Roll No. x……®……ΔEÚ SS—40—PHY. No. of Questions — 30 No. of Printed Pages — 11 =SS… ®……v™… ®…EÚ {…Æ˙“I……, 2013 SENIOR SECONDARY EXAMINATION, 2013 ¶……Ë i…EÚ ¥…Y……x… PHYSICS ∫…®…™… : 3 {…⁄h……»EÚ 1 P…h]‰ı 4 : 56 {…Æ˙“I……Ãl…™…… E‰Ú ±…B ∫……®……x™… x…nÊ˘∂… : GENERAL INSTRUCTIONS TO THE EXAMINEES : 1. {…Æ˙“I……l…‘ ∫…¥…«|…l…®… +{…x…‰ |…∂x… {…j… {…Æ˙ x……®……ΔEÚ + x…¥……™…«i…: ±…J…Â* Candidate must write first his / her Roll No. on the question paper compulsorily. 2. ∫…¶…“ |…∂x… EÚÆ˙x…‰ + x…¥……™…« ΩÈ˛* All the questions are compulsory. 3. |…i™…‰EÚ |…∂x… EÚ… =k…Æ˙ n˘“ M…<« =k…Æ˙-{…÷Œ∫i…EÚ… ®…Â Ω˛“ ±…J…Â* Write the answer to each question in the given answer-book only. 4. V…x… |…∂x…… ®… +…xi… Æ˙EÚ J…hb˜ ΩÈ˛, =x… ∫…¶…“ E‰Ú =k…Æ˙ BEÚ ∫……l… Ω˛“ ±…J…Â* For questions having more than one part, the answers to those parts are to be written together in continuity. 5. |…∂x… {…j… E‰Ú Ω˛xn˘“ ¥… +ΔO…‰V…“ ∞¸{……xi…Æ˙ ®… EÚ∫…“ |…EÚ…Æ˙ EÚ“ j…÷ ]ı / +xi…Æ˙ / ¥…Æ˙…‰v……¶……∫… Ω˛…‰x…‰ {…Æ˙ Ω˛xn˘“ ¶……π…… E‰Ú |…∂x… EÚ…‰ ∫…Ω˛“ ®……x…Â* If there is any error / difference / contradiction in Hindi & English versions of the question paper, the question of Hindi version should be treated valid. SS—40—Phy. SS–580 [ Turn over 2 6. 7. |…∂x… ∫…ΔJ™…… +ΔEÚ |…i™…‰EÚ |…∂x… 1 – 13 1 14 – 24 2 25 – 27 3 28 – 30 4 Q. Nos. Marks per question 1 – 13 1 14 – 24 2 25 – 27 3 28 – 30 4 |…∂x… ∫…ΔJ™…… 21 i…l…… 27 ∫…‰ 30 ®… +…xi… Æ˙EÚ ¥…EÚ±{… ΩÈ˛* There are internal choices in Q. Nos. 21 and 27 to 30. 8. {…Æ˙“I…… ®… EËÚ±…E÷Ú±…‰]ıÆ˙ E‰Ú ={…™……‰M… EÚ“ +x…÷®… i… x…Ω˛” Ω˲* Use of calculator is not allowed in the examination. 1. ¥…t÷i… I…‰j… Ɖ˙J……+… E‰Ú EÚ…‰<« n˘…‰ M…÷h… ±… J…B* Write any two properties of electric field lines. 2. 1 ∫…®… ¥…¶…¥… {…fiπ`ˆ EÚ…‰ {… Æ˙¶…… π…i… EÚ“ V…B* Define equipotential surface. 3. 