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The second revolution was in Nuclear Physics. During and after the Second

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: : Editor. Hony.Idvisor : ' I r' ,:','], : : r' . 'Managing

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revolution of the conquest of space technology with the advances in computer technology.'

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Our government and the HRD minister in particular are very keen about the new lndian revolution of unification of lndia in a new way, Holding common exams, common cuniculum right from CBSE and the constant emphasis on the importance of Research are all connected to each other.

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Full length Practice Paper IITJEE 2011 10 Best Problems

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As the Hon. Minister put it, "a college is a teaching institution and university is an institution where you create knowledge".

a

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Learn Fast Nuclear Physics

,ti

We had been emphasising all these years that we have to develop the art of questioning. The teachers of ancient lndia taught the disciples by asking questions and the students were equally keen to lebrn by askingthe fundamental questions.This is the interaction between teachers and students. lt was the collective search for truth.

Ihought Provoking ProblemsSimple tlarmonic Motion & Waves

27 'i

Neither today nor in the yesteryears, a fresh graduate is ready to start research on his own 0r take part in production work. The key for the solution to the problem is inhouse training. Graduates, post-graduates and doctorate holders have to put in a lot of work with the seniors in the field helping them initially. 0nce they are ready to take on their own, the time and energy spent in training them is never regretted. Universities are certainly the place for research. This is

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one institution where no questions are asked why research is necessary. Research needs no justification. Teaching is part of research, where thestudents are subtly prodded to ask questions, to understand the present limitations of knowledge and then try to solve the internal contradictions by boldly going on their own. There is only one way that any nation can become a superpower. That is research and more research in every field agriculture to medicine,

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PHYsrcs FoR You I reanurnv 'r

r

3

ffiww-#ffiffi

2

11Exam Date1Oth

April

2011

Usefuldata:GasconstanlR:8.314 IK-lmol-l,Molecularmass:H2SO4=9g&C2HpH=46gr =94 g, 1 F = 95500 C; Atomic Numbers: H= t Li= 3, B-= 5;i= 6, N =7; O = 8, F = 9, Na =,11, P: 15, S = 16, Cl = 17, Ar= 18, K = L9, V 23, Cr= 24 Mn = 25, Fe ='Zi, = Co=27, Ni=2& Cu=29,2n=30,Ge=32,8r=35, Ag=47,I=53,Xe =54,pt=78,Hs.=g0, pb=g2H2O = 18 g, C6H5OH

Straight Objective

llpe3.

(a) (c)a is

34 mL 30 mL

(b) 10 mL (d) 50 mL

This section contains 9 multiple choice questions numbered 1 to 9. Each question has 4 choices (a), (b), k) and (d), out

of which ONLY ONE is conect.

1.

'

During the test of halogens, sodium extract is first boiled with nitric acid as to(a) decompose NaCN and NarS (b) make silver halide'insoluble (c) :rrcrease the solubility of AgNO3

For a perfectly crystalline solid Co.-. = aTB, wherc constant. If C,- is 0.{2IlK-mol at 10 K, molar entropy at 10 K is (a) 0.42llK-mol O) 0.1-1llK-mol (d) zero @) a.zllK-mol

Which of the following statement is notcorrect?

(d) ,.{issolve AgCN

2.

The system shown in the diagram is at equilibrium at 27"C and volume of the bulb is 150 mL. At this temperature, the vapour pressure of water is

(a) H2O2 oxidises Fe(II) to Fe(Itr). (b) HrO2 can be obtained by' electrolysis of dil. H2so4. (c) HrO2 reduce Mn(Vtr) to Mn(II). (d) H2O, is a weak base.:

28 millimetres of mercury..

If the bulb containsliquid water is

Fluorine does not show highest oxidation stateopposite to other halogens, because (a) it is most electronegative. (b) it has no d-orbital. (c) its atomic radius is very small. (d) F- ion is stable and isoelectronic with neon.

