;x=0.04\,\mathrm m\)

Required Determine the stated quantity.

Formula / Substitution / Calculation \(a=-\omega^2x=-25(0.04)=-1.0\,\mathrm{m\,s^{-2}}\)

Final Answer \(\boxed{-1.0\,\mathrm{m\,s^{-2}}}\)

PREPMODE PRACTICE • NOT OFFICIAL BIEK QUESTIONS

1. In SHM, kinetic energy is maximum at the:

Answer: mean position
Speed is maximum at equilibrium.

2. The displacement-time graph of ideal SHM is a:

Answer: sine/cosine curve
SHM is sinusoidal.

3. If mass of a spring oscillator doubles, its period becomes:

Answer: \(\sqrt2T\)
Since \(T\propto\sqrt m\).

4. Resonance occurs when driving frequency is close to the system's:

Answer: natural frequency
Forced amplitude peaks near natural frequency, with damping limiting the peak.

5. In SHM acceleration is opposite to:

Answer: displacement
a=-ω²x.

6. At an extreme position, SHM speed is:

Answer: zero
The object reverses direction.

7. Total energy of ideal SHM is proportional to:

Answer: \(A^2\)
E=1/2 kA².

8. Increasing spring constant makes period:

Answer: smaller
T∝1/√k.

9. Pendulum period for small oscillations is independent of bob:

Answer: mass
T=2π√(L/g).

10. Resonance is strongest when driving frequency is near:

Answer: natural frequency
Energy transfer is most effective near resonance.

Acoustics studies sound and mechanical waves. Wave speed is related to frequency and wavelength by

\[v=f\lambda\]

The Doppler effect is the apparent change in frequency caused by relative motion between source and observer.

Stationary waves form through interference of two waves of equal frequency and amplitude traveling in opposite directions.

Coverage check

  • Sound as a longitudinal mechanical wave.
  • Speed of sound and its temperature dependence.
  • Intensity, intensity level and decibel scale.
  • Doppler effect, beats and stationary waves.
\[\beta=10\log_{10}\left(\frac{I}{I_0}\right)\,\mathrm{dB}\]

Complete chapter revision

Use this as the chapter-level revision pass before moving to derivations and numericals.

Sound waves

Sound in air is a longitudinal mechanical wave involving compressions and rarefactions. Wave speed obeys \(v=f\lambda\).

Intensity and loudness

Intensity is power per area and falls approximately as inverse square of distance for a point source in free space.

Doppler effect

Observed frequency changes when source and observer move relative to the medium; approaching generally raises observed frequency.

Beats

Beats arise from interference of two close frequencies and occur at \(f_b=|f_1-f_2|\).

Stationary waves

Nodes have zero displacement amplitude and antinodes maximum amplitude. Allowed frequencies depend on boundary conditions.

Temperature

Sound speed in a gas increases approximately as the square root of absolute temperature.

Standing waves on a stretched string: for a string fixed at both ends, allowed wavelengths satisfy

\[L=\frac{n\lambda_n}{2}\]

so

\[\lambda_n=\frac{2L}{n}\]

Using \(v=f\lambda\),

\[\boxed{f_n=\frac{nv}{2L}}\]

Doppler relation

For motion along the line joining source and observer, a useful sign-aware form is

\[\boxed{f'=f\frac{v\pm v_o}{v\mp v_s}}\]

Choose signs so motion toward each other increases observed frequency and motion apart decreases it.

Derivation bank

Revise the starting relation, the key transformation and the final result for each.

1

String harmonics

\[L=\frac{n\lambda_n}{2},\;f_n=\frac{v}{\lambda_n}\Rightarrow\boxed{f_n=\frac{nv}{2L}}\]
2

Beat frequency

\[\boxed{f_b=|f_1-f_2|}\]
3

Temperature dependence of sound speed

\[v\propto\sqrt{T}\Rightarrow\boxed{\frac{v_2}{v_1}=\sqrt{\frac{T_2}{T_1}}}\]

Given Speed of sound at \(27^\circ\mathrm C\) is \(345\,\mathrm{m\,s^{-1}}\). Estimate speed at \(127^\circ\mathrm C\), using \(v\propto\sqrt T\).

Required \(v_2\)

Formula \(v_2/v_1=\sqrt{T_2/T_1}\)

Substitution \(v_2=345\sqrt{400/300}\)

Calculation \(v_2\approx398\,\mathrm{m\,s^{-1}}\)

Final Answer \(\boxed{398\,\mathrm{m\,s^{-1}}}\)

Second worked example

Given Two tones: \(256\,\mathrm{Hz}\) and \(260\,\mathrm{Hz}\).

