NEET Exam > NEET Notes > Additional Study Material for NEET > Past Year Paper, Physics (Set - 1), Outside Delhi, 2015, Class 12, Physics
Past Year Paper, Physics (Set - 1), Outside Delhi, 2015, Class 12, Physics | Additional Study Material for NEET PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
Page 1
55/1 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-†ú ´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, ¯ÖÏŸµÖêÛ 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/C
55/1
• 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 [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-†ú ´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, ¯ÖÏŸµÖêÛ 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/C
55/1
• 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 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 [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-†ú ´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, ¯ÖÏŸµÖêÛ 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/C
55/1
• 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 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 3 [P.T.O.
ÜÖÞ›ü – †
Section – A
1. ×ÛúÃÖß “ÖÖ»ÖÛú Ûêú ÃÖ´Ö×¾Ö³Ö¾Ö ¯Öéšü Ûêú ×ÛúÃÖß ×²Ö®¤ãü ¯Ö¸ü ×¾ÖªãŸÖ õÖê¡Ö ¸êüÜÖÖ‹Ñ ¯Öéšü Ûêú »Ö´²Ö¾ÖŸÖ ŒµÖÖë ÆüÖêŸÖß Æïü ? 1
Why are electric field lines perpendicular at a point on an equipotential surface of a
conductor ?
2. ×ÛúÃÖß ÃÖÓ¬ÖÖ׸ü¡Ö ÛúÖê ×ÛúÃÖß ¯Ö׸ü¾ÖŸÖá †Ö¾Öé×¢Ö Ûêú AC ÄÖÖêŸÖ ÃÖê ÃÖÓµÖÖê×•ÖŸÖ ×ÛúµÖÖ ÝÖµÖÖ Æîü … µÖפü AC ÄÖÖêŸÖ Ûúß †Ö¾Öé×¢Ö
‘Ö™üÖ ¤üß •ÖÖ‹, ŸÖÖê ŒµÖÖ ×¾ÖãÖÖ¯Ö®Ö ¬ÖÖ¸üÖ ¯Ö׸ü¾ÖÙŸÖŸÖ ÆüÖê •ÖÖ‹ÝÖß ? 1
A variable frequency AC source is connected to a capacitor. Will the displacement
current change if the frequency of the AC source is decreased ?
3. NAND ÝÖê™ü ÛúÖ ŸÖÛÔú ¯ÖÏŸÖßÛú ÜÖàד֋ †Öî¸ü ‡ÃÖÛúß ÃÖŸµÖ´ÖÖ®Ö ÃÖÖ¸üÞÖß ¤üßו֋ … 1
Draw the logic symbol of NAND gate and give its Truth Table.
4. AC ÄÖÖêŸÖ Ûúß †Ö¾Öé×¢Ö ´Öë ¯Ö׸ü¾ÖŸÖÔ®Ö Ûêú ÃÖÖ£Ö ÃÖÓ¬ÖÖ׸ü¡Ö ¯ÖÏן֑ÖÖŸÖ ´Öë ×¾Ö“Ö¸üÞÖ ÛúÖê ¤ü¿ÖÖÔ®Öê Ûêú ×»Ö‹ ÝÖÏÖ±ú ÜÖàד֋ … 1
Plot a graph showing variation of capacitive reactance with the change in the frequency
of the AC source.
5. †ÖµÖÖ´Ö ´ÖÖò›ãü»Ö®Ö †Öî¸ü †Ö¾Öé×¢Ö ´ÖÖò›ãü»Ö®Ö Ûêú ²Öß“Ö ×¾Ö³Öê¤ü®Ö Ûúßו֋ … 1
Distinguish between amplitude modulation and frequency modulation.
ÜÖÞ›ü – ²Ö
Section – B
6. ‘×¾ÖªãŸÖ °»ÖŒÃÖ’ Ûúß ¯Ö׸ü³ÖÖÂÖÖ †Öî¸ü ‡ÃÖÛúÖ SI ´ÖÖ¡ÖÛú ×»Ö×ÜÖ‹ … ×¾ÖªãŸÖ õÖê¡Ö
?
E = 3 × 10
3
^
i N/C Ûêú ÛúÖ¸üÞÖ
×ÛúÃÖß 10 cm ³Öã•ÖÖ ¾ÖÖ»Öê ¾ÖÝÖÔ ÃÖê ÝÖã•Ö¸ü®Öê ¾ÖÖ»ÖÖ °»ÖŒÃÖ ×ÛúŸÖ®ÖÖ Æîü, •Ö²Ö×Ûú ‡ÃÖê
?
E Ûêú †×³Ö»Ö´²Ö¾ÖŸÖË ¸üÜÖÖ ÝÖµÖÖ Æîü … 2
Define the term ‘electric flux’. Write its SI units. What is the flux due to electric field
?
