NEET Exam > NEET Notes > Additional Study Material for NEET > Past Year Paper - Solutions, Physics (Set - 1), Delhi, 2014, Class 12, Physics
Past Year Paper - Solutions, Physics (Set - 1), Delhi, 2014, 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
Delhi SET I Page 1 of 20 Final Draft 12/3/2013 2: 00 pm
MARKING SCHEME
SET 55/1/1
Q. No. Expected Answer / Value Points Marks Total
Marks
1. Magnitude of the drift velocity of charge carrier per unit Electric field is
called mobility.
Alternatively, µ
=
E
v
d
| |
or
m
et
SI unit = m
2
/ (volt second) or ms
-1
N
-1
C
½
½
1
2.
Modulation Index =
c
m
a
a
= 1/2 = 0.5
½
½
1
3. If Electric field is not normal, it will have non-zero component along the
surface. In that case, work would be done in moving a charge on an
equipotential surface.
1
1
4. Glass.
In glass there is no effect of electromagnetic induction, due to presence of
Earth’s magnetic field, unlike in the case of metallic ball.
½
½
1
5.
1
1
6. 20cm 1 1
7.
Perpendicular to the plane formed by
/
[Note: Give full credit for writing the expression.]
½
½
1
8. X: Channel
It connects the Transmitter to the Receiver
½
½
1
Page 2
Delhi SET I Page 1 of 20 Final Draft 12/3/2013 2: 00 pm
MARKING SCHEME
SET 55/1/1
Q. No. Expected Answer / Value Points Marks Total
Marks
1. Magnitude of the drift velocity of charge carrier per unit Electric field is
called mobility.
Alternatively, µ
=
E
v
d
| |
or
m
et
SI unit = m
2
/ (volt second) or ms
-1
N
-1
C
½
½
1
2.
Modulation Index =
c
m
a
a
= 1/2 = 0.5
½
½
1
3. If Electric field is not normal, it will have non-zero component along the
surface. In that case, work would be done in moving a charge on an
equipotential surface.
1
1
4. Glass.
In glass there is no effect of electromagnetic induction, due to presence of
Earth’s magnetic field, unlike in the case of metallic ball.
½
½
1
5.
1
1
6. 20cm 1 1
7.
Perpendicular to the plane formed by
/
[Note: Give full credit for writing the expression.]
½
½
1
8. X: Channel
It connects the Transmitter to the Receiver
½
½
1
Delhi SET I Page 2 of 20 Final Draft 12/3/2013 2: 00 pm
9.
A: Paramagnetic
B: Diamagnetic
Susceptibility
For A: positive
For B: negative
½
½
½
½
2
10.
cos
= 5 10
10
cos0
!
50
!
=5 10
10
cos60
!
= 25 N
!
½
½
½
½
2
11.
In the first case, the overlapping of the contributions of the wavelets from
two halves of a single slit produces a minimum because corresponding
wavelets from two halves have a path difference of
2
?
.
In the second case, the overlapping of the wavefronts from the two slits
produces first maximum because these wavefronts have the path difference
of #.
(Alternatively, if a student writes the conditions given below, give full
credit.)
Condition for first minimum in single slit diffraction is , ˜ ? / a,
Whereas in case of two narrow slits separated by distance a, first maximum
occurs at angle ˜ ? / a
[Note: Award 1 mark even if the candidate attempts this question partly.]
1
1
2
Identification of magnetic material ½ + ½
Susceptibility ½ + ½
Finding flux in two cases 1+1
Explanation of the given statement 1 + 1
Page 3
Delhi SET I Page 1 of 20 Final Draft 12/3/2013 2: 00 pm
MARKING SCHEME
SET 55/1/1
Q. No. Expected Answer / Value Points Marks Total
Marks
1. Magnitude of the drift velocity of charge carrier per unit Electric field is
called mobility.
Alternatively, µ
=
E
v
d
| |
or
m
et
SI unit = m
2
/ (volt second) or ms
-1
N
-1
C
½
½
1
2.
Modulation Index =
c
m
a
a
= 1/2 = 0.5
½
½
1
3. If Electric field is not normal, it will have non-zero component along the
surface. In that case, work would be done in moving a charge on an
equipotential surface.
1
1
4. Glass.
In glass there is no effect of electromagnetic induction, due to presence of
Earth’s magnetic field, unlike in the case of metallic ball.
½
½
1
5.
1
1
6. 20cm 1 1
7.
Perpendicular to the plane formed by
/
[Note: Give full credit for writing the expression.]
