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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

1

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

2

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): 0.06 9 0 0 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. 4.237 g mass  Density = , Density = 1.6948 = 1.7 g cm–3 volume 2.5 cm 3 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 = 16 3.62 cm2 = 164 cm2 (in significant figures) Dl Db DA  = l b A 1.01  1.62 .1 .1 2.63 DA   = 16.2 10.1 16.2  10.1 16.2  10.1 A 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]  [ML2 T 1 ] = [T 1 ] Dimensions of h and L are equal. (c) Tension = Force = [MLT–2] Force [MLT 2 ] Surface tension =   [ML0 T 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 0.5 0.01 0.05  0.025 DA DB Dx    = = A B 2.5 0.10 2.5  0.10 x 0.075 Dx = , Dx = 0.075  0.08 0.25 x (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  1.414 m. 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   2.0  1.0   1  0.2    0.1    2  1.0 2.0   1.0  2.0  0.3  1.414 0.3 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  ai ) must be minimum for more accuracy. Ans. (a): Error or absolute error Da1 = 5  4.9  0.1 cm , Da2 = 5  4.805  0.195 cm =

0.25 cm , Da4 = 5  5.4  Da3 = 5  5.25  0.4 cm Da1 is minimum. Hence verifies option (a). 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  10 5 dynes  or Y = 1 m2 10 4 cm 2 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

Area A = [L2] Time T = [T1] Energy = F.s = [MLT–2L] = [ML2T–2] [ML2T–2] = [MLT–1]a[L2]b [T]c [M1L2T–2] = [Ma La + 2b T–a+c] Comparing the powers a = 1 a + 2b = 2 – a + c = – 2 1 + 2b = 2 – 1 + c = – 2 2b = 2 – 1 c = – 2 + 1 1 b = c = – 1 2 \ Dimensional formula of energy is [P1A1/2T–1]. Verifies the option (d). MULTIPLE CHOICE QUESTIONS-II Q2.13. On the basis of dimensions, decide which of the following relations for the displacement of particle undergoing simple harmonic motion is not correct.  2 t  (a) y = a sin (b) y = a sin vt  T 

\ \

a 2 t 2 t  t  sin    cos (d) y = a 2  sin  a  T T  T Main concept used: Displacement y and amplitude a has same dimensions, angle of sin and cos is dimensionless. So dimensions in both side must be equal by the principle of homogeneity. Ans. (b, c): In (a) and (d) option the dimensions of y and a in L.H.S. and R.H.S. are equal to L and angles of sin, cos are dimensionless. In option (b) angle is v.t (where v is velocity) \ dimension of v.t is [LT–1][T] = [L]. So sin vt is not dimensionless so option (b) is wrong. In option (c) in a [L] R.H.S. dimension of amplitude   [LT 1 ] which not equal to the T [T] t [T] dimension of y i.e., L and angle  [L1T] is not dimensionless. a [L] Hence, verifies the option (b) and (c). Q2.14. P, Q, R are physical quantities, having different dimensions, which of the following combinations can never be a meaningful quantity? PR – Q 2 (P  Q) (R + Q) PQ (a)  (b) (PQ – R) (c) (d) (e) R R P R Main concept used: Addition or subtraction may be possible of the same physical quantity. But multiplication can be possible for different physical quantities or different dimensions. After multiplication or (c) y =