E = 3 × 10
3
^
i N/C through a square of side 10 cm, when it is held normal to
?
E ?
7. ×“Ö¡Ö ´Öë ¤ü¿ÖÖÔ‹ ÝÖ‹ ¯ÖÏןָüÖê¬ÖÛúÖë Ûêú ®Öê™ü¾ÖÛÔú «üÖ¸üÖ ²Öî™ü¸üß ÃÖê »Öß ÝÖµÖß ¬ÖÖ¸üÖ ¯Ö׸üÛú×»ÖŸÖ Ûúßו֋ … 2
A
1 O
B
D
C
5 O 4 O
2 O
2 O
4 V
Calculate the current drawn from the battery by the network of resistors shown in the
figure.
A
1 O
B
D
C
5 O 4 O
2 O
2 O
4 V
Page 4
55/1 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-†ú ´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, ¯ÖÏŸµÖêÛ 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/C
55/1
• 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 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 3 [P.T.O.
ÜÖÞ›ü – †
Section – A
1. ×ÛúÃÖß “ÖÖ»ÖÛú Ûêú ÃÖ´Ö×¾Ö³Ö¾Ö ¯Öéšü Ûêú ×ÛúÃÖß ×²Ö®¤ãü ¯Ö¸ü ×¾ÖªãŸÖ õÖê¡Ö ¸êüÜÖÖ‹Ñ ¯Öéšü Ûêú »Ö´²Ö¾ÖŸÖ ŒµÖÖë ÆüÖêŸÖß Æïü ? 1
Why are electric field lines perpendicular at a point on an equipotential surface of a
conductor ?
2. ×ÛúÃÖß ÃÖÓ¬ÖÖ׸ü¡Ö ÛúÖê ×ÛúÃÖß ¯Ö׸ü¾ÖŸÖá †Ö¾Öé×¢Ö Ûêú AC ÄÖÖêŸÖ ÃÖê ÃÖÓµÖÖê×•ÖŸÖ ×ÛúµÖÖ ÝÖµÖÖ Æîü … µÖפü AC ÄÖÖêŸÖ Ûúß †Ö¾Öé×¢Ö
‘Ö™üÖ ¤üß •ÖÖ‹, ŸÖÖê ŒµÖÖ ×¾ÖãÖÖ¯Ö®Ö ¬ÖÖ¸üÖ ¯Ö׸ü¾ÖÙŸÖŸÖ ÆüÖê •ÖÖ‹ÝÖß ? 1
A variable frequency AC source is connected to a capacitor. Will the displacement
current change if the frequency of the AC source is decreased ?
3. NAND ÝÖê™ü ÛúÖ ŸÖÛÔú ¯ÖÏŸÖßÛú ÜÖàד֋ †Öî¸ü ‡ÃÖÛúß ÃÖŸµÖ´ÖÖ®Ö ÃÖÖ¸üÞÖß ¤üßו֋ … 1
Draw the logic symbol of NAND gate and give its Truth Table.
4. AC ÄÖÖêŸÖ Ûúß †Ö¾Öé×¢Ö ´Öë ¯Ö׸ü¾ÖŸÖÔ®Ö Ûêú ÃÖÖ£Ö ÃÖÓ¬ÖÖ׸ü¡Ö ¯ÖÏן֑ÖÖŸÖ ´Öë ×¾Ö“Ö¸üÞÖ ÛúÖê ¤ü¿ÖÖÔ®Öê Ûêú ×»Ö‹ ÝÖÏÖ±ú ÜÖàד֋ … 1
Plot a graph showing variation of capacitive reactance with the change in the frequency
of the AC source.
5. †ÖµÖÖ´Ö ´ÖÖò›ãü»Ö®Ö †Öî¸ü †Ö¾Öé×¢Ö ´ÖÖò›ãü»Ö®Ö Ûêú ²Öß“Ö ×¾Ö³Öê¤ü®Ö Ûúßו֋ … 1
Distinguish between amplitude modulation and frequency modulation.
ÜÖÞ›ü – ²Ö
Section – B
6. ‘×¾ÖªãŸÖ °»ÖŒÃÖ’ Ûúß ¯Ö׸ü³ÖÖÂÖÖ †Öî¸ü ‡ÃÖÛúÖ SI ´ÖÖ¡ÖÛú ×»Ö×ÜÖ‹ … ×¾ÖªãŸÖ õÖê¡Ö
?
E = 3 × 10
3
^
i N/C Ûêú ÛúÖ¸üÞÖ
×ÛúÃÖß 10 cm ³Öã•ÖÖ ¾ÖÖ»Öê ¾ÖÝÖÔ ÃÖê ÝÖã•Ö¸ü®Öê ¾ÖÖ»ÖÖ °»ÖŒÃÖ ×ÛúŸÖ®ÖÖ Æîü, •Ö²Ö×Ûú ‡ÃÖê
?
