EASY Marks NCERT Exemplar Problems Solutions Physics Class 11 for NEET AIPMT IIT JEE Main and Advanced ( PDFDrive

345 Pages·2016·8.43 MB·English

11
NCERT EXEMPLAR
PROBLEMS-SOLUTIONS
Physics
(Progressive Educational Publishers)
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NEW EDITION
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Note from the Publisher
National Council of Educational Research and Training (NCERT) developed
Exemplar Problems in Science and Mathematics. The prime objective is to
provide the students with number of quality problems to facilitate the concept
of learning.
Easy Marks NCERT Solutions to Exemplar Problems Physics-XI is
mainly based on the idea to present the considerable requirements of the
Exemplar Problems in a simple and detailed manner.
Salient features of the book:
● Scientific and methodological solutions to the textual questions are
provided.
● Multiple Choice Questions (MCQs) with explanation for understanding
the concept better.
● The explanation of the answers are provided with diagram, wherever
needed.
● Very Short, Short and Long Answer Type Questions are given to provide
students with more practical problems.
It has always been our endeavour to provide better quality material to the
students. If there are any suggestions for the betterment of the book, we will
certainly try to incorporate them.
CONTENTS
1. Introduction ...................................................................................... 5
2. Units and Measurement ................................................................... 6
3. Motion in a Straight Line ................................................................. 26
4. Motion in a Plane ............................................................................. 44
5. Laws of Motion ................................................................................ 74
6. Work, Energy and Power ................................................................. 96
7. System of Particles and Rotational Motion ..................................... 120
8. Gravitation ....................................................................................... 137
9. Mechanical Properties of Solids ...................................................... 156
10. Mechanical Properties of Fluids ...................................................... 174
11. Thermal Properties of Matter ........................................................... 186
12. Thermodynamics .............................................................................. 198
13. Kinetic Theory ................................................................................. 214
14. Oscillations ...................................................................................... 230
15. Waves ............................................................................................... 256
• Sample Question Paper-I ................................................................. 274
• Sample Question Paper-II ................................................................ 301
• Formulae .......................................................................................... 324
11
Introduction
A NOTE TO STUDENTS
A good number of problems have been provided in this book. Some
are easy, some are of average difficult level, some difficult and some
problems will challenge even the best amongst you. It is advised that
you first master the concepts covered in your textbook, solve the
examples and exercises provided in your textbook and then attempt to
solve the problems given in this book. There is no single prescription
which can help you in solving each and every problem in physics but
still researches in physics education show that most of the problems can
be attempted if you follow certain steps in a sequence. The following
prescription due to Dan Styer presents one such set of steps :
1. Strategy design
(a) Classify the problem by its method of solution.
(b) Summarise the situation with a diagram.
(c) Keep the goal in sight (perhaps by writing it down).
2. Execution tactics
(a) Work with symbols.
(b) Keep packets of related variables together.
(c) Be neat and organised.
(d) Keep it simple.
3. Answer checking
(a) Dimensionally consistent?
(b) Numerically reasonable (including sign)?
(c) Algebraically possible? (Example: no imaginary or infinite
answers)
(d) Functionally reasonable? (Example: greater range with
greater initial speed)
(e) Check special cases and symmetry.
(f ) Report numbers with units specified and with reasonable
significant figures.
We would like to emphasise that the problems in this book
should be used to improve the quality of teaching-learning process
of physics. Some can be directly adopted for evaluation purpose but
most of them should be suitably adapted according to the time/marks
assigned. Most of the problems included under SA and LA can be
used to generate more problems of VSA or SA categories, respectively.
qqq
5
22 Units and
Measurement
MULTIPLE CHOICE QUESTIONS-I
Q2.1. The number of significant figures in 0.06900 is:
(a) 5 (b) 4 (c) 2 (d) 3
Main concept used: In a number less than one (i.e., decimal).
The zer oes on the left of non zero number are not significant figures,
and zeroes of right side of non zero number are significant figures.
Ans. (b): 0.06900 two zeroes before six are not significant figure and
two zero on right side of 9 are significant figures. Significant figures
are underlined figures so verifies option (b).
Q2.2. The sum of the numbers 436.32, 227.2 and 0.301 in appropriate
significant figures is
(a) 663.821 (b) 664 (c) 663.8 (d) 663.82
Main concept used: In addition the result will be in least number
of places after decimal and minimum number of significant figure.
Ans. (c): On adding the given numbers result is 663.821, but in given
number the minimum number of places is one so result in significant
figure is 663.8. (upto one place of decimal after rounding off).
Q2.3. The mass and volume of a body are 4.237 g and 2.5 cm3,
respectively. The density of the material of the body in correct
significant figures is:
(a) 1.6048 g cm–3 (b) 1.69 g cm–3 (c) 1.7 g cm–3 (d) 1.695 g cm–3
Main concept used: The final result in either division or multiplication
retain as many minimum numbers of significant figures (after
rounding off) as there in the original numbers.
Ans. (c): The significant figures in given numbers 4.237 g and 2.5 cm3 are
four and two respectively so result must have only two significant figures.
mass 4.237g
Density =  , Density = 1.6948 = 1.7 g cm–3
volume 2.5cm3
rounding off upto 2 significant figures.
Q2.4. The numbers 2.745 and 2.735 on rounding off to 3 significant
figures will give.
(a) 2.75 and 2.74 (b) 2.74 and 2.73
(c) 2.75 and 2.73 (d) 2.74 and 2.74
Main concept used: (i) If the preceding digit of dropped out digit 5
is even the no change in rounding off.
(ii) If the preceding digit of dropped out digit 5 is odd then preceding
digit is increased by one.
6
Ans. (d): (i) In given number 2.745 it is round off upto 3 significant
figure, IVth digit is 5 and its preceding is even, so no change in 4 i.e.,
answer is 2.74.
(ii) Given figure 2.735 is round off upto 3 significant figure here IVth
i.e., next digit is 5 and its preceding digit is 3 (odd). So 3 is increased
by 1 and answer becomes 2.74.
Hence, verifies the option (d).
Q2.5. The length and breadth of a rectangular sheet are 16.2 cm and
10.1 cm respectively. The area of the sheet in appropriate significant
figures and error is
(a) 164 ± 3 cm2 (b) 163.62 ± 2.6 cm2
(c) 163.6 ± 2.6 cm2 (d) 163.62 ± 3 cm2
Main concept used: (i) Significant figures in the result multiplication
(or division) is the minimum number of significant figures in given
number.
Dx
(ii) If Dx is error in quantity x then relative error or error is .
x
Ans. (a): l = 16.2 cm Dl = 0.1
b = 10.1 cm Db = 0.1
l = 16.2 ± .1
b = 10.1 ± 0.1
A = Area = l  b = 16.2  10.1 = 163.62 cm2
= 164 cm2 (in significant figures)
DA Dl Db
= 
A l b
DA .1 .1 1.011.62 2.63
=   
A 16.2 10.1 16.210.1 16.210.1
DA = 2.63 cm2.
Now rounding off upto significant figures in Dl and Db i.e., one
DA = 3 cm2
\ A = (164 ± 3) cm2. Hence, verifies the option (a).
Q2.6. Which of the following pairs of physical quantities does not have
same dimensional formula?
(a) Work and torque
(b) Angular momentum and Planck’s constant
(c) Tension and surface tension
(d) Impulse and linear momentum.
Main concept used: In a formula dimensions of each term are same.
Ans. (a): Work = Force  displacement = [MLT–2][L] = [ML2T–2]
Torque = Force  distance = [MLT–2][L] = [ML2T–2]
R.H.S has same dimensions.
(b) Angular momentum L = mvr = [M][LT–1][L] = [ML2T–1]
E F.s
Planck’s constant h =  ( E = hn)
n n
Units and MeasUreMent n 7
[MLT2][L]
= [ML2T1]
[T1]
Dimensions of h and L are equal.
(c) Tension = Force = [MLT–2]
Force [MLT2]
Surface tension =  [ML0T2]
l [L]
(d) Impulse = F  t = [MLT–2][T] = [MLT–1]
Momentum = mv = [MLT–1]
R.H.S. has same dimensions.
Hence, verify the option (c).
Q2.7. Measure of two quantities along with the precision of respective
measuring instrument is A = 2.5 ms– 1 ± 0.5 ms–1 and B = 0.10 s ± 0.01 s.
The value of AB will be:
(a) (0.25 ± 0.08) m (b) (0.25 ± 0.5) m
(c) (0.25 ± 0.05) m (d) (0.25 ± 0.135) m
Main concept used: Rules of significant figure in multiplication
and addition.
Ans. (a): A = (2.5 ± 0.5) ms–1
B = (0.10 ± 0.01) s
x = AB = 2.5  0.10 = 0.25 m
Dx DA DB 0.5 0.01 0.050.025
=  =  
x A B 2.5 0.10 2.50.10
Dx 0.075
= , Dx = 0.075  0.08
x 0.25
(Rounding off upto 2 significant figures)
\ AB = (0.25 ± 0.08) m.
Hence, verifies the option (a).
Q2.8. You measure two quantities as A = (1.0 ± 0.2) m, B = 2.0 m ± 0.2 m.
We report correct value of AB as:
(a) 1.4 m ± 0.4 m (b) 1.41 m ± 0.15 m
(c) 1.4 m ± 0.3 m (d) 1.4 m ± 0.2 m
Main concept used: In significant figures of measured quantities
zeroes are included in significant figures.
Ans. (d): Quantities A and B are measured quantities so number of
significant figures in 1.0 m and 2.0 m are two.
AB = 1.02.0  2 1.414m.
rounding off upto minimum numbers of significant figure in 1.0 and
2.0 result must be in 2 significant figures
x = AB 1.4
Dx 1DA DB 10.2 0.2
=   
   
x 2 A B  21.0 2.0
8 n NCERT exeMplar probleMs physics–XI
1  1 1  2.01.0
= 0.2  0.1
   
2 1.0 2.0 1.02.0
0.3 0.31.414
Dx = x 0.2121
1.02.0 1.02.0
Dx = 0.2 m rounding off upto 1 place of decimal.
\ AB = (1.4 ± 0.2) m
Verifies the option (d).
Q2.9. Which of the following measurements is most precise?
(a) 5.00 mm (b) 5.00 cm (c) 5.00 m (d) 5.00 km
Main concept used: In these problems unit must be least and in
digits number of digits including zeroes after decimal must be zero.
Ans. (a): All the measurements are upto two places of decimal, least
unit is mm. So 5.00 mm measurement is most precise.
Hence, verifies answer (a).
Q2.10. The mean length of an object is 5 cm. Which of the following
measurements is most accurate?
(a) 4.9 cm (b) 4.805 cm (c) 5.25 cm (d) 5.4 cm
Main concept used: Absolute error (a a ) must be minimum for
i
more accuracy.
Ans. (a): Error or absolute error
Da1 = 54.9 0.1cm, Da2 = 54.805 0.195cm
Da = 55.25 0.25cm, Da = 55.4 0.4cm
3 4
Da is minimum. Hence verifies option (a).
1
Q2.11. Young’s modulus of steel is 1.9  1011 N/m2. When expressed
in CGS units of dynes/cm2. It will be equal to (1 N = 105 dynes, and
1 m2 = 104 cm2).
(a) 1.9  1010 (b) 1.9  1011 (c) 1.9  1012 (d) 1.9  1013
Ans. (c): Y = 1.9  1011 N/m2
1.91011 N 1.91011 105 dynes
or Y = 
1m2 104 cm2
Y = 1.9  1011 + 5 – 4
Y = 1.9  1012 dyne/cm2
Verifies the option (c).
Q2.12. If the momentum (P), area (A) and time (T) are taken to be
fundamental quantities, then energy has dimensional formula.
(a) [P1A–1T1] (b) [P2A1T1] (c) [P1A–1/2T1] (d) [P1A1/2T–1]
Ans. (d): Let the dimensional formula for energy in fundamental
quantities P, A and T is [Pa Ab Tc].
Dimensional formula of E = [Pa Ab Tc] dimensional formula of
momentum P = mv = [MLT–1].
Units and MeasUreMent n 9