Required Beat frequency.

Formula \(f_b=|f_2-f_1|\)

Calculation \(f_b=4\,\mathrm{Hz}\).

Final Answer \(\boxed{4\,\mathrm{beats/s}}\)

Additional revision examples

These PrepMode examples reinforce the same chapter relationships and are not labelled as official BIEK questions.

Revision Example 1: Wavelength

Given \(v=340\,\mathrm{m\,s^{-1}},\;f=680\,\mathrm{Hz}\)

Required Determine the stated quantity.

Formula / Substitution / Calculation \(\lambda=v/f=340/680=0.50\,\mathrm m\)

Final Answer \(\boxed{0.50\,\mathrm m}\)

Revision Example 2: Beat frequency

Given \(f_1=256\,\mathrm{Hz},\;f_2=260\,\mathrm{Hz}\)

Required Determine the stated quantity.

Formula / Substitution / Calculation \(f_b=|260-256|=4\,\mathrm{Hz}\)

Final Answer \(\boxed{4\,\mathrm{beats\,s^{-1}}}\)

PREPMODE PRACTICE • NOT OFFICIAL BIEK QUESTIONS

1. The Doppler effect is used in medical:

Answer: ultrasound
Doppler ultrasound measures motion such as blood flow.

2. For a fixed string, the fundamental frequency is:

Answer: \(v/2L\)
For \(n=1\), \(f_1=v/(2L)\).

3. At nodes of a stationary wave, displacement amplitude is:

Answer: zero
Nodes remain at zero displacement.

4. Beat frequency equals:

Answer: absolute difference of frequencies
Beats occur at \(|f_1-f_2|\).

5. Sound cannot travel through:

Answer: vacuum
Mechanical waves require a medium.

6. Beat frequency is:

Answer: absolute frequency difference
fb=|f1-f2|.

7. In a standing wave, antinodes have:

Answer: maximum amplitude
Antinodes are points of greatest displacement amplitude.

8. Doppler shift occurs because of:

Answer: relative motion of source and observer
Relative motion changes arrival rate of wavefronts.

9. Sound speed in ideal gas rises when absolute temperature:

Answer: rises
v∝√T.

10. Wave relation is:

Answer: \(v=f\lambda\)
Wave speed equals frequency times wavelength.

Physical optics treats light as a wave and explains interference, diffraction and polarization.

For double-slit interference, the path difference controls whether waves reinforce or cancel.

\[\Delta=m\lambda\quad\text{(bright)}\]
\[\Delta=\left(m+\frac12\right)\lambda\quad\text{(dark)}\]

Newton's rings are circular interference fringes produced by the thin air film between a curved lens and a flat glass plate.

Coverage check

  • Wave nature of light and coherent sources.
  • Young's double-slit interference and fringe spacing.
  • Diffraction and resolving effects.
  • Newton's rings.
  • Polarization as evidence of transverse wave nature.
\[\beta_{fringe}=\frac{\lambda D}{d}\]

Complete chapter revision

Use this as the chapter-level revision pass before moving to derivations and numericals.

Wave nature

Interference, diffraction and polarization are wave phenomena. Stable interference requires coherent sources with a constant phase relationship.

Young interference

For slit separation \(d\) and screen distance \(D\), fringe spacing is \(\beta=\lambda D/d\) under small-angle conditions.

Diffraction

Diffraction is significant when aperture size is comparable with wavelength and causes spreading into the geometrical shadow.

Polarization

Polarization restricts the vibration direction and confirms the transverse nature of light.

Newton's rings

Thin-film interference between a lens and glass plate produces concentric bright and dark rings.

Exam focus

State conditions for maxima/minima carefully and distinguish path difference from phase difference.

Newton's rings, reflected light: for a plano-convex lens of radius of curvature \(R\), film thickness at radius \(r\) satisfies

\[t\approx\frac{r^2}{2R}\]

For dark rings in reflected light, \(2t=n\lambda\). Therefore

\[\frac{r_n^2}{R}=n\lambda\]
\[\boxed{r_n=\sqrt{n\lambda R}}\]

Double-slit fringe spacing

For small angle \(\theta\), path difference \(d\sin\theta\approx dy/D\). Bright fringes satisfy \(dy/D=m\lambda\), hence

\[y_m=\frac{m\lambda D}{d}\]
\[\boxed{\beta=y_{m+1}-y_m=\frac{\lambda D}{d}}\]

Derivation bank

Revise the starting relation, the key transformation and the final result for each.