E Ûêú †×³Ö»Ö´²Ö¾ÖŸÖË ¸üÜÖÖ ÝÖµÖÖ Æîü … 2
Define the term ‘electric flux’. Write its SI units. What is the flux due to electric field
?
E = 3 × 10
3
^
i N/C through a square of side 10 cm, when it is held normal to
?
E ?
7. ×“Ö¡Ö ´Öë ¤ü¿ÖÖÔ‹ ÝÖ‹ ¯ÖÏןָüÖê¬ÖÛúÖë Ûêú ®Öê™ü¾ÖÛÔú «üÖ¸üÖ ²Öî™ü¸üß ÃÖê »Öß ÝÖµÖß ¬ÖÖ¸üÖ ¯Ö׸üÛú×»ÖŸÖ Ûúßו֋ … 2
A
1 O
B
D
C
5 O 4 O
2 O
2 O
4 V
Calculate the current drawn from the battery by the network of resistors shown in the
figure.
A
1 O
B
D
C
5 O 4 O
2 O
2 O
4 V
55/1 4
8. 20 cm ³Öã•ÖÖ ¾ÖÖ»Öê ×ÛúÃÖß ¾ÖÝÖÖÔÛúÖ¸ü »Öæ¯Ö וÖÃÖÃÖê 1A ¬ÖÖ¸üÖ ¯ÖϾÖÖ×ÆüŸÖ ÆüÖê ¸üÆüß Æîü, ÛúÖê ×ÛúÃÖß †®Ö®ŸÖ »Ö´²ÖÖ‡Ô Ûêú ÃÖ߬Öê
ŸÖÖ¸ü וÖÃÖÃÖê 2A ¬ÖÖ¸üÖ ¯ÖϾÖÖ×ÆüŸÖ ÆüÖê ¸üÆüß Æîü Ûêú ×®ÖÛú™ü ×“Ö¡Ö ´Öë ¤ü¿ÖÖÔ‹ †®ÖãÃÖÖ¸ü ÃÖ´ÖÖ®Ö ŸÖ»Ö ´Öë ¸üÜÖÖ ÝÖµÖÖ Æîü … 2
¬ÖÖ¸üÖ¾ÖÖÆüß “ÖÖ»ÖÛú Ûêú ÛúÖ¸üÞÖ »Öæ¯Ö ¯Ö¸ü †Ö¸üÖê×¯ÖŸÖ ®Öê™ü ²Ö»Ö ÛúÖ ¯Ö׸ü´ÖÖÞÖ †Öî¸ü פü¿ÖÖ ¯Ö׸üÛú×»ÖŸÖ Ûúßו֋ …
†£Ö¾ÖÖ
200 ±êú¸üÖë †Öî¸ü 100 cm
2
õÖê¡Ö±ú»Ö Ûúß ×ÛúÃÖß ¾ÖÝÖÖÔÛúÖ¸ü ÃÖ´ÖŸÖ»Ö ÛãúÞ›ü»Öß ÃÖê 5A †¯Ö׸ü¾ÖŸÖá ¬ÖÖ¸üÖ ¯ÖϾÖÖ×ÆüŸÖ ÆüÖê ¸üÆüß
Æîü … µÖÆü ÛãúÞ›ü»Öß 0.2 T Ûêú ‹êÃÖê ‹ÛúÃÖ´ÖÖ®Ö “Öã´²ÖÛúßµÖ õÖê¡Ö ´Öë ×Ã£ÖŸÖ Æîü, וÖÃÖÛúß ×¤ü¿ÖÖ ÛãúÞ›ü»Öß Ûêú ŸÖ»Ö Ûêú
»Ö´²Ö¾ÖŸÖ Æîü … •Ö²Ö ‡ÃÖ ÛãúÞ›ü»Öß ÛúÖ ŸÖ»Ö “Öã´²ÖÛúßµÖ õÖê¡Ö ÃÖê 60° ÛúÖ ÛúÖêÞÖ ²Ö®ÖÖŸÖÖ Æîü ŸÖ²Ö ˆÃÖ ×ãÖ×ŸÖ ´Öë ÛãúÞ›ü»Öß
¯Ö¸ü »ÖÝÖÖ ²Ö»Ö-†Ö‘ÖæÞÖÔ ¯Ö׸üÛú×»ÖŸÖ Ûúßו֋ … ×ÛúÃÖ ×¾Ö®µÖÖÃÖ ´Öë µÖÆü ÛãúÞ›ü»Öß Ã£ÖÖµÖß ÃÖÖ´µÖÖ¾ÖãÖÖ ´Öë ÆüÖêÝÖß ?