10 n NCERT Exemplar Problems Physics–XI

division of two quantities their dimension may be equal to IIIrd and can be added or subtracted. Ans. (a, e): In option (a) and (e) there is term (P – Q) and (R + Q) as different physical quantities can never be added or subtracted so option (a) and (e) can never be meaningful. In option (b), the dimension of PQ may be equal to dimension of R so option (b) can be possible. Similarly dimensions of PR and Q2 may be equal and gives the possibility of option (d). In option (c), there is no addition subtraction gives the possibilities of option (c). Hence, verifies the right option (a) and (e). Q2.15. Photon is quantum of radiation with energy E = hn, where n is frequency and h is Planck‘s constant. The dimensions of h are the same as that of: (a) Linear impulse (b) Angular impulse (c) Linear momentum (d) Angular momentum Main concept used: Dimensional formulae. Ans. (b, d):  E = hn E [ML2 T 2 ]   [ML2 T 1 ] h = 1 n [T ] dp . dt = dp Linear impulse = F.t = dt = mv = [MLT–1] dL . dt = dL = mvr Angular impulse = t.dt = dt = [MLT–1L] = [ML2T–1] Linear momentum = mv = [MLT–1] Angular momentum L = mvr = [ML2T–1]. So the dimensional formulae of h, Angular impulse and Angular momentum are same. Hence, verifies the option (b) and (d). Q2.16. If the Planck’s constant (h) and the speed of light in vacuum (c) are taken as two fundamental quantities, which one of the following can, in addition, be taken to express length, mass and time in terms of the three chosen fundamental quantities. (a) Mass of electron (me) (b) Universal gravitational constant (G) (c) Charge of electron (e) (d) Mass of proton (mp) Ans. (a, b, and d): dimension of E [ML2 T 2 ] [ML2 T 1 ]   h = n [T 1 ] s 1 c =  [LT ] t Fr 2 [ML3 T 2 ]   [M 1L3 T 2 ] G = M1 M 2 [M][M]



Units and Measurement  n 11

hc = [ML2T–1]  [LT–1] = [ML3T–2] [ML3 T 2 ] hc  [M 2 ] = G M 1L3 T 2  hc  [h1/ 2 c1/ 2 G –1/ 2 ] M = G h [ML2 T 1 ] hc [ML] = L =  -1 c G [LT ]



h G Gh   3/2  [G1/ 2 h1/ 2 c 3/ 2 ] c hc c c = [LT–1] = [G1/2 h1/2 c–3/2 T–1]



L =



3

1

T = [G1/ 2 h1/ 2 c 2 ] = [G1/2 h1/2 c–5/2] Hence, physical quantities (a, b and d) can be used to represent L, M, T in terms of the chosen fundamental quantities. Q2.17. Which of the following ratios express pressure? (a) Force/Area (b) Energy/Volume (c) Energy/Area (d) Force/Volume Main concept used: Ratio can be express as pressure (P) if the dimension of ratio is same as P. F [MLT 2 ] Ans. (a, b): Dimension of pressure P =  A [L2 ] Dimension of pressure P = [ML–1T–2] [MLT 2 ] Force  [ML1T 2 ] = dimension of P (a) Dimension of = 2 Area [L ] 2 2 E [ML T ] (b) Dimension =  [ML1T 2 ] = dimension of P 3 V [L ]

(c) Dimension of of P (d) Dimension of

E [ML2 T 2 ] [M1L0 T 2 ]  [ML1T 2 ] dimension =  A [L2 ] F [MLT 2 ] = [ML2 T 2 ]  [ML1T 2 ] dimension V [L3 ]

of P Hence, required option are (a) and (b). Q2.18. Which of the following are not unit of time? (a) Second (b) Parsec (c) Year (d) Light year Main concept used: Dimension of time is [T]. Ans. (b, d): The second, and year measures the time so their dimension is [M0L0T1] or unit of time.

12 n NCERT Exemplar Problems Physics–XI

But Parsec and light year measures the distance and dimension of distance [L] is not equal to the dimension of Time [T]. Hence, (b) and (d) are not unit of time. VERY SHORT ANSWER TYPE QUESTIONS Q2.19. Why do we have different units for same physical quantity? Ans. Same physical quantity measures the different order of same physical quantity e.g., velocity is physical quantity it measures the speed of light 108 m/s, speed of a person 100 m/s, speed of bus 101 m/s speed of a, e, p particle measured in terms of speed of light. So, we have the different units of same physical quantity. So we have different units of velocity cm/s, m/s, km/hr velocity of light. Q2.20. The radius of atom is of the order of 1 Å and radius of nucleus is of the order fermi. How many magnitudes higher is the volume of atom as compared to the volume of nucleus? Ans. Radius (R) of atom = 1 Å = 10–10 m Radius (r) of nucleus = 1 fermi = 10–15 m Ratio of volume of atom to nucleus 4 3 R 3 R3  10 10  3 5 3  ) 1015 .  (10  = 4 3 =   10 15  3 r r 3 Q2.21. Name the device used for measuring the mass of atoms and molecules. Ans. Deflection of a charge particle or ionised atom or molecule depends on the magnitude of either magnetic or electric field, mass and charge of particle by using this principle spectrograph or spectrometer measures the mass of atom and molecules. Q2.22. Express unified atomic mass unit in kg. 1 Ans. Unified atomic mass unit (amu) or (u) = the mass of one 12 carbon (C12) atom. By Avogadro’s number 6.023  1023 we know that mass of 6.023  1023 atoms of carbon 6C12 is equal to 12 gm. 12 gm So mass of one atom of 6C12 = 6.023  10 23 1 12 gm 1 amu = mass of one 6C12 atom = 12 12  6.023  10 23 = 1.67  10–24 gm 1 amu = 1.67  10–27 kg. Q2.23. A function f(q) is defined as: q2 q3 q4 f(q) = 1  q     ... 2! 3! 4! Why is it necessary for f(q) to be a dimensionless quantity? Units and Measurement  n 13