1

Young fringe spacing

\[\Delta=d\sin\theta\approx d\frac{y}{D}=m\lambda\Rightarrow y_m=\frac{m\lambda D}{d},\quad\boxed{\beta=\frac{\lambda D}{d}}\]
2

Newton dark-ring radius

\[t\approx\frac{r^2}{2R},\;2t=n\lambda\Rightarrow\boxed{r_n=\sqrt{n\lambda R}}\]
3

Single-slit first minimum

\[\boxed{a\sin\theta=\lambda}\quad\text{for the first diffraction minimum}\]

Given \(\lambda=600\,\mathrm{nm}\), \(R=1.0\,\mathrm m\), \(n=4\)

Required Radius of 4th dark ring.

Formula \(r_n=\sqrt{n\lambda R}\)

Substitution \(r_4=\sqrt{4(600\times10^{-9})(1)}\)

Calculation \(r_4\approx1.55\times10^{-3}\,\mathrm m\)

Final Answer \(\boxed{r_4\approx1.55\,\mathrm{mm}}\)

Second worked example

Given \(\lambda=600\,\mathrm{nm},D=2.0\,\mathrm m,d=0.50\,\mathrm{mm}\).

Required Fringe spacing.

Formula \(\beta=\lambda D/d\)

Calculation \(\beta=2.4\times10^{-3}\,\mathrm m\).

Final Answer \(\boxed{2.4\,\mathrm{mm}}\)

Additional revision examples

These PrepMode examples reinforce the same chapter relationships and are not labelled as official BIEK questions.

Revision Example 1: Young fringe spacing

Given \(\lambda=600\,\mathrm{nm},\;D=2.0\,\mathrm m,\;d=0.50\,\mathrm{mm}\)

Required Determine the stated quantity.

Formula / Substitution / Calculation \(\beta=\lambda D/d=600\times10^{-9}(2)/(5\times10^{-4})=2.4\times10^{-3}\,\mathrm m\)

Final Answer \(\boxed{2.4\,\mathrm{mm}}\)

Revision Example 2: Diffraction angle

Given \(a=0.20\,\mathrm{mm},\;\lambda=500\,\mathrm{nm}\)

Required Determine the stated quantity.

Formula / Substitution / Calculation \(\sin\theta=\lambda/a=2.5\times10^{-3}\Rightarrow\theta\approx0.143^\circ\)

Final Answer \(\boxed{0.143^\circ}\)

PREPMODE PRACTICE • NOT OFFICIAL BIEK QUESTIONS

1. Interference is a consequence of:

Answer: superposition
Overlapping coherent waves add according to superposition.

2. For constructive interference, path difference is:

Answer: \(m\lambda\)
Whole-wavelength path differences give constructive interference.

3. Polarization demonstrates that light is:

Answer: transverse
Only transverse waves can be polarized.

4. Polarization is possible for:

Answer: transverse waves
Polarization selects vibration direction and is a transverse-wave property.

5. Coherent sources have constant:

Answer: phase difference
Stable fringes require fixed phase relation.

6. Young fringe spacing increases with:

Answer: wavelength
β=λD/d.

7. Polarization is possible for:

Answer: transverse waves
Polarization selects transverse vibration direction.

8. Diffraction becomes pronounced when aperture is comparable to:

Answer: wavelength
Wave spreading is strongest at comparable scales.

9. In reflected Newton rings, the center is commonly:

Answer: dark
A phase reversal causes the central dark condition.

10. For destructive two-source interference, path difference may be:

Answer: \((m+1/2)\lambda\)
Half-integer wavelength difference gives cancellation.

Communication systems transfer information from a source to a destination. A basic system includes transmitter, channel and receiver.

Modulation places an information signal onto a higher-frequency carrier for efficient transmission.

\[\text{signal}+\text{carrier}\;\xrightarrow{\text{modulation}}\;\text{modulated carrier}\]

Electromagnetic waves and optical fibers are important communication channels.

Coverage check

  • Elements of a communication system: source, transmitter, channel, receiver and destination.
  • Carrier waves, modulation and demodulation.
  • Amplitude and frequency modulation at introductory level.
  • Communication channels including radio and optical fiber.
  • Electromagnetic spectrum and bandwidth idea.

Complete chapter revision

Use this as the chapter-level revision pass before moving to derivations and numericals.

System blocks

A communication system includes information source, transducer, transmitter, channel, receiver and output transducer.

Modulation

Modulation transfers information to a high-frequency carrier. AM varies carrier amplitude; FM varies carrier frequency.

Why carriers

High-frequency carriers enable practical antenna dimensions, channel separation, efficient radiation and multiplexing.

Bandwidth

A modulated signal occupies a range of frequencies; bandwidth determines channel requirements.

EM spectrum

Different frequency bands serve radio, microwave, infrared and optical communication according to propagation and bandwidth needs.

Optical fiber

Total internal reflection guides light through a higher-index core surrounded by lower-index cladding, giving low-loss high-bandwidth transmission.

Why modulation is used: low-frequency information signals are generally inefficient for direct radiation. Shifting information to a high-frequency carrier permits practical antennas, channel allocation and long-range transmission.

At the receiver, demodulation recovers the information signal from the carrier.

AM sidebands

If a carrier of frequency \(f_c\) is amplitude-modulated by a single tone \(f_m\), the spectrum contains:

\[\boxed{f_c-f_m,\quad f_c,\quad f_c+f_m}\]

The occupied bandwidth is \(2f_m\) for a single maximum modulating frequency.

Derivation bank

Revise the starting relation, the key transformation and the final result for each.

1

AM sidebands

\[f_{USB}=f_c+f_m,\qquad f_{LSB}=f_c-f_m\]
2

AM bandwidth

\[\boxed{BW=2f_{m,\max}}\]
3

Critical angle for optical fiber

\[n_1\sin\theta_c=n_2\Rightarrow\boxed{\sin\theta_c=\frac{n_2}{n_1}}\quad(n_1>n_2)\]

Given Carrier frequency \(f_c=1.0\,\mathrm{MHz}\), audio signal \(f_m=5.0\,\mathrm{kHz}\)

Required AM sideband frequencies.

Formula \(f_{USB}=f_c+f_m\), \(f_{LSB}=f_c-f_m\)

Calculation \(f_{USB}=1.005\,\mathrm{MHz}\), \(f_{LSB}=0.995\,\mathrm{MHz}\)

Final Answer \(\boxed{0.995\,\mathrm{MHz},\;1.005\,\mathrm{MHz}}\)

Second worked example

Given AM highest audio frequency \(f_m=5\,\mathrm{kHz}\).

Required Required bandwidth.

Formula \(B=2f_m\)

Calculation \(B=10\,\mathrm{kHz}\).

Final Answer \(\boxed{10\,\mathrm{kHz}}\)

Additional revision examples

These PrepMode examples reinforce the same chapter relationships and are not labelled as official BIEK questions.

Revision Example 1: AM bandwidth

Given \(f_{m,\max}=5\,\mathrm{kHz}\)

Required Determine the stated quantity.

Formula / Substitution / Calculation \(BW=2f_m=10\,\mathrm{kHz}\)

Final Answer \(\boxed{10\,\mathrm{kHz}}\)

Revision Example 2: Critical angle

Given \(n_1=1.50,\;n_2=1.00\)

Required Determine the stated quantity.

Formula / Substitution / Calculation \(\sin\theta_c=n_2/n_1=0.667\Rightarrow\theta_c=41.8^\circ\)

Final Answer \(\boxed{41.8^\circ}\)

PREPMODE PRACTICE • NOT OFFICIAL BIEK QUESTIONS

1. Superposition of a signal wave on a carrier is called:

Answer: modulation
Modulation varies a carrier according to the information signal.

2. The shortest wavelength among radio, microwave, infrared and ultraviolet is:

Answer: ultraviolet
Within these options, ultraviolet has the highest frequency and shortest wavelength.

3. A device that recovers information from a modulated carrier performs:

Answer: demodulation
The receiver demodulates the incoming carrier.

4. Demodulation is performed mainly at the:

Answer: receiver
The receiver extracts the information signal from the carrier.

5. Modulation uses a high-frequency:

Answer: carrier
Information is imposed on a carrier.

6. AM changes carrier:

Answer: amplitude
AM = amplitude modulation.

7. FM changes carrier:

Answer: frequency
FM = frequency modulation.

8. Demodulation occurs mainly in the:

Answer: receiver
It recovers information.

9. Optical fiber relies on:

Answer: total internal reflection
Core-cladding geometry traps light.

10. AM bandwidth for highest modulating frequency fm is:

Answer: \(2f_m\)
Upper and lower sidebands each extend fm.

QUICK REVISION

High-value relationships

\[v=u+at\]
\[F=ma\]
\[F_c=\frac{mv^2}{r}\]
\[W=\Delta K\]
\[P=P_0+\rho gh\]
\[P+\frac12\rho v^2+\rho gh=\text{constant}\]
\[E=\frac{F}{q}\]
\[C=\frac{Q}{V}\]
\[V=IR\]
\[T=2\pi\sqrt{\frac{m}{k}}\]
\[v=f\lambda\]
\[r_n=\sqrt{n\lambda R}\]