A square loop of side 20 cm carrying current of 1A is kept near an infinite long
straight wire carrying a current of 2A in the same plane as shown in the figure.
Calculate the magnitude and direction of the net force exerted on the loop due to the
current carrying conductor.
OR
A square shaped plane coil of area 100 cm
2
of 200 turns carries a steady current of 5A.
It is placed in a uniform magnetic field of 0.2 T acting perpendicular to the plane of
the coil. Calculate the torque on the coil when its plane makes an angle of 60° with the
direction of the field. In which orientation will the coil be in stable equilibrium ?
9. ˆ®Ö ¾ÖîªãŸÖ “Öã´²ÖÛúßµÖ ×¾Ö×Ûú¸üÞÖÖë ÛúÖ ®ÖÖ´Ö ×»Ö×ÜÖ‹ (i) ו֮ÖÛúÖ ˆ¯ÖµÖÖêÝÖ ÛïúÃÖ¸ü Ûúß ÛúÖê׿ÖÛúÖ†Öë ÛúÖê ®Ö™ü Ûú¸ü®Öê ´Öë
×ÛúµÖÖ •ÖÖŸÖÖ Æîü, (ii) ו֮ÖÃÖê Ÿ¾Ö“ÖÖ ŸÖÖ´ÖÏ ¸ÓüÝÖ Ûúß ÆüÖê •ÖÖŸÖß Æîü, (iii) ¯Ö飾Öß Ûúß ˆÂÞÖŸÖÖ ²Ö®ÖÖ‹ ¸üÜÖŸÖê Æïü … 2
‡®Ö´Öë ÃÖê ×ÛúÃÖß ‹Ûú ¯ÖÏÛúÖ¸ü Ûúß ŸÖ¸ÓüÝÖÖë ÛúÖê ˆŸ¯Ö®®Ö Ûú¸ü®Öê Ûúß ×¾Ö×¬Ö ÛúÖ ÃÖÓõÖê¯Ö ´Öë ¾ÖÞÖÔ®Ö Ûúßו֋ …
Name the types of e.m. radiations which (i) are used in destroying cancer cells,
(ii) cause tanning of the skin and (iii) maintain the earth’s warmth.
Write briefly a method of producing any one of these waves.
Page 5
55/1 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-†ú ´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, ¯ÖÏŸµÖêÛ 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/C
55/1
• 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 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 3 [P.T.O.
ÜÖÞ›ü – †
Section – A
1. ×ÛúÃÖß “ÖÖ»ÖÛú Ûêú ÃÖ´Ö×¾Ö³Ö¾Ö ¯Öéšü Ûêú ×ÛúÃÖß ×²Ö®¤ãü ¯Ö¸ü ×¾ÖªãŸÖ õÖê¡Ö ¸êüÜÖÖ‹Ñ ¯Öéšü Ûêú »Ö´²Ö¾ÖŸÖ ŒµÖÖë ÆüÖêŸÖß Æïü ? 1
Why are electric field lines perpendicular at a point on an equipotential surface of a
conductor ?
2. ×ÛúÃÖß ÃÖÓ¬ÖÖ׸ü¡Ö ÛúÖê ×ÛúÃÖß ¯Ö׸ü¾ÖŸÖá †Ö¾Öé×¢Ö Ûêú AC ÄÖÖêŸÖ ÃÖê ÃÖÓµÖÖê×•ÖŸÖ ×ÛúµÖÖ ÝÖµÖÖ Æîü … µÖפü AC ÄÖÖêŸÖ Ûúß †Ö¾Öé×¢Ö
‘Ö™üÖ ¤üß •ÖÖ‹, ŸÖÖê ŒµÖÖ ×¾ÖãÖÖ¯Ö®Ö ¬ÖÖ¸üÖ ¯Ö׸ü¾ÖÙŸÖŸÖ ÆüÖê •ÖÖ‹ÝÖß ? 1
A variable frequency AC source is connected to a capacitor. Will the displacement
current change if the frequency of the AC source is decreased ?
3. NAND ÝÖê™ü ÛúÖ ŸÖÛÔú ¯ÖÏŸÖßÛú ÜÖàד֋ †Öî¸ü ‡ÃÖÛúß ÃÖŸµÖ´ÖÖ®Ö ÃÖÖ¸üÞÖß ¤üßו֋ … 1
Draw the logic symbol of NAND gate and give its Truth Table.
4. AC ÄÖÖêŸÖ Ûúß †Ö¾Öé×¢Ö ´Öë ¯Ö׸ü¾ÖŸÖÔ®Ö Ûêú ÃÖÖ£Ö ÃÖÓ¬ÖÖ׸ü¡Ö ¯ÖÏן֑ÖÖŸÖ ´Öë ×¾Ö“Ö¸üÞÖ ÛúÖê ¤ü¿ÖÖÔ®Öê Ûêú ×»Ö‹ ÝÖÏÖ±ú ÜÖàד֋ … 1
Plot a graph showing variation of capacitive reactance with the change in the frequency
of the AC source.