½
½
1
8. X: Channel
It connects the Transmitter to the Receiver
½
½
1
Delhi SET I Page 2 of 20 Final Draft 12/3/2013 2: 00 pm
9.
A: Paramagnetic
B: Diamagnetic
Susceptibility
For A: positive
For B: negative
½
½
½
½
2
10.
cos
= 5 10
10
cos0
!
50
!
=5 10
10
cos60
!
= 25 N
!
½
½
½
½
2
11.
In the first case, the overlapping of the contributions of the wavelets from
two halves of a single slit produces a minimum because corresponding
wavelets from two halves have a path difference of
2
?
.
In the second case, the overlapping of the wavefronts from the two slits
produces first maximum because these wavefronts have the path difference
of #.
(Alternatively, if a student writes the conditions given below, give full
credit.)
Condition for first minimum in single slit diffraction is , ˜ ? / a,
Whereas in case of two narrow slits separated by distance a, first maximum
occurs at angle ˜ ? / a
[Note: Award 1 mark even if the candidate attempts this question partly.]
1
1
2
Identification of magnetic material ½ + ½
Susceptibility ½ + ½
Finding flux in two cases 1+1
Explanation of the given statement 1 + 1
Delhi SET I Page 3 of 20 Final Draft 12/3/2013 2: 00 pm
12.
Truth Table
Input Output
A B Y’ Y
0 0 0 0
0 1 1 0
1 0 1 1
1 1 1 1
Gate R: OR Gate
S: AND Gate
OR
P: NAND Gate
Q: OR Gate
Truth Table
Input Output
A B X
0 0 1
1 0 1
0 1 1
1 1 1
1
½
½
½
½
1
2
2
13.
Junction rule: At any junction, the sum of the currents entering the junction
is equal to the sum of currents leaving the junction.
Alternatively, ? i =0
½
Statements of two Laws ½ + ½
Justification ½ + ½
Truth Table 1
Names of gates used ½ + ½
Identification 1
Truth Table 1
Page 4
Delhi SET I Page 1 of 20 Final Draft 12/3/2013 2: 00 pm
MARKING SCHEME
SET 55/1/1
Q. No. Expected Answer / Value Points Marks Total
Marks
1. Magnitude of the drift velocity of charge carrier per unit Electric field is
called mobility.
Alternatively, µ
=
E
v
d
| |
or
m
et
SI unit = m
2
/ (volt second) or ms
-1
N
-1
C
½
½
1
2.
Modulation Index =
c
m
a
a
= 1/2 = 0.5
½
½
1
3. If Electric field is not normal, it will have non-zero component along the
surface. In that case, work would be done in moving a charge on an
equipotential surface.
1
1
4. Glass.
In glass there is no effect of electromagnetic induction, due to presence of
Earth’s magnetic field, unlike in the case of metallic ball.
½
½
1
5.
1
1
6. 20cm 1 1
7.
Perpendicular to the plane formed by
/
[Note: Give full credit for writing the expression.]
½
½
1
8. X: Channel
It connects the Transmitter to the Receiver
½
½
1
Delhi SET I Page 2 of 20 Final Draft 12/3/2013 2: 00 pm
9.
A: Paramagnetic
B: Diamagnetic
Susceptibility
For A: positive
For B: negative
½
½
½
½
2
10.
cos
= 5 10
10
cos0
!
50
!
=5 10
10
cos60
!
= 25 N
!
½
½
½
½
2
11.
In the first case, the overlapping of the contributions of the wavelets from
two halves of a single slit produces a minimum because corresponding
wavelets from two halves have a path difference of
2
?
.
In the second case, the overlapping of the wavefronts from the two slits
produces first maximum because these wavefronts have the path difference
of #.
(Alternatively, if a student writes the conditions given below, give full
credit.)
Condition for first minimum in single slit diffraction is , ˜ ? / a,
Whereas in case of two narrow slits separated by distance a, first maximum
occurs at angle ˜ ? / a
[Note: Award 1 mark even if the candidate attempts this question partly.]
1
1
2
Identification of magnetic material ½ + ½
Susceptibility ½ + ½
Finding flux in two cases 1+1
Explanation of the given statement 1 + 1
Delhi SET I Page 3 of 20 Final Draft 12/3/2013 2: 00 pm
12.