0.001 mol of Orn, volume of the

approijmately

Calcium imide on hydrolysis gives gas (B) which on oxidation by bleaching powder gives gas(C). Gas(C) on reaction with magnesium give compound (D) which on hydrolysis gives again gas(B).Identify (B),(C) and (D). (a) NH3, N2 Mg3N, (b) N" NH3, MgNH

(c) N, N2O5, Mg(NO3)2 (d) NH3, NO" Mg(NOj,By Momentum : JABALPUR : (076r) 2400022, NAGPUR : (071,2) 225291,1, GWALIOR : (07s7) 243L6'10.

4

pHystcs FoR you rernunnv,r I

r

7.

(Sirou)2'-anion is obtained when (a) no iixygen of a SiOf-tetrahedron is shared with another SiOf- tetrahedron (b) one oxygen of a SiOf- tetrahedron is shared with another SiOt- tetrahedron (c) two oxygen of a SiOf- tetrahedron are shared with another SiOf- tetrahedron

Statement-2 : In an adiabatic process, no heatexdrange between the system and tlre zurroundings

takes place.

.i,,'

,,r

,

Linked,Comprehension rype3 paragraphs Prr-rr Pr"r, and Prr-rr.

This section contains

(d)

three oxygen of a Siot-tetrahedron are shared

Based upon each paragraph,3 multiple choice questions have

with another SiOf-tetrahedronPlatinum-saltmethod is used for the determination of molecular weight of the organic

to be answered. Each question has 4 chokes (a), (b), k) and(d), out of which ONLY ONE is correct.

Paragraph for Question Nos. 13 to 15The pronounced change from non-metallic behaviour

(a) bases (b) acids (c) phenolic compounds (d) all are correctWhich is/are the structure of XeFa ?

as also the increase in basicity of oxides from N, R As, Sb io Bi is principally due to the increasing size of the atoms. The ionization potential indicate that it is much more difficult to pull electrons of small nitrogen atom than the larger bismuth atom. It is interesting to

(a)

(b)

note that nitrogen obtained from the decomposition of.compounds such as NHaNO2 is of lower density than the residual gas obtained from the atmosphere by removal of oxygery carbon dioxide and water.

13. \Alhich of the following oxides is most acidic

?

(a) (c)

As2O3Sb2O3

(b) Bio: (d) POr

14. Which of the following hydrides would be most

':, . r,,,rAsgertion ReaSon ltpe,r,';: r,.

:,

basic

?

This section contains 3 multiple choice questions numbered 10 to 12. Each question contains statement-l (Assertion) and

(a) PH3 (c) NHs

(b) tuH. (d) sbH3

statement-2 (Reason). Each question has 4 choices (a), (b), (c) and (d) out of which ONLY ONE is correct.

15. The residual nitrogen obtained from air after removal of oxygen, carbon dioxide and watervapour has a greater densilv than that obtained from chemical compounds because (a) it is an allotrophic modification of nitrogen. (b) it is mixed with some heavier gas. (c) it is rich in heavier isotopes of nitrogen.

(a) (b) (c) (d)

Statement-1 is True, Statement-2 is True; Statement-2' is a correct explanation for Statement-1.

Statement-l is True, Statement-2 is True; Statement-2 is not a correct explanation for Statement-1 Statement-l is True, Statement-2 is False. Statement-l is False, Statement-2 is True. is slower

10. Statement-l : Chemical reaction of H2Othan D2O.

(d) nitrogen obtained from chemical sources contains some light gas. Paragraph for Question Nos. 16 to 18An organic compound on analysis gave the followingdata.

Statement-2 : Heavier isotope(deuterium) is less reactive and bond energy of O - H bond is lesser

thanO-Dbond.11.

(i)

0.4 gmof organic compoundoncomplete combustion

gave 0.44 gm CO2 and 0.18 gm H2O.

Statement-l : Li2SOa does not form alum.Statement-2 : The size of Li* is very small and it can not show co-ordination of six H2O molecules.