1 B¥…Δ Y = 48 × 10−8 Ω -m E‰Ú S……±…EÚ…Â EÚ“ ±…®§……<« +…v…“ EÚÆ˙x…‰ {…Æ˙ X ¥… ±…B ∫…ΔM…i… ®……x… ±… J…B* X = 4Ω Y E‰Ú Write the corresponding values of X and Y for which the lengths of conductors X = 4Ω and Y = 48 × 10 −8 Ω -m are reduced to half. 4. n˘B M…™…‰ S…j…… ®… §…xn÷ P {…Æ˙ =i{…z… S…÷®§…EÚ“™… I…‰j… EÚ“ n˘∂…… ⊗ ¥… ☼ E‰Ú ∞¸{… ®… ±… J…B* S…j… A SS—40—Phy. S…j… B SS–580 1 3 In given diagrams write the direction of magnetic field produced at 1 point P in form of ⊗ and ☼. Fig. A 5. Fig. B °ËÚÆ˙…b‰˜ EÚ… ¥…Ët÷i…S…÷®§…EÚ“™… |…‰Æ˙h… EÚ… x…™…®… ±… J…B* Write Faraday's law of electromagnetic induction. 6. 1 EÚ∫…“ |…i™……¥…i…‘ {… Æ˙{…l… ®… +…Æ˙…‰ {…i… ¥……‰±]ıi…… 220V Ω˲* ™… n˘ R = 8Ω, i……‰ x…®x… EÚ… ®……x… ±… J…B : a) ¥……‰±]ıi…… EÚ… ¥…M…« ®……v™… ®…⁄±… ( rms ) ®……x… b) {… Æ˙{…l… EÚ“ |… i…§……v……* X L = X C = 6Ω Ω˲ In an alternating circuit applied voltage is 220V. If R = 8Ω, X L = X C = 6Ω then write the values of the following : 1 7. a) b) a) b) a) b) 8. Root mean square value of voltage Impedance of circuit. S…÷®§…EÚi¥… E‰Ú ±…B M……>∫… x…™…®… EÚ…‰ ®…ËC∫…¥…‰±… ∫…®…“EÚÆ˙h… E‰Ú ∞¸{… ®… ±… J…B* 1 μ0ε0 EÚ… ®……x… ±… J…B* Write Gauss' law for magnetism in the form of Maxwell equation. 1 Write the value of . 1 μ0ε0 +À§…n÷˘EÚi…… n˘…‰π… EÚ…‰ ∫…Δ∂……‰ v…i… EÚÆ˙x…‰ E‰Ú ±…™…‰ EÚ∫… |…EÚ…Æ˙ E‰Ú ±…Â∫… EÚ… |…™……‰M… EÚÆ˙i…‰ ΩÈ˛ ? Which type of lens is used to correct astigmatism ? 9. {…⁄h…« +…xi… Æ˙EÚ {…Æ˙…¥…i…«x… EÚ“ P…]ıx…… EÚ…‰ {… Æ˙¶…… π…i… EÚ“ V…B* 10. Define the phenomenon of total internal reflection. EÚ…‰<« i…i¥… A x…®x… n˘…‰ S…Æ˙h… |… GÚ™……+… u˘…Æ˙… i…i¥… C ®… ¥…P… ]ıi… Ω˛…‰i…… 1 1 Ω˲ : A ⎯⎯→ 2He4 + B B ⎯⎯→ 2 ( −1e 0 ) + C i…i¥…… A, B ¥… C ®… ∫…‰ ∫…®…∫l…… x…EÚ ™…÷M®… UÙ…Δ ]ıB* SS—40—Phy. SS–580 [ Turn over 4 An element A decays into element C by the following two step processes : A ⎯⎯→ 2He4 + B B ⎯⎯→ 2 ( −1e 0 ) + C Select pair for isobars from elements A, B and C. 1 11. S…j… : P A B Y Y' 0 0 0 1 0 1 1 0 1 0 1 0 1 1 1 0 ∫……Æ˙h…“ : Q S…j… P B¥…Δ ∫……Æ˙h…“ Q ∫…‰ ∫…Δ§…Δ v…i… i……ÃEÚEÚ u˘…Æ˙… E‰Ú x……®… ±… J…B* Figure : P 12. A B Y Y' 0 0 0 1 0 1 1 0 1 0 1 0 1 1 1 0 Table : Q Write the names of logic gates related to figure P and table Q. {…fil¥…“ ∫…i…Ω˛ ∫…‰ |…‰π…h… B¥…Δ + ¶…O……Ω˛“ BÂ]ı“x…… EÚ“ >ƒS……<™……ƒ GÚ®…∂…: hT ¥… hR ΩÈ˛* n˘…‰x…… 1 BÂ]ı“x……+… E‰Ú §…“S… EÚ“ + v…EÚi…®… o˘Œπ]ıƉ˙J…“™… n⁄˘Æ˙“ (LOS) E‰Ú ±…B ∫…Δ§…Δv… ±… J…B* Heights of transmitting and receiving antennas from earth surface are hT and h R respectively. Write the relation for maximum line-of-sight (LOS) distance between two antennas. 13. 1 ®……Ïb÷˜±…x… C™…… Ω˲ ? +…™……®… ®……Ïb÷˜ ±…i… i…ÆΔ˙M… E‰Ú ∫…Δ∫…⁄S…x…Ú E‰Ú ±…B |…n˘k… + ¶…O……Ω˛“ §±……ÏEÚ-+…Ɖ˙J… ®… ¶……M… X EÚ… x……®… ±… J…B* SS—40—Phy. SS–580 5 What is modulation ? In the given block diagram of a receiver for detection of amplitude modulated wave, write the name of part X. 1 14. a) ®…“]ıÆ˙ ∫…‰i…÷ EÚ“ ∫…xi…÷±…x… +¥…∫l…… ®… n˘B M…™…‰ {… Æ˙{…l… S…j… ®… +Y……i… |… i…Æ˙…‰v… S EÚ… ®……x… Y……i… EÚ“ V…B* b) n˘B M…™…‰ {… Æ˙{…l… ®… X ¥… Y E‰Ú ®…v™… {… Æ˙h……®…“ |… i…Æ˙…‰v… EÚ… ®……x… ±… J…B ™… n˘ E÷ΔÚV…“ K i) J…÷±…“ Ω˛…‰ ii) §…xn˘ Ω˛…‰* a) In the given circuit diagram, find the value of unknown resistance S, in the balancing condition of meter bridge. SS—40—Phy. SS–580 [ Turn over 6 b) In the given circuit write the value of resultant resistance in between X and Y when key K is i) opened ii) closed. 1 1 1+2 +2 =2 15. EÚÆ˙J……‰°Ú EÚ… |…l…®… x…™…®… (∫…Δ v… x…™…®…) ±… J…B* ¥Ω˛“]ı∫]ı…‰x… ∫…‰i…÷ EÚ… {… Æ˙{…l… S…j… §…x……EÚÆ˙ ∫…‰i…÷ ®… ∂…⁄x™… ¥…I…‰{… E‰Ú ±…B |… i…§…xv… EÚ“ ¥™…÷i{… k… EÚ“ V…B* Write Kirchhoff's first rule (law of junction). Drawing a circuit diagram of Wheatstone bridge, derive condition for zero deflection in the bridge. 1 2 16. 1 +2 +1=2 → b) n˘E¬ÚEÚ…±… (®…÷HÚ +…EÚ…∂…) ®… EÚ∫…“ §…xn÷ {…Æ˙ ¥…t÷i…-I…‰j… ∫… n˘∂… ( E ) EÚ… {… Æ˙®……h… → 9·3 V/m Ω˲* <∫… §…xn÷ {…Æ˙ S…÷®§…EÚ“™… I…‰j… ∫… n˘∂… ( B ) EÚ… {… Æ˙®……h… Y……i… EÚ“ V…B* {…Æ˙…§…ÈM…x…“, +¥…Æ˙HÚ i…l…… X- EÚÆ˙h…… ®… ∫…‰ EÚ∫…EÚ“ i…ÆΔ˙M…nˢP™…« + v…EÚi…®… Ω˛…‰i…“ Ω˲ ? a) In free space at a point magnitude of electric field vector ( E ) is 9·3 b) V/m. Find the magnitude of magnetic field vector ( B ) at the point. Out of ultraviolet, infrared and X-rays whose wavelength is a) → → 1 2 maximum ? 17. 1 ∫…÷®…‰ ±…i… EÚ“ V…B : Eڅϱ…®… I μ = tan i p P Eڅϱ…®… II ∫x…˱… EÚ… x…™…®… 1 sin ic Q •…÷∫]ıÆ˙ EÚ… x…™…®… C sin i μ= sin r R |…W®… D ⎛ A + Dm sin ⎜⎜ 2 ⎝ μ= A sin 2 S {…⁄h…« +…xi… Æ˙EÚ {…Æ˙…¥…i…«x… A B μ= SS—40—Phy. ⎞ ⎟ ⎟ ⎠ SS–580 1 1 +2 +2 +2 =2 7 Match the following : Column I μ = tan i p A P Column II Snell's law B μ= 1 sin ic Q Brewster's law C μ= sin i sin r R Prism D ⎛ A + Dm sin ⎜⎜ 2 ⎝ μ= A sin 2 S Total internal reflection ⎞ ⎟ ⎟ ⎠ 1 4×2 =2 18. |…EÚ…∂… ¥…t÷i… |…¶……¥… EÚ“ P…]ıx…… ®… x…®x… EÚ…‰ {… Æ˙¶…… π…i… EÚ“ V…B : a) i) EÚ…™…« °Ú±…x… ii) x…Æ˙…‰v…“ ¥…¶…¥… (+xi…EÚ ¥…¶…¥…) b) 3·31 Å i…ÆΔ˙M…nˢP™…« E‰Ú °Ú…‰]ı…Ïx… EÚ“ >V……« EÚ“ M…h…x…… EÚ“ V…B* Define the following in photoelectric effect phenomenon : a) i) Work function ii) Stopping potential (cut-off potential) b) Calculate energy of photon of wavelength 3·31 Å. 1 2 1 +2 +1=2 19. 400V 20. Calculate de Broglie wavelength associated with an electron, accelerated through a potential difference of 400V. 2 P-N ∫…Δ v… b˜…™……‰b˜ E‰Ú +O… n˘ ∂…EÚ §……™…∫… (+ ¶…x… i…) ®… V-I + ¶…±……I… h…EÚ ¥…GÚ |……{i… EÚÆ˙x…‰ ¥…¶…¥……xi…Æ˙ ∫…‰ i¥… Æ˙i… <±…‰C]≈ı…Ïx… ∫…‰ ∫…®§…r˘ n‰˘ •……ÏM±…“ i…ÆΔ˙M…nˢP™…« EÚ“ M…h…x…… EÚ“ V…B* EÚ“ EÚ…™…« ¥… v… EÚ… ¥…h…«x… EÚ“ V…B* |……™……‰ M…EÚ ¥™…¥…∫l…… EÚ… {… Æ˙{…l… S…j… §…x……∫… x…™…®… EÚ… EÚl…x… ±… J…B* S…j… §…x……EÚÆ˙ BEÚ ∫…®……x… +…¥…‰ ∂…i… +x…xi… ∫…®…i…±… S……n˘Æ˙ E‰Ú EÚ…Æ˙h… <∫…E‰Ú x…V…n˘“EÚ EÚ∫…“ §…xn÷ {…Æ˙ ¥…t÷i… I…‰j… E‰Ú ±…B ¥™…ΔV…EÚ ¥™…÷i{…z… EÚ“ V…B* n˘B M…™…‰ S…j… ®…Â, {…fiπ`ˆ ∫…‰ x…M…«i… ¥…t÷i… }±…C∫… EÚ… ®……x… ±… J…B : • • q1 = 2μc q 2 = −1μc +l…¥…… SS—40—Phy. SS–580 [ Turn over 10 ∫…Δv…… Æ˙j… EÚ…‰ {… Æ˙¶…… π…i… EÚ“ V…B* {… Æ˙{…l… S…j… §…x……EÚÆ˙ ∫…Δv…… Æ˙j…… E‰Ú ∏…‰h…“ ∫…Δ™……‰V…x… ®… i…÷±™… v…… Æ˙i…… EÚ… ∫…Δ§…Δv… |……{i… EÚ“ V…B* n˘B M…™…‰ {… Æ˙{…l… ®… ¥……‰±]ıi…… V1 EÚ… ®……x… ±… J…B* Write the statement of Gauss' law for electrostatics. Draw a diagram and derive an expression for electric field due to a uniformly charged infinite plane sheet at a point near the sheet. In the given diagram write the value of electric flux passing from the surface. • q = 2μc 1 • q 2 = −1μc 1 1 1+ 2 +2+2 =4 OR Define capacitor. Draw a circuit diagram and obtain a relation for equivalent capacitance in the series combination of capacitors. In given circuit diagram write the value of V1 . 1 1 1+ 2 +2+2 =4 29. |…i™……¥…i…‘ ¥……‰±]ıi…… ª……‰i… E‰Ú ∫……l… ∫…®§…r˘ ∏…‰h…“ LCR {… Æ˙{…l… E‰Ú ±…B °‰ÚV…Æ˙ S…j… J…”S…i…‰ Ω÷˛B {… Æ˙{…l… EÚ“ |… i…§……v…… EÚ… ¥™…ΔV…EÚ Y……i… EÚÆÂ˙* +l…¥…… |…i™……¥…i…‘ v……Æ˙… V… x…j… EÚ… S…j… §…x……EÚÆ˙ ¥…h…«x… EÚ“ V…B * |…‰ Æ˙i… ¥…t÷i… ¥……Ω˛EÚ §…±… E‰Ú i……iI… h…EÚ ®……x… E‰Ú ¥™…ΔV…EÚ EÚ“ ¥™…÷i{… k… EÚ“ V…B* Draw phasor diagram for a series LCR circuit with alternating voltage source, determine the expression for the impedance of the circuit. 1+3=4 OR Draw a diagram of AC generator and describe it. Derive an expression for instantaneous value of induced emf. 1+3=4 SS—40—Phy. SS–580 11 30. ¥™… i…EÚÆ˙h… À£ÚV… |… i…∞¸{… =i{…z… EÚÆ˙x…‰ E‰Ú ±…™…‰ ™…ΔM… u-˘Œ∫±…]ı |…™……‰M… EÚ… EÚÆ˙h… S…j… §…x……<™…‰* |…n˘“{i… À£ÚV…… E‰Ú ±…™…‰ À£ÚV… S……Ëc˜…<« EÚ… ¥™…ΔV…EÚ ¥™…÷i{…z… EÚ“ V…B* ™… n˘ |…n˘“{i… À£ÚV…… E‰Ú ±…™…‰ À£ÚV… S……Ëc˜…<« 2 mm Ω˛…‰ i……‰ +n˘“{i… À£ÚV…… E‰Ú ±…B À£ÚV… S……Ëc˜…<« ±… J…B* ¥™… i…EÚÆ˙h… À£ÚV… |… i…∞¸{… EÚ… E‰Úxp˘“™… §…xn÷˘ S…®…EÚ“±…… Ω˛…‰M…… ™…… EÚ…±…… ? ∫{…π]ı EÚ“ V…B* +l…¥…… +¥…i…±… n˘{…«h… E‰Ú ±…™…‰ EÚÆ˙h… S…j… §…x……EÚÆ˙ ∫…r˘ EÚ“ V…B EÚ n˘{…«h… EÚ“ °Ú…‰EÚ∫… n⁄˘Æ˙“ =∫…EÚ“ ¥…GÚi…… j…V™…… EÚ“ +…v…“ Ω˛…‰i…“ Ω˲* +¥…i…±… n˘{…«h… E‰Ú ±…™…‰ n˘{…«h… ∫…®…“EÚÆ˙h… ±… J…™…‰* |…EÚ…∂… V…§… ¥…Æ˙±… ∫…‰ ∫…P…x… ®……v™…®… ®… |…¥…‰∂… EÚÆ˙i…… Ω˲ i……‰ =∫…EÚ“ i…ÆΔ˙M…nˢP™…« B¥…Δ +…¥…fi k… {…Æ˙ C™…… |…¶……¥… {…c˜i…… Ω˲ ? To produce interference fringe pattern, draw a ray diagram of Young's double slit experiment. Derive an expression of fringe width for bright interference fringes. If fringe width for bright fringes is 2 mm, write the fringe width for dark fringes. Whether the centre point of interference 1 2 fringe pattern is bright or dark ? Explain it. 1 +3+ 2 =4 OR Draw a ray diagram for a concave mirror and prove that focal length of the mirror is half of its radius of curvature. Write mirror equation for a concave mirror. When light enters from rare to denser media, what effect occurs on the values of wavelength and frequency of it ? 1 2 SS—40—Phy. SS–580 1 + 22 + 1 = 4 [ Turn over