Truth Table
Input Output
A B Y’ Y
0 0 0 0
0 1 1 0
1 0 1 1
1 1 1 1
Gate R: OR Gate
S: AND Gate
OR
P: NAND Gate
Q: OR Gate
Truth Table
Input Output
A B X
0 0 1
1 0 1
0 1 1
1 1 1
1
½
½
½
½
1
2
2
13.
Junction rule: At any junction, the sum of the currents entering the junction
is equal to the sum of currents leaving the junction.
Alternatively, ? i =0
½
Statements of two Laws ½ + ½
Justification ½ + ½
Truth Table 1
Names of gates used ½ + ½
Identification 1
Truth Table 1
Delhi SET I Page 4 of 20 Final Draft 12/3/2013 2: 00 pm
Justification : Conservation of charge
Loop rule: The Algebraic sum of changes in the potential around any closed
loop involving resistors and cells in the loop is zero.
Alternatively, ? ? V =0 , where ? V is the changes in potential
Justification : Conservation of energy
½
½
½
2
14.
(i) Reactance of the capacitor will decrease, resulting in increase of the
current in the circuit. Therefore the bulb will glow brighter.
(ii) Increased resistance will decrease the current in the circuit, which
will decrease glow of the bulb.
[Note : Do not deduct any mark for not giving the reasons.]
1
1
2
15.
It makes use of the principle that the energy of the charged particles / ions
can be made to increase in presence of crossed Electric and magnetic fields.
A normal Magnetic field acts on the charged particle and makes them move
in a circular path .While moving from one dee to another; particle is acted
upon by the alternating electric field, and is accelerated by this field, which
increases the energy of the particle.
1
1
2
16.
tpEsin?
4v3 = pEsin60
0
v
? pE = 8
Potential energy
½
½
½
Calculation of Potential energy of the dipole 2
Underlying principle 1
Brief working 1
Effect on glow of bulb in Part (i) 1
Part (ii) 1
Page 5
Delhi SET I Page 1 of 20 Final Draft 12/3/2013 2: 00 pm
MARKING SCHEME
SET 55/1/1
Q. No. Expected Answer / Value Points Marks Total
Marks
1. Magnitude of the drift velocity of charge carrier per unit Electric field is
called mobility.
Alternatively, µ
=
E
v
d
| |
or
m
et
SI unit = m
2
/ (volt second) or ms
-1
N
-1
C
½
½
1
2.
Modulation Index =
c
m
a
a
= 1/2 = 0.5
½
½
1
3. If Electric field is not normal, it will have non-zero component along the
surface. In that case, work would be done in moving a charge on an
equipotential surface.
1
1
4. Glass.
In glass there is no effect of electromagnetic induction, due to presence of
Earth’s magnetic field, unlike in the case of metallic ball.
½
½
1
5.
1
1
6. 20cm 1 1
7.
Perpendicular to the plane formed by
/
[Note: Give full credit for writing the expression.]
½
½
1
8. X: Channel
It connects the Transmitter to the Receiver
½
½
1
Delhi SET I Page 2 of 20 Final Draft 12/3/2013 2: 00 pm
9.
A: Paramagnetic
B: Diamagnetic
Susceptibility
For A: positive
For B: negative
½
½
½
½
2
10.
cos
= 5 10
10
cos0
!
50
!
=5 10
10
cos60
!
= 25 N
!
½
½
½
½
2
11.
In the first case, the overlapping of the contributions of the wavelets from
two halves of a single slit produces a minimum because corresponding
wavelets from two halves have a path difference of
2
?
.
In the second case, the overlapping of the wavefronts from the two slits
produces first maximum because these wavefronts have the path difference
of #.
(Alternatively, if a student writes the conditions given below, give full
credit.)
Condition for first minimum in single slit diffraction is , ˜ ? / a,
Whereas in case of two narrow slits separated by distance a, first maximum
occurs at angle ˜ ? / a
[Note: Award 1 mark even if the candidate attempts this question partly.]
1
1
2
Identification of magnetic material ½ + ½
Susceptibility ½ + ½
Finding flux in two cases 1+1
Explanation of the given statement 1 + 1
Delhi SET I Page 3 of 20 Final Draft 12/3/2013 2: 00 pm
12.
Truth Table
Input Output
A B Y’ Y
0 0 0 0
0 1 1 0
1 0 1 1
1 1 1 1
Gate R: OR Gate
S: AND Gate
OR
P: NAND Gate
Q: OR Gate
Truth Table
Input Output
A B X
0 0 1
1 0 1
0 1 1
1 1 1
1
½
½
½
½
1
2
2
13.
Junction rule: At any junction, the sum of the currents entering the junction
is equal to the sum of currents leaving the junction.