5. †ÖµÖÖ´Ö ´ÖÖò›ãü»Ö®Ö †Öî¸ü †Ö¾Öé×¢Ö ´ÖÖò›ãü»Ö®Ö Ûêú ²Öß“Ö ×¾Ö³Öê¤ü®Ö Ûúßו֋ … 1
Distinguish between amplitude modulation and frequency modulation.
ÜÖÞ›ü – ²Ö
Section – B
6. ‘×¾ÖªãŸÖ °»ÖŒÃÖ’ Ûúß ¯Ö׸ü³ÖÖÂÖÖ †Öî¸ü ‡ÃÖÛúÖ SI ´ÖÖ¡ÖÛú ×»Ö×ÜÖ‹ … ×¾ÖªãŸÖ õÖê¡Ö
?
E = 3 × 10
3
^
i N/C Ûêú ÛúÖ¸üÞÖ
×ÛúÃÖß 10 cm ³Öã•ÖÖ ¾ÖÖ»Öê ¾ÖÝÖÔ ÃÖê ÝÖã•Ö¸ü®Öê ¾ÖÖ»ÖÖ °»ÖŒÃÖ ×ÛúŸÖ®ÖÖ Æîü, •Ö²Ö×Ûú ‡ÃÖê
?
E Ûêú †×³Ö»Ö´²Ö¾ÖŸÖË ¸üÜÖÖ ÝÖµÖÖ Æîü … 2
Define the term ‘electric flux’. Write its SI units. What is the flux due to electric field
?
E = 3 × 10
3
^
i N/C through a square of side 10 cm, when it is held normal to
?
E ?
7. ×“Ö¡Ö ´Öë ¤ü¿ÖÖÔ‹ ÝÖ‹ ¯ÖÏןָüÖê¬ÖÛúÖë Ûêú ®Öê™ü¾ÖÛÔú «üÖ¸üÖ ²Öî™ü¸üß ÃÖê »Öß ÝÖµÖß ¬ÖÖ¸üÖ ¯Ö׸üÛú×»ÖŸÖ Ûúßו֋ … 2
A
1 O
B
D
C
5 O 4 O
2 O
2 O
4 V
Calculate the current drawn from the battery by the network of resistors shown in the
figure.
A
1 O
B
D
C
5 O 4 O
2 O
2 O
4 V
55/1 4
8. 20 cm ³Öã•ÖÖ ¾ÖÖ»Öê ×ÛúÃÖß ¾ÖÝÖÖÔÛúÖ¸ü »Öæ¯Ö וÖÃÖÃÖê 1A ¬ÖÖ¸üÖ ¯ÖϾÖÖ×ÆüŸÖ ÆüÖê ¸üÆüß Æîü, ÛúÖê ×ÛúÃÖß †®Ö®ŸÖ »Ö´²ÖÖ‡Ô Ûêú ÃÖ߬Öê
ŸÖÖ¸ü וÖÃÖÃÖê 2A ¬ÖÖ¸üÖ ¯ÖϾÖÖ×ÆüŸÖ ÆüÖê ¸üÆüß Æîü Ûêú ×®ÖÛú™ü ×“Ö¡Ö ´Öë ¤ü¿ÖÖÔ‹ †®ÖãÃÖÖ¸ü ÃÖ´ÖÖ®Ö ŸÖ»Ö ´Öë ¸üÜÖÖ ÝÖµÖÖ Æîü … 2
¬ÖÖ¸üÖ¾ÖÖÆüß “ÖÖ»ÖÛú Ûêú ÛúÖ¸üÞÖ »Öæ¯Ö ¯Ö¸ü †Ö¸üÖê×¯ÖŸÖ ®Öê™ü ²Ö»Ö ÛúÖ ¯Ö׸ü´ÖÖÞÖ †Öî¸ü פü¿ÖÖ ¯Ö׸üÛú×»ÖŸÖ Ûúßו֋ …
†£Ö¾ÖÖ
200 ±êú¸üÖë †Öî¸ü 100 cm
2
õÖê¡Ö±ú»Ö Ûúß ×ÛúÃÖß ¾ÖÝÖÖÔÛúÖ¸ü ÃÖ´ÖŸÖ»Ö ÛãúÞ›ü»Öß ÃÖê 5A †¯Ö׸ü¾ÖŸÖá ¬ÖÖ¸üÖ ¯ÖϾÖÖ×ÆüŸÖ ÆüÖê ¸üÆüß
Æîü … µÖÆü ÛãúÞ›ü»Öß 0.2 T Ûêú ‹êÃÖê ‹ÛúÃÖ´ÖÖ®Ö “Öã´²ÖÛúßµÖ õÖê¡Ö ´Öë ×Ã£ÖŸÖ Æîü, וÖÃÖÛúß ×¤ü¿ÖÖ ÛãúÞ›ü»Öß Ûêú ŸÖ»Ö Ûêú
»Ö´²Ö¾ÖŸÖ Æîü … •Ö²Ö ‡ÃÖ ÛãúÞ›ü»Öß ÛúÖ ŸÖ»Ö “Öã´²ÖÛúßµÖ õÖê¡Ö ÃÖê 60° ÛúÖ ÛúÖêÞÖ ²Ö®ÖÖŸÖÖ Æîü ŸÖ²Ö ˆÃÖ ×ãÖ×ŸÖ ´Öë ÛãúÞ›ü»Öß
¯Ö¸ü »ÖÝÖÖ ²Ö»Ö-†Ö‘ÖæÞÖÔ ¯Ö׸üÛú×»ÖŸÖ Ûúßו֋ … ×ÛúÃÖ ×¾Ö®µÖÖÃÖ ´Öë µÖÆü ÛãúÞ›ü»Öß Ã£ÖÖµÖß ÃÖÖ´µÖÖ¾ÖãÖÖ ´Öë ÆüÖêÝÖß ?