(ii) 0.4 gm of the compound on analysis by Duma's method gave 11.2 ml nitrogen gas at STP. 16. % of carbon in the compound is

12. Statement-l : In an adiabatic process, entropy of the system remains constant.

(a)

(c) 60%

30%

b)

40Y'

(d) s0%

5

PHYSICS FOR YOU

I

rranUnnv',r r

L7. % of hydrogen in the comPound is

(a) 10% (b)

s%

(c)

20o/" (d) 15%This section contains

18. % of nitrogen in the organic compound is (b) 45'/" (a) 40o/"

2 questions.A

(c)

35%

(d)

30%

Each question contains statements given in two columns which have be matched. statements (A,B,C,D) Column I have to be matched statements (p,q,r,s) in Column

pqrs

Paragraph for Question Nos. 19 to 21 For a non-ideal gas, the compressibility factor (Z) is defined asZ

answers to these questions have to

t@re-L@l to in B l@@@ol with c l@@o@l ll.The D l@@o@l

=Pv* : vRT

=

Molar volume

Compressibility of an unknown gas at 600 K and 1.0 atm was found to be 1.2. Also this gas was foundto effuse 1.58 times slower than the pure methane gas

be appropriately bubbled as illustrbted in the following example. lf the correct matches,are A-p, A-s, B-q, B-r, C-p, C-q and D-s, then the correctly bubbled 4 x 4 matrix shouldbe as shown.

under identical conditions. t!. Density of the gas in the above mentioned experimental condition is

22. Match the co}umn

olumn,:I(A)(B)

Column(p)(q)

-'II

Heavy water Nuclear reactionD2

DrOColourlEss

1ai o.ra g

rr

(c)

1.02 g

L{

@) @) La7 gLaof the gas in.the given experimental

0.58 g

Lr

20. Molar volumeconditions is

(c)

(r)(s)

Low boiling pointDiatornic

(D) Elz

(a) (c)

40.8 L 59.8 L

b) 3e.2L (d\ 27.2L

23. Match the column.

21. The value of the virial coefficient"B" in the virial equation is (Ignore the higher terms from theequation during calculations) virial equation :

olumn-

I(p)

Coltifin *rII

(A) CO2G; + Cc", -+ 2CO [AHi: CO2 = -394 and

as>0

CO:

-220 kl/mol,

z=1*B*C+D+.--.-v^ "-'' i* ' ' '"""'nu,] u,1@)

respectivelyl(B) SOCl2nf -+ SO2kl,+,Ch1*,

(q) AH>AE (r)(s)

(a) 8.16 L mol-l (c) 11..76 L mol-l

(d)

7.84 L mol-l 5.44 L mol-l

(c) CO + ClzcltD) Ctrs6y-r

-r

COCl26y

AH q" (b) < q"37.

(c)

q"

(d)

q

Triangle ABC is right angle at C. Inradius and circumradius of triangle ABC with sides a, b, c

o(x>(a+b)\_

\x2a)

are r and (a)4r21

R,

then

>*r,t"(Lf9)=(b) 4R2

(a) P(x = b) (c) q38.

(b) P(x > b)

(d)

1

(c)

4R

(d)

1. -+rR

1.

o(

\ x>a )

x=

a+b\_(b) P(x > b)

(a) P(x = b) (c) o

(d)

't

This section contains 3 multiple choice questions numbered 33 to 35. Each question contains Statement-l (Assertion) and

Statement-2 (Reason). Each question has 4 choices (a), (b), (c) and (d) out of which ONLY ONE is conect.

