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55/1/2/D 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) ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë Ûãú»Ö 26 ¯ÖÏ¿®Ö Æïü … ÃÖ³Öß ¯ÖÏ¿®Ö †×®Ö¾ÖÖµÖÔ Æïü …  
 (ii) ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö Ûêú 5 ³ÖÖÝÖ Æïü : ÜÖÞ›ü-Ûú, ÜÖÞ›ü-ÜÖ, ÜÖÞ›ü-ÝÖ, ÜÖÞ›ü-‘Ö †Öî¸ü ÜÖÞ›ü-’û …  
 (iii) ÜÖÞ›ü-Ûú ´Öë 5 ¯ÖÏ¿®Ö ¯ÖÏŸµÖêÛú 1 †ÓÛú ÛúÖ, ÜÖÞ›ü-ÜÖ ´Öë 5 ¯ÖÏ¿®Ö ¯ÖÏŸµÖêÛú 2 †ÓÛú Ûêú, ÜÖÞ›ü-ÝÖ ´Öë 12 ¯ÖÏ¿®Ö 
¯ÖÏŸµÖêÛú 3 †ÓÛú Ûêú, ÜÖÞ›ü-‘Ö ´Öë 4 †ÓÛú ÛúÖ ‹Ûú ´Ö滵ÖÖ¬ÖÖ׸üŸÖ ¯ÖÏ¿®Ö †Öî¸ü ÜÖÞ›ü-’û ´Öë 3 ¯ÖÏ¿®Ö ¯ÖÏŸµÖêÛú 5 †ÓÛú 
Ûêú פü‹ ÝÖ‹ Æïü …  
 (iv) ÃÖ´ÖÝÖÏ ¯Ö¸ü ÛúÖê‡Ô ×¾ÖÛú»¯Ö ®ÖÆüà Æîü … ×±ú¸ü ³Öß 2 †ÓÛú Ûêú 1 ¯ÖÏ¿®Ö, 3 †ÓÛú Ûêú 1 ¯ÖÏ¿®Ö †Öî¸ü 5 †ÓÛúÖë Ûêú 3 ¯ÖÏ¿®ÖÖë 
´Öë ³Öߟָüß ×¾ÖÛú»¯Ö פü‹ ÝÖ‹ Æïü … ‹êÃÖê ¯ÖÏ¿®ÖÖë ´Öë †Ö¯ÖÛúÖê ×¾ÖÛú»¯ÖÖë ´Öë ÃÖê ‹Ûú ÛúÖê Æü»Ö Ûú¸ü®ÖÖ Æîü …  
 Series : ONS/1 
55/1/2/D 
• Ûéú¯ÖµÖÖ •ÖÖÑ“Ö Ûú¸ü »Öë ×Ûú ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë ´ÖãצüŸÖ ¯Öéšü 15 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 15 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 minute 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 – 2 
Page 2


55/1/2/D 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) ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë Ûãú»Ö 26 ¯ÖÏ¿®Ö Æïü … ÃÖ³Öß ¯ÖÏ¿®Ö †×®Ö¾ÖÖµÖÔ Æïü …  
 (ii) ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö Ûêú 5 ³ÖÖÝÖ Æïü : ÜÖÞ›ü-Ûú, ÜÖÞ›ü-ÜÖ, ÜÖÞ›ü-ÝÖ, ÜÖÞ›ü-‘Ö †Öî¸ü ÜÖÞ›ü-’û …  
 (iii) ÜÖÞ›ü-Ûú ´Öë 5 ¯ÖÏ¿®Ö ¯ÖÏŸµÖêÛú 1 †ÓÛú ÛúÖ, ÜÖÞ›ü-ÜÖ ´Öë 5 ¯ÖÏ¿®Ö ¯ÖÏŸµÖêÛú 2 †ÓÛú Ûêú, ÜÖÞ›ü-ÝÖ ´Öë 12 ¯ÖÏ¿®Ö 
¯ÖÏŸµÖêÛú 3 †ÓÛú Ûêú, ÜÖÞ›ü-‘Ö ´Öë 4 †ÓÛú ÛúÖ ‹Ûú ´Ö滵ÖÖ¬ÖÖ׸üŸÖ ¯ÖÏ¿®Ö †Öî¸ü ÜÖÞ›ü-’û ´Öë 3 ¯ÖÏ¿®Ö ¯ÖÏŸµÖêÛú 5 †ÓÛú 
Ûêú פü‹ ÝÖ‹ Æïü …  
 (iv) ÃÖ´ÖÝÖÏ ¯Ö¸ü ÛúÖê‡Ô ×¾ÖÛú»¯Ö ®ÖÆüà Æîü … ×±ú¸ü ³Öß 2 †ÓÛú Ûêú 1 ¯ÖÏ¿®Ö, 3 †ÓÛú Ûêú 1 ¯ÖÏ¿®Ö †Öî¸ü 5 †ÓÛúÖë Ûêú 3 ¯ÖÏ¿®ÖÖë 
´Öë ³Öߟָüß ×¾ÖÛú»¯Ö פü‹ ÝÖ‹ Æïü … ‹êÃÖê ¯ÖÏ¿®ÖÖë ´Öë †Ö¯ÖÛúÖê ×¾ÖÛú»¯ÖÖë ´Öë ÃÖê ‹Ûú ÛúÖê Æü»Ö Ûú¸ü®ÖÖ Æîü …  
 Series : ONS/1 
55/1/2/D 
• Ûéú¯ÖµÖÖ •ÖÖÑ“Ö Ûú¸ü »Öë ×Ûú ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë ´ÖãצüŸÖ ¯Öéšü 15 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 15 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 minute 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 – 2 
55/1/2/D 2 
 (v) •ÖÆüÖÑ †Ö¾Ö¿µÖÛú ÆüÖê, ¾ÖÆüÖÑ †Ö¯Ö ³ÖÖîןÖÛ  †“Ö¸üÖë Ûêú ×®Ö´®Ö×»Ö×ÜÖŸÖ ´Ö滵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 
  ‡»ÖꌙÒüÖê®Ö ÛúÖ ¦ü¾µÖ´ÖÖ®Ö = 9.1 × 10
–31
 kg 
  ®µÖæ™ÒüÖò®Ö ÛúÖ ¦ü¾µÖ´ÖÖ®Ö = 1.675 × 10
–27
 kg 
  ¯ÖÏÖê™üÖê®Ö ÛúÖ ¦ü¾µÖ´ÖÖ®Ö = 1.673 × 10
–27
 kg 
  ‹ê¾ÖÖêÝÖê›ÒüÖê ÃÖÓܵÖÖ = 6.023 × 10
23
 ¯ÖÏ×ŸÖ ÝÖÏÖ´Ö ´ÖÖê»Ö  
  ²ÖÖò»™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 
  Mass of electron = 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


55/1/2/D 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) ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë Ûãú»Ö 26 ¯ÖÏ¿®Ö Æïü … ÃÖ³Öß ¯ÖÏ¿®Ö †×®Ö¾ÖÖµÖÔ Æïü …  
 (ii) ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö Ûêú 5 ³ÖÖÝÖ Æïü : ÜÖÞ›ü-Ûú, ÜÖÞ›ü-ÜÖ, ÜÖÞ›ü-ÝÖ, ÜÖÞ›ü-‘Ö †Öî¸ü ÜÖÞ›ü-’û …  
 (iii) ÜÖÞ›ü-Ûú ´Öë 5 ¯ÖÏ¿®Ö ¯ÖÏŸµÖêÛú 1 †ÓÛú ÛúÖ, ÜÖÞ›ü-ÜÖ ´Öë 5 ¯ÖÏ¿®Ö ¯ÖÏŸµÖêÛú 2 †ÓÛú Ûêú, ÜÖÞ›ü-ÝÖ ´Öë 12 ¯ÖÏ¿®Ö 
¯ÖÏŸµÖêÛú 3 †ÓÛú Ûêú, ÜÖÞ›ü-‘Ö ´Öë 4 †ÓÛú ÛúÖ ‹Ûú ´Ö滵ÖÖ¬ÖÖ׸üŸÖ ¯ÖÏ¿®Ö †Öî¸ü ÜÖÞ›ü-’û ´Öë 3 ¯ÖÏ¿®Ö ¯ÖÏŸµÖêÛú 5 †ÓÛú 
Ûêú פü‹ ÝÖ‹ Æïü …  
 (iv) ÃÖ´ÖÝÖÏ ¯Ö¸ü ÛúÖê‡Ô ×¾ÖÛú»¯Ö ®ÖÆüà Æîü … ×±ú¸ü ³Öß 2 †ÓÛú Ûêú 1 ¯ÖÏ¿®Ö, 3 †ÓÛú Ûêú 1 ¯ÖÏ¿®Ö †Öî¸ü 5 †ÓÛúÖë Ûêú 3 ¯ÖÏ¿®ÖÖë 
´Öë ³Öߟָüß ×¾ÖÛú»¯Ö פü‹ ÝÖ‹ Æïü … ‹êÃÖê ¯ÖÏ¿®ÖÖë ´Öë †Ö¯ÖÛúÖê ×¾ÖÛú»¯ÖÖë ´Öë ÃÖê ‹Ûú ÛúÖê Æü»Ö Ûú¸ü®ÖÖ Æîü …  
 Series : ONS/1 
55/1/2/D 
• Ûéú¯ÖµÖÖ •ÖÖÑ“Ö Ûú¸ü »Öë ×Ûú ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë ´ÖãצüŸÖ ¯Öéšü 15 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 15 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 minute 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 – 2 
55/1/2/D 2 
 (v) •ÖÆüÖÑ †Ö¾Ö¿µÖÛú ÆüÖê, ¾ÖÆüÖÑ †Ö¯Ö ³ÖÖîןÖÛ  †“Ö¸üÖë Ûêú ×®Ö´®Ö×»Ö×ÜÖŸÖ ´Ö滵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 
  ‡»ÖꌙÒüÖê®Ö ÛúÖ ¦ü¾µÖ´ÖÖ®Ö = 9.1 × 10
–31
 kg 
  ®µÖæ™ÒüÖò®Ö ÛúÖ ¦ü¾µÖ´ÖÖ®Ö = 1.675 × 10
–27
 kg 
  ¯ÖÏÖê™üÖê®Ö ÛúÖ ¦ü¾µÖ´ÖÖ®Ö = 1.673 × 10
–27
 kg 
  ‹ê¾ÖÖêÝÖê›ÒüÖê ÃÖÓܵÖÖ = 6.023 × 10
23
 ¯ÖÏ×ŸÖ ÝÖÏÖ´Ö ´ÖÖê»Ö  
  ²ÖÖò»™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 
  Mass of electron = 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
 
55/1/2/D 3 [P.T.O. 
ÜÖÞ›ü – Ûú 
SECTION – A 
 
1. “Ö»Ö ÛãúÞ›ü»Öß ÝÖÖê®ÖÖê´Öß™ü¸ü ÛúÖ †Ö¬ÖÖ׸üŸÖ ×ÃÖ¨üÖ®ŸÖ ×»Ö×ÜÖ‹ … 1 
 Write the underlying principle of a moving coil galvanometer. 
 
2. ×¾Ö´ÖÖ®Ö ÃÖÓ“ÖÖ»Ö®Ö Ûúß ¸ü›üÖ¸ü ¯ÖÏÞÖÖ»Öß Ûêú ×»Ö‹ ÃÖæõ´Ö ŸÖ¸ÓüÝÖÖë ÛúÖê ˆ¯ÖµÖãŒŸÖ ŒµÖÖë ´ÖÖ®ÖÖ •ÖÖŸÖÖ Æîü ? 1 
 Why are microwaves considered suitable for radar systems used in aircraft          
navigation ? 
 
3. ÁÖêÞÖß LCR ¯Ö׸ü¯Ö£Ö ´Öë †®Öã®ÖÖ¤ü Ûêú ‘ÝÖãÞÖŸÖÖ ÛúÖ¸üÛú’ Ûúß ¯Ö׸ü³ÖÖÂÖÖ ×»Ö×ÜÖ‹ … ‡ÃÖÛúÖ SI ´ÖÖ¡ÖÛú ŒµÖÖ Æîü ? 1 
 Define ‘quality factor’ of resonance in series LCR circuit. What is its SI unit ? 
 
4.  ×“Ö¡Ö ´Öë ¤ü¿ÖÖÔ‹ †®ÖãÃÖÖ¸ü ÛúÖê‡Ô ײ֮¤ãü×ÛúŸÖ †Ö¾Öê¿Ö +Q ×ÛúÃÖß ×²Ö®¤ãü O ¯Ö¸ü ×Ã£ÖŸÖ Æîü … ˆ»»ÖêÜÖ Ûúßו֋ ×Ûú ŒµÖÖ 
×¾Ö³Ö¾ÖÖ®ŸÖ¸ü V
A
 – V
B
 ¬Ö®ÖÖŸ´ÖÛú, ŠúÞÖÖŸ´ÖÛú †£Ö¾ÖÖ ¿Öæ®µÖ Æîü …  1  
  +Q•----------------------------•------------• 
       O   A B 
 A point charge +Q is placed at point O as shown in the figure. Is the potential 
difference V
A
 – V
B
 positive, negative or zero ? 
  +Q•----------------------------•------------• 
       O   A B 
  
5. µÖפü ×ÛúÃÖß ÝÖÖê»ÖßµÖ ÝÖÖÃÖßµÖ ¯Öéšü Ûúß ×¡Ö•µÖÖ ´Öë ¾Öéרü Ûú¸ü ¤üß •ÖÖ‹, ŸÖÖê ˆÃÖ´Öë ¯Ö׸ü²Ö¨ü ×ÛúÃÖß ×²Ö®¤ãü×ÛúŸÖ †Ö¾Öê¿Ö Ûêú 
ÛúÖ¸üÞÖ ×¾ÖªãŸÖ õÖê¡Ö ´Öë ŒµÖÖ ¯Ö׸ü¾ÖŸÖÔ®Ö ÆüÖêÝÖÖ ?  1 
 How does the electric flux due to a point charge enclosed by a spherical Gaussian 
surface get affected when its radius is increased ? 
Page 4


55/1/2/D 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) ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë Ûãú»Ö 26 ¯ÖÏ¿®Ö Æïü … ÃÖ³Öß ¯ÖÏ¿®Ö †×®Ö¾ÖÖµÖÔ Æïü …  
 (ii) ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö Ûêú 5 ³ÖÖÝÖ Æïü : ÜÖÞ›ü-Ûú, ÜÖÞ›ü-ÜÖ, ÜÖÞ›ü-ÝÖ, ÜÖÞ›ü-‘Ö †Öî¸ü ÜÖÞ›ü-’û …  
 (iii) ÜÖÞ›ü-Ûú ´Öë 5 ¯ÖÏ¿®Ö ¯ÖÏŸµÖêÛú 1 †ÓÛú ÛúÖ, ÜÖÞ›ü-ÜÖ ´Öë 5 ¯ÖÏ¿®Ö ¯ÖÏŸµÖêÛú 2 †ÓÛú Ûêú, ÜÖÞ›ü-ÝÖ ´Öë 12 ¯ÖÏ¿®Ö 
¯ÖÏŸµÖêÛú 3 †ÓÛú Ûêú, ÜÖÞ›ü-‘Ö ´Öë 4 †ÓÛú ÛúÖ ‹Ûú ´Ö滵ÖÖ¬ÖÖ׸üŸÖ ¯ÖÏ¿®Ö †Öî¸ü ÜÖÞ›ü-’û ´Öë 3 ¯ÖÏ¿®Ö ¯ÖÏŸµÖêÛú 5 †ÓÛú 
Ûêú פü‹ ÝÖ‹ Æïü …  
 (iv) ÃÖ´ÖÝÖÏ ¯Ö¸ü ÛúÖê‡Ô ×¾ÖÛú»¯Ö ®ÖÆüà Æîü … ×±ú¸ü ³Öß 2 †ÓÛú Ûêú 1 ¯ÖÏ¿®Ö, 3 †ÓÛú Ûêú 1 ¯ÖÏ¿®Ö †Öî¸ü 5 †ÓÛúÖë Ûêú 3 ¯ÖÏ¿®ÖÖë 
´Öë ³Öߟָüß ×¾ÖÛú»¯Ö פü‹ ÝÖ‹ Æïü … ‹êÃÖê ¯ÖÏ¿®ÖÖë ´Öë †Ö¯ÖÛúÖê ×¾ÖÛú»¯ÖÖë ´Öë ÃÖê ‹Ûú ÛúÖê Æü»Ö Ûú¸ü®ÖÖ Æîü …  
 Series : ONS/1 
55/1/2/D 
• Ûéú¯ÖµÖÖ •ÖÖÑ“Ö Ûú¸ü »Öë ×Ûú ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë ´ÖãצüŸÖ ¯Öéšü 15 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 15 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 minute 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 – 2 
55/1/2/D 2 
 (v) •ÖÆüÖÑ †Ö¾Ö¿µÖÛú ÆüÖê, ¾ÖÆüÖÑ †Ö¯Ö ³ÖÖîןÖÛ  †“Ö¸üÖë Ûêú ×®Ö´®Ö×»Ö×ÜÖŸÖ ´Ö滵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 
  ‡»ÖꌙÒüÖê®Ö ÛúÖ ¦ü¾µÖ´ÖÖ®Ö = 9.1 × 10
–31
 kg 
  ®µÖæ™ÒüÖò®Ö ÛúÖ ¦ü¾µÖ´ÖÖ®Ö = 1.675 × 10
–27
 kg 
  ¯ÖÏÖê™üÖê®Ö ÛúÖ ¦ü¾µÖ´ÖÖ®Ö = 1.673 × 10
–27
 kg 
  ‹ê¾ÖÖêÝÖê›ÒüÖê ÃÖÓܵÖÖ = 6.023 × 10
23
 ¯ÖÏ×ŸÖ ÝÖÏÖ´Ö ´ÖÖê»Ö  
  ²ÖÖò»™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 
  Mass of electron = 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
 
55/1/2/D 3 [P.T.O. 
ÜÖÞ›ü – Ûú 
SECTION – A 
 
1. “Ö»Ö ÛãúÞ›ü»Öß ÝÖÖê®ÖÖê´Öß™ü¸ü ÛúÖ †Ö¬ÖÖ׸üŸÖ ×ÃÖ¨üÖ®ŸÖ ×»Ö×ÜÖ‹ … 1 
 Write the underlying principle of a moving coil galvanometer. 
 
2. ×¾Ö´ÖÖ®Ö ÃÖÓ“ÖÖ»Ö®Ö Ûúß ¸ü›üÖ¸ü ¯ÖÏÞÖÖ»Öß Ûêú ×»Ö‹ ÃÖæõ´Ö ŸÖ¸ÓüÝÖÖë ÛúÖê ˆ¯ÖµÖãŒŸÖ ŒµÖÖë ´ÖÖ®ÖÖ •ÖÖŸÖÖ Æîü ? 1 
 Why are microwaves considered suitable for radar systems used in aircraft          
navigation ? 
 
3. ÁÖêÞÖß LCR ¯Ö׸ü¯Ö£Ö ´Öë †®Öã®ÖÖ¤ü Ûêú ‘ÝÖãÞÖŸÖÖ ÛúÖ¸üÛú’ Ûúß ¯Ö׸ü³ÖÖÂÖÖ ×»Ö×ÜÖ‹ … ‡ÃÖÛúÖ SI ´ÖÖ¡ÖÛú ŒµÖÖ Æîü ? 1 
 Define ‘quality factor’ of resonance in series LCR circuit. What is its SI unit ? 
 
4.  ×“Ö¡Ö ´Öë ¤ü¿ÖÖÔ‹ †®ÖãÃÖÖ¸ü ÛúÖê‡Ô ײ֮¤ãü×ÛúŸÖ †Ö¾Öê¿Ö +Q ×ÛúÃÖß ×²Ö®¤ãü O ¯Ö¸ü ×Ã£ÖŸÖ Æîü … ˆ»»ÖêÜÖ Ûúßו֋ ×Ûú ŒµÖÖ 
×¾Ö³Ö¾ÖÖ®ŸÖ¸ü V
A
 – V
B
 ¬Ö®ÖÖŸ´ÖÛú, ŠúÞÖÖŸ´ÖÛú †£Ö¾ÖÖ ¿Öæ®µÖ Æîü …  1  
  +Q•----------------------------•------------• 
       O   A B 
 A point charge +Q is placed at point O as shown in the figure. Is the potential 
difference V
A
 – V
B
 positive, negative or zero ? 
  +Q•----------------------------•------------• 
       O   A B 
  
5. µÖפü ×ÛúÃÖß ÝÖÖê»ÖßµÖ ÝÖÖÃÖßµÖ ¯Öéšü Ûúß ×¡Ö•µÖÖ ´Öë ¾Öéרü Ûú¸ü ¤üß •ÖÖ‹, ŸÖÖê ˆÃÖ´Öë ¯Ö׸ü²Ö¨ü ×ÛúÃÖß ×²Ö®¤ãü×ÛúŸÖ †Ö¾Öê¿Ö Ûêú 
ÛúÖ¸üÞÖ ×¾ÖªãŸÖ õÖê¡Ö ´Öë ŒµÖÖ ¯Ö׸ü¾ÖŸÖÔ®Ö ÆüÖêÝÖÖ ?  1 
 How does the electric flux due to a point charge enclosed by a spherical Gaussian 
surface get affected when its radius is increased ? 
55/1/2/D 4 
ÜÖÞ›ü – ÜÖ  
SECTION – B 
 
6. ¦ü¾µÖ´ÖÖ®Ö ÃÖÓܵÖÖ A = 240 ŸÖ£ÖÖ ²Ö®¬Ö®Ö ‰ú•ÖÖÔ ¯ÖÏ×ŸÖ ®µÖã׌»Ö†Öò®Ö BE/A = 7.6 MeV ÛúÖ ÛúÖê‡Ô ®ÖÖ׳ÖÛú ¤üÖê ™ãüÛú›ÌüÖë 
´Öë ×¾ÖÜÖ×Þ›üŸÖ ÆüÖêŸÖÖ Æîü וִ֮Öë ¯ÖÏŸµÖêÛú Ûêú ×»Ö‹ A = 120 †Öî¸ü BE/A = 8.5 MeV Æîü … ´Ö㌟Ö-‰ú•ÖÖÔ ¯Ö׸üÛú×»ÖŸÖ 
Ûúßו֋ …   2  
†£Ö¾ÖÖ 
 ÃÖÓ»ÖµÖ®Ö †×³Ö×ÛÎúµÖÖ 
2
1
H + 
2
1
H ?? 
3
2
He + n, •Ö²Ö×Ûú, ²ÖÓ¬Ö®Ö ‰ú•ÖÖÔ (BE) 
2
1
H Ûúß 2.23 MeV ŸÖ£ÖÖ 
3
2
He Ûúß 
7.73 MeV Æîü, ´Öë ‰ú•ÖÖÔ ¯Ö׸üÛú×»ÖŸÖ Ûúßו֋ … 
 A nucleus with mass number A = 240 and BE/A = 7.6 MeV breaks into two fragments 
each of A = 120 with BE/A = 8.5 MeV. Calculate the released energy. 
OR 
 Calculate the energy in fusion reaction :  
 
2
1
H + 
2
1
H ?? 
3
2
He + n, where BE of  
2
1
H = 2.23 MeV and of 
3
2
He = 7.73 MeV. 
 
7. ¤üÖê ÃÖê»Ö, ו֮ÖÛúß emf 1.5 V †Öî¸ü 2.0 V ŸÖ£ÖÖ †Ö®ŸÖ׸üÛú ¯ÖÏןָüÖê¬Ö ÛÎú´Ö¿Ö: 0.2 O ŸÖ£ÖÖ 0.3 O Æïü, ¯ÖÖ¿¾ÖÔ ´Öë 
ÃÖÓµÖÖê×•ÖŸÖ Æïü … ‡®ÖÛêú ŸÖã»µÖ ÃÖê»Ö Ûúß emf †Öî¸ü †Ö®ŸÖ׸üÛú ¯ÖÏןָüÖê¬Ö ¯Ö׸üÛú×»ÖŸÖ Ûúßו֋ … 2 
 Two cells of emfs 1.5 V and 2.0 V having internal resistances 0.2 O  and 0.3 O 
respectively are connected in parallel. Calculate the emf and internal resistance of the 
equivalent cell. 
 
8. ²ÖÎæÙü¸ü ×®ÖµÖ´Ö ×»Ö×ÜÖ‹ …  
 ×¾Ö׳֮®Ö ¾ÖÞÖÖí Ûêú ¯ÖÏÛúÖ¿Ö Ûêú ×»Ö‹ ¯ÖÖ¸ü¤ü¿Öá ´ÖÖ¬µÖ´Ö Ûêú ²ÖÎæÙü¸ü ÛúÖêÞÖ ÛúÖ ´ÖÖ®Ö ×³Ö®®Ö-׳֮®Ö ÆüÖêŸÖÖ Æîü … ÛúÖ¸üÞÖ ¤üßו֋ … 2 
 State Brewster’s law. 
 The value of Brewster angle for a transparent medium is different for light of different 
colours. Give reason.  
 
9. ÃÖÓ“ÖÖ¸ü ¯ÖÏÞÖÖ»Öß ´Öë ˆ¯ÖµÖÖêÝÖ ÆüÖê®Öê ¾ÖÖ»Öê ¯Ö¤üÖë (i) ÃÖÓÛúßÞÖÔ®Ö (õÖßÞÖŸÖÖ) (ii) ×¾Ö´ÖÖò›ãü»Ö®Ö Ûúß ¾µÖÖܵÖÖ Ûúßו֋ … 2  
 Explain the terms (i) Attenuation and (ii) Demodulation used in Communication 
System.   
Page 5


55/1/2/D 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) ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë Ûãú»Ö 26 ¯ÖÏ¿®Ö Æïü … ÃÖ³Öß ¯ÖÏ¿®Ö †×®Ö¾ÖÖµÖÔ Æïü …  
 (ii) ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö Ûêú 5 ³ÖÖÝÖ Æïü : ÜÖÞ›ü-Ûú, ÜÖÞ›ü-ÜÖ, ÜÖÞ›ü-ÝÖ, ÜÖÞ›ü-‘Ö †Öî¸ü ÜÖÞ›ü-’û …  
 (iii) ÜÖÞ›ü-Ûú ´Öë 5 ¯ÖÏ¿®Ö ¯ÖÏŸµÖêÛú 1 †ÓÛú ÛúÖ, ÜÖÞ›ü-ÜÖ ´Öë 5 ¯ÖÏ¿®Ö ¯ÖÏŸµÖêÛú 2 †ÓÛú Ûêú, ÜÖÞ›ü-ÝÖ ´Öë 12 ¯ÖÏ¿®Ö 
¯ÖÏŸµÖêÛú 3 †ÓÛú Ûêú, ÜÖÞ›ü-‘Ö ´Öë 4 †ÓÛú ÛúÖ ‹Ûú ´Ö滵ÖÖ¬ÖÖ׸üŸÖ ¯ÖÏ¿®Ö †Öî¸ü ÜÖÞ›ü-’û ´Öë 3 ¯ÖÏ¿®Ö ¯ÖÏŸµÖêÛú 5 †ÓÛú 
Ûêú פü‹ ÝÖ‹ Æïü …  
 (iv) ÃÖ´ÖÝÖÏ ¯Ö¸ü ÛúÖê‡Ô ×¾ÖÛú»¯Ö ®ÖÆüà Æîü … ×±ú¸ü ³Öß 2 †ÓÛú Ûêú 1 ¯ÖÏ¿®Ö, 3 †ÓÛú Ûêú 1 ¯ÖÏ¿®Ö †Öî¸ü 5 †ÓÛúÖë Ûêú 3 ¯ÖÏ¿®ÖÖë 
´Öë ³Öߟָüß ×¾ÖÛú»¯Ö פü‹ ÝÖ‹ Æïü … ‹êÃÖê ¯ÖÏ¿®ÖÖë ´Öë †Ö¯ÖÛúÖê ×¾ÖÛú»¯ÖÖë ´Öë ÃÖê ‹Ûú ÛúÖê Æü»Ö Ûú¸ü®ÖÖ Æîü …  
 Series : ONS/1 
55/1/2/D 
• Ûéú¯ÖµÖÖ •ÖÖÑ“Ö Ûú¸ü »Öë ×Ûú ‡ÃÖ ¯ÖÏ¿®Ö-¯Ö¡Ö ´Öë ´ÖãצüŸÖ ¯Öéšü 15 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 15 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 minute 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 – 2 
55/1/2/D 2 
 (v) •ÖÆüÖÑ †Ö¾Ö¿µÖÛú ÆüÖê, ¾ÖÆüÖÑ †Ö¯Ö ³ÖÖîןÖÛ  †“Ö¸üÖë Ûêú ×®Ö´®Ö×»Ö×ÜÖŸÖ ´Ö滵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 
  ‡»ÖꌙÒüÖê®Ö ÛúÖ ¦ü¾µÖ´ÖÖ®Ö = 9.1 × 10
–31
 kg 
  ®µÖæ™ÒüÖò®Ö ÛúÖ ¦ü¾µÖ´ÖÖ®Ö = 1.675 × 10
–27
 kg 
  ¯ÖÏÖê™üÖê®Ö ÛúÖ ¦ü¾µÖ´ÖÖ®Ö = 1.673 × 10
–27
 kg 
  ‹ê¾ÖÖêÝÖê›ÒüÖê ÃÖÓܵÖÖ = 6.023 × 10
23
 ¯ÖÏ×ŸÖ ÝÖÏÖ´Ö ´ÖÖê»Ö  
  ²ÖÖò»™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 
  Mass of electron = 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
 
55/1/2/D 3 [P.T.O. 
ÜÖÞ›ü – Ûú 
SECTION – A 
 
1. “Ö»Ö ÛãúÞ›ü»Öß ÝÖÖê®ÖÖê´Öß™ü¸ü ÛúÖ †Ö¬ÖÖ׸üŸÖ ×ÃÖ¨üÖ®ŸÖ ×»Ö×ÜÖ‹ … 1 
 Write the underlying principle of a moving coil galvanometer. 
 
2. ×¾Ö´ÖÖ®Ö ÃÖÓ“ÖÖ»Ö®Ö Ûúß ¸ü›üÖ¸ü ¯ÖÏÞÖÖ»Öß Ûêú ×»Ö‹ ÃÖæõ´Ö ŸÖ¸ÓüÝÖÖë ÛúÖê ˆ¯ÖµÖãŒŸÖ ŒµÖÖë ´ÖÖ®ÖÖ •ÖÖŸÖÖ Æîü ? 1 
 Why are microwaves considered suitable for radar systems used in aircraft          
navigation ? 
 
3. ÁÖêÞÖß LCR ¯Ö׸ü¯Ö£Ö ´Öë †®Öã®ÖÖ¤ü Ûêú ‘ÝÖãÞÖŸÖÖ ÛúÖ¸üÛú’ Ûúß ¯Ö׸ü³ÖÖÂÖÖ ×»Ö×ÜÖ‹ … ‡ÃÖÛúÖ SI ´ÖÖ¡ÖÛú ŒµÖÖ Æîü ? 1 
 Define ‘quality factor’ of resonance in series LCR circuit. What is its SI unit ? 
 
4.  ×“Ö¡Ö ´Öë ¤ü¿ÖÖÔ‹ †®ÖãÃÖÖ¸ü ÛúÖê‡Ô ײ֮¤ãü×ÛúŸÖ †Ö¾Öê¿Ö +Q ×ÛúÃÖß ×²Ö®¤ãü O ¯Ö¸ü ×Ã£ÖŸÖ Æîü … ˆ»»ÖêÜÖ Ûúßו֋ ×Ûú ŒµÖÖ 
×¾Ö³Ö¾ÖÖ®ŸÖ¸ü V
A
 – V
B
 ¬Ö®ÖÖŸ´ÖÛú, ŠúÞÖÖŸ´ÖÛú †£Ö¾ÖÖ ¿Öæ®µÖ Æîü …  1  
  +Q•----------------------------•------------• 
       O   A B 
 A point charge +Q is placed at point O as shown in the figure. Is the potential 
difference V
A
 – V
B
 positive, negative or zero ? 
  +Q•----------------------------•------------• 
       O   A B 
  
5. µÖפü ×ÛúÃÖß ÝÖÖê»ÖßµÖ ÝÖÖÃÖßµÖ ¯Öéšü Ûúß ×¡Ö•µÖÖ ´Öë ¾Öéרü Ûú¸ü ¤üß •ÖÖ‹, ŸÖÖê ˆÃÖ´Öë ¯Ö׸ü²Ö¨ü ×ÛúÃÖß ×²Ö®¤ãü×ÛúŸÖ †Ö¾Öê¿Ö Ûêú 
ÛúÖ¸üÞÖ ×¾ÖªãŸÖ õÖê¡Ö ´Öë ŒµÖÖ ¯Ö׸ü¾ÖŸÖÔ®Ö ÆüÖêÝÖÖ ?  1 
 How does the electric flux due to a point charge enclosed by a spherical Gaussian 
surface get affected when its radius is increased ? 
55/1/2/D 4 
ÜÖÞ›ü – ÜÖ  
SECTION – B 
 
6. ¦ü¾µÖ´ÖÖ®Ö ÃÖÓܵÖÖ A = 240 ŸÖ£ÖÖ ²Ö®¬Ö®Ö ‰ú•ÖÖÔ ¯ÖÏ×ŸÖ ®µÖã׌»Ö†Öò®Ö BE/A = 7.6 MeV ÛúÖ ÛúÖê‡Ô ®ÖÖ׳ÖÛú ¤üÖê ™ãüÛú›ÌüÖë 
´Öë ×¾ÖÜÖ×Þ›üŸÖ ÆüÖêŸÖÖ Æîü וִ֮Öë ¯ÖÏŸµÖêÛú Ûêú ×»Ö‹ A = 120 †Öî¸ü BE/A = 8.5 MeV Æîü … ´Ö㌟Ö-‰ú•ÖÖÔ ¯Ö׸üÛú×»ÖŸÖ 
Ûúßו֋ …   2  
†£Ö¾ÖÖ 
 ÃÖÓ»ÖµÖ®Ö †×³Ö×ÛÎúµÖÖ 
2
1
H + 
2
1
H ?? 
3
2
He + n, •Ö²Ö×Ûú, ²ÖÓ¬Ö®Ö ‰ú•ÖÖÔ (BE) 
2
1
H Ûúß 2.23 MeV ŸÖ£ÖÖ 
3
2
He Ûúß 
7.73 MeV Æîü, ´Öë ‰ú•ÖÖÔ ¯Ö׸üÛú×»ÖŸÖ Ûúßו֋ … 
 A nucleus with mass number A = 240 and BE/A = 7.6 MeV breaks into two fragments 
each of A = 120 with BE/A = 8.5 MeV. Calculate the released energy. 
OR 
 Calculate the energy in fusion reaction :  
 
2
1
H + 
2
1
H ?? 
3
2
He + n, where BE of  
2
1
H = 2.23 MeV and of 
3
2
He = 7.73 MeV. 
 
7. ¤üÖê ÃÖê»Ö, ו֮ÖÛúß emf 1.5 V †Öî¸ü 2.0 V ŸÖ£ÖÖ †Ö®ŸÖ׸üÛú ¯ÖÏןָüÖê¬Ö ÛÎú´Ö¿Ö: 0.2 O ŸÖ£ÖÖ 0.3 O Æïü, ¯ÖÖ¿¾ÖÔ ´Öë 
ÃÖÓµÖÖê×•ÖŸÖ Æïü … ‡®ÖÛêú ŸÖã»µÖ ÃÖê»Ö Ûúß emf †Öî¸ü †Ö®ŸÖ׸üÛú ¯ÖÏןָüÖê¬Ö ¯Ö׸üÛú×»ÖŸÖ Ûúßו֋ … 2 
 Two cells of emfs 1.5 V and 2.0 V having internal resistances 0.2 O  and 0.3 O 
respectively are connected in parallel. Calculate the emf and internal resistance of the 
equivalent cell. 
 
8. ²ÖÎæÙü¸ü ×®ÖµÖ´Ö ×»Ö×ÜÖ‹ …  
 ×¾Ö׳֮®Ö ¾ÖÞÖÖí Ûêú ¯ÖÏÛúÖ¿Ö Ûêú ×»Ö‹ ¯ÖÖ¸ü¤ü¿Öá ´ÖÖ¬µÖ´Ö Ûêú ²ÖÎæÙü¸ü ÛúÖêÞÖ ÛúÖ ´ÖÖ®Ö ×³Ö®®Ö-׳֮®Ö ÆüÖêŸÖÖ Æîü … ÛúÖ¸üÞÖ ¤üßו֋ … 2 
 State Brewster’s law. 
 The value of Brewster angle for a transparent medium is different for light of different 
colours. Give reason.  
 
9. ÃÖÓ“ÖÖ¸ü ¯ÖÏÞÖÖ»Öß ´Öë ˆ¯ÖµÖÖêÝÖ ÆüÖê®Öê ¾ÖÖ»Öê ¯Ö¤üÖë (i) ÃÖÓÛúßÞÖÔ®Ö (õÖßÞÖŸÖÖ) (ii) ×¾Ö´ÖÖò›ãü»Ö®Ö Ûúß ¾µÖÖܵÖÖ Ûúßו֋ … 2  
 Explain the terms (i) Attenuation and (ii) Demodulation used in Communication 
System.   
55/1/2/D 5 [P.T.O. 
10. ÃÖ´ÖÖ®Ö †Ö¾Öê¿Ö ¯Ö¸ü®ŸÖã ×¾Ö׳֮®Ö ¦ü¾µÖ´ÖÖ®ÖÖë m
1
 , m
2
 (m
1
 > m
2
) Ûêú ¤üÖê ÛúÞÖÖë A †Öî¸ü B Ûêú 
1
V
  †Öî¸ü            
¤êü-²ÖÎÖòÝ»Öß ŸÖ¸ÓüÝÖ¤îü‘µÖÔ ? Ûêú ²Öß“Ö ×¾Ö“Ö¸üÞÖ ÛúÖê ¤ü¿ÖÖÔ®Öê Ûêú ×»Ö‹ ÝÖÏÖ±ú ÜÖàד֋ … µÖפü V Ÿ¾Ö¸üÛú ×¾Ö³Ö¾Ö ÛúÖê ×®Öºþ×¯ÖŸÖ 
Ûú¸üŸÖÖ Æîü, ŸÖÖê ‡®Ö ¤üÖê®ÖÖë ´Öë ÃÖê ÛúÖî®Ö ”ûÖê™êü ¦ü¾µÖ´ÖÖ®Ö ÛúÖê ¯ÖϤüÙ¿ÖŸÖ Ûú¸üŸÖÖ Æîü ? ÛúÖ¸üÞÖ ¤üßו֋ … 2 
 Plot a graph showing variation of de-Broglie wavelength ? versus 
1
V
 , where V is 
accelerating potential for two particles A and B carrying same charge but of masses 
m
1
, m
2
 (m
1
 > m
2
). Which one of the two represents a particle of smaller mass and   
why ? 
 
ÜÖÞ›ü-ÝÖ 
SECTION – C 
 
11. (i) †®µÖÖê®µÖ ¯ÖÏê¸üÞÖ Ûúß ¯Ö׸ü³ÖÖÂÖÖ ×»Ö×ÜÖ‹ … 
 (ii) ÃÖÓ»ÖÝ®Ö ÛãúÞ›ü×»ÖµÖÖë Ûêú ×ÛúÃÖß µÖãÝÖ»Ö ÛúÖ †®µÖÖê®µÖ ¯ÖÏê¸üÞÖ 1.5 H Æîü … µÖפü ‹Ûú ÛãúÞ›ü»Öß ´Öë 0.5 s ´Öë ¬ÖÖ¸üÖ      
0 ÃÖê 20 A ÆüÖê •ÖÖŸÖß Æîü, ŸÖÖê †®µÖ ÛãúÞ›ü»Öß ´Öë °»ÖŒÃÖ-ÝÖÏÓ×£ÖÛúÖ ÛúÖ ¯Ö׸ü¾ÖŸÖÔ®Ö ×ÛúŸÖ®ÖÖ ÆüÖêÝÖÖ ? 3 
 (i) Define mutual inductance. 
 (ii) A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one 
coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the 
other coil ? 
 
12. ¤üÖê ÃÖ´ÖÖ®ŸÖ¸ü ¯Ö×¼üÛúÖ ÃÖÓ¬ÖÖ׸ü¡ÖÖë X ŸÖ£ÖÖ Y Ûúß ¯Ö×¼üÛúÖ†Öë Ûêú õÖê¡Ö ÃÖ´ÖÖ®Ö Æïü †Öî¸ü ˆ®ÖÛêú ²Öß“Ö ¯Öé£ÖÛËú®Ö ³Öß ÃÖ´ÖÖ®Ö Æïü … 
X Ûúß ¯Ö×¼üÛúÖ†Öë Ûêú ²Öß“Ö ¾ÖÖµÖã Æîü, •Ö²Ö×Ûú Y ´Öë e
r
 = 4 ÛúÖ ¯Ö¸üÖ¾ÖîªãŸÖ ´ÖÖ¬µÖ´Ö Æîü … 
 
 (i) µÖפü ÃÖÓµÖÖê•Ö®Ö Ûúß ŸÖã»µÖ ¬ÖÖ׸üŸÖÖ 4 µF Æîü, ŸÖÖê ¯ÖÏŸµÖêÛú ÃÖÓ¬ÖÖ׸ü¡Ö Ûúß ¬ÖÖ׸üŸÖÖ ¯Ö׸üÛú×»ÖŸÖ Ûúßו֋ … 
 (ii) X ŸÖ£ÖÖ Y Ûúß ¯Ö×¼üÛúÖ†Öë Ûêú ²Öß“Ö ×¾Ö³Ö¾ÖÖ®ŸÖ¸ü ¯Ö׸üÛú×»ÖŸÖ Ûúßו֋ … 
 (iii) X ŸÖ£ÖÖ Y ´Öë ÃÖÓ×“ÖŸÖ ×ãָü ×¾ÖªãŸÖ ‰ú•ÖÖÔ Ûêú †®Öã¯ÖÖŸÖ ÛúÖ †®Öã´ÖÖ®Ö »ÖÝÖÖ‡‹ … 3 
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FAQs on Past Year Paper, Physics (Set - 2), Delhi, 2016, Class 12, Physics - Additional Study Material for NEET

1. How can I access the past year paper for Class 12 Physics Delhi Board, 2016?
Ans. You can access the past year paper for Class 12 Physics Delhi Board, 2016 by searching for it on educational websites, online forums, or by asking your school or teachers if they have a copy.
2. What are the benefits of solving past year papers for Class 12 Physics?
Ans. Solving past year papers for Class 12 Physics can help you understand the exam pattern, identify important topics, and practice time management. It also gives you an idea of the types of questions asked in the exam and helps in improving your overall exam preparation.
3. How can solving past year papers help in improving my performance in the Class 12 Physics exam?
Ans. Solving past year papers helps in improving your performance in the Class 12 Physics exam by familiarizing you with the exam format, allowing you to practice answering different types of questions, and helping you identify your strengths and weaknesses. It also helps in boosting your confidence and reducing exam stress.
4. Are the questions in the past year paper for Class 12 Physics Delhi Board, 2016 still relevant for the current syllabus?
Ans. Yes, the questions in the past year paper for Class 12 Physics Delhi Board, 2016 are still relevant for the current syllabus. Although the syllabus may have undergone minor changes, the core concepts and topics covered in the past year paper are still applicable.
5. Can I rely solely on solving past year papers for Class 12 Physics to prepare for the exam?
Ans. While solving past year papers for Class 12 Physics is an important part of exam preparation, it is not sufficient on its own. It is recommended to also study the textbook, refer to other study materials, and practice additional sample papers to get a comprehensive understanding of the subject and perform well in the exam.
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