A square loop of side 20 cm carrying current of 1A is kept near an infinite long
straight wire carrying a current of 2A in the same plane as shown in the figure.
Calculate the magnitude and direction of the net force exerted on the loop due to the
current carrying conductor.
OR
A square shaped plane coil of area 100 cm
2
of 200 turns carries a steady current of 5A.
It is placed in a uniform magnetic field of 0.2 T acting perpendicular to the plane of
the coil. Calculate the torque on the coil when its plane makes an angle of 60° with the
direction of the field. In which orientation will the coil be in stable equilibrium ?
9. ˆ®Ö ¾ÖîªãŸÖ “Öã´²ÖÛúßµÖ ×¾Ö×Ûú¸üÞÖÖë ÛúÖ ®ÖÖ´Ö ×»Ö×ÜÖ‹ (i) ו֮ÖÛúÖ ˆ¯ÖµÖÖêÝÖ ÛïúÃÖ¸ü Ûúß ÛúÖê׿ÖÛúÖ†Öë ÛúÖê ®Ö™ü Ûú¸ü®Öê ´Öë
×ÛúµÖÖ •ÖÖŸÖÖ Æîü, (ii) ו֮ÖÃÖê Ÿ¾Ö“ÖÖ ŸÖÖ´ÖÏ ¸ÓüÝÖ Ûúß ÆüÖê •ÖÖŸÖß Æîü, (iii) ¯Ö飾Öß Ûúß ˆÂÞÖŸÖÖ ²Ö®ÖÖ‹ ¸üÜÖŸÖê Æïü … 2
‡®Ö´Öë ÃÖê ×ÛúÃÖß ‹Ûú ¯ÖÏÛúÖ¸ü Ûúß ŸÖ¸ÓüÝÖÖë ÛúÖê ˆŸ¯Ö®®Ö Ûú¸ü®Öê Ûúß ×¾Ö×¬Ö ÛúÖ ÃÖÓõÖê¯Ö ´Öë ¾ÖÞÖÔ®Ö Ûúßו֋ …
Name the types of e.m. radiations which (i) are used in destroying cancer cells,
(ii) cause tanning of the skin and (iii) maintain the earth’s warmth.
Write briefly a method of producing any one of these waves.
55/1 5 [P.T.O.
10. ×“Ö¡Ö ´Öë ÆüÖ‡›ÒüÖê•Ö®Ö ¯Ö¸ü´ÖÖÞÖã ÛúÖ ‰ú•ÖÖÔ ÃŸÖ¸ü †Ö¸êüÜÖ ¤ü¿ÖÖÔµÖÖ ÝÖµÖÖ Æîü : 2
(a) ¾ÖÆü ÃÖÓÛÎú´ÖÞÖ –ÖÖŸÖ Ûúßו֋ וÖÃÖ´Öë 496 nm ŸÖ¸ÓüÝÖ¤îü¬µÖÔ Ûêú ±úÖê™üÖò®Ö ÛúÖ ˆŸÃÖ•ÖÔ®Ö ÆüÖêŸÖÖ Æîü …
n = 4
n = 3
n = 2
n = 1
(b) ×ÛúÃÖ ÃÖÓÛÎú´ÖÞÖ Ûêú ÃÖÓÝÖŸÖ †×¬ÖÛúŸÖ´Ö ŸÖ¸ÓüÝÖ¤îü¬µÖÔ Ûêú ×¾Ö×Ûú¸üÞÖ ˆŸÃÖÙ•ÖŸÖ ÆüÖêŸÖê Æïü ? †¯Ö®Öê ˆ¢Ö¸ü Ûúß ¯Öã×™ü
Ûúßו֋ …
The figure shows energy level diagram of hydrogen atom.
(a) Find out the transition which results in the emission of a photon of wavelength
496 nm.
n = 4
n = 3
n = 2
n = 1
(b) Which transition corresponds to the emission of radiation of maximum
wavelength ? Justify your answer.
ÜÖÞ›ü – ÃÖ
Section – C
11. (a) ×ÛúÃÖß “ÖÖ»ÖÛú ´Öë ¯ÖϾÖÖ×ÆüŸÖ ¬ÖÖ¸üÖ ‘I’ †Öî¸ü ‡»ÖꌙÒüÖò®Ö Ûêú †¯Ö¾ÖÖÆü ¾ÖêÝÖ
?
?
d
Ûêú ²Öß“Ö ÃÖÓ²ÖÓ¬Ö ¾µÖ㟯֮®Ö Ûúßו֋ … 3
(b) ×“Ö¡Ö ´Öë ×ÛúÃÖß ŸÖÖ¸ü Ûúß †®Öã¯ÖÏÃ£Ö ÛúÖ™ü ÃÖê ¯ÖϾÖÖ×ÆüŸÖ ¬ÖÖ¸üÖ ‘I’ †Öî¸ü ÃÖ´ÖµÖ ‘t’ Ûêú ²Öß“Ö ÝÖÏÖ±ú ¤ü¿ÖÖÔµÖÖ ÝÖµÖÖ Æîü …
‡ÃÖ ÝÖÏÖ±ú ÛúÖ ˆ¯ÖµÖÖêÝÖ Ûú¸üÛêú ŸÖÖ¸ü ÃÖê 10s ´Öë ¯ÖϾÖÖ×ÆüŸÖ †Ö¾Öê¿Ö –ÖÖŸÖ Ûúßו֋ …
Read More
|
26 videos|287 docs|64 tests
|
FAQs on Past Year Paper, Physics (Set - 1), Outside Delhi, 2015, Class 12, Physics - Additional Study Material for NEET
| 1. What are the important topics to study in Class 12 Physics for the Outside Delhi exam? |  |
Ans. The important topics to study for the Class 12 Physics exam in Outside Delhi include electromagnetism, optics, modern physics, nuclear physics, and semiconductor devices. It is crucial to have a strong understanding of these topics to score well in the exam.
| 2. How can I effectively prepare for the Class 12 Physics exam in Outside Delhi? | |
Ans. To prepare effectively for the Class 12 Physics exam in Outside Delhi, it is recommended to thoroughly understand the concepts and principles of each topic. Practice solving numerical problems, solve past year papers, and take mock tests to improve your problem-solving skills. Additionally, create a study schedule and allocate dedicated time for each topic to ensure comprehensive preparation.
| 3. Can you suggest some useful resources for studying Class 12 Physics for the Outside Delhi exam? | |
Ans. Yes, there are several useful resources for studying Class 12 Physics for the Outside Delhi exam. Some recommended resources include NCERT textbooks, reference books like HC Verma's Concepts of Physics, previous year question papers, online learning platforms, and video tutorials. These resources provide comprehensive explanations, examples, and practice questions to enhance your understanding and preparation.
| 4. What are the key formulas and equations I should focus on for the Class 12 Physics exam in Outside Delhi? | |
Ans. Some key formulas and equations to focus on for the Class 12 Physics exam in Outside Delhi include Ohm's law (V = IR), Newton's laws of motion, Coulomb's law, electromagnetic induction equation (ε = -dΦ/dt), Snell's law, lens formula (1/f = 1/v - 1/u), and the equation of motion (v = u + at). It is important to understand the derivation and application of these formulas to solve problems effectively.
| 5. How can I improve my conceptual understanding of Class 12 Physics for the Outside Delhi exam? | |
Ans. To improve your conceptual understanding of Class 12 Physics for the Outside Delhi exam, focus on building a strong foundation by thoroughly studying the fundamental concepts of each topic. Practice visualizing and understanding the principles behind the phenomena discussed in the syllabus. Additionally, engage in active learning by discussing concepts with peers, solving conceptual questions, and seeking clarification from teachers or online forums when needed. Regular revision and solving challenging problems will also aid in strengthening your conceptual understanding.
About this Document
|
4.88/5 Rating |
|
Dec 22, 2024 Last updated |
Document Description: Past Year Paper, Physics (Set - 1), Outside Delhi, 2015, Class 12, Physics for NEET 2024 is part of Additional Study Material for NEET preparation.
The notes and questions for Past Year Paper, Physics (Set - 1), Outside Delhi, 2015, Class 12, Physics have been prepared according to the NEET exam syllabus. Information about Past Year Paper, Physics (Set - 1), Outside Delhi, 2015, Class 12, Physics covers topics
like and Past Year Paper, Physics (Set - 1), Outside Delhi, 2015, Class 12, Physics Example, for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Past Year Paper, Physics (Set - 1), Outside Delhi, 2015, Class 12, Physics.
Introduction of Past Year Paper, Physics (Set - 1), Outside Delhi, 2015, Class 12, Physics in English is available as part of our Additional Study Material for NEET
for NEET & Past Year Paper, Physics (Set - 1), Outside Delhi, 2015, Class 12, Physics in Hindi for Additional Study Material for NEET course.
Download more important topics related with notes, lectures and mock test series for NEET
Exam by signing up for free. NEET: Past Year Paper, Physics (Set - 1), Outside Delhi, 2015, Class 12, Physics | Additional Study Material for NEET
Description
Full syllabus notes, lecture & questions for Past Year Paper, Physics (Set - 1), Outside Delhi, 2015, Class 12, Physics | Additional Study Material for NEET - NEET | Plus excerises question with solution to help you revise complete syllabus for Additional Study Material for NEET | Best notes, free PDF download
Information about Past Year Paper, Physics (Set - 1), Outside Delhi, 2015, Class 12, Physics
In this doc you can find the meaning of Past Year Paper, Physics (Set - 1), Outside Delhi, 2015, Class 12, Physics defined & explained in the simplest way possible. Besides explaining types of
Past Year Paper, Physics (Set - 1), Outside Delhi, 2015, Class 12, Physics theory, EduRev gives you an ample number of questions to practice Past Year Paper, Physics (Set - 1), Outside Delhi, 2015, Class 12, Physics tests, examples and also practice NEET
tests
|
Explore Courses for NEET exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
Physics | Additional Study Material for NEET
,Physics | Additional Study Material for NEET
,Physics (Set - 1)
,2015
,Semester Notes
,practice quizzes
,Past Year Paper
,ppt
,Outside Delhi
,Physics | Additional Study Material for NEET
,Past Year Paper
,2015
,Extra Questions
,Physics (Set - 1)
,Class 12
,2015
,mock tests for examination
,Class 12
,shortcuts and tricks
,past year papers
,Important questions
,Outside Delhi
,MCQs
,Physics (Set - 1)
,Summary
,Past Year Paper
,Exam
,video lectures
,Sample Paper
,Class 12
,Outside Delhi
,Previous Year Questions with Solutions
,Viva Questions
,Free
,Objective type Questions
,study material
;
Additional Information about Past Year Paper, Physics (Set - 1), Outside Delhi, 2015, Class 12, Physics for NEET Preparation
Past Year Paper, Physics (Set - 1), Outside Delhi, 2015, Class 12, Physics Free PDF Download
The Past Year Paper, Physics (Set - 1), Outside Delhi, 2015, Class 12, Physics is an invaluable resource that delves deep into the core of the NEET exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the Past Year Paper, Physics (Set - 1), Outside Delhi, 2015, Class 12, Physics now and kickstart your journey towards success in the NEET exam.
Importance of Past Year Paper, Physics (Set - 1), Outside Delhi, 2015, Class 12, Physics
The importance of Past Year Paper, Physics (Set - 1), Outside Delhi, 2015, Class 12, Physics cannot be overstated, especially for NEET aspirants.
This document holds the key to success in the NEET exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
Past Year Paper, Physics (Set - 1), Outside Delhi, 2015, Class 12, Physics Notes
Past Year Paper, Physics (Set - 1), Outside Delhi, 2015, Class 12, Physics Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to Past Year Paper, Physics (Set - 1), Outside Delhi, 2015, Class 12, Physics.
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, Past Year Paper, Physics (Set - 1), Outside Delhi, 2015, Class 12, Physics Notes on EduRev are your ultimate resource for success.
Past Year Paper, Physics (Set - 1), Outside Delhi, 2015, Class 12, Physics NEET Questions
The "Past Year Paper, Physics (Set - 1), Outside Delhi, 2015, Class 12, Physics NEET Questions" guide is a valuable resource for all aspiring students preparing for the
NEET exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study Past Year Paper, Physics (Set - 1), Outside Delhi, 2015, Class 12, Physics on the App
Students of NEET can study Past Year Paper, Physics (Set - 1), Outside Delhi, 2015, Class 12, Physics alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the Past Year Paper, Physics (Set - 1), Outside Delhi, 2015, Class 12, Physics,
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of Past Year Paper, Physics (Set - 1), Outside Delhi, 2015, Class 12, Physics is prepared as per the latest NEET syllabus.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up
within 7 days!
within 7 days!
Takes less than 10 seconds to signup