Alternatively, ? i =0
½
Statements of two Laws ½ + ½
Justification ½ + ½
Truth Table 1
Names of gates used ½ + ½
Identification 1
Truth Table 1
Delhi SET I Page 4 of 20 Final Draft 12/3/2013 2: 00 pm
Justification : Conservation of charge
Loop rule: The Algebraic sum of changes in the potential around any closed
loop involving resistors and cells in the loop is zero.
Alternatively, ? ? V =0 , where ? V is the changes in potential
Justification : Conservation of energy
½
½
½
2
14.
(i) Reactance of the capacitor will decrease, resulting in increase of the
current in the circuit. Therefore the bulb will glow brighter.
(ii) Increased resistance will decrease the current in the circuit, which
will decrease glow of the bulb.
[Note : Do not deduct any mark for not giving the reasons.]
1
1
2
15.
It makes use of the principle that the energy of the charged particles / ions
can be made to increase in presence of crossed Electric and magnetic fields.
A normal Magnetic field acts on the charged particle and makes them move
in a circular path .While moving from one dee to another; particle is acted
upon by the alternating electric field, and is accelerated by this field, which
increases the energy of the particle.
1
1
2
16.
tpEsin?
4v3 = pEsin60
0
v
? pE = 8
Potential energy
½
½
½
Calculation of Potential energy of the dipole 2
Underlying principle 1
Brief working 1
Effect on glow of bulb in Part (i) 1
Part (ii) 1
Delhi SET I Page 5 of 20 Final Draft 12/3/2013 2: 00 pm
U = -pE cos
= -8 x cos 60
0
= -4J
[Give full credit to alternative methods of finding Potential energy.]
½
2
17.
(a) de Broglie wavelength is given by
?=
2
3 456
As mass of proton < mass of deuteron and q
p
= q
d
and v is same
=>
?p > ?
d
for same accelerating potential.
(b) Momentum =
2
7
? ?
p
> ?
d
? momentum of proton will be less , than that of deuteron
½
½
½
½
2
18.
(a) Power =
h, where n = no. of photons per second
2.0 x 10
-3
=
× 6.6 × 10
<
× 6 × 10
<
n =
. =
>?
@.@ ×
>?A
×@×
BA
= 0.050 x 10
17
= 5x 10
15
photons / second
[Note: Even if the student doesn’t write the formula but calculates correctly,
give full credit to this part]
(b)
½
½
Part (a) and reason ½ + ½
Part (b) and reason ½ + ½
(a) Estimation of no. of photons per second 1
(b) Plot showing the variation 1
Read More
|
26 videos|287 docs|64 tests
|
FAQs on Past Year Paper - Solutions, Physics (Set - 1), Delhi, 2014, Class 12, Physics - Additional Study Material for NEET
| 1. What is the importance of solving past year papers in preparing for the Class 12 Physics exam? |  |
Ans. Solving past year papers is important as it helps students familiarize themselves with the exam pattern, types of questions asked, and the level of difficulty. It allows them to practice time management and identify their strengths and weaknesses in different topics of the Physics syllabus. Additionally, it helps students gain confidence and reduces exam anxiety.
| 2. How can solving past year papers help in improving my physics problem-solving skills? | |
Ans. Solving past year papers requires applying concepts and solving problems from various topics of Physics. By regularly practicing these papers, students can enhance their problem-solving skills, develop a deeper understanding of concepts, and learn different approaches to tackle complex problems. It also enables students to identify common mistakes and improve their overall accuracy.
| 3. Can solving past year papers help me predict the important topics for the Class 12 Physics exam? | |
Ans. Yes, solving past year papers can provide insights into the important topics for the Class 12 Physics exam. By analyzing the frequency of questions from different topics in previous years' papers, students can identify the areas that are frequently tested and allocate their study time accordingly. However, it is important to note that the exam pattern may change, and all topics should be thoroughly covered.
| 4. How should I use the solutions of past year papers to maximize my preparation for the Class 12 Physics exam? | |
Ans. To maximize preparation, students should solve the past year papers first without referring to the solutions. After completing the paper, they can cross-check their answers with the provided solutions. If any mistakes are made, they should understand the correct approach and rectify their errors. It is important to review the solutions in detail, especially for questions that were incorrectly answered or took a significant amount of time to solve.
| 5. Are the solutions provided for past year papers in the same language as the article title? | |
Ans. Yes, the solutions provided for past year papers are in the same language as the article title, which is Delhi 2014 Class 12 Physics paper. This ensures that students can easily understand and follow the solutions while preparing for the exam. The language used in the solutions is kept at a level similar to the complexity of the exam and the text.
