CBSE Class 12 Physics: sample paper 2023-24

Section A

MCQ 1-12, assertion-reason 13-16

1. Q1 · 1 mark · MCQElectrostatic Potential and Capacitance

   Which of the following is not the property of an equipotential surface?

   1. (A)They do not cross each other.
   2. (B)The work done in carrying a charge from one point to another on an equipotential surface is zero.
   3. (C)For a uniform electric field, they are concentric spheres.
   4. (D)They can be imaginary spheres.

   Try it first, then show our solutionSolution

   1. For a uniform electric field, equipotential surfaces are parallel planes perpendicular to the field, not concentric spheres (concentric spheres arise for a point charge).

      1 mark

   Answer: Option (C)

   Where the marks are

   • Identifies that equipotentials of a uniform field are planes, so (C) is not a property (1 mark)

2. Q2 · 1 mark · MCQElectric Charges and Fields

   An electric dipole placed in an electric field of intensity $2\times 10^5$ N/C at an angle of 30° experiences a torque equal to 4 Nm. The charge on the dipole of dipole length 2 cm is

   1. (A)7 µC
   2. (B)8 mC
   3. (C)2 mC
   4. (D)5 mC

   Try it first, then show our solutionSolution

   1. Torque $\tau =pE\sin \theta =q(2a)E\sin \theta$, so $q=\frac{\tau }{(2a)E\sin \theta }$.

      ½ mark

   2. $q=\frac{4}{2\times 10^{-2}\times 2\times 10^5\times \sin 30^{\circ }}=\frac{4}{2000}=2\times 10^{-3}$ C $=2$ mC.

      ½ mark

   Answer: 2 mC, option (C)

   Where the marks are

   • Uses $\tau =q(2a)E\sin \theta$ (½ mark)
   • Gets $q=2$ mC (½ mark)

   Open Q2 on its own page

3. Q3 · 1 mark · MCQDual Nature of Radiation and Matter

   A metallic plate exposed to white light emits electrons. For which of the following colours of light, the stopping potential will be maximum?

   1. (A)Blue
   2. (B)Yellow
   3. (C)Red
   4. (D)Violet

   Try it first, then show our solutionSolution

   1. $e{V}_0=h\nu -{\varphi }_0$, so stopping potential increases with frequency.

      ½ mark

   2. Among the given colours violet has the highest frequency, so its stopping potential is maximum.

      ½ mark

   Answer: Violet, option (D)

   Where the marks are

   • Relates stopping potential to frequency via Einstein's equation and picks violet (1 mark)

   Open Q3 on its own page

4. Q4 · 1 mark · MCQAtoms

   When alpha particles are sent through a thin gold foil, most of them go straight through the foil, because

   1. (A)alpha particles are positively charged
   2. (B)the mass of an alpha particle is more than the mass of an electron
   3. (C)most of the part of an atom is empty space
   4. (D)alpha particles move with high velocity

   Try it first, then show our solutionSolution

   1. Most alpha particles pass undeflected because the atom is mostly empty space; the positive charge and mass are concentrated in a tiny nucleus that few particles approach.

      1 mark

   Answer: Option (C)

   Where the marks are

   • States atom is mostly empty space (1 mark)

5. Q5 · 1 mark · MCQMoving Charges and Magnetism

   An electron is moving along positive x-axis in a magnetic field which is parallel to the positive y-axis. In what direction will the magnetic force be acting on the electron?

   1. (A)Along -x axis
   2. (B)Along -z axis
   3. (C)Along +z axis
   4. (D)Along -y axis

   Try it first, then show our solutionSolution

   1. $\vec{F}=q(\vec{v}\times \vec{B})$ with $q=-e$: $\hat{i}\times \hat{j}=\hat{k}$.

      ½ mark

   2. Negative charge reverses it, so the force is along $-z$ axis.

      ½ mark

   Answer: Along −z axis, option (B)

   Where the marks are

   • Applies $q(\vec{v}\times \vec{B})$ with negative charge to get −z (1 mark)

6. Q6 · 1 mark · MCQMagnetism and Matter

   The relative magnetic permeability of a substance X is slightly less than unity and that of substance Y is slightly more than unity, then

   1. (A)X is paramagnetic and Y is ferromagnetic
   2. (B)X is diamagnetic and Y is ferromagnetic
   3. (C)X and Y both are paramagnetic
   4. (D)X is diamagnetic and Y is paramagnetic

   Try it first, then show our solutionSolution

   1. Diamagnetic: ${\mu }_r$ slightly less than 1; paramagnetic: ${\mu }_r$ slightly more than 1; ferromagnetic: ${\mu }_r\gg 1$. Hence X is diamagnetic and Y is paramagnetic.

      1 mark

   Answer: Option (D)

   Where the marks are

   • Classifies by relative permeability correctly (1 mark)

7. Q7 · 1 mark · MCQMoving Charges and Magnetism

   An ammeter of resistance 0.81 ohm reads up to 1 A. The value of the required shunt to increase the range to 10 A is

   1. (A)0.9 ohm
   2. (B)0.09 ohm
   3. (C)0.03 ohm
   4. (D)0.3 ohm

   Try it first, then show our solutionSolution

   1. Shunt carries $10-1=9$ A; same p.d.: $I_{g}G=(I-I_g)S$.

      ½ mark

   2. $1\times 0.81=9S\Rightarrow S=0.09\Omega$.

      ½ mark

   Answer: 0.09 Ω, option (B)

   Where the marks are

   • Equates p.d. across meter and shunt (½ mark)
   • S = 0.09 Ω (½ mark)

8. Q8 · 1 mark · MCQAtoms

   An electron with angular momentum L moving around the nucleus has a magnetic moment given by

   1. (A)eL/2m
   2. (B)eL/3m
   3. (C)eL/4m
   4. (D)eL/m

   Try it first, then show our solutionSolution

   1. $\mu =IA=\frac{ev}{2\pi r}\pi r^2=\frac{evr}{2}=\frac{e}{2m}(mvr)=\frac{eL}{2m}$.

      1 mark

   Answer: eL/2m, option (A)

   Where the marks are

   • Derives $\mu =eL/2m$ (1 mark)

9. Q9 · 1 mark · MCQAlternating Current

   The large scale transmission of electrical energy over long distances is done with the use of transformers. The voltage output of the generator is stepped-up because of

   1. (A)reduction of current
   2. (B)reduction of current and voltage both
   3. (C)power loss is cut down
   4. (D)(a) and (c) both

   Try it first, then show our solutionSolution

   1. For the same power $P=VI$, stepping up V reduces current I.

      ½ mark

   2. Line loss $I^{2}R$ then falls, so both (a) and (c) are correct.

      ½ mark

   Answer: Option (D)

   Where the marks are

   • Links higher voltage to lower current and lower $I^{2}R$ loss (1 mark)

10. Q10 · 1 mark · MCQElectromagnetic Waves

    The diagram below shows the electric field (E) and magnetic field (B) components of an electromagnetic wave at a certain time and location.

    The direction of the propagation of the electromagnetic wave is

    1. (A)perpendicular to E and B and out of plane of the paper
    2. (B)perpendicular to E and B and into the plane of the paper
    3. (C)parallel and in the same direction as E
    4. (D)parallel and in the same direction as B

    Figure in the paper: Vector E drawn pointing downward and vector B drawn pointing to the right, both starting from a common point in the plane of the paper. See it in the official paper.

    Try it first, then show our solutionSolution

    1. Propagation is along $\vec{E}\times \vec{B}$. With E downward ($-\hat{j}$) and B to the right ($\hat{i}$): $(-\hat{j})\times \hat{i}=\hat{k}$, i.e. out of the paper.

       1 mark

    Answer: Perpendicular to E and B, out of the paper, option (A)

    Where the marks are

    • Uses $\vec{E}\times \vec{B}$ to get out of the plane (1 mark)

11. Q11 · 1 mark · MCQElectromagnetic Induction

    In a coil of resistance 100 $\Omega$ a current is induced by changing the magnetic flux through it. The variation of current with time is as shown in the figure. The magnitude of change in flux through coil is

    1. (A)200 Wb
    2. (B)275 Wb
    3. (C)225 Wb
    4. (D)250 Wb

    Figure in the paper: Graph of current I (vertical axis) versus time t (horizontal axis): a straight line falling from I = 10 A at t = 0 to zero at t = 0.5 s. See it in the official paper.

    Try it first, then show our solutionSolution

    1. $I=\frac{1}{R}\frac{d\varphi }{dt}\Rightarrow \Delta \varphi =R\int Idt=R\times$ (area under I–t graph).

       ½ mark

    2. Area $=\frac{1}{2}\times 10\times 0.5=2.5$ C, so $\Delta \varphi =100\times 2.5=250$ Wb.

       ½ mark

    Answer: 250 Wb, option (D)

    Where the marks are

    • Δφ = R × area under I–t graph (½ mark)
    • Computes 250 Wb (½ mark)

    Open Q11 on its own page

12. Q12 · 1 mark · MCQAtoms

    The energy of an electron in nth orbit of hydrogen atom is $E_n=-13.6/n^2$ eV. The negative sign of energy indicates that

    1. (A)electron is free to move.
    2. (B)electron is bound to the nucleus.
    3. (C)kinetic energy of electron is equal to potential energy of electron.
    4. (D)atom is radiating energy.

    Try it first, then show our solutionSolution

    1. Zero energy corresponds to a free electron at infinity; negative total energy means energy must be supplied to remove it, i.e. the electron is bound to the nucleus.

       1 mark

    Answer: Option (B)

    Where the marks are

    • Interprets negative energy as bound state (1 mark)

13. Q13 · 1 mark · Assertion-reasonCompetency-basedDual Nature of Radiation and Matter

    Assertion (A): For the radiation of a frequency greater than the threshold frequency, photoelectric current is proportional to the intensity of the radiation.
    Reason (R): Greater the number of energy quanta available, greater is the number of electrons absorbing the energy quanta and greater is number of electrons coming out of the metal.

    1. (A)If both Assertion and Reason are true and Reason is correct explanation of Assertion.
    2. (B)If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
    3. (C)If Assertion is true but Reason is false.
    4. (D)If both Assertion and Reason are false.

    Try it first, then show our solutionSolution

    1. Above threshold, higher intensity means more photons per second, so more electrons are emitted and photocurrent ∝ intensity: A is true.

       ½ mark

    2. R states exactly this photon picture, so it correctly explains A.

       ½ mark

    Answer: Option (A)

    Where the marks are

    • Both true, R explains A (1 mark)

    Open Q13 on its own page

14. Q14 · 1 mark · Assertion-reasonCompetency-basedSemiconductor Electronics: Materials, Devices and Simple Circuits

    Assertion (A): Putting p type semiconductor slab directly in physical contact with n type semiconductor slab cannot form the pn junction.
    Reason (R): The roughness at contact will be much more than inter atomic crystal spacing and continuous flow of charge carriers is not possible.

    1. (A)If both Assertion and Reason are true and Reason is correct explanation of Assertion.
    2. (B)If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
    3. (C)If Assertion is true but Reason is false.
    4. (D)If both Assertion and Reason are false.

    Try it first, then show our solutionSolution

    1. A p-n junction cannot be made by pressing two slabs together: the surface roughness (~µm) far exceeds interatomic spacing (~Å), creating a discontinuity that blocks continuous carrier flow. Both true and R explains A.

       1 mark

    Answer: Option (A)

    Where the marks are

    • Both true with R as correct explanation (1 mark)

15. Q15 · 1 mark · Assertion-reasonCompetency-basedElectrostatic Potential and Capacitance

    Assertion (A): An electron has a higher potential energy when it is at a location associated with a negative value of potential and has a lower potential energy when at a location associated with a positive potential.
    Reason (R): Electrons move from a region of higher potential to a region of lower potential.

    1. (A)If both Assertion and Reason are true and Reason is correct explanation of Assertion.
    2. (B)If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
    3. (C)If Assertion is true but Reason is false.
    4. (D)If both Assertion and Reason are false.

    Try it first, then show our solutionSolution

    1. $U=qV=-eV$: negative V gives positive (higher) U and positive V gives lower U, so A is true.

       ½ mark

    2. Electrons, being negative, move from lower to higher potential, so R is false.

       ½ mark

    Answer: Option (C)

    Where the marks are

    • A true via $U=-eV$, R false (1 mark)

    Open Q15 on its own page

16. Q16 · 1 mark · Assertion-reasonCompetency-basedRay Optics and Optical Instruments

    Assertion (A): Propagation of light through an optical fibre is due to total internal reflection taking place at the core-cladding interface.
    Reason (R): Refractive index of the material of the cladding of the optical fibre is greater than that of the core.

    1. (A)If both Assertion and Reason are true and Reason is correct explanation of Assertion.
    2. (B)If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
    3. (C)If Assertion is true but Reason is false.
    4. (D)If both Assertion and Reason are false.

    Try it first, then show our solutionSolution

    1. Light is guided by total internal reflection at the core–cladding interface: A true.

       ½ mark

    2. TIR needs light going from denser to rarer medium, so the cladding has a lower refractive index than the core: R false.

       ½ mark

    Answer: Option (C)

    Where the marks are

    • A true, R false since cladding index is lower (1 mark)

    Open Q16 on its own page

Section B

1. Q17 · 2 marks · Very short answerSemiconductor Electronics: Materials, Devices and Simple Circuits

   (a) Name the device which utilizes unilateral action of a pn diode to convert ac into dc.
   (b) Draw the circuit diagram of full wave rectifier.

   Try it first, then show our solutionSolution

   1. (a) The device is a rectifier. It uses the fact that a pn junction diode conducts only when forward biased, so it converts ac into dc.

      1 mark

   2. (b) Full wave rectifier circuit (described): The ac input goes to the primary of a centre-tap transformer. The two ends A and B of the secondary connect to the p-sides of diodes $D_1$ and $D_2$. The n-sides of both diodes are joined together and connected to one end of the load resistance $R_L$. The other end of $R_L$ goes to the centre tap of the secondary. The output is taken across $R_L$. In one half cycle $D_1$ conducts and in the other $D_2$ conducts, so current through $R_L$ always flows in the same direction.

      1 mark

   Answer: (a) Rectifier; (b) centre-tap full wave rectifier circuit with two diodes and load $R_L$.

   Where the marks are

   • Names the device as rectifier (1 mark)
   • Correct, labelled circuit of full wave rectifier: centre-tap transformer, two diodes, load resistance, output (1 mark)

   Open Q17 on its own page

2. Q18 · 2 marks · Very short answerDual Nature of Radiation and Matter

   The wavelength λ of a photon and the de Broglie wavelength of an electron of mass m have the same value. Show that the energy of the photon is 2λmc/h times the kinetic energy of the electron, where c and h have their usual meanings.

   Try it first, then show our solutionSolution

   1. de Broglie wavelength of the electron: $\lambda =\frac{h}{mv}$, so $v=\frac{h}{m\lambda }$ ...(i)

      ½ mark

   2. Energy of a photon of wavelength $\lambda$: $E=\frac{hc}{\lambda }$ ...(ii)

      ½ mark

   3. Kinetic energy of the electron: $K=\frac{1}{2}m{v}^2=\frac{1}{2}m\frac{h^2}{m^2{\lambda }^2}=\frac{h^2}{2m{\lambda }^2}$ ...(iii)

      ½ mark

   4. Dividing (ii) by (iii): $\frac{E}{K}=\frac{hc}{\lambda }\cdot \frac{2m{\lambda }^2}{h^2}=\frac{2\lambda mc}{h}$. Hence $E=\frac{2\lambda mc}{h}K$.

      ½ mark

   Answer: $E/K=2\lambda mc/h$ (shown).

   Where the marks are

   • Writes $v=h/(m\lambda )$ from de Broglie relation (½ mark)
   • Writes photon energy $E=hc/\lambda$ (½ mark)
   • Expresses $K=h^2/(2m{\lambda }^2)$ (½ mark)
   • Takes the ratio to get $2\lambda mc/h$ (½ mark)

   Open Q18 on its own page

3. Q19 · 2 marks · Very short answerRay Optics and Optical Instruments

   A ray of monochromatic light passes through an equilateral glass prism in such a way that the angle of incidence is equal to the angle of emergence and each of these angles is 3/4 times the angle of the prism. Determine the angle of deviation and the refractive index of the glass prism.

   Try it first, then show our solutionSolution

   1. For an equilateral prism $A=60^{\circ }$. Since $i=e$, the ray passes symmetrically and the deviation is minimum.
      $i=e=\frac{3}{4}A=\frac{3}{4}\times 60^{\circ }=45^{\circ }$.
      Using $i+e=A+{\delta }_m$: ${\delta }_m=2i-A=90^{\circ }-60^{\circ }=30^{\circ }$.

      1 mark

   2. Refractive index: $n=\frac{\sin \frac{A+{\delta }_m}{2}}{\sin \frac{A}{2}}=\frac{\sin 45^{\circ }}{\sin 30^{\circ }}=\frac{1/\sqrt{2}}{1/2}=\sqrt{2}\approx 1.414$

      1 mark

   Answer: Angle of deviation = 30° (minimum deviation); refractive index = √2 ≈ 1.41

   Where the marks are

   • Uses $i=e$ ⇒ minimum deviation, finds $i=45^{\circ }$ and ${\delta }_m=30^{\circ }$ (1 mark)
   • Uses prism formula to get $n=\sqrt{2}$ (1 mark)

   Open Q19 on its own page

4. Q20 · 2 marks · Very short answerCurrent Electricity

   A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds to a steady value of 2.8 A. What is the steady temperature of the heating element if the room temperature is 27.0 °C and the temperature coefficient of resistance of nichrome is $1.70\times 10^{-4}$ °C${}^{-1}$ ?

   Try it first, then show our solutionSolution

   1. Resistance varies with temperature as $R=R_0(1+\alpha \Delta T)$, where $R_0$ is the resistance at room temperature 27.0 °C.

      ½ mark

   2. $R_0=\frac{V}{I_0}=\frac{230}{3.2}=71.9\Omega$ and $R=\frac{V}{I}=\frac{230}{2.8}=82.1\Omega$. So $\frac{R}{R_0}=\frac{I_0}{I}=1+\alpha \Delta T$.

      ½ mark

   3. $\Delta T=\frac{I_0/I-1}{\alpha }=\frac{3.2/2.8-1}{1.70\times 10^{-4}}=\frac{0.1429}{1.70\times 10^{-4}}\approx 840{}^{\circ }\text{C}$.
      Steady temperature $T=27+840=867{}^{\circ }\text{C}$.

      1 mark

   Answer: Steady temperature ≈ 867 °C

   Where the marks are

   • Writes $R=R_0(1+\alpha \Delta T)$ (½ mark)
   • Expresses $R_0=V/I_0$ and $R=V/I$ (½ mark)
   • Obtains $\Delta T\approx 840$ °C and $T\approx 867$ °C (1 mark)

   Open Q20 on its own page

5. Answer one of these 2 alternatives.

   Q21(a) · 2 marks · Very short answerRay Optics and Optical Instruments

   Show that the least possible distance between an object and its real image in a convex lens is 4f, where f is the focal length of the lens.

   Try it first, then show our solutionSolution

   1. Let $d$ be the distance between the object and its real image, with the object at distance $x$ from the lens. Then $u=-x$, $v=+(d-x)$. Lens formula: $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$.

      ½ mark

   2. $\frac{1}{f}=\frac{1}{d-x}+\frac{1}{x}=\frac{d}{x(d-x)}$

      ½ mark

   3. $\Rightarrow xd-x^2=fd\Rightarrow x^2-dx+fd=0\Rightarrow x=\frac{d\pm \sqrt{d^2-4fd}}{2}$

      ½ mark

   4. For a real image, $x$ must be real, so $d^2-4fd\ge 0\Rightarrow d\ge 4f$. Hence the least distance between object and its real image is $4f$.

      ½ mark

   Answer: $d_{min}=4f$ (shown)

   Where the marks are

   • Takes $u=-x$, $v=d-x$ with lens formula (½ mark)
   • Gets $1/f=d/[x(d-x)]$ (½ mark)
   • Forms the quadratic $x^2-dx+fd=0$ and solves it (½ mark)
   • Uses real-root condition $d^2-4fd\ge 0$ to get $d\ge 4f$ (½ mark)

   Open Q21(a) on its own page

   OR

   Q21(b) · 2 marks · Very short answerRay Optics and Optical Instruments

   In an astronomical telescope in normal adjustment a straight black line of length L is drawn on the objective lens. The eyepiece forms a real image of this line whose length is $l$. What is the angular magnification of the telescope?

   Try it first, then show our solutionSolution

   1. In normal adjustment the separation of objective and eyepiece is $f_o+f_e$. The line on the objective acts as an object for the eyepiece: $u=-(f_o+f_e)$, $f=f_e$.
      $\frac{1}{v}-\frac{1}{-(f_o+f_e)}=\frac{1}{f_e}\Rightarrow \frac{1}{v}=\frac{1}{f_e}-\frac{1}{f_o+f_e}=\frac{f_o}{f_e(f_o+f_e)}\Rightarrow v=\frac{f_e(f_o+f_e)}{f_o}$

      1 mark

   2. Linear magnification by the eyepiece: $\left|\frac{v}{u}\right|=\frac{f_e}{f_o}=\frac{\text{image size}}{\text{object size}}=\frac{l}{L}$

      ½ mark

   3. Angular magnification of the telescope: $M=\frac{f_o}{f_e}=\frac{L}{l}$

      ½ mark

   Answer: $M=f_o/f_e=L/l$

   Where the marks are

   • Applies lens formula for eyepiece with $u=-(f_o+f_e)$ and finds $v$ (1 mark)
   • Finds linear magnification $f_e/f_o=l/L$ (½ mark)
   • Concludes $M=f_o/f_e=L/l$ (½ mark)

   Open Q21(b) on its own page

Section C

1. Q22 · 3 marks · Short answerNuclei

   A given coin has a mass of 3.0 g. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity assume that the coin is entirely made of ${}_{29}^{63}Cu$ atoms (of mass 62.92960 u).
   Given $m_p$ = 1.007825u and $m_n$ = 1.008665u.

   Try it first, then show our solutionSolution

   1. Number of Cu atoms in 3.0 g: $N=\frac{6.023\times 10^{23}\times 3}{63}=2.87\times 10^{22}$

      ½ mark

   2. Each ${}_{29}^{63}$Cu nucleus has 29 protons and $63-29=34$ neutrons.
      Mass defect: $\Delta m=29(1.007825)+34(1.008665)-62.92960$
      $=29.226925+34.294610-62.92960=0.591935\text{u}$

      1 mark

   3. Binding energy per nucleus: $E_1=0.591935\times 931.5\text{MeV}=551.4\text{MeV}$

      ½ mark

   4. Energy required for the whole coin: $E=N{E}_1=2.87\times 10^{22}\times 551.4\text{MeV}\approx 1.58\times 10^{25}\text{MeV}$ $(\approx 2.5\times 10^{12}\text{J})$

      1 mark

   Answer: ≈ 1.58 × 10^25 MeV (≈ 2.5 × 10^12 J)

   Where the marks are

   • Number of atoms in 3 g ≈ $2.87\times 10^{22}$ (½ mark)
   • Mass defect per atom ≈ 0.5919 u using 29 protons and 34 neutrons (1 mark)
   • Energy per atom = Δm × 931.5 MeV (½ mark)
   • Total energy ≈ $1.58\times 10^{25}$ MeV (1 mark)

   Open Q22 on its own page

2. Q23 · 3 marks · Short answerElectrostatic Potential and Capacitance

   Charges (+q) and (–q) are placed at the points A and B respectively which are a distance 2L apart. C is the midpoint between A and B. What is the work done in moving a charge +Q along the semicircle CRD.

   Figure in the paper: Points A, C, B, D lie on a horizontal line in that order (A leftmost, D rightmost). A semicircle CRD is drawn above the line from C to D with top point R, with B at its centre (C and D on either side of B). See it in the official paper.

   Try it first, then show our solutionSolution

   1. C is the midpoint of AB, so AC = BC = L. Potential at C:
      $V_C=\frac{1}{4\pi {\epsilon }_0}\left(\frac{q}{L}-\frac{q}{L}\right)=0$

      1 mark

   2. B is the centre of the semicircle of radius BC = L, so BD = L and AD = 3L. Potential at D:
      $V_D=\frac{1}{4\pi {\epsilon }_0}\left(\frac{q}{3L}-\frac{q}{L}\right)=-\frac{2q}{3(4\pi {\epsilon }_0)L}=-\frac{q}{6\pi {\epsilon }_{0}L}$

      1 mark

   3. Work done does not depend on the path (electrostatic field is conservative):
      $W=Q(V_D-V_C)=-\frac{Qq}{6\pi {\epsilon }_{0}L}$

      1 mark

   Answer: $W=-\frac{Qq}{6\pi {\epsilon }_{0}L}$

   Where the marks are

   • $V_C=0$ (1 mark)
   • $V_D=-q/(6\pi {\epsilon }_{0}L)$ (1 mark)
   • $W=Q(V_D-V_C)=-Qq/(6\pi {\epsilon }_{0}L)$ (1 mark)

   Open Q23 on its own page

3. Q24 · 3 marks · Short answerAtoms

   The total energy of an electron in the first excited state of the hydrogen atom is about –3.4 eV.
   a. What is the kinetic energy of the electron in this state?
   b. What is the potential energy of the electron in this state?
   c. Which of the answers above would change if the choice of the zero of potential energy is changed?

   Try it first, then show our solutionSolution

   1. For an electron in a Bohr orbit: $K=-E$ and $U=-2K$ (so $E=K+U=-K$).

      1 mark

   2. (a) $K=-(-3.4)=3.4\text{eV}$
      (b) $U=-2K=-6.8\text{eV}$

      1 mark

   3. (c) The kinetic energy does not change. The potential energy, and hence the total energy, depend on the chosen zero of potential energy, so their values would change.

      1 mark

   Answer: (a) 3.4 eV; (b) −6.8 eV; (c) K stays the same; U and E change

   Where the marks are

   • Uses $K=-E$, $U=-2K$ (1 mark)
   • K = 3.4 eV and U = −6.8 eV (1 mark)
   • States K unchanged, U and total energy change (1 mark)

   Open Q24 on its own page

4. Q25 · 3 marks · Short answerCurrent Electricity

   A wire of uniform cross-section and resistance 4 ohm is bent in the shape of square ABCD. Point A is connected to a point P on DC by a wire AP of resistance 1 ohm. When a potential difference is applied between A and C, the points B and P are seen to be at the same potential. What is the resistance of the part DP?

   Figure in the paper: Square ABCD with A top-left, B top-right, C bottom-right, D bottom-left; point P on side DC; a wire AP drawn diagonally from A to P. See it in the official paper.

   Try it first, then show our solutionSolution

   1. Total resistance 4 Ω is shared equally by the four sides, so each side has 1 Ω. Let resistance of DP = $x$, then PC = $(1-x)$.
      Between A and P there are two parallel paths: wire AP (1 Ω) and AD + DP $(1+x)$ Ω, so $R_{AP}=\frac{1\cdot (1+x)}{2+x}$.
      Since B and P are at the same potential, the bridge ABCP is balanced: $\frac{R_{AB}}{R_{BC}}=\frac{R_{AP}}{R_{PC}}$.

      1½ marks

   2. $\frac{1}{1}=\frac{(1+x)/(2+x)}{1-x}\Rightarrow 1+x=(2+x)(1-x)=2-x-x^2$
      $\Rightarrow x^2+2x-1=0\Rightarrow x=-1+\sqrt{2}$ (positive root)
      Resistance of DP $=(\sqrt{2}-1)\Omega \approx 0.414\Omega$

      1½ marks

   Answer: $R_{DP}=(\sqrt{2}-1)\Omega \approx 0.41\Omega$

   Where the marks are

   • Each side 1 Ω, sets DP = x, PC = 1 − x, combines AP with AD+DP in parallel and writes balanced bridge condition (1½ marks)
   • Solves $x^2+2x-1=0$ to get $x=(\sqrt{2}-1)$ Ω (1½ marks)

   Open Q25 on its own page

5. Q26 · 3 marks · Short answerMoving Charges and Magnetism

   The given figure shows a long straight wire of a circular cross-section (radius a) carrying steady current I. The current I is uniformly distributed across this cross-section. Calculate the magnetic field in the region r < a and r > a.

   Figure in the paper: Long horizontal wire carrying current I to the right, with circular cross-section of radius a shown. Amperian loop 1 is a circle of radius r1 inside the wire (r<a); loop 2 is a larger circle of radius r outside the wire, with point P on it. See it in the official paper.

   Try it first, then show our solutionSolution

   1. Region r > a: Take a circular Amperian loop (loop 2) of radius r, concentric with the wire. By symmetry B is tangential and of the same magnitude everywhere on it, length $L=2\pi r$. Whole current I is enclosed.
      Ampere's circuital law: $B(2\pi r)={\mu }_{0}I\Rightarrow B=\frac{{\mu }_{0}I}{2\pi r}$, so $B\propto \frac{1}{r}$.

      1½ marks

   2. Region r < a: Take loop 1 of radius r inside the wire. Current is uniform, so the enclosed current is $I_e=I\frac{\pi r^2}{\pi a^2}=\frac{I{r}^2}{a^2}$.
      $B(2\pi r)={\mu }_0\frac{I{r}^2}{a^2}\Rightarrow B=\frac{{\mu }_{0}I}{2\pi a^2}r$, so $B\propto r$.

      1½ marks

   Answer: $r>a$: $B={\mu }_{0}I/(2\pi r)$; $r<a$: $B={\mu }_{0}Ir/(2\pi a^2)$

   Where the marks are

   • Applies Ampere's law to outer loop to get $B={\mu }_{0}I/2\pi r$ (1½ marks)
   • Finds enclosed current $I{r}^2/a^2$ and gets $B={\mu }_{0}Ir/2\pi a^2$ (1½ marks)

   Open Q26 on its own page

6. Q27 · 3 marks · Short answerElectromagnetic Waves

   Identify the part of the electromagnetic spectrum which:
   a) produces heating effect,
   b) is absorbed by the ozone layer in the atmosphere,
   c) is used for studying crystal structure.

   Write any one method of the production of each of the above radiations.

   Try it first, then show our solutionSolution

   1. (a) Infrared waves produce heating effect. Produced by hot bodies and by vibrations of atoms and molecules (e.g. an electric heater or a hot iron).

      1 mark

   2. (b) Ultraviolet rays are absorbed by the ozone layer. Produced by very hot bodies such as the sun and by special lamps like mercury vapour / arc lamps.

      1 mark

   3. (c) X-rays are used to study crystal structure. Produced when a metal target (e.g. tungsten) is bombarded with high-energy electrons, as in a Coolidge tube.

      1 mark

   Answer: (a) Infrared; (b) Ultraviolet; (c) X-rays

   Where the marks are

   • Identifies infrared (½ mark) and gives its production method (½ mark)
   • Identifies ultraviolet (½ mark) and gives its production method (½ mark)
   • Identifies X-rays (½ mark) and gives its production method (½ mark)

   Open Q27 on its own page

7. Answer one of these 2 alternatives.

   Q28(a) · 3 marks · Short answerElectromagnetic Induction

   a. Define mutual inductance and write its SI unit.
   b. Two circular loops, one of small radius r and other of larger radius R, such that R >> r, are placed coaxially with centres coinciding. Obtain the mutual inductance of the arrangement.

   Try it first, then show our solutionSolution

   1. Mutual inductance of two coils is the magnetic flux linked with one coil per unit current in the other (${\Phi }_2=M{I}_1$). Equivalently, it equals the emf induced in one coil per unit rate of change of current in the other.

      ½ mark

   2. SI unit: henry (H).

      ½ mark

   3. Let current $I$ flow in the large loop of radius R. Field at its centre: $B=\frac{{\mu }_{0}I}{2R}$.

      ½ mark

   4. Since $r\ll R$, this field is nearly uniform over the small loop. Flux through small loop: $\Phi =B\pi r^2=\frac{{\mu }_{0}I\pi r^2}{2R}$.

      ½ mark

   5. Also $\Phi =MI$.

      ½ mark

   6. $\therefore M=\frac{\Phi }{I}=\frac{{\mu }_0\pi r^2}{2R}$

      ½ mark

   Answer: Unit: henry; $M={\mu }_0\pi r^2/(2R)$

   Where the marks are

   • Correct definition of mutual inductance (½ mark)
   • SI unit henry (½ mark)
   • Field at centre of large loop ${\mu }_{0}I/2R$ (½ mark)
   • Flux through small loop $B\pi r^2$ (½ mark)
   • Uses $\Phi =MI$ (½ mark)
   • Obtains $M={\mu }_0\pi r^2/2R$ (½ mark)

   Open Q28(a) on its own page

   OR

   Q28(b) · 3 marks · Short answerMoving Charges and Magnetism

   Two long straight parallel current carrying conductors are kept 'a' distant apart in air. The direction of current in both the conductors is same. Find the magnitude of force per unit length and direction of the force between them. Hence define one ampere.

   Try it first, then show our solutionSolution

   1. Field of conductor 1 (current $I_1$) at conductor 2, distance a away: $B_1=\frac{{\mu }_0{I}_1}{2\pi a}$

      ½ mark

   2. Force on length $l$ of conductor 2: $F_2=I_{2}l{B}_1\sin 90^{\circ }=\frac{{\mu }_0{I}_1{I}_{2}l}{2\pi a}$, so $\frac{F_2}{l}=\frac{{\mu }_0{I}_1{I}_2}{2\pi a}$, directed towards conductor 1.

      ½ mark

   3. Similarly, the field of conductor 2 at conductor 1: $B_2=\frac{{\mu }_0{I}_2}{2\pi a}$

      ½ mark

   4. Force per unit length on conductor 1: $\frac{F_1}{l}=\frac{{\mu }_0{I}_1{I}_2}{2\pi a}$, directed towards conductor 2. Equal and opposite forces; for currents in the same direction the force is attractive.

      ½ mark

   5. Putting $I_1=I_2=1$ A, $a=1$ m: $\frac{F}{l}=\frac{4\pi \times 10^{-7}\times 1\times 1}{2\pi \times 1}=2\times 10^{-7}$ N/m.
      One ampere is that steady current which, flowing in each of two infinitely long, straight, parallel conductors placed 1 m apart in vacuum, makes each conductor experience a force of $2\times 10^{-7}$ N per metre of its length.

      1 mark

   Answer: $F/l={\mu }_0{I}_1{I}_2/(2\pi a)$, attractive; 1 A gives $2\times 10^{-7}$ N/m at 1 m separation

   Where the marks are

   • Field due to first conductor at second (½ mark)
   • Force per unit length on second conductor with direction (½ mark)
   • Field due to second conductor at first (½ mark)
   • Force on first conductor; forces attractive (½ mark)
   • Definition of one ampere using $2\times 10^{-7}$ N/m (1 mark)

   Open Q28(b) on its own page

Section D

case study based

1. Q29 · 4 marks · Case-basedCompetency-basedSemiconductor Electronics: Materials, Devices and Simple Circuits

   Read the following paragraph and answer the questions that follow.
   A semiconductor diode is basically a pn junction with metallic contacts provided at the ends for the application of an external voltage. It is a two terminal device. When an external voltage is applied across a semiconductor diode such that p-side is connected to the positive terminal of the battery and n-side to the negative terminal, it is said to be forward biased. When an external voltage is applied across the diode such that n-side is positive and p-side is negative, it is said to be reverse biased. An ideal diode is one whose resistance in forward biasing is zero and the resistance is infinite in reverse biasing. When the diode is forward biased, it is found that beyond forward voltage called knee voltage, the conductivity is very high. When the biasing voltage is more than the knee voltage the potential barrier is overcome and the current increases rapidly with increase in forward voltage. When the diode is reverse biased, the reverse bias voltage produces a very small current about a few microamperes which almost remains constant with bias. This small current is reverse saturation current.

   1. i.

      In the given figure, a diode D is connected to an external resistance R = 100 $\Omega$ and an emf of 3.5 V. If the barrier potential developed across the diode is 0.5 V, the current in the circuit will be:
      (a) 40 mA (b) 20 mA (c) 35 mA (d) 30 mA

      1. (A)40 mA
      2. (B)20 mA
      3. (C)35 mA
      4. (D)30 mA

      1 mark
   2. ii.

      In which of the following figures, the pn diode is reverse biased?

      1. (A)Diode with p-side at -12 V and resistor R end at -5 V
      2. (B)Diode p-side earthed, resistor R end at -10 V
      3. (C)Diode with resistor R end at +5 V and other end earthed
      4. (D)Diode p-side at +5 V, resistor R end at +10 V

      1 mark
   3. iii.

      Based on the V-I characteristics of the diode, we can classify diode as

      1. (A)bilateral device
      2. (B)ohmic device
      3. (C)non-ohmic device
      4. (D)passive element

      1 mark
   4. iv.

      Two identical PN junctions can be connected in series by three different methods as shown in the figure. If the potential difference in the junctions is the same, then the correct connections will be
      (1) P N N P, (2) P N P N, (3) N P N P, each connected across a battery with + and − terminals.

      1. (A)in the circuits (1) and (2)
      2. (B)in the circuits (2) and (3)
      3. (C)in the circuits (1) and (3)
      4. (D)only in the circuit (1)

      1 mark

   5. OR

      iv.

      The V-I characteristic of a diode is shown in the figure. The ratio of the resistance of the diode at I = 15 mA to the resistance at V = -10 V is

      1. (A)100
      2. (B)$10^6$
      3. (C)10
      4. (D)$10^{-6}$

      1 mark

   Figure in the paper: Circuit: diode D in series with R = 100 Ω and a 3.5 V cell. Part ii figures: four diode-resistor circuits with the given potentials. Part iv (OR) figure: three series PN junction connections with battery. Final V-I graph for silicon diode: forward current I (mA) axis marked 10, 15, 20, 30; voltage axis marked 0.5, 0.7, 0.8 V; steep rise near 0.7 V; reverse side at -10 V shows current about 1 µA. See it in the official paper.

   Try it first, then show our solutionSolution

   1. (i) Voltage across R $=3.5-0.5=3.0$ V, so $I=\frac{3.0}{100}=0.03$ A $=30$ mA. Option (d).

      1 mark

   2. (ii) A diode is reverse biased when its n-side is at a higher potential than its p-side. In circuit (c) the n-side is held at +5 V through R while the p-side is earthed (0 V), so it is reverse biased. Option (c).

      1 mark

   3. (iii) The V-I graph of a diode is not a straight line (current does not vary linearly with voltage), so the diode is a non-ohmic device. Option (c).

      1 mark

   4. (iv)(a) For the same potential difference across both junctions, both must be biased the same way, i.e. connected in the same orientation. In (2) PNPN and (3) NPNP both junctions face the same way; in (1) PNNP one is forward and one reverse biased. Option (b).

      1 mark

   5. (iv)(b) From the graph, near $I=15$ mA: $R=\frac{\Delta V}{\Delta I}=\frac{0.8-0.7}{(20-10)\times 10^{-3}}=10\Omega$. At $V=-10$ V, $I=1\mu$A: $R=\frac{10}{10^{-6}}=10^7\Omega$. Ratio $=\frac{10}{10^7}=10^{-6}$. Option (d).

   Answer: (i) (d) 30 mA; (ii) (c); (iii) (c) non-ohmic; (iv) (b) circuits (2) and (3) OR (d) $10^{-6}$

   Where the marks are

   • (i) 30 mA, option (d) (1 mark)
   • (ii) option (c), n-side at higher potential (1 mark)
   • (iii) non-ohmic device, option (c) (1 mark)
   • (iv) circuits (2) and (3), option (b); OR ratio $10^{-6}$, option (d) (1 mark)

   Open Q29 on its own page

2. Q30 · 4 marks · Case-basedCompetency-basedRay Optics and Optical Instruments

   Read the following paragraph and answer the questions that follow.
   Types of Lenses and their combination
   A convex or converging lens is thicker at the centre than at the edges. It converges a beam of light on refraction through it. It has a real focus. Convex lens is of three types: Double convex lens, Plano convex lens and Concavo-convex lens.
   Concave lens is thinner at the centre than at the edges. It diverges a beam of light on refraction through it. It has a virtual focus. Concave lenses are of three types: Double concave lens, Plano concave lens and Convexo-concave lens.
   When two thin lenses of focal lengths $f_1$ and $f_2$ are placed in contact with each other along their common principal axis, then the two lens system is regarded as a single lens of focal length f and
   $$\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}$$
   If several thin lenses of focal length $f_1,f_2,...f_n$ are placed in contact, then the effective focal length of the combination is given by
   $$\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}+.....+\frac{1}{f_n}$$
   and in terms of power, we can write
   P = P1 + P2 + .... + Pn
   The value of focal length and power of a lens must be used with proper sign consideration.

   1. i.

      Two thin lenses are kept coaxially in contact with each other and the focal length of the combination is 80 cm. If the focal length of one lens is 20 cm, the focal length of the other would be

      1. (A)-26.7cm
      2. (B)60cm
      3. (C)80cm
      4. (D)30cm

      1 mark
   2. ii.

      A spherical air bubble is embedded in a piece of glass. For a ray of light passing through the bubble, it behaves like a

      1. (A)converging lens
      2. (B)diverging lens
      3. (C)mirror
      4. (D)thin plane sheet of glass

      1 mark
   3. iii.

      Lens generally used in magnifying glass is

      1. (A)single concave lens
      2. (B)single convex lens
      3. (C)combination of convex lens of lower power and concave lens of lower focal length
      4. (D)Planoconcave lens

      1 mark
   4. iv.

      The magnification of an image by a convex lens is positive only when the object is placed

      1. (A)at its focus F
      2. (B)between F and 2F
      3. (C)at 2F
      4. (D)between F and optical centre

      1 mark

   5. OR

      iv.

      A convex lens of 20 cm focal length forms a real image which is three times magnified. The distance of the object from the lens is

      1. (A)13.33 cm
      2. (B)14 cm
      3. (C)26.66 cm
      4. (D)25 cm

      1 mark

   Try it first, then show our solutionSolution

   1. (i) $\frac{1}{f_2}=\frac{1}{f}-\frac{1}{f_1}=\frac{1}{80}-\frac{1}{20}=-\frac{3}{80}\Rightarrow f_2=-26.7$ cm. Option (a).

      1 mark

   2. (ii) The air bubble is a biconvex shape of a rarer medium (air) inside a denser medium (glass), so it spreads light out and acts as a diverging lens. Option (b).

      1 mark

   3. (iii) A single convex lens with the object placed within its focal length gives an erect, magnified, virtual image, so it is used as a magnifying glass. Option (b).

      1 mark

   4. (iv)(a) Positive magnification means an erect, virtual image, which a convex lens forms only when the object is between F and the optical centre. Option (d).

      1 mark

   5. (iv)(b) Real image, $m=-3\Rightarrow v=-3u$. With $u=-x$, $v=3x$: $\frac{1}{3x}+\frac{1}{x}=\frac{1}{20}\Rightarrow \frac{4}{3x}=\frac{1}{20}\Rightarrow x=26.66$ cm. Option (c).

   Answer: (i) (a) −26.7 cm; (ii) (b); (iii) (b); (iv) (d) OR (c) 26.66 cm

   Where the marks are

   • (i) −26.7 cm, option (a) (1 mark)
   • (ii) diverging lens, option (b) (1 mark)
   • (iii) single convex lens, option (b) (1 mark)
   • (iv) between F and optical centre, option (d); OR 26.66 cm, option (c) (1 mark)

   Open Q30 on its own page

Section E

long answer

1. Answer one of these 2 alternatives.

   Q31(a) · 5 marks · Long answerRay Optics and Optical Instruments

   i. Draw a ray diagram for the formation of image of a point object by a thin double convex lens having radii of curvature $R_1$ and $R_2$. Hence derive lens maker's formula.
   ii. A converging lens has a focal length of 10 cm in air. It is made of a material of refractive index 1.6. If it is immersed in a liquid of refractive index 1.3, find its new focal length.

   Try it first, then show our solutionSolution

   1. Ray diagram (described): A thin double convex lens of refractive index $n_2$ is placed in a medium of refractive index $n_1$. Point object O on the principal axis. A ray from O strikes the first surface ABC (centre $C_1$, radius $R_1$) at A and bends towards the axis, heading to $I_1$ (image by the first surface). Before reaching $I_1$ it meets the second surface ADC (centre $C_2$, radius $R_2$) and is refracted to meet the axis at the final image I. B and D are the poles of the two surfaces.

      1 mark

   2. Refraction at the first surface (rarer to denser), image at $I_1$:
      $\frac{n_2}{B{I}_1}+\frac{n_1}{OB}=\frac{n_2-n_1}{B{C}_1}$ ...(i)

      ½ mark

   3. $I_1$ acts as a virtual object for the second surface (denser to rarer), final image at I:
      $\frac{n_1}{DI}-\frac{n_2}{D{I}_1}=\frac{n_2-n_1}{D{C}_2}$ ...(ii)

      ½ mark

   4. For a thin lens $B{I}_1\approx D{I}_1$. Adding (i) and (ii):
      $\frac{n_1}{OB}+\frac{n_1}{DI}=(n_2-n_1)\left(\frac{1}{B{C}_1}+\frac{1}{D{C}_2}\right)$ ...(iii)
      If the object is at infinity ($OB\to \infty$), the image forms at the focus: $DI=f$:
      $\frac{n_1}{f}=(n_2-n_1)\left(\frac{1}{B{C}_1}+\frac{1}{D{C}_2}\right)$

      ½ mark

   5. Sign convention: $B{C}_1=+R_1$, $D{C}_2=-R_2$. Hence
      $\frac{1}{f}=\left(\frac{n_2}{n_1}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)=(n_{21}-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$ — the lens maker's formula.

      ½ mark

   6. (ii) In air: $\frac{1}{f_a}=(1.6-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)=0.6\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$ ...(1)

      1 mark

   7. In liquid: $\frac{1}{f_l}=\left(\frac{1.6}{1.3}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)=\frac{0.3}{1.3}\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$ ...(2)
      Dividing (1) by (2): $\frac{f_l}{f_a}=\frac{0.6\times 1.3}{0.3}=2.6\Rightarrow f_l=2.6\times 10=26$ cm.

      1 mark

   Answer: $\frac{1}{f}=(n_{21}-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$; new focal length = 26 cm

   Where the marks are

   • Labelled ray diagram of refraction through double convex lens (1 mark)
   • Refraction equation at first surface (½ mark)
   • Refraction equation at second surface with $I_1$ as virtual object (½ mark)
   • Adds equations, takes object at infinity so image at focus (½ mark)
   • Applies sign convention $B{C}_1=R_1$, $D{C}_2=-R_2$ to get lens maker's formula (½ mark)
   • Lens maker's equation in air with 0.6 factor (1 mark)
   • Equation in liquid and $f_l=26$ cm (1 mark)

   Open Q31(a) on its own page

   OR

   Q31(b) · 5 marks · Long answerWave Optics

   i. Define a wavefront. How is it different from a ray?
   ii. Using Huygens's construction of secondary wavelets draw a diagram showing the passage of a plane wavefront from a denser to a rarer medium. Using it verify Snell's law.
   iii. In a double slit experiment using light of wavelength 600nm and the angular width of the fringe formed on a distant screen is 0.1°. Find the spacing between the two slits.
   iv. Write two differences between interference pattern and diffraction pattern.

   Try it first, then show our solutionSolution

   1. (i) A wavefront is the locus of all points of the medium vibrating in the same phase (a surface of constant phase). A ray shows the direction in which the wave travels, whereas a wavefront is a surface; the ray at any point is always normal to the wavefront at that point.

      1 mark

   2. (ii) Diagram (described): A plane wavefront AB in the denser medium 1 (speed $v_1$) meets the boundary XY obliquely at angle of incidence i, end A touching first. In time t, the wavelet from B reaches C on the boundary: $BC=v_{1}t$. In the same time, a secondary wavelet from A spreads into the rarer medium 2 with radius $AE=v_{2}t$ (larger, since $v_2>v_1$). The tangent CE drawn from C to this wavelet is the refracted wavefront; the refracted ray bends away from the normal ($r>i$).

      1 mark

   3. From triangles ABC and AEC: $\sin i=\frac{BC}{AC}$, $\sin r=\frac{AE}{AC}$
      $\Rightarrow \frac{\sin i}{\sin r}=\frac{BC}{AE}=\frac{v_{1}t}{v_{2}t}=\frac{v_1}{v_2}=$ constant. With $n=c/v$, this gives $n_1\sin i=n_2\sin r$, which is Snell's law.

      1 mark

   4. (iii) Angular fringe width $\theta =\frac{\lambda }{d}$. $\theta =0.1^{\circ }=0.1\times \frac{\pi }{180}=1.745\times 10^{-3}$ rad.
      $d=\frac{\lambda }{\theta }=\frac{6\times 10^{-7}}{1.745\times 10^{-3}}=3.4\times 10^{-4}$ m (≈ 0.34 mm).

      1 mark

   5. (iv) Differences: (1) Interference results from superposition of waves from two coherent sources; diffraction results from superposition of wavelets from different parts of the same wavefront. (2) In interference all bright fringes have nearly equal width and intensity; in diffraction the central maximum is twice as wide as the others and intensity of successive maxima falls sharply.

      1 mark

   Answer: (iii) Slit spacing ≈ $3.4\times 10^{-4}$ m; Snell's law $\sin i/\sin r=v_1/v_2$ verified

   Where the marks are

   • Definition of wavefront and its difference from a ray (1 mark)
   • Huygens construction diagram for denser to rarer refraction (1 mark)
   • Shows $\sin i/\sin r=v_1/v_2$ = constant (1 mark)
   • Uses $\theta =\lambda /d$ with θ in radians to get $d=3.4\times 10^{-4}$ m (1 mark)
   • Two correct differences between interference and diffraction patterns (1 mark)

   Open Q31(b) on its own page

2. Answer one of these 2 alternatives.

   Q32(a) · 5 marks · Long answerElectrostatic Potential and Capacitance

   i. Derive an expression for the capacitance of a parallel plate capacitor with air present between the two plates.
   ii. Obtain the equivalent capacitance of the network shown in figure. For a 300 V supply, determine the charge on each capacitor.

   Figure in the paper: Network across a 300 V supply: C1 = 100 pF in the top branch; C2 = 200 pF and C3 = 200 pF in series in the middle branch; C4 = 100 pF in the bottom branch; the three branches are in parallel across the 300 V source. See it in the official paper.

   Try it first, then show our solutionSolution

   1. (i) Two parallel plates, each of area A, separated by distance d, carry charges +Q and −Q, with surface charge density $\sigma =Q/A$. Each sheet produces field $\frac{\sigma }{2{\epsilon }_0}$. Between the plates the two fields are in the same direction and add; outside they cancel. So the field between the plates is $E=\frac{\sigma }{{\epsilon }_0}=\frac{Q}{A{\epsilon }_0}$.

      1 mark

   2. Potential difference: $V=Ed=\frac{Qd}{A{\epsilon }_0}$. Capacitance: $C=\frac{Q}{V}=\frac{{\epsilon }_{0}A}{d}$.

      1 mark

   3. (ii) $C_2$ and $C_3$ in series: $\frac{200\times 200}{400}=100$ pF. This is in parallel with $C_1$: $100+100=200$ pF. This combination is in series with $C_4$:
      $C_{eq}=\frac{200\times 100}{300}=\frac{200}{3}$ pF ≈ 66.7 pF

      1 mark

   4. Charge drawn from supply = charge on $C_4$: $Q_4=\frac{200}{3}\times 10^{-12}\times 300=2\times 10^{-8}$ C. P.d. across $C_4=\frac{2\times 10^{-8}}{100\times 10^{-12}}=200$ V.

      ½ mark

   5. P.d. across $C_1$ = $300-200=100$ V, so $Q_1=100\times 10^{-12}\times 100=1\times 10^{-8}$ C.

      ½ mark

   6. The series pair $C_2$, $C_3$ also has 100 V across it, i.e. 50 V across each (equal capacitances).
      $Q_2=Q_3=200\times 10^{-12}\times 50=1\times 10^{-8}$ C.

      1 mark

   Answer: $C={\epsilon }_{0}A/d$; $C_{eq}=200/3$ pF; $Q_1=10^{-8}$ C, $Q_2=Q_3=10^{-8}$ C, $Q_4=2\times 10^{-8}$ C

   Where the marks are

   • Field between plates $E=\sigma /{\epsilon }_0=Q/A{\epsilon }_0$ (1 mark)
   • $V=Ed$ and $C={\epsilon }_{0}A/d$ (1 mark)
   • Equivalent capacitance 200/3 pF (1 mark)
   • Charge on $C_4=2\times 10^{-8}$ C (½ mark)
   • Charge on $C_1=1\times 10^{-8}$ C (½ mark)
   • Charges on $C_2$ and $C_3=1\times 10^{-8}$ C each (½ + ½ mark)

   Open Q32(a) on its own page

   OR

   Q32(b) · 5 marks · Long answerElectrostatic Potential and Capacitance

   i. A dielectric slab of thickness 't' is kept between the plates of a parallel plate capacitor with plate separation 'd' (t < d). Derive the expression for the capacitance of the capacitor.
   ii. A capacitor of capacity $C_1$ is charged to the potential of $V_o$. On disconnecting with the battery, it is connected with an uncharged capacitor of capacity $C_2$ as shown in the adjoining figure. Find the ratio of energies before and after the connection of switch S.

   Figure in the paper: Circuit with capacitor C1 (charged to V0) and uncharged capacitor C2 connected in parallel through a switch S in the top wire. See it in the official paper.

   Try it first, then show our solutionSolution

   1. (i) Let each plate have area $A$ and charge density $\sigma =Q/A$. In the air gap (total thickness $d-t$) the field is $E_0=\frac{\sigma }{{\epsilon }_0}=\frac{Q}{A{\epsilon }_0}$.

      ½ mark

   2. Inside the dielectric slab of constant $K$, polarisation reduces the field to $E=\frac{E_0}{K}=\frac{Q}{KA{\epsilon }_0}$.

      1 mark

   3. Potential difference between the plates: $V=E_0(d-t)+Et=\frac{Q}{A{\epsilon }_0}\left(d-t+\frac{t}{K}\right)$.

      1 mark

   4. Capacitance: $C=\frac{Q}{V}=\frac{{\epsilon }_{0}A}{d-t+\frac{t}{K}}$. This is greater than $\frac{{\epsilon }_{0}A}{d}$ because $K>1$.

      ½ mark

   5. (ii) Before S is closed, only $C_1$ is charged: $U_i=\frac{1}{2}{C}_1{V}_0^2$.

      ½ mark

   6. Charge is conserved when S is closed. With the capacitors in parallel, the common potential is $V=\frac{C_1{V}_0}{C_1+C_2}$.

      ½ mark

   7. Final energy: $U_f=\frac{1}{2}(C_1+C_2)V^2=\frac{C_1^2{V}_0^2}{2(C_1+C_2)}$.

      ½ mark

   8. Ratio: $\frac{U_f}{U_i}=\frac{C_1}{C_1+C_2}$, i.e. $U_i:U_f=(C_1+C_2):C_1$. The final energy is less; the difference is lost as heat and radiation during charge redistribution.

      ½ mark

   Answer: (i) $C=\frac{{\epsilon }_{0}A}{d-t+t/K}$; (ii) $U_i:U_f=(C_1+C_2):C_1$ (i.e. $U_f/U_i=C_1/(C_1+C_2)$)

   Where the marks are

   • Writes the field in the air gap $E_0=\sigma /{\epsilon }_0$ (½ mark)
   • Writes the field in the dielectric $E_0/K$ (1 mark)
   • Gives $V=E_0(d-t)+E_{0}t/K$ (1 mark)
   • Obtains $C={\epsilon }_{0}A/(d-t+t/K)$ (½ mark)
   • Initial energy $\frac{1}{2}{C}_1{V}_0^2$ (½ mark)
   • Common potential $C_1{V}_0/(C_1+C_2)$ (½ mark)
   • Final energy $C_1^2{V}_0^2/2(C_1+C_2)$ (½ mark)
   • Ratio $U_f:U_i=C_1:(C_1+C_2)$ (½ mark)

   Open Q32(b) on its own page

3. Answer one of these 2 alternatives.

   Q33(a) · 5 marks · Long answerAlternating Current

   a. Draw graphs showing the variations of inductive reactance and capacitive reactance with frequency of applied ac source.
   b. Draw the phasor diagram for a series LRC circuit connected to an AC source.
   c. When an alternating voltage of 220V is applied across a device X, a current of 0.25A flows which lags behind the applied voltage in phase by π/2 radian. If the same voltage is applied across another device Y, the same current flows but now it is in phase with the applied voltage.
   (i) Name the devices X and Y.
   (ii) Calculate the current flowing in the circuit when the same voltage is applied across the series combination of X and Y.

   Try it first, then show our solutionSolution

   1. a. $X_L=2\pi fL$, so the graph of $X_L$ against $f$ is a straight line through the origin. $X_C=\frac{1}{2\pi fC}$, so the graph of $X_C$ against $f$ is a rectangular hyperbola: it is very large at low $f$ and approaches the $f$-axis as $f$ increases. Both are drawn on the same $f$-axis, with the curves labelled $X_L$ and $X_C$.

      1 mark

   2. b. Phasor diagram:

      • Draw $I$ along the reference (x) axis.
      • Draw $V_R=IR$ along $I$.
      • Draw $V_L=I{X}_L$ vertically up, leading $I$ by $90^{\circ }$.
      • Draw $V_C=I{X}_C$ vertically down, lagging $I$ by $90^{\circ }$.

      The resultant $V$ is the hypotenuse of a triangle with sides $V_R$ and $(V_L-V_C)$. It makes an angle $\varphi$ with $I$, where $\tan \varphi =\frac{X_L-X_C}{R}$.

      1 mark

   3. c(i). In X, the current lags the voltage by $\pi /2$, so X is a pure inductor. In Y, the current is in phase with the voltage, so Y is a pure resistor.

      1 mark

   4. c(ii). $X_L=\frac{V}{I}=\frac{220}{0.25}=880\Omega$ and $R=\frac{220}{0.25}=880\Omega$.

      1 mark

   5. For the series combination, $Z=\sqrt{R^2+X_L^2}=880\sqrt{2}\approx 1244.5\Omega$.

      So $I=\frac{V}{Z}=\frac{220}{880\sqrt{2}}=\frac{1}{4\sqrt{2}}\approx 0.177\text{A}$.

      1 mark

   Answer: (c)(i) X = inductor, Y = resistor; (ii) $Z=880\sqrt{2}\Omega$, $I\approx 0.177$ A

   Where the marks are

   • Graphs: $X_L$ straight line through origin, $X_C$ hyperbola (½ + ½ mark)
   • Correct phasor diagram for series LCR with $V_L$ leading and $V_C$ lagging $I$ (1 mark)
   • Identifies X as inductor and Y as resistor (½ + ½ mark)
   • Finds $X_L=R=880\Omega$ (1 mark)
   • Finds $Z=880\sqrt{2}\Omega$ and $I=0.177$ A (1 mark)

   Open Q33(a) on its own page

   OR

   Q33(b) · 5 marks · Long answerAlternating Current

   a. A series LCR circuit is connected to an ac source. Using the phasor diagram, derive the expression for the impedance of the circuit.
   b. Plot a graph to show the variation of current with frequency of the ac source, explaining the nature of its variation for two different resistances $R_1$ and $R_2$ ($R_1<R_2$) at resonance.

   Try it first, then show our solutionSolution

   1. a. Let $v=v_m\sin \omega t$ be applied to L, C and R in series. The same current $i=i_m\sin (\omega t+\varphi )$ flows through all three elements. The voltage amplitudes are $V_R=i_{m}R$, $V_L=i_m{X}_L$ and $V_C=i_m{X}_C$.

      1 mark

   2. Phasor diagram (described): $I$ and $V_R$ lie along the same line. $V_L$ is perpendicular to $I$ and leads it by $90^{\circ }$. $V_C$ is opposite to $V_L$ and lags $I$ by $90^{\circ }$. Taking $V_L>V_C$, the net reactive voltage is $(V_L-V_C)$, perpendicular to $V_R$.

      1 mark

   3. The source voltage is the phasor sum of these, so by Pythagoras:

      $v_m^2=V_R^2+(V_L-V_C{)}^2=i_m^2\left[R^2+(X_L-X_C{)}^2\right]$

      1 mark

   4. Hence $i_m=\frac{v_m}{\sqrt{R^2+(X_L-X_C{)}^2}}=\frac{v_m}{Z}$.

      The impedance is $Z=\sqrt{R^2+(X_L-X_C{)}^2}$. The phase angle is $\tan \varphi =\frac{X_L-X_C}{R}$.

      1 mark

   5. b. Graph of $i_m$ against $\omega$ for both resistances:

      • Each curve rises to a peak at the resonant frequency ${\omega }_0=1/\sqrt{LC}$, where $X_L=X_C$ and $Z=R$, so $i_{max}=v_m/R$. On either side of ${\omega }_0$ the current falls.
      • Curve (i), for $R_1$, has the higher and sharper peak, because $R_1<R_2$.
      • Curve (ii), for $R_2$, has a lower and broader peak.
      • Both peaks occur at the same ${\omega }_0$, since ${\omega }_0$ does not depend on $R$.
      1 mark

   Answer: $Z=\sqrt{R^2+(X_L-X_C{)}^2}$; at resonance ${\omega }_0=1/\sqrt{LC}$, the $R_1$ curve has a higher, sharper peak and the $R_2$ curve a lower, broader peak.

   Where the marks are

   • States the same current through all elements and the voltage amplitudes $V_R$, $V_L$, $V_C$ (1 mark)
   • Correct phasor diagram with the phase relations (1 mark)
   • Applies Pythagoras: $v_m^2=i_m^2[R^2+(X_L-X_C{)}^2]$ (1 mark)
   • Obtains $Z=\sqrt{R^2+(X_L-X_C{)}^2}$ (and $\tan \varphi$) (1 mark)
   • Resonance curves: same ${\omega }_0$; $R_1$ gives a higher, sharper peak than $R_2$ (1 mark)

   Open Q33(b) on its own page