Ans. q is represented by angle which is equal to

arc so angle q is radius

dimensionless physical quantity. First term is 1 which is dimensionless, next term contain only powers of q, as q is dimensionless so their power will be dimensionless. Hence, each term in R.H.S. expression are dimensionless i.e. R.H.S is dimensionless so left hand side f(q) must be dimensionless. Q2.24. Why length, mass and time are chosen as base quantities in mechanics? Ans. The length, mass and time cannot be derived from any other physical quantity and all physical quantities of mechanic can be represented in terms of only length, mass and time. So length, mass and time are chosen as the base quantities in mechanics. SHORT ANSWER TYPE QUESTIONS Q2.25. (a) The earth-moon distance is about 60 earth radius. What will be the diameter of earth (approximately in degrees) as seen from the moon. (b) Moon is seen to be of (1/2)0 diameter from the earth. What must be the relative size of compared to the earth. (c) From parallax measurement, the sun is found to be at a distance of about 400 times of earth-moon distance. Estimate the ratio of sun-earth diameters. Re arc 1 rad. Ans. (a): q =   radius 60 R e 60

The angle from the moon to the diameter of earth 1 180 6  2q = 2.  60   6  2  3.14 (b) As moon is seen from earth (diametrically) angle = If earth is seen from moon the angle = 2° size of moon 1/2 1   \ size (diameter) of earth 2 4 1 \ Size of moon is the size (diameter) of earth. 4

14 n NCERT Exemplar Problems Physics–XI

1 2

(c) Let the distance between earth and moon rem = x m Then the distance between earth and sun rse = 400x m On the complete solar eclipse sun disappear or covered by moon completely. So the angle formed by the diameters of moon and sun on earth are equal \ qm = qs. Where angle from earth to moon = q°m. Angle from earth to sun = q°s qm = qs ⇒ Dm = Ds rem rse Dm Ds = x 400 x Ds 400 x = Dm x D  4Dm = De or Dm = e 4 Ds 4Ds \ = 400    400 De /4 De 400 Ds D \ =    s  100 4 De De Ds = 100De Q2.26. Which of the following time measuring devices is most precise? (a) A wall clock (b) A stop watch (c) A digital watch (d) An atomic clock Give reason for your answer. Ans. (d): The least count of a wall clock, stop watch, digital watch and 1 1 1 atomic clock are 1 sec, sec, sec and 13 sec . So atomic clock 10 10 100 is most precise. Q2.27. The distance of galaxy is of the order 1025 m. Calculate the order of magnitude of time taken by light to reach us from the galaxy. Ans. Distance travelled by light from galaxy to earth = 1025 m Speed of light = 3  108 m/s distance 10 25 1  sec.  1017 seconds So , Required time = 8 speed 3 3  10 10  1016  3.33  1016 seconds. Required time = 3 Q2.28. The vernier scale of travelling microscope has 50 divisions which coincide with 49 divisions of main scale divisions. If each main scale division is 0.5 mm, calculate the minimum inaccuracy in measurement of distance.



Units and Measurement  n 15

Full Marks Pvt Ltd has prepared NCERT Exemplar Problems-Solutions series with an objective to provide the students with foolproof solutions to questions for the book NCERT Exemplar Problems. NCERT has developed Exemplar Problems in Science and Mathematics, whose prime objective is to provide the students with a number of quality problems to facilitate the concept of learning. It has always been our endeavour to provide better quality material to the students.

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