About this Document
|
4.2K Views |
|
4.93/5 Rating |
|
Dec 22, 2024 Last updated |
Document Description: Past Year Paper - Solutions, Physics (Set - 1), Delhi, 2014, Class 12, Physics for NEET 2024 is part of Additional Study Material for NEET preparation.
The notes and questions for Past Year Paper - Solutions, Physics (Set - 1), Delhi, 2014, Class 12, Physics have been prepared according to the NEET exam syllabus. Information about Past Year Paper - Solutions, Physics (Set - 1), Delhi, 2014, Class 12, Physics covers topics
like and Past Year Paper - Solutions, Physics (Set - 1), Delhi, 2014, Class 12, Physics Example, for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Past Year Paper - Solutions, Physics (Set - 1), Delhi, 2014, Class 12, Physics.
Introduction of Past Year Paper - Solutions, Physics (Set - 1), Delhi, 2014, Class 12, Physics in English is available as part of our Additional Study Material for NEET
for NEET & Past Year Paper - Solutions, Physics (Set - 1), Delhi, 2014, 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 - Solutions, Physics (Set - 1), Delhi, 2014, Class 12, Physics | Additional Study Material for NEET
Description
Full syllabus notes, lecture & questions for Past Year Paper - Solutions, Physics (Set - 1), Delhi, 2014, 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 - Solutions, Physics (Set - 1), Delhi, 2014, Class 12, Physics
In this doc you can find the meaning of Past Year Paper - Solutions, Physics (Set - 1), Delhi, 2014, Class 12, Physics defined & explained in the simplest way possible. Besides explaining types of
Past Year Paper - Solutions, Physics (Set - 1), Delhi, 2014, Class 12, Physics theory, EduRev gives you an ample number of questions to practice Past Year Paper - Solutions, Physics (Set - 1), Delhi, 2014, 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
video lectures
,Delhi
,Previous Year Questions with Solutions
,Semester Notes
,Important questions
,2014
,Physics | Additional Study Material for NEET
,2014
,mock tests for examination
,Past Year Paper - Solutions
,Past Year Paper - Solutions
,practice quizzes
,Free
,Summary
,Physics | Additional Study Material for NEET
,Viva Questions
,MCQs
,Delhi
,Physics (Set - 1)
,shortcuts and tricks
,ppt
,Physics (Set - 1)
,Class 12
,study material
,Exam
,Extra Questions
,Physics (Set - 1)
,Class 12
,Sample Paper
,Past Year Paper - Solutions
,2014
,past year papers
,Physics | Additional Study Material for NEET
,Class 12
,Delhi
,Objective type Questions
;
Additional Information about Past Year Paper - Solutions, Physics (Set - 1), Delhi, 2014, Class 12, Physics for NEET Preparation
Past Year Paper - Solutions, Physics (Set - 1), Delhi, 2014, Class 12, Physics Free PDF Download
The Past Year Paper - Solutions, Physics (Set - 1), Delhi, 2014, 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 - Solutions, Physics (Set - 1), Delhi, 2014, Class 12, Physics now and kickstart your journey towards success in the NEET exam.
Importance of Past Year Paper - Solutions, Physics (Set - 1), Delhi, 2014, Class 12, Physics
The importance of Past Year Paper - Solutions, Physics (Set - 1), Delhi, 2014, 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 - Solutions, Physics (Set - 1), Delhi, 2014, Class 12, Physics Notes
Past Year Paper - Solutions, Physics (Set - 1), Delhi, 2014, 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 - Solutions, Physics (Set - 1), Delhi, 2014, 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 - Solutions, Physics (Set - 1), Delhi, 2014, Class 12, Physics Notes on EduRev are your ultimate resource for success.
Past Year Paper - Solutions, Physics (Set - 1), Delhi, 2014, Class 12, Physics NEET Questions
The "Past Year Paper - Solutions, Physics (Set - 1), Delhi, 2014, 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 - Solutions, Physics (Set - 1), Delhi, 2014, Class 12, Physics on the App
Students of NEET can study Past Year Paper - Solutions, Physics (Set - 1), Delhi, 2014, 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 - Solutions, Physics (Set - 1), Delhi, 2014, 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 - Solutions, Physics (Set - 1), Delhi, 2014, Class 12, Physics is prepared as per the latest NEET syllabus.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup