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55/1/1 1 [P.T.O. 
 
 
 
 
¸üÖê»Ö ®ÖÓ. 
Roll No.  
 
 
 
³ÖÖîןÖÛú ×¾Ö–ÖÖ®Ö (ÃÖî¨üÖ×®ŸÖÛú) 
PHYSICS (Theory)  
×®Ö¬ÖÖÔ׸üŸÖ ÃÖ´ÖµÖ  : 3 ‘ÖÓ™êü ] [ †×¬ÖÛúŸÖ´Ö †ÓÛú  : 70 
Time allowed : 3 hours ] [ Maximum Marks : 70 
 
ÃÖÖ´ÖÖ®µÖ ×®Ö¤ìü¿Ö    : : : :   
    (i) ÃÖ³Öß ¯ÖÏ¿®Ö †×®Ö¾ÖÖµÖÔ Æïü … ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë Ûãú»Ö 26 ¯ÖÏ¿®Ö Æïü … 
 (ii) ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö Ûêú 5 ³ÖÖÝÖ Æïü : ÜÖÞ›ü-†, ÜÖÞ›ü-²Ö, ÜÖÞ›ü-ÃÖ, ÜÖÞ›ü-¤ü †Öî¸ü ÜÖÞ›ü-µÖ …  
 (iii) ÜÖÞ›ü-† ´Öë 5 ¯ÖÏ¿®Ö Æïü, ¯ÖÏŸµÖêÛú ÛúÖ 1 †ÓÛú Æîü … ÜÖÞ›ü-²Ö ´Öë 5 ¯ÖÏ¿®Ö Æïü, ¯ÖÏŸµÖêÛú Ûêú 2 †ÓÛú Æïü … ÜÖÞ›ü-ÃÖ ´Öë 
12 ¯ÖÏ¿®Ö Æïü, ¯ÖÏŸµÖêÛú Ûêú 3 †ÓÛú Æïü … ÜÖÞ›ü-¤ü ´Öë 4 †ÓÛú ÛúÖ ‹Ûú ´Ö滵ÖÖ¬ÖÖ׸üŸÖ ¯ÖÏ¿®Ö Æîü †Öî¸ü ÜÖÞ›ü-µÖ ´Öë 3 
¯ÖÏ¿®Ö Æïü, ¯ÖÏŸµÖêÛú Ûêú 5 †ÓÛú Æïü …  
 (iv) ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë ÃÖ´ÖÝÖÏ ¯Ö¸ü ÛúÖê‡Ô ×¾ÖÛú»¯Ö ®ÖÆüà Æîü … ŸÖ£ÖÖׯÖ, ¤üÖê †ÓÛúÖë ¾ÖÖ»Öê ‹Ûú ¯ÖÏ¿®Ö ´Öë, ŸÖß®Ö †ÓÛúÖë ¾ÖÖ»Öê ‹Ûú 
¯ÖÏ¿®Ö ´Öë †Öî¸ü ¯ÖÖÑ“Ö †ÓÛúÖë ¾ÖÖ»Öê ŸÖß®ÖÖë ¯ÖÏ¿®ÖÖë ´Öë †Ö®ŸÖ׸üÛú “ÖµÖ®Ö ¯ÖϤüÖ®Ö ×ÛúµÖÖ ÝÖµÖÖ Æîü … ‹êÃÖê ¯ÖÏ¿®ÖÖë ´Öë †Ö¯ÖÛúÖê 
פü‹ ÝÖ‹ “ÖµÖ®Ö ´Öë ÃÖê Ûêú¾Ö»Ö ‹Ûú ¯ÖÏ¿®Ö Æüß Ûú¸ü®ÖÖ Æîü …   
 Series : SSO/1/C 
55/1/1
• Ûéú¯ÖµÖÖ •ÖÖÑ“Ö Ûú¸ü »Öë ×Ûú ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë ´ÖãצüŸÖ ¯Öéšü 12  Æïü … 
• ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë ¤üÖ×Æü®Öê ÆüÖ£Ö Ûúß †Öê¸ü פü‹ ÝÖ‹ ÛúÖê›ü ®Ö´²Ö¸ü ÛúÖê ”ûÖ¡Ö ˆ¢Ö¸ü-¯Öã×ßÖÛúÖ Ûêú ´ÖãÜÖ-¯Öéšü ¯Ö¸ü ×»ÖÜÖë …  
• Ûéú¯ÖµÖÖ •ÖÖÑ“Ö Ûú¸ü »Öë ×Ûú ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë 26 ¯ÖÏ¿®Ö Æïü …  
• Ûéú¯ÖµÖÖ ¯ÖÏ¿®Ö ÛúÖ ˆ¢Ö¸ü ×»ÖÜÖ®ÖÖ ¿Öãºþ Ûú¸ü®Öê ÃÖê ¯ÖÆü»Öê, ¯ÖÏ¿®Ö ÛúÖ ÛÎú´ÖÖÓÛú †¾Ö¿µÖ ×»ÖÜÖë …  
• ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö ÛúÖê ¯ÖœÌü®Öê Ûêú ×»Ö‹ 15 ×´Ö®Ö™ü ÛúÖ ÃÖ´ÖµÖ ×¤üµÖÖ ÝÖµÖÖ Æîü … ¯ÖÏ¿®Ö-¯Ö¡Ö ÛúÖ ×¾ÖŸÖ¸üÞÖ ¯Öæ¾ÖÖÔÆËü®Ö ´Öë 10.15 ²Ö•Öê 
×ÛúµÖÖ •ÖÖµÖêÝÖÖ … 10.15 ²Ö•Öê ÃÖê 10.30 ²Ö•Öê ŸÖÛú ”ûÖ¡Ö Ûêú¾Ö»Ö ¯ÖÏ¿®Ö-¯Ö¡Ö ÛúÖê ¯ÖœÌëüÝÖê †Öî¸ü ‡ÃÖ †¾Ö×¬Ö Ûêú ¤üÖî¸üÖ®Ö ¾Öê 
ˆ¢Ö¸ü-¯Öã×ßÖÛúÖ ¯Ö¸ü ÛúÖê‡Ô ˆ¢Ö¸ü ®ÖÆüà ×»ÖÜÖëÝÖê …  
• Please check that this question paper contains 12 printed pages. 
• Code number given on the right hand side of the question paper should be written on the 
title page of the answer-book by the candidate. 
• Please check that this question paper contains 26 questions. 
• Please write down the Serial Number of the question before attempting it. 
• 15 minutes time has been allotted to read this question paper. The question paper will be 
distributed at 10.15 a.m. From 10.15 a.m. to 10.30 a.m., the students will read the 
question paper only and will not write any answer on the answer-book during this period. 
ÛúÖê›ü ®ÖÓ. 
Code No.  
    
¯Ö¸üßõÖÖ£Öá ÛúÖê›ü ÛúÖê ˆ¢Ö¸ü-¯Öã×ßÖÛúÖ Ûêú ´ÖãÜÖ-¯Öéšü 
¯Ö¸ü †¾Ö¿µÖ ×»ÖÜÖë … 
Candidates must write the Code on 
the title page of the answer-book. 
 
SET – 1 
Page 2


55/1/1 1 [P.T.O. 
 
 
 
 
¸üÖê»Ö ®ÖÓ. 
Roll No.  
 
 
 
³ÖÖîןÖÛú ×¾Ö–ÖÖ®Ö (ÃÖî¨üÖ×®ŸÖÛú) 
PHYSICS (Theory)  
×®Ö¬ÖÖÔ׸üŸÖ ÃÖ´ÖµÖ  : 3 ‘ÖÓ™êü ] [ †×¬ÖÛúŸÖ´Ö †ÓÛú  : 70 
Time allowed : 3 hours ] [ Maximum Marks : 70 
 
ÃÖÖ´ÖÖ®µÖ ×®Ö¤ìü¿Ö    : : : :   
    (i) ÃÖ³Öß ¯ÖÏ¿®Ö †×®Ö¾ÖÖµÖÔ Æïü … ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë Ûãú»Ö 26 ¯ÖÏ¿®Ö Æïü … 
 (ii) ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö Ûêú 5 ³ÖÖÝÖ Æïü : ÜÖÞ›ü-†, ÜÖÞ›ü-²Ö, ÜÖÞ›ü-ÃÖ, ÜÖÞ›ü-¤ü †Öî¸ü ÜÖÞ›ü-µÖ …  
 (iii) ÜÖÞ›ü-† ´Öë 5 ¯ÖÏ¿®Ö Æïü, ¯ÖÏŸµÖêÛú ÛúÖ 1 †ÓÛú Æîü … ÜÖÞ›ü-²Ö ´Öë 5 ¯ÖÏ¿®Ö Æïü, ¯ÖÏŸµÖêÛú Ûêú 2 †ÓÛú Æïü … ÜÖÞ›ü-ÃÖ ´Öë 
12 ¯ÖÏ¿®Ö Æïü, ¯ÖÏŸµÖêÛú Ûêú 3 †ÓÛú Æïü … ÜÖÞ›ü-¤ü ´Öë 4 †ÓÛú ÛúÖ ‹Ûú ´Ö滵ÖÖ¬ÖÖ׸üŸÖ ¯ÖÏ¿®Ö Æîü †Öî¸ü ÜÖÞ›ü-µÖ ´Öë 3 
¯ÖÏ¿®Ö Æïü, ¯ÖÏŸµÖêÛú Ûêú 5 †ÓÛú Æïü …  
 (iv) ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë ÃÖ´ÖÝÖÏ ¯Ö¸ü ÛúÖê‡Ô ×¾ÖÛú»¯Ö ®ÖÆüà Æîü … ŸÖ£ÖÖׯÖ, ¤üÖê †ÓÛúÖë ¾ÖÖ»Öê ‹Ûú ¯ÖÏ¿®Ö ´Öë, ŸÖß®Ö †ÓÛúÖë ¾ÖÖ»Öê ‹Ûú 
¯ÖÏ¿®Ö ´Öë †Öî¸ü ¯ÖÖÑ“Ö †ÓÛúÖë ¾ÖÖ»Öê ŸÖß®ÖÖë ¯ÖÏ¿®ÖÖë ´Öë †Ö®ŸÖ׸üÛú “ÖµÖ®Ö ¯ÖϤüÖ®Ö ×ÛúµÖÖ ÝÖµÖÖ Æîü … ‹êÃÖê ¯ÖÏ¿®ÖÖë ´Öë †Ö¯ÖÛúÖê 
פü‹ ÝÖ‹ “ÖµÖ®Ö ´Öë ÃÖê Ûêú¾Ö»Ö ‹Ûú ¯ÖÏ¿®Ö Æüß Ûú¸ü®ÖÖ Æîü …   
 Series : SSO/1/C 
55/1/1
• Ûéú¯ÖµÖÖ •ÖÖÑ“Ö Ûú¸ü »Öë ×Ûú ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë ´ÖãצüŸÖ ¯Öéšü 12  Æïü … 
• ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë ¤üÖ×Æü®Öê ÆüÖ£Ö Ûúß †Öê¸ü פü‹ ÝÖ‹ ÛúÖê›ü ®Ö´²Ö¸ü ÛúÖê ”ûÖ¡Ö ˆ¢Ö¸ü-¯Öã×ßÖÛúÖ Ûêú ´ÖãÜÖ-¯Öéšü ¯Ö¸ü ×»ÖÜÖë …  
• Ûéú¯ÖµÖÖ •ÖÖÑ“Ö Ûú¸ü »Öë ×Ûú ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë 26 ¯ÖÏ¿®Ö Æïü …  
• Ûéú¯ÖµÖÖ ¯ÖÏ¿®Ö ÛúÖ ˆ¢Ö¸ü ×»ÖÜÖ®ÖÖ ¿Öãºþ Ûú¸ü®Öê ÃÖê ¯ÖÆü»Öê, ¯ÖÏ¿®Ö ÛúÖ ÛÎú´ÖÖÓÛú †¾Ö¿µÖ ×»ÖÜÖë …  
• ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö ÛúÖê ¯ÖœÌü®Öê Ûêú ×»Ö‹ 15 ×´Ö®Ö™ü ÛúÖ ÃÖ´ÖµÖ ×¤üµÖÖ ÝÖµÖÖ Æîü … ¯ÖÏ¿®Ö-¯Ö¡Ö ÛúÖ ×¾ÖŸÖ¸üÞÖ ¯Öæ¾ÖÖÔÆËü®Ö ´Öë 10.15 ²Ö•Öê 
×ÛúµÖÖ •ÖÖµÖêÝÖÖ … 10.15 ²Ö•Öê ÃÖê 10.30 ²Ö•Öê ŸÖÛú ”ûÖ¡Ö Ûêú¾Ö»Ö ¯ÖÏ¿®Ö-¯Ö¡Ö ÛúÖê ¯ÖœÌëüÝÖê †Öî¸ü ‡ÃÖ †¾Ö×¬Ö Ûêú ¤üÖî¸üÖ®Ö ¾Öê 
ˆ¢Ö¸ü-¯Öã×ßÖÛúÖ ¯Ö¸ü ÛúÖê‡Ô ˆ¢Ö¸ü ®ÖÆüà ×»ÖÜÖëÝÖê …  
• Please check that this question paper contains 12 printed pages. 
• Code number given on the right hand side of the question paper should be written on the 
title page of the answer-book by the candidate. 
• Please check that this question paper contains 26 questions. 
• Please write down the Serial Number of the question before attempting it. 
• 15 minutes time has been allotted to read this question paper. The question paper will be 
distributed at 10.15 a.m. From 10.15 a.m. to 10.30 a.m., the students will read the 
question paper only and will not write any answer on the answer-book during this period. 
ÛúÖê›ü ®ÖÓ. 
Code No.  
    
¯Ö¸üßõÖÖ£Öá ÛúÖê›ü ÛúÖê ˆ¢Ö¸ü-¯Öã×ßÖÛúÖ Ûêú ´ÖãÜÖ-¯Öéšü 
¯Ö¸ü †¾Ö¿µÖ ×»ÖÜÖë … 
Candidates must write the Code on 
the title page of the answer-book. 
 
SET – 1 
55/1/1 2 
 (v) •ÖÆüÖÑ †Ö¾Ö¿µÖÛú ÆüÖê †Ö¯Ö ×®Ö´®Ö×»Ö×ÜÖŸÖ ³ÖÖîןÖÛ  ×®ÖµÖŸÖÖÓÛúÖë Ûêú ´ÖÖ®ÖÖë ÛúÖ ˆ¯ÖµÖÖêÝÖ Ûú¸ü ÃÖÛúŸÖê Æïü : 
  c = 3 × 10
8
 m/s 
  h = 6.63 × 10
–34
 Js 
  e = 1.6 × 10
–19
 C 
  µ
0
 = 4p × 10
–7
 T m A
–1 
  
e
0
 = 8.854 × 10
–12
 C
2
 N
–1
 m
–2 
 
  
1
4pe
0
 = 9 × 10
9
 N m
2
 C
–2 
  m
e
 = 9.1 × 10
–31
 kg 
  ®µÖæ™ÒüÖò®Ö ÛúÖ ¦ü¾µÖ´ÖÖ®Ö = 1.675 × 10
–27
 kg 
  ¯ÖÏÖê™üÖò®Ö ÛúÖ ¦ü¾µÖ´ÖÖ®Ö = 1.673 × 10
–27
 kg 
  †Ö¾ÖÖêÝÖÖ¦üÖê ÃÖÓܵÖÖ = 6.023 × 10
23
 ¯ÖÏ×ŸÖ ÝÖÏÖ´Ö ´ÖÖê»Ö  
  ²ÖÖê»™Ëü•ÖÌ´ÖÖ®Ö ×®ÖµÖŸÖÖÓÛú = 1.38 × 10
–23
 JK
–1
 
 
General Instructions :   
 (i) All questions are compulsory. There are 26 questions in all. 
 (ii) This question paper has five sections : Section A, Section B, Section C, Section D 
and Section E. 
 (iii) Section A contains five questions of one mark each, Section B contains five 
questions of two marks each, Section C contains twelve questions of three marks 
each, Section D contains one value based question of four marks and Section E 
contains three questions of five marks each. 
 (iv) There is no overall choice. However, an internal choice has been provided in one 
question of two marks, one question of three marks and all the three questions of 
five marks weightage. You have to attempt only one of the choices in such 
questions. 
 (v) You may use the following values of physical constants wherever necessary : 
  c = 3 × 10
8
 m/s 
  h = 6.63 × 10
–34
 Js 
  e = 1.6 × 10
–19
 C 
  µ
0
 = 4p × 10
–7
 T m A
–1 
  
e
0
 = 8.854 × 10
–12
 C
2
 N
–1
 m
–2 
 
  
1
4pe
0
 = 9 × 10
9
 N m
2
 C
–2 
  m
e
 = 9.1 × 10
–31
 kg 
  Mass of neutron = 1.675 × 10
–27
 kg 
  Mass of proton = 1.673 × 10
–27
 kg 
  Avogadro’s number = 6.023 × 10
23
 per gram mole 
  Boltzmann constant = 1.38 × 10
–23
 JK
–1
 
Page 3


55/1/1 1 [P.T.O. 
 
 
 
 
¸üÖê»Ö ®ÖÓ. 
Roll No.  
 
 
 
³ÖÖîןÖÛú ×¾Ö–ÖÖ®Ö (ÃÖî¨üÖ×®ŸÖÛú) 
PHYSICS (Theory)  
×®Ö¬ÖÖÔ׸üŸÖ ÃÖ´ÖµÖ  : 3 ‘ÖÓ™êü ] [ †×¬ÖÛúŸÖ´Ö †ÓÛú  : 70 
Time allowed : 3 hours ] [ Maximum Marks : 70 
 
ÃÖÖ´ÖÖ®µÖ ×®Ö¤ìü¿Ö    : : : :   
    (i) ÃÖ³Öß ¯ÖÏ¿®Ö †×®Ö¾ÖÖµÖÔ Æïü … ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë Ûãú»Ö 26 ¯ÖÏ¿®Ö Æïü … 
 (ii) ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö Ûêú 5 ³ÖÖÝÖ Æïü : ÜÖÞ›ü-†, ÜÖÞ›ü-²Ö, ÜÖÞ›ü-ÃÖ, ÜÖÞ›ü-¤ü †Öî¸ü ÜÖÞ›ü-µÖ …  
 (iii) ÜÖÞ›ü-† ´Öë 5 ¯ÖÏ¿®Ö Æïü, ¯ÖÏŸµÖêÛú ÛúÖ 1 †ÓÛú Æîü … ÜÖÞ›ü-²Ö ´Öë 5 ¯ÖÏ¿®Ö Æïü, ¯ÖÏŸµÖêÛú Ûêú 2 †ÓÛú Æïü … ÜÖÞ›ü-ÃÖ ´Öë 
12 ¯ÖÏ¿®Ö Æïü, ¯ÖÏŸµÖêÛú Ûêú 3 †ÓÛú Æïü … ÜÖÞ›ü-¤ü ´Öë 4 †ÓÛú ÛúÖ ‹Ûú ´Ö滵ÖÖ¬ÖÖ׸üŸÖ ¯ÖÏ¿®Ö Æîü †Öî¸ü ÜÖÞ›ü-µÖ ´Öë 3 
¯ÖÏ¿®Ö Æïü, ¯ÖÏŸµÖêÛú Ûêú 5 †ÓÛú Æïü …  
 (iv) ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë ÃÖ´ÖÝÖÏ ¯Ö¸ü ÛúÖê‡Ô ×¾ÖÛú»¯Ö ®ÖÆüà Æîü … ŸÖ£ÖÖׯÖ, ¤üÖê †ÓÛúÖë ¾ÖÖ»Öê ‹Ûú ¯ÖÏ¿®Ö ´Öë, ŸÖß®Ö †ÓÛúÖë ¾ÖÖ»Öê ‹Ûú 
¯ÖÏ¿®Ö ´Öë †Öî¸ü ¯ÖÖÑ“Ö †ÓÛúÖë ¾ÖÖ»Öê ŸÖß®ÖÖë ¯ÖÏ¿®ÖÖë ´Öë †Ö®ŸÖ׸üÛú “ÖµÖ®Ö ¯ÖϤüÖ®Ö ×ÛúµÖÖ ÝÖµÖÖ Æîü … ‹êÃÖê ¯ÖÏ¿®ÖÖë ´Öë †Ö¯ÖÛúÖê 
פü‹ ÝÖ‹ “ÖµÖ®Ö ´Öë ÃÖê Ûêú¾Ö»Ö ‹Ûú ¯ÖÏ¿®Ö Æüß Ûú¸ü®ÖÖ Æîü …   
 Series : SSO/1/C 
55/1/1
• Ûéú¯ÖµÖÖ •ÖÖÑ“Ö Ûú¸ü »Öë ×Ûú ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë ´ÖãצüŸÖ ¯Öéšü 12  Æïü … 
• ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë ¤üÖ×Æü®Öê ÆüÖ£Ö Ûúß †Öê¸ü פü‹ ÝÖ‹ ÛúÖê›ü ®Ö´²Ö¸ü ÛúÖê ”ûÖ¡Ö ˆ¢Ö¸ü-¯Öã×ßÖÛúÖ Ûêú ´ÖãÜÖ-¯Öéšü ¯Ö¸ü ×»ÖÜÖë …  
• Ûéú¯ÖµÖÖ •ÖÖÑ“Ö Ûú¸ü »Öë ×Ûú ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë 26 ¯ÖÏ¿®Ö Æïü …  
• Ûéú¯ÖµÖÖ ¯ÖÏ¿®Ö ÛúÖ ˆ¢Ö¸ü ×»ÖÜÖ®ÖÖ ¿Öãºþ Ûú¸ü®Öê ÃÖê ¯ÖÆü»Öê, ¯ÖÏ¿®Ö ÛúÖ ÛÎú´ÖÖÓÛú †¾Ö¿µÖ ×»ÖÜÖë …  
• ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö ÛúÖê ¯ÖœÌü®Öê Ûêú ×»Ö‹ 15 ×´Ö®Ö™ü ÛúÖ ÃÖ´ÖµÖ ×¤üµÖÖ ÝÖµÖÖ Æîü … ¯ÖÏ¿®Ö-¯Ö¡Ö ÛúÖ ×¾ÖŸÖ¸üÞÖ ¯Öæ¾ÖÖÔÆËü®Ö ´Öë 10.15 ²Ö•Öê 
×ÛúµÖÖ •ÖÖµÖêÝÖÖ … 10.15 ²Ö•Öê ÃÖê 10.30 ²Ö•Öê ŸÖÛú ”ûÖ¡Ö Ûêú¾Ö»Ö ¯ÖÏ¿®Ö-¯Ö¡Ö ÛúÖê ¯ÖœÌëüÝÖê †Öî¸ü ‡ÃÖ †¾Ö×¬Ö Ûêú ¤üÖî¸üÖ®Ö ¾Öê 
ˆ¢Ö¸ü-¯Öã×ßÖÛúÖ ¯Ö¸ü ÛúÖê‡Ô ˆ¢Ö¸ü ®ÖÆüà ×»ÖÜÖëÝÖê …  
• Please check that this question paper contains 12 printed pages. 
• Code number given on the right hand side of the question paper should be written on the 
title page of the answer-book by the candidate. 
• Please check that this question paper contains 26 questions. 
• Please write down the Serial Number of the question before attempting it. 
• 15 minutes time has been allotted to read this question paper. The question paper will be 
distributed at 10.15 a.m. From 10.15 a.m. to 10.30 a.m., the students will read the 
question paper only and will not write any answer on the answer-book during this period. 
ÛúÖê›ü ®ÖÓ. 
Code No.  
    
¯Ö¸üßõÖÖ£Öá ÛúÖê›ü ÛúÖê ˆ¢Ö¸ü-¯Öã×ßÖÛúÖ Ûêú ´ÖãÜÖ-¯Öéšü 
¯Ö¸ü †¾Ö¿µÖ ×»ÖÜÖë … 
Candidates must write the Code on 
the title page of the answer-book. 
 
SET – 1 
55/1/1 2 
 (v) •ÖÆüÖÑ †Ö¾Ö¿µÖÛú ÆüÖê †Ö¯Ö ×®Ö´®Ö×»Ö×ÜÖŸÖ ³ÖÖîןÖÛ  ×®ÖµÖŸÖÖÓÛúÖë Ûêú ´ÖÖ®ÖÖë ÛúÖ ˆ¯ÖµÖÖêÝÖ Ûú¸ü ÃÖÛúŸÖê Æïü : 
  c = 3 × 10
8
 m/s 
  h = 6.63 × 10
–34
 Js 
  e = 1.6 × 10
–19
 C 
  µ
0
 = 4p × 10
–7
 T m A
–1 
  
e
0
 = 8.854 × 10
–12
 C
2
 N
–1
 m
–2 
 
  
1
4pe
0
 = 9 × 10
9
 N m
2
 C
–2 
  m
e
 = 9.1 × 10
–31
 kg 
  ®µÖæ™ÒüÖò®Ö ÛúÖ ¦ü¾µÖ´ÖÖ®Ö = 1.675 × 10
–27
 kg 
  ¯ÖÏÖê™üÖò®Ö ÛúÖ ¦ü¾µÖ´ÖÖ®Ö = 1.673 × 10
–27
 kg 
  †Ö¾ÖÖêÝÖÖ¦üÖê ÃÖÓܵÖÖ = 6.023 × 10
23
 ¯ÖÏ×ŸÖ ÝÖÏÖ´Ö ´ÖÖê»Ö  
  ²ÖÖê»™Ëü•ÖÌ´ÖÖ®Ö ×®ÖµÖŸÖÖÓÛú = 1.38 × 10
–23
 JK
–1
 
 
General Instructions :   
 (i) All questions are compulsory. There are 26 questions in all. 
 (ii) This question paper has five sections : Section A, Section B, Section C, Section D 
and Section E. 
 (iii) Section A contains five questions of one mark each, Section B contains five 
questions of two marks each, Section C contains twelve questions of three marks 
each, Section D contains one value based question of four marks and Section E 
contains three questions of five marks each. 
 (iv) There is no overall choice. However, an internal choice has been provided in one 
question of two marks, one question of three marks and all the three questions of 
five marks weightage. You have to attempt only one of the choices in such 
questions. 
 (v) You may use the following values of physical constants wherever necessary : 
  c = 3 × 10
8
 m/s 
  h = 6.63 × 10
–34
 Js 
  e = 1.6 × 10
–19
 C 
  µ
0
 = 4p × 10
–7
 T m A
–1 
  
e
0
 = 8.854 × 10
–12
 C
2
 N
–1
 m
–2 
 
  
1
4pe
0
 = 9 × 10
9
 N m
2
 C
–2 
  m
e
 = 9.1 × 10
–31
 kg 
  Mass of neutron = 1.675 × 10
–27
 kg 
  Mass of proton = 1.673 × 10
–27
 kg 
  Avogadro’s number = 6.023 × 10
23
 per gram mole 
  Boltzmann constant = 1.38 × 10
–23
 JK
–1
 
55/1/1 3 [P.T.O. 
ÜÖÞ›ü – † 
Section – A 
 
1. ¦ü¾µÖ´ÖÖ®Ö ‘m’ †Öî¸ü †Ö¾Öê¿Ö ‘q’ ÛúÖ ÛúÖê‡Ô ÛúÞÖ ‘v’ ¾ÖêÝÖ ÃÖê ×ÛúÃÖß ‹ÛúÃÖ´ÖÖ®Ö “Öã´²ÖÛúßµÖ õÖê¡Ö, •ÖÖê ÛúÞÖ Ûúß ÝÖ×ŸÖ Ûúß 
פü¿ÖÖ Ûêú »Ö´²Ö¾ÖŸÖ Æîü, ´Öë ¯ÖϾÖê¿Ö Ûú¸üŸÖÖ Æîü, ˆÃÖÛúß ÝÖ×ŸÖ•Ö ‰ú•ÖÖÔ ×ÛúÃÖ ¯ÖÏÛúÖ¸ü ¯ÖϳÖÖ×¾ÖŸÖ ÆüÖêŸÖß Æîü ? 1 
 A particle of mass ‘m’ and charge ‘q’ moving with velocity ‘v’ enters the region of 
uniform magnetic field at right angle to the direction of its motion. How does its 
kinetic energy get affected ? 
 
2.  ×“Ö¡Ö ´Öë ×ÛúÃÖß ¬ÖÖ¸üÖ¾ÖÖÆüß ¯Ö׸ü®ÖÖ×»ÖÛúÖ ÛúÖê ×ÛúÃÖß “ÖÖ»ÖÛú »Öæ¯Ö (¯ÖÖ¿Ö) Ûúß †Öê¸ü ÝÖןִÖÖ®Ö ¤ü¿ÖÖÔµÖÖ ÝÖµÖÖ Æîü … »Öæ¯Ö ´Öë 
¯ÖÏê׸üŸÖ ¬ÖÖ¸üÖ Ûúß ×¤ü¿ÖÖ ²ÖŸÖÖ‡‹ …  1 
 
 Figure shows a current carrying solenoid moving towards a conducting loop. Find the 
direction of the current induced in the loop. 
 
 
3.  •Ö²Ö ×ÛúÃÖß ×²Ö´²Ö ÛúÖê ×ÛúÃÖß †¾ÖŸÖ»Ö ¤ü¯ÖÔÞÖ Ûêú f †Öî¸ü 2f Ûêú ²Öß“Ö ¸üÜÖÖ •ÖÖŸÖÖ Æîü, ŸÖ²Ö ²Ö®Ö®Öê ¾ÖÖ»ÖÖ ¯ÖÏןÖײִ²Ö ÛîúÃÖÖ 
ÆüÖêŸÖÖ Æîü – (i) ¾ÖÖßÖ×¾ÖÛú †£Ö¾ÖÖ †Ö³ÖÖÃÖß †Öî¸ü (ii) ”ûÖê™üÖ †£Ö¾ÖÖ ×¾Ö¾ÖÙ¬ÖŸÖ ?  1 
 When an object is placed between f and 2f of a concave mirror, would the image 
formed be (i) real or virtual and (ii) diminished or magnified ? 
 
4.  ‡»ÖꌙÒüÖò®Ö Ûúß ÝÖ×ŸÖ•Ö ‰ú•ÖÖÔ Ûêú ±ú»Ö®Ö Ûêú ºþ¯Ö ´Öë ‡ÃÖÛúß ¤êü-²ÖÎÖÝ»Öß ŸÖ¸ÓüÝÖ¤îü¬µÖÔ Ûêú ×¾Ö“Ö¸üÞÖ ÛúÖê ÝÖÏÖ±ú ÜÖà“ÖÛú¸ü   
¤ü¿ÖÖÔ‡‹ …    1 
 Draw a plot showing the variation of de Broglie wavelength of electron as a function 
of its K. E. 
 
5.  ´ÖÖê²ÖÖ‡»Ö ±úÖê®Ö ´Öë †ÖÝÖ´Öß ×ÃÖݮֻÖÖë †Öî¸ü ×®ÖÝÖÔ´Öß ×ÃÖݮֻÖÖë Ûúß †Ö¾Öé×¢ÖµÖÖÑ ×³Ö®®Ö ŒµÖÖë ÆüÖêŸÖß Æîü ?  1 
 Why is the frequency of outgoing and incoming signals different in a mobile phone ? 
Page 4


55/1/1 1 [P.T.O. 
 
 
 
 
¸üÖê»Ö ®ÖÓ. 
Roll No.  
 
 
 
³ÖÖîןÖÛú ×¾Ö–ÖÖ®Ö (ÃÖî¨üÖ×®ŸÖÛú) 
PHYSICS (Theory)  
×®Ö¬ÖÖÔ׸üŸÖ ÃÖ´ÖµÖ  : 3 ‘ÖÓ™êü ] [ †×¬ÖÛúŸÖ´Ö †ÓÛú  : 70 
Time allowed : 3 hours ] [ Maximum Marks : 70 
 
ÃÖÖ´ÖÖ®µÖ ×®Ö¤ìü¿Ö    : : : :   
    (i) ÃÖ³Öß ¯ÖÏ¿®Ö †×®Ö¾ÖÖµÖÔ Æïü … ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë Ûãú»Ö 26 ¯ÖÏ¿®Ö Æïü … 
 (ii) ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö Ûêú 5 ³ÖÖÝÖ Æïü : ÜÖÞ›ü-†, ÜÖÞ›ü-²Ö, ÜÖÞ›ü-ÃÖ, ÜÖÞ›ü-¤ü †Öî¸ü ÜÖÞ›ü-µÖ …  
 (iii) ÜÖÞ›ü-† ´Öë 5 ¯ÖÏ¿®Ö Æïü, ¯ÖÏŸµÖêÛú ÛúÖ 1 †ÓÛú Æîü … ÜÖÞ›ü-²Ö ´Öë 5 ¯ÖÏ¿®Ö Æïü, ¯ÖÏŸµÖêÛú Ûêú 2 †ÓÛú Æïü … ÜÖÞ›ü-ÃÖ ´Öë 
12 ¯ÖÏ¿®Ö Æïü, ¯ÖÏŸµÖêÛú Ûêú 3 †ÓÛú Æïü … ÜÖÞ›ü-¤ü ´Öë 4 †ÓÛú ÛúÖ ‹Ûú ´Ö滵ÖÖ¬ÖÖ׸üŸÖ ¯ÖÏ¿®Ö Æîü †Öî¸ü ÜÖÞ›ü-µÖ ´Öë 3 
¯ÖÏ¿®Ö Æïü, ¯ÖÏŸµÖêÛú Ûêú 5 †ÓÛú Æïü …  
 (iv) ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë ÃÖ´ÖÝÖÏ ¯Ö¸ü ÛúÖê‡Ô ×¾ÖÛú»¯Ö ®ÖÆüà Æîü … ŸÖ£ÖÖׯÖ, ¤üÖê †ÓÛúÖë ¾ÖÖ»Öê ‹Ûú ¯ÖÏ¿®Ö ´Öë, ŸÖß®Ö †ÓÛúÖë ¾ÖÖ»Öê ‹Ûú 
¯ÖÏ¿®Ö ´Öë †Öî¸ü ¯ÖÖÑ“Ö †ÓÛúÖë ¾ÖÖ»Öê ŸÖß®ÖÖë ¯ÖÏ¿®ÖÖë ´Öë †Ö®ŸÖ׸üÛú “ÖµÖ®Ö ¯ÖϤüÖ®Ö ×ÛúµÖÖ ÝÖµÖÖ Æîü … ‹êÃÖê ¯ÖÏ¿®ÖÖë ´Öë †Ö¯ÖÛúÖê 
פü‹ ÝÖ‹ “ÖµÖ®Ö ´Öë ÃÖê Ûêú¾Ö»Ö ‹Ûú ¯ÖÏ¿®Ö Æüß Ûú¸ü®ÖÖ Æîü …   
 Series : SSO/1/C 
55/1/1
• Ûéú¯ÖµÖÖ •ÖÖÑ“Ö Ûú¸ü »Öë ×Ûú ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë ´ÖãצüŸÖ ¯Öéšü 12  Æïü … 
• ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë ¤üÖ×Æü®Öê ÆüÖ£Ö Ûúß †Öê¸ü פü‹ ÝÖ‹ ÛúÖê›ü ®Ö´²Ö¸ü ÛúÖê ”ûÖ¡Ö ˆ¢Ö¸ü-¯Öã×ßÖÛúÖ Ûêú ´ÖãÜÖ-¯Öéšü ¯Ö¸ü ×»ÖÜÖë …  
• Ûéú¯ÖµÖÖ •ÖÖÑ“Ö Ûú¸ü »Öë ×Ûú ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë 26 ¯ÖÏ¿®Ö Æïü …  
• Ûéú¯ÖµÖÖ ¯ÖÏ¿®Ö ÛúÖ ˆ¢Ö¸ü ×»ÖÜÖ®ÖÖ ¿Öãºþ Ûú¸ü®Öê ÃÖê ¯ÖÆü»Öê, ¯ÖÏ¿®Ö ÛúÖ ÛÎú´ÖÖÓÛú †¾Ö¿µÖ ×»ÖÜÖë …  
• ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö ÛúÖê ¯ÖœÌü®Öê Ûêú ×»Ö‹ 15 ×´Ö®Ö™ü ÛúÖ ÃÖ´ÖµÖ ×¤üµÖÖ ÝÖµÖÖ Æîü … ¯ÖÏ¿®Ö-¯Ö¡Ö ÛúÖ ×¾ÖŸÖ¸üÞÖ ¯Öæ¾ÖÖÔÆËü®Ö ´Öë 10.15 ²Ö•Öê 
×ÛúµÖÖ •ÖÖµÖêÝÖÖ … 10.15 ²Ö•Öê ÃÖê 10.30 ²Ö•Öê ŸÖÛú ”ûÖ¡Ö Ûêú¾Ö»Ö ¯ÖÏ¿®Ö-¯Ö¡Ö ÛúÖê ¯ÖœÌëüÝÖê †Öî¸ü ‡ÃÖ †¾Ö×¬Ö Ûêú ¤üÖî¸üÖ®Ö ¾Öê 
ˆ¢Ö¸ü-¯Öã×ßÖÛúÖ ¯Ö¸ü ÛúÖê‡Ô ˆ¢Ö¸ü ®ÖÆüà ×»ÖÜÖëÝÖê …  
• Please check that this question paper contains 12 printed pages. 
• Code number given on the right hand side of the question paper should be written on the 
title page of the answer-book by the candidate. 
• Please check that this question paper contains 26 questions. 
• Please write down the Serial Number of the question before attempting it. 
• 15 minutes time has been allotted to read this question paper. The question paper will be 
distributed at 10.15 a.m. From 10.15 a.m. to 10.30 a.m., the students will read the 
question paper only and will not write any answer on the answer-book during this period. 
ÛúÖê›ü ®ÖÓ. 
Code No.  
    
¯Ö¸üßõÖÖ£Öá ÛúÖê›ü ÛúÖê ˆ¢Ö¸ü-¯Öã×ßÖÛúÖ Ûêú ´ÖãÜÖ-¯Öéšü 
¯Ö¸ü †¾Ö¿µÖ ×»ÖÜÖë … 
Candidates must write the Code on 
the title page of the answer-book. 
 
SET – 1 
55/1/1 2 
 (v) •ÖÆüÖÑ †Ö¾Ö¿µÖÛú ÆüÖê †Ö¯Ö ×®Ö´®Ö×»Ö×ÜÖŸÖ ³ÖÖîןÖÛ  ×®ÖµÖŸÖÖÓÛúÖë Ûêú ´ÖÖ®ÖÖë ÛúÖ ˆ¯ÖµÖÖêÝÖ Ûú¸ü ÃÖÛúŸÖê Æïü : 
  c = 3 × 10
8
 m/s 
  h = 6.63 × 10
–34
 Js 
  e = 1.6 × 10
–19
 C 
  µ
0
 = 4p × 10
–7
 T m A
–1 
  
e
0
 = 8.854 × 10
–12
 C
2
 N
–1
 m
–2 
 
  
1
4pe
0
 = 9 × 10
9
 N m
2
 C
–2 
  m
e
 = 9.1 × 10
–31
 kg 
  ®µÖæ™ÒüÖò®Ö ÛúÖ ¦ü¾µÖ´ÖÖ®Ö = 1.675 × 10
–27
 kg 
  ¯ÖÏÖê™üÖò®Ö ÛúÖ ¦ü¾µÖ´ÖÖ®Ö = 1.673 × 10
–27
 kg 
  †Ö¾ÖÖêÝÖÖ¦üÖê ÃÖÓܵÖÖ = 6.023 × 10
23
 ¯ÖÏ×ŸÖ ÝÖÏÖ´Ö ´ÖÖê»Ö  
  ²ÖÖê»™Ëü•ÖÌ´ÖÖ®Ö ×®ÖµÖŸÖÖÓÛú = 1.38 × 10
–23
 JK
–1
 
 
General Instructions :   
 (i) All questions are compulsory. There are 26 questions in all. 
 (ii) This question paper has five sections : Section A, Section B, Section C, Section D 
and Section E. 
 (iii) Section A contains five questions of one mark each, Section B contains five 
questions of two marks each, Section C contains twelve questions of three marks 
each, Section D contains one value based question of four marks and Section E 
contains three questions of five marks each. 
 (iv) There is no overall choice. However, an internal choice has been provided in one 
question of two marks, one question of three marks and all the three questions of 
five marks weightage. You have to attempt only one of the choices in such 
questions. 
 (v) You may use the following values of physical constants wherever necessary : 
  c = 3 × 10
8
 m/s 
  h = 6.63 × 10
–34
 Js 
  e = 1.6 × 10
–19
 C 
  µ
0
 = 4p × 10
–7
 T m A
–1 
  
e
0
 = 8.854 × 10
–12
 C
2
 N
–1
 m
–2 
 
  
1
4pe
0
 = 9 × 10
9
 N m
2
 C
–2 
  m
e
 = 9.1 × 10
–31
 kg 
  Mass of neutron = 1.675 × 10
–27
 kg 
  Mass of proton = 1.673 × 10
–27
 kg 
  Avogadro’s number = 6.023 × 10
23
 per gram mole 
  Boltzmann constant = 1.38 × 10
–23
 JK
–1
 
55/1/1 3 [P.T.O. 
ÜÖÞ›ü – † 
Section – A 
 
1. ¦ü¾µÖ´ÖÖ®Ö ‘m’ †Öî¸ü †Ö¾Öê¿Ö ‘q’ ÛúÖ ÛúÖê‡Ô ÛúÞÖ ‘v’ ¾ÖêÝÖ ÃÖê ×ÛúÃÖß ‹ÛúÃÖ´ÖÖ®Ö “Öã´²ÖÛúßµÖ õÖê¡Ö, •ÖÖê ÛúÞÖ Ûúß ÝÖ×ŸÖ Ûúß 
פü¿ÖÖ Ûêú »Ö´²Ö¾ÖŸÖ Æîü, ´Öë ¯ÖϾÖê¿Ö Ûú¸üŸÖÖ Æîü, ˆÃÖÛúß ÝÖ×ŸÖ•Ö ‰ú•ÖÖÔ ×ÛúÃÖ ¯ÖÏÛúÖ¸ü ¯ÖϳÖÖ×¾ÖŸÖ ÆüÖêŸÖß Æîü ? 1 
 A particle of mass ‘m’ and charge ‘q’ moving with velocity ‘v’ enters the region of 
uniform magnetic field at right angle to the direction of its motion. How does its 
kinetic energy get affected ? 
 
2.  ×“Ö¡Ö ´Öë ×ÛúÃÖß ¬ÖÖ¸üÖ¾ÖÖÆüß ¯Ö׸ü®ÖÖ×»ÖÛúÖ ÛúÖê ×ÛúÃÖß “ÖÖ»ÖÛú »Öæ¯Ö (¯ÖÖ¿Ö) Ûúß †Öê¸ü ÝÖןִÖÖ®Ö ¤ü¿ÖÖÔµÖÖ ÝÖµÖÖ Æîü … »Öæ¯Ö ´Öë 
¯ÖÏê׸üŸÖ ¬ÖÖ¸üÖ Ûúß ×¤ü¿ÖÖ ²ÖŸÖÖ‡‹ …  1 
 
 Figure shows a current carrying solenoid moving towards a conducting loop. Find the 
direction of the current induced in the loop. 
 
 
3.  •Ö²Ö ×ÛúÃÖß ×²Ö´²Ö ÛúÖê ×ÛúÃÖß †¾ÖŸÖ»Ö ¤ü¯ÖÔÞÖ Ûêú f †Öî¸ü 2f Ûêú ²Öß“Ö ¸üÜÖÖ •ÖÖŸÖÖ Æîü, ŸÖ²Ö ²Ö®Ö®Öê ¾ÖÖ»ÖÖ ¯ÖÏןÖײִ²Ö ÛîúÃÖÖ 
ÆüÖêŸÖÖ Æîü – (i) ¾ÖÖßÖ×¾ÖÛú †£Ö¾ÖÖ †Ö³ÖÖÃÖß †Öî¸ü (ii) ”ûÖê™üÖ †£Ö¾ÖÖ ×¾Ö¾ÖÙ¬ÖŸÖ ?  1 
 When an object is placed between f and 2f of a concave mirror, would the image 
formed be (i) real or virtual and (ii) diminished or magnified ? 
 
4.  ‡»ÖꌙÒüÖò®Ö Ûúß ÝÖ×ŸÖ•Ö ‰ú•ÖÖÔ Ûêú ±ú»Ö®Ö Ûêú ºþ¯Ö ´Öë ‡ÃÖÛúß ¤êü-²ÖÎÖÝ»Öß ŸÖ¸ÓüÝÖ¤îü¬µÖÔ Ûêú ×¾Ö“Ö¸üÞÖ ÛúÖê ÝÖÏÖ±ú ÜÖà“ÖÛú¸ü   
¤ü¿ÖÖÔ‡‹ …    1 
 Draw a plot showing the variation of de Broglie wavelength of electron as a function 
of its K. E. 
 
5.  ´ÖÖê²ÖÖ‡»Ö ±úÖê®Ö ´Öë †ÖÝÖ´Öß ×ÃÖݮֻÖÖë †Öî¸ü ×®ÖÝÖÔ´Öß ×ÃÖݮֻÖÖë Ûúß †Ö¾Öé×¢ÖµÖÖÑ ×³Ö®®Ö ŒµÖÖë ÆüÖêŸÖß Æîü ?  1 
 Why is the frequency of outgoing and incoming signals different in a mobile phone ? 
55/1/1 4 
ÜÖÞ›ü – ²Ö 
Section – B 
6.  ×ÛúÃÖß “ÖÖ»ÖÛú ´Öë †Ö¾Öê¿Ö ¾ÖÖÆüÛúÖë Ûêú †¯Ö¾ÖÖÆü ¾ÖêÝÖ Ûúß †¾Ö¬ÖÖ¸üÞÖÖ ÛúÖ ˆ¯ÖµÖÖêÝÖ Ûú¸üÛêú ˆÃÖ “ÖÖ»ÖÛú Ûúß ¯ÖÏןָüÖê¬ÖÛúŸÖÖ 
†Öî¸ü ¬ÖÖ¸üÖ ‘Ö®ÖŸ¾Ö Ûêú ²Öß“Ö ÃÖÓ²ÖÓ¬Ö ¾µÖ㟯֮®Ö Ûúßו֋ …  2 
 Using the concept of drift velocity of charge carriers in a conductor, deduce the 
relationship between current density and resistivity of the conductor. 
 
7.  †¬ÖÐã×¾ÖŸÖ ¯ÖÏÛúÖ¿Ö †Öî¸ü ¸îü×ÜÖÛúŸÖ: ¬ÖÐã×¾ÖŸÖ ¯ÖÏÛúÖ¿Ö Ûêú ²Öß“Ö ×¾Ö³Öê¤ü®Ö Ûúßו֋ … †Ö¸êüÜÖ Ûúß ÃÖÆüÖµÖŸÖÖ ÃÖê ¾ÖÞÖÔ®Ö Ûúßו֋ 
×Ûú ×ÛúÃÖ ¯ÖÏÛúÖ¸ü ¯ÖÏÛúßÞÖÔ®Ö «üÖ¸üÖ †¬ÖÐã×¾ÖŸÖ ¯ÖÏÛúÖ¿Ö ¸îü×ÜÖÛúŸÖ: ¬ÖÐã×¾ÖŸÖ ÆüÖê •ÖÖŸÖÖ Æîü …  2 
 Distinguish between unpolarised and a linearly polarised light. Describe, with the help 
of a diagram, how unpolarised light gets linearly polarised by scattering. 
 
8.  ¿¾ÖêŸÖ ¯ÖÏÛúÖ¿Ö ×ÛúÃÖß ÛúÖÑ“Ö Ûêú ׯÖÏ•ûÌ´Ö ÃÖê ÝÖã•Ö¸ü®Öê ¯Ö¸ü ¯Ö׸üõÖê×¯ÖŸÖ ÆüÖê •ÖÖŸÖÖ Æîü …  
  »ÖëÃÖ ´ÖîÛú¸ü ÃÖæ¡Ö ÛúÖ ˆ¯ÖµÖÖêÝÖ Ûú¸üÛêú µÖÆü ¤ü¿ÖÖÔ‡‹ ×Ûú ×ÛúÃÖß ×¤ü‹ ÝÖ‹ »ÖëÃÖ Ûúß ±úÖêÛúÃÖ ¤æü¸üß ˆÃÖ ¯Ö¸ü †Ö¯Ö×ŸÖŸÖ 
¯ÖÏÛúÖ¿Ö Ûêú ¾ÖÞÖÔ (¸ÓüÝÖ) ¯Ö¸ü ×®Ö³ÖÔ¸ü Ûú¸üŸÖß Æîü …  2 
 Why does white light disperse when passed through a glass prism ? 
 Using lens maker’s formula, show how the focal length of a given lens depends upon 
the colour of light incident on it. 
 
9.  ×“Ö¡Ö ´Öë ¤ü¿ÖÖÔ‹ ÝÖ‹ ×®Ö¸üÖê¬Öß ×¾Ö³Ö¾Ö †Öî¸ü ±úÖê™üÖò®Ö Ûúß †Ö¯ÖŸÖ®Ö †Ö¾Öé×¢Ö Ûêú ²Öß“Ö ÝÖÏÖ±ú ÛúÖ ˆ¯ÖµÖÖêÝÖ Ûú¸üÛêú ¯»ÖÖÓÛú 
×®ÖµÖŸÖÖÓÛú ¯Ö׸üÛú×»ÖŸÖ Ûúßו֋ …  2 
 
 Using the graph shown in the figure for stopping potential V/s the incident frequency 
of photons, calculate Planck’s constant. 
 
 
Page 5


55/1/1 1 [P.T.O. 
 
 
 
 
¸üÖê»Ö ®ÖÓ. 
Roll No.  
 
 
 
³ÖÖîןÖÛú ×¾Ö–ÖÖ®Ö (ÃÖî¨üÖ×®ŸÖÛú) 
PHYSICS (Theory)  
×®Ö¬ÖÖÔ׸üŸÖ ÃÖ´ÖµÖ  : 3 ‘ÖÓ™êü ] [ †×¬ÖÛúŸÖ´Ö †ÓÛú  : 70 
Time allowed : 3 hours ] [ Maximum Marks : 70 
 
ÃÖÖ´ÖÖ®µÖ ×®Ö¤ìü¿Ö    : : : :   
    (i) ÃÖ³Öß ¯ÖÏ¿®Ö †×®Ö¾ÖÖµÖÔ Æïü … ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë Ûãú»Ö 26 ¯ÖÏ¿®Ö Æïü … 
 (ii) ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö Ûêú 5 ³ÖÖÝÖ Æïü : ÜÖÞ›ü-†, ÜÖÞ›ü-²Ö, ÜÖÞ›ü-ÃÖ, ÜÖÞ›ü-¤ü †Öî¸ü ÜÖÞ›ü-µÖ …  
 (iii) ÜÖÞ›ü-† ´Öë 5 ¯ÖÏ¿®Ö Æïü, ¯ÖÏŸµÖêÛú ÛúÖ 1 †ÓÛú Æîü … ÜÖÞ›ü-²Ö ´Öë 5 ¯ÖÏ¿®Ö Æïü, ¯ÖÏŸµÖêÛú Ûêú 2 †ÓÛú Æïü … ÜÖÞ›ü-ÃÖ ´Öë 
12 ¯ÖÏ¿®Ö Æïü, ¯ÖÏŸµÖêÛú Ûêú 3 †ÓÛú Æïü … ÜÖÞ›ü-¤ü ´Öë 4 †ÓÛú ÛúÖ ‹Ûú ´Ö滵ÖÖ¬ÖÖ׸üŸÖ ¯ÖÏ¿®Ö Æîü †Öî¸ü ÜÖÞ›ü-µÖ ´Öë 3 
¯ÖÏ¿®Ö Æïü, ¯ÖÏŸµÖêÛú Ûêú 5 †ÓÛú Æïü …  
 (iv) ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë ÃÖ´ÖÝÖÏ ¯Ö¸ü ÛúÖê‡Ô ×¾ÖÛú»¯Ö ®ÖÆüà Æîü … ŸÖ£ÖÖׯÖ, ¤üÖê †ÓÛúÖë ¾ÖÖ»Öê ‹Ûú ¯ÖÏ¿®Ö ´Öë, ŸÖß®Ö †ÓÛúÖë ¾ÖÖ»Öê ‹Ûú 
¯ÖÏ¿®Ö ´Öë †Öî¸ü ¯ÖÖÑ“Ö †ÓÛúÖë ¾ÖÖ»Öê ŸÖß®ÖÖë ¯ÖÏ¿®ÖÖë ´Öë †Ö®ŸÖ׸üÛú “ÖµÖ®Ö ¯ÖϤüÖ®Ö ×ÛúµÖÖ ÝÖµÖÖ Æîü … ‹êÃÖê ¯ÖÏ¿®ÖÖë ´Öë †Ö¯ÖÛúÖê 
פü‹ ÝÖ‹ “ÖµÖ®Ö ´Öë ÃÖê Ûêú¾Ö»Ö ‹Ûú ¯ÖÏ¿®Ö Æüß Ûú¸ü®ÖÖ Æîü …   
 Series : SSO/1/C 
55/1/1
• Ûéú¯ÖµÖÖ •ÖÖÑ“Ö Ûú¸ü »Öë ×Ûú ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë ´ÖãצüŸÖ ¯Öéšü 12  Æïü … 
• ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë ¤üÖ×Æü®Öê ÆüÖ£Ö Ûúß †Öê¸ü פü‹ ÝÖ‹ ÛúÖê›ü ®Ö´²Ö¸ü ÛúÖê ”ûÖ¡Ö ˆ¢Ö¸ü-¯Öã×ßÖÛúÖ Ûêú ´ÖãÜÖ-¯Öéšü ¯Ö¸ü ×»ÖÜÖë …  
• Ûéú¯ÖµÖÖ •ÖÖÑ“Ö Ûú¸ü »Öë ×Ûú ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë 26 ¯ÖÏ¿®Ö Æïü …  
• Ûéú¯ÖµÖÖ ¯ÖÏ¿®Ö ÛúÖ ˆ¢Ö¸ü ×»ÖÜÖ®ÖÖ ¿Öãºþ Ûú¸ü®Öê ÃÖê ¯ÖÆü»Öê, ¯ÖÏ¿®Ö ÛúÖ ÛÎú´ÖÖÓÛú †¾Ö¿µÖ ×»ÖÜÖë …  
• ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö ÛúÖê ¯ÖœÌü®Öê Ûêú ×»Ö‹ 15 ×´Ö®Ö™ü ÛúÖ ÃÖ´ÖµÖ ×¤üµÖÖ ÝÖµÖÖ Æîü … ¯ÖÏ¿®Ö-¯Ö¡Ö ÛúÖ ×¾ÖŸÖ¸üÞÖ ¯Öæ¾ÖÖÔÆËü®Ö ´Öë 10.15 ²Ö•Öê 
×ÛúµÖÖ •ÖÖµÖêÝÖÖ … 10.15 ²Ö•Öê ÃÖê 10.30 ²Ö•Öê ŸÖÛú ”ûÖ¡Ö Ûêú¾Ö»Ö ¯ÖÏ¿®Ö-¯Ö¡Ö ÛúÖê ¯ÖœÌëüÝÖê †Öî¸ü ‡ÃÖ †¾Ö×¬Ö Ûêú ¤üÖî¸üÖ®Ö ¾Öê 
ˆ¢Ö¸ü-¯Öã×ßÖÛúÖ ¯Ö¸ü ÛúÖê‡Ô ˆ¢Ö¸ü ®ÖÆüà ×»ÖÜÖëÝÖê …  
• Please check that this question paper contains 12 printed pages. 
• Code number given on the right hand side of the question paper should be written on the 
title page of the answer-book by the candidate. 
• Please check that this question paper contains 26 questions. 
• Please write down the Serial Number of the question before attempting it. 
• 15 minutes time has been allotted to read this question paper. The question paper will be 
distributed at 10.15 a.m. From 10.15 a.m. to 10.30 a.m., the students will read the 
question paper only and will not write any answer on the answer-book during this period. 
ÛúÖê›ü ®ÖÓ. 
Code No.  
    
¯Ö¸üßõÖÖ£Öá ÛúÖê›ü ÛúÖê ˆ¢Ö¸ü-¯Öã×ßÖÛúÖ Ûêú ´ÖãÜÖ-¯Öéšü 
¯Ö¸ü †¾Ö¿µÖ ×»ÖÜÖë … 
Candidates must write the Code on 
the title page of the answer-book. 
 
SET – 1 
55/1/1 2 
 (v) •ÖÆüÖÑ †Ö¾Ö¿µÖÛú ÆüÖê †Ö¯Ö ×®Ö´®Ö×»Ö×ÜÖŸÖ ³ÖÖîןÖÛ  ×®ÖµÖŸÖÖÓÛúÖë Ûêú ´ÖÖ®ÖÖë ÛúÖ ˆ¯ÖµÖÖêÝÖ Ûú¸ü ÃÖÛúŸÖê Æïü : 
  c = 3 × 10
8
 m/s 
  h = 6.63 × 10
–34
 Js 
  e = 1.6 × 10
–19
 C 
  µ
0
 = 4p × 10
–7
 T m A
–1 
  
e
0
 = 8.854 × 10
–12
 C
2
 N
–1
 m
–2 
 
  
1
4pe
0
 = 9 × 10
9
 N m
2
 C
–2 
  m
e
 = 9.1 × 10
–31
 kg 
  ®µÖæ™ÒüÖò®Ö ÛúÖ ¦ü¾µÖ´ÖÖ®Ö = 1.675 × 10
–27
 kg 
  ¯ÖÏÖê™üÖò®Ö ÛúÖ ¦ü¾µÖ´ÖÖ®Ö = 1.673 × 10
–27
 kg 
  †Ö¾ÖÖêÝÖÖ¦üÖê ÃÖÓܵÖÖ = 6.023 × 10
23
 ¯ÖÏ×ŸÖ ÝÖÏÖ´Ö ´ÖÖê»Ö  
  ²ÖÖê»™Ëü•ÖÌ´ÖÖ®Ö ×®ÖµÖŸÖÖÓÛú = 1.38 × 10
–23
 JK
–1
 
 
General Instructions :   
 (i) All questions are compulsory. There are 26 questions in all. 
 (ii) This question paper has five sections : Section A, Section B, Section C, Section D 
and Section E. 
 (iii) Section A contains five questions of one mark each, Section B contains five 
questions of two marks each, Section C contains twelve questions of three marks 
each, Section D contains one value based question of four marks and Section E 
contains three questions of five marks each. 
 (iv) There is no overall choice. However, an internal choice has been provided in one 
question of two marks, one question of three marks and all the three questions of 
five marks weightage. You have to attempt only one of the choices in such 
questions. 
 (v) You may use the following values of physical constants wherever necessary : 
  c = 3 × 10
8
 m/s 
  h = 6.63 × 10
–34
 Js 
  e = 1.6 × 10
–19
 C 
  µ
0
 = 4p × 10
–7
 T m A
–1 
  
e
0
 = 8.854 × 10
–12
 C
2
 N
–1
 m
–2 
 
  
1
4pe
0
 = 9 × 10
9
 N m
2
 C
–2 
  m
e
 = 9.1 × 10
–31
 kg 
  Mass of neutron = 1.675 × 10
–27
 kg 
  Mass of proton = 1.673 × 10
–27
 kg 
  Avogadro’s number = 6.023 × 10
23
 per gram mole 
  Boltzmann constant = 1.38 × 10
–23
 JK
–1
 
55/1/1 3 [P.T.O. 
ÜÖÞ›ü – † 
Section – A 
 
1. ¦ü¾µÖ´ÖÖ®Ö ‘m’ †Öî¸ü †Ö¾Öê¿Ö ‘q’ ÛúÖ ÛúÖê‡Ô ÛúÞÖ ‘v’ ¾ÖêÝÖ ÃÖê ×ÛúÃÖß ‹ÛúÃÖ´ÖÖ®Ö “Öã´²ÖÛúßµÖ õÖê¡Ö, •ÖÖê ÛúÞÖ Ûúß ÝÖ×ŸÖ Ûúß 
פü¿ÖÖ Ûêú »Ö´²Ö¾ÖŸÖ Æîü, ´Öë ¯ÖϾÖê¿Ö Ûú¸üŸÖÖ Æîü, ˆÃÖÛúß ÝÖ×ŸÖ•Ö ‰ú•ÖÖÔ ×ÛúÃÖ ¯ÖÏÛúÖ¸ü ¯ÖϳÖÖ×¾ÖŸÖ ÆüÖêŸÖß Æîü ? 1 
 A particle of mass ‘m’ and charge ‘q’ moving with velocity ‘v’ enters the region of 
uniform magnetic field at right angle to the direction of its motion. How does its 
kinetic energy get affected ? 
 
2.  ×“Ö¡Ö ´Öë ×ÛúÃÖß ¬ÖÖ¸üÖ¾ÖÖÆüß ¯Ö׸ü®ÖÖ×»ÖÛúÖ ÛúÖê ×ÛúÃÖß “ÖÖ»ÖÛú »Öæ¯Ö (¯ÖÖ¿Ö) Ûúß †Öê¸ü ÝÖןִÖÖ®Ö ¤ü¿ÖÖÔµÖÖ ÝÖµÖÖ Æîü … »Öæ¯Ö ´Öë 
¯ÖÏê׸üŸÖ ¬ÖÖ¸üÖ Ûúß ×¤ü¿ÖÖ ²ÖŸÖÖ‡‹ …  1 
 
 Figure shows a current carrying solenoid moving towards a conducting loop. Find the 
direction of the current induced in the loop. 
 
 
3.  •Ö²Ö ×ÛúÃÖß ×²Ö´²Ö ÛúÖê ×ÛúÃÖß †¾ÖŸÖ»Ö ¤ü¯ÖÔÞÖ Ûêú f †Öî¸ü 2f Ûêú ²Öß“Ö ¸üÜÖÖ •ÖÖŸÖÖ Æîü, ŸÖ²Ö ²Ö®Ö®Öê ¾ÖÖ»ÖÖ ¯ÖÏןÖײִ²Ö ÛîúÃÖÖ 
ÆüÖêŸÖÖ Æîü – (i) ¾ÖÖßÖ×¾ÖÛú †£Ö¾ÖÖ †Ö³ÖÖÃÖß †Öî¸ü (ii) ”ûÖê™üÖ †£Ö¾ÖÖ ×¾Ö¾ÖÙ¬ÖŸÖ ?  1 
 When an object is placed between f and 2f of a concave mirror, would the image 
formed be (i) real or virtual and (ii) diminished or magnified ? 
 
4.  ‡»ÖꌙÒüÖò®Ö Ûúß ÝÖ×ŸÖ•Ö ‰ú•ÖÖÔ Ûêú ±ú»Ö®Ö Ûêú ºþ¯Ö ´Öë ‡ÃÖÛúß ¤êü-²ÖÎÖÝ»Öß ŸÖ¸ÓüÝÖ¤îü¬µÖÔ Ûêú ×¾Ö“Ö¸üÞÖ ÛúÖê ÝÖÏÖ±ú ÜÖà“ÖÛú¸ü   
¤ü¿ÖÖÔ‡‹ …    1 
 Draw a plot showing the variation of de Broglie wavelength of electron as a function 
of its K. E. 
 
5.  ´ÖÖê²ÖÖ‡»Ö ±úÖê®Ö ´Öë †ÖÝÖ´Öß ×ÃÖݮֻÖÖë †Öî¸ü ×®ÖÝÖÔ´Öß ×ÃÖݮֻÖÖë Ûúß †Ö¾Öé×¢ÖµÖÖÑ ×³Ö®®Ö ŒµÖÖë ÆüÖêŸÖß Æîü ?  1 
 Why is the frequency of outgoing and incoming signals different in a mobile phone ? 
55/1/1 4 
ÜÖÞ›ü – ²Ö 
Section – B 
6.  ×ÛúÃÖß “ÖÖ»ÖÛú ´Öë †Ö¾Öê¿Ö ¾ÖÖÆüÛúÖë Ûêú †¯Ö¾ÖÖÆü ¾ÖêÝÖ Ûúß †¾Ö¬ÖÖ¸üÞÖÖ ÛúÖ ˆ¯ÖµÖÖêÝÖ Ûú¸üÛêú ˆÃÖ “ÖÖ»ÖÛú Ûúß ¯ÖÏןָüÖê¬ÖÛúŸÖÖ 
†Öî¸ü ¬ÖÖ¸üÖ ‘Ö®ÖŸ¾Ö Ûêú ²Öß“Ö ÃÖÓ²ÖÓ¬Ö ¾µÖ㟯֮®Ö Ûúßו֋ …  2 
 Using the concept of drift velocity of charge carriers in a conductor, deduce the 
relationship between current density and resistivity of the conductor. 
 
7.  †¬ÖÐã×¾ÖŸÖ ¯ÖÏÛúÖ¿Ö †Öî¸ü ¸îü×ÜÖÛúŸÖ: ¬ÖÐã×¾ÖŸÖ ¯ÖÏÛúÖ¿Ö Ûêú ²Öß“Ö ×¾Ö³Öê¤ü®Ö Ûúßו֋ … †Ö¸êüÜÖ Ûúß ÃÖÆüÖµÖŸÖÖ ÃÖê ¾ÖÞÖÔ®Ö Ûúßו֋ 
×Ûú ×ÛúÃÖ ¯ÖÏÛúÖ¸ü ¯ÖÏÛúßÞÖÔ®Ö «üÖ¸üÖ †¬ÖÐã×¾ÖŸÖ ¯ÖÏÛúÖ¿Ö ¸îü×ÜÖÛúŸÖ: ¬ÖÐã×¾ÖŸÖ ÆüÖê •ÖÖŸÖÖ Æîü …  2 
 Distinguish between unpolarised and a linearly polarised light. Describe, with the help 
of a diagram, how unpolarised light gets linearly polarised by scattering. 
 
8.  ¿¾ÖêŸÖ ¯ÖÏÛúÖ¿Ö ×ÛúÃÖß ÛúÖÑ“Ö Ûêú ׯÖÏ•ûÌ´Ö ÃÖê ÝÖã•Ö¸ü®Öê ¯Ö¸ü ¯Ö׸üõÖê×¯ÖŸÖ ÆüÖê •ÖÖŸÖÖ Æîü …  
  »ÖëÃÖ ´ÖîÛú¸ü ÃÖæ¡Ö ÛúÖ ˆ¯ÖµÖÖêÝÖ Ûú¸üÛêú µÖÆü ¤ü¿ÖÖÔ‡‹ ×Ûú ×ÛúÃÖß ×¤ü‹ ÝÖ‹ »ÖëÃÖ Ûúß ±úÖêÛúÃÖ ¤æü¸üß ˆÃÖ ¯Ö¸ü †Ö¯Ö×ŸÖŸÖ 
¯ÖÏÛúÖ¿Ö Ûêú ¾ÖÞÖÔ (¸ÓüÝÖ) ¯Ö¸ü ×®Ö³ÖÔ¸ü Ûú¸üŸÖß Æîü …  2 
 Why does white light disperse when passed through a glass prism ? 
 Using lens maker’s formula, show how the focal length of a given lens depends upon 
the colour of light incident on it. 
 
9.  ×“Ö¡Ö ´Öë ¤ü¿ÖÖÔ‹ ÝÖ‹ ×®Ö¸üÖê¬Öß ×¾Ö³Ö¾Ö †Öî¸ü ±úÖê™üÖò®Ö Ûúß †Ö¯ÖŸÖ®Ö †Ö¾Öé×¢Ö Ûêú ²Öß“Ö ÝÖÏÖ±ú ÛúÖ ˆ¯ÖµÖÖêÝÖ Ûú¸üÛêú ¯»ÖÖÓÛú 
×®ÖµÖŸÖÖÓÛú ¯Ö׸üÛú×»ÖŸÖ Ûúßו֋ …  2 
 
 Using the graph shown in the figure for stopping potential V/s the incident frequency 
of photons, calculate Planck’s constant. 
 
 
55/1/1 5 [P.T.O. 
10.  ®Öß“Öê ¤üß ÝÖµÖß ®ÖÖ׳ÖÛúßµÖ †×³Ö×ÛÎúµÖÖ ÛúÖê ¯Öæ¸üÖ Ûúßו֋ :  2 
 (a) 
10
5
B + 
1
0
n ?? 
4
2
He + …… 
 (b) 
94
42
Mo + 
2
1
H ?? 
95
43
Te + …… 
    †£Ö¾ÖÖ  
 µÖפü ×ÛúÃÖß ®ÖÖ׳ÖÛúßµÖ †×³Ö×ÛÎúµÖÖ ´Öë ¯ÖÏÖê™üÖò®ÖÖë †Öî¸ü ®µÖæ™ÒüÖò®ÖÖë ¤üÖê®ÖÖë Ûúß ÃÖÓܵÖÖ ÃÖÓ¸ü×õÖŸÖ ¸üÆüŸÖß Æîü, ŸÖ²Ö ×ÛúÃÖ ¯ÖÏÛúÖ¸ü 
¦ü¾µÖ´ÖÖ®Ö ÛúÖ ºþ¯ÖÖ®ŸÖ¸üÞÖ ‰ú•ÖÖÔ ´Öë (†£Ö¾ÖÖ ‡ÃÖÛúÖ ¾µÖãŸÛÎú´Ö) ÆüÖêŸÖÖ Æîü ?  ‹Ûú ˆ¤üÖÆü¸üÞÖ ÃÖ×ÆüŸÖ ¾µÖÖܵÖÖ Ûúßו֋ …  
 Complete the following nuclear reactions : 
 (a) 
10
5
B + 
1
0
n ?? 
4
2
He + …… 
 (b) 
94
42
Mo + 
2
1
H ?? 
95
43
Te + …… 
    OR 
 If both the number of protons and neutrons in a nuclear reaction is conserved, in what 
way is mass converted into energy (or vice verse) ? Explain giving one example. 
 
ÜÖÞ›ü – ÃÖ 
Section – C 
 
11. ׫ü¬ÖÐã¾Ö †Ö‘ÖæÞÖÔ 
?
p Ûêú ×ÛúÃÖß ¾ÖîªãŸÖ ׫ü¬ÖÐã¾Ö ÛúÖê ×ÛúÃÖß ‹ÛúÃÖ´ÖÖ®Ö ×¾ÖªãŸÖ õÖê¡Ö 
?
E ´Öë ¸üÜÖÖ ÝÖµÖÖ Æîü … ‡ÃÖ ×«ü¬ÖÐã¾Ö «üÖ¸üÖ 
†®Öã³Ö¾Ö ×Ûú‹ •ÖÖ®Öê ¾ÖÖ»Öê ²Ö»Ö †Ö‘ÖæÞÖÔ 
?
t Ûêú ×»Ö‹ ¾µÖÓ•ÖÛú ¯ÖÏÖ¯ŸÖ Ûúßו֋ … ‡ÃÖ ¾µÖÓ•ÖÛú ´Öë »Ö´²Ö¾ÖŸÖ ÃÖפü¿ÖÖë Ûêú ¤üÖê 
µÖãÝ´ÖÖë ÛúÖê ¯ÖÆü“ÖÖ×®Ö‹ …   3 
 An electric dipole of dipole moment 
?
p is placed in a uniform electric field 
?
E. Obtain 
the expression for the torque 
?
t experienced by the dipole. Identify two pairs of 
perpendicular vectors in the expression. 
 
12.  (a) R
1
 †Öî¸ü R
2
 (R
2
 > R
1
) ס֕µÖÖ†Öë Ûêú ¤üÖê ÝÖÖê»ÖßµÖ “ÖÖ»ÖÛúÖë ÛúÖê †Ö¾Öê×¿ÖŸÖ ×ÛúµÖÖ ÝÖµÖÖ Æîü … µÖפü ‡®Æëü ×ÛúÃÖß 
“ÖÖ»ÖÛú ŸÖÖ¸ü ÃÖê ÃÖÓµÖÖê×•ÖŸÖ ×ÛúµÖÖ •ÖÖŸÖÖ Æîü, ŸÖÖê ‡®ÖÛêú ¯ÖéšüßµÖ †Ö¾Öê¿Ö ‘Ö®ÖŸ¾ÖÖë ÛúÖ †®Öã¯ÖÖŸÖ –ÖÖŸÖ Ûúßו֋ …   
 (b) ×ÛúÃÖß †ÃÖ´ÖÖ®Ö †®Öã¯ÖÏÃ£Ö ÛúÖ™ü Ûêú ¬ÖÖן¾ÖÛú “ÖÖ»ÖÛú ÃÖê ÛúÖê‡Ô ãÖÖµÖß ¬ÖÖ¸üÖ ¯ÖϾÖÖ×ÆüŸÖ ÆüÖê ¸üÆüß Æîü … ‡ÃÖ “ÖÖ»ÖÛú 
Ûêú †®Öãפü¿Ö ÛúÖî®Ö ÃÖß ¸üÖ×¿Ö ×®ÖµÖŸÖ Æîü : ×¾ÖªãŸÖ ¬ÖÖ¸üÖ, ¬ÖÖ¸üÖ ‘Ö®ÖŸ¾Ö, ×¾ÖªãŸÖ õÖê¡Ö, †¯Ö¾ÖÖÆü “ÖÖ»Ö ?  3 
 (a) Two spherical conductors of radii R
1
 and R
2
 (R
2
 > R
1
) are charged. If they are 
connected by a conducting wire, find out the ratio of the surface charge densities 
on them. 
 (b) A steady current flows in a metallic conductor of non-uniform cross-section. 
Which of these quantities is constant along the conductor : current, current 
density, electric field, drift speed ?  
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FAQs on Past Year Paper, Physics (Set - 4), Delhi, 2015, Class 12, Physics - Additional Study Material for NEET

1. What is the importance of solving past year papers for the Class 12 Physics exam?
Ans. Solving past year papers for the Class 12 Physics exam is important as it helps students understand the exam pattern, time management, and the types of questions asked in the actual exam. It allows students to assess their preparation level and identify their strengths and weaknesses in different topics of Physics.
2. How can solving past year papers help in improving exam performance?
Ans. Solving past year papers helps in improving exam performance by familiarizing students with the format and style of questions asked in the Class 12 Physics exam. It also helps in enhancing problem-solving skills, time management, and accuracy. Regular practice of past year papers enables students to gain confidence and reduces exam anxiety.
3. Are the questions in the Delhi 2015 Class 12 Physics exam similar to the ones in the past year papers?
Ans. Yes, the questions in the Delhi 2015 Class 12 Physics exam are likely to be similar to the ones in the past year papers. However, the specific questions may vary in terms of values, figures, or contextual scenarios. It is important to practice a variety of past year papers to get a comprehensive understanding of the different types of questions that can be asked in the exam.
4. How can I utilize the solutions of past year papers to improve my understanding of Class 12 Physics concepts?
Ans. Utilizing the solutions of past year papers can help improve your understanding of Class 12 Physics concepts by providing step-by-step explanations and logical reasoning behind the solutions. By comparing your own answers with the solutions, you can identify any gaps in your knowledge and clarify doubts. It is important to thoroughly analyze the solutions and seek help from teachers or peers for any difficult concepts.
5. What should be the ideal approach to solve past year papers for the Class 12 Physics exam?
Ans. The ideal approach to solve past year papers for the Class 12 Physics exam is to first read the question carefully and understand the given information. Then, apply the relevant concepts and formulas to solve the problem. It is important to show all the steps and calculations neatly. After solving the question, cross-check the answer with the provided solution and rectify any mistakes. Finally, practice solving papers within the allotted time to improve speed and accuracy.
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