Paragraph for Question Nos. 39 to 41 B be a function defined by y =/(x) such that / is bijective then there exists a unique function g : B -+ A such that f(x)=y e Cg)= x,YxeA and

Letf': A -r

(a) (b) (c) (d)

Statement-l is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-l. Statement-l is True, Statement-2 is True; Statement-2 is not a correct explanation for Statement-l. Statement-'l is True, Statement-2 is False. Statement-l is False, Statement-2 is True.PHYsrcs FoR

yeB. Here., g is said to be inverse of / and g = f-t : B'-+ A = llf(x), x\: [x, f(x)lef ;. ir we consider two inverse functions sin-1x and cofl r withdomain [-1, U &

[-*, -]and

rai'rge

respectively. Then

l-;,;l&

(0,

r)

-"]=r ir"

g

you I

reaRumv'r

,t

o,[',#](.)40. If

(b)

[0,

uThis

section contains 2 {uestions.

t#,']

-I

< x < 0 then sin-1 .r =

",[;,']

(a) (c)41.

cot-ltqz)"t+

"*-'[q7] ", -..r(qt)n (d) none of these=O

to in Column I have to be matched with statements (p,q,r,s) in Column ll. The

Each question contains statements A given in two columns which have B be matched. Statements (A,B,C,D) _

u

D

answers to these questions have to be appropriately bubbled

and [.] de-notes the greatest integer function then complete set of values of r is

ff real numbei

If [cot-r

- ri"-'

as illustrated in the following example. lf the correci matches A-s; B-q, B-r, C-p, C-q and D-s, then the correctly Irl.4-p: bubbled 4 x 4 matrix should be as shown.

"l

wherexisnonnegative

45. Match the column.

I12

iColumn II= (p)1

(a) (cos1, cotl) (c) (cot1, 1l

(b)

(cos1,1l

(d) none of these

(A)

312

r=0

> "rrr'(-+l

Paragraph for Question Nos.42 to 44 Inradius and circumradius of triangle ABC with sides a, b; c are r and R respectively. Incircle of the triangle touches the sides a, b, c at At, Br, C1 respectively. Similarly the incircle of triangle ArBlCt touches its sides at 42, 82, C2 respectively and so on. 42. ln MrBrC, the value of Z.B2A2C2= 5n- A (a) 3n - A(c)

(B)

54eis

remainder

divided by 7 then (q) is

3

If m:20101 then the value of(c)the expressiorl

(t) 1 -....- 1 , 1 -,log3m log2s1nm = Iogzm(s)

5

/r-c \ 43' cosf(a) (c)

8 5n+A 1.6

,o, '-'

1.6

n+A4 B,C,

(D) solutions of xp2xs = 30 is 4 )" + 3 then the value of l, is

Number of positive integral6

46. Match the column. Column(A) lcosr I = sinxcosx

f irc =) ) (b) (o)

I

Column

II

-""(f

(p)(q)

znn+f,

't"(+)

(B)

*'(fR 2R

= lsinr

I

_*,(+)(c)cos3x

zrn+&4

r( Areaof ' 44. Areaof ABC\ 'A1B,C1

-3costsin2x = cos3x

(r)

^3n znlt-r '4znn-!4

(a)

(c)

(d) none of these

0)4 2

(D)

sec2/+cosec2r

(secr+cosecr)+4=0

- 2Ji

(s)

least count of the vernier.This section contains 9 multiple choice questions numbered 47 to 55. Each question has 4 choices (a), (b), k) and (d), out of which 0NLY ONE is conect.

(a) (*)

*,(#)

(c)

(il

(d) Can't be determined

47, If. nth division of main scale coincides with (n + 1)th divisions of vernier scale. Given onemain scale division is equal to a units. Find theI rranunnv,rr

48. A long capillary tube of radius r is initially just vertically completely immersed inside a liquid . of angle of contact 0o. If the tube is slowly

lo

PHysrcs FoRyou

raised then relation between radius of curvature of meniscus(R) inside the capillary tube and displacement (h) of tube can be repiesented by

51.

Figure shows a parabolic graph between T and1

7

for a mixture of a gas undergoinganadiabatic

'

process. What is the ratio of ur^" and speed of(a)

sound in the mixfure?

(c)

(b) 3 (d) None of these 52. An irregular shaped body of rr.ass m, density o is falling with a terminal speed z' in a viscous1

(a)

(c) J,

medium of density p and viscosifr q. The viscous drag force acting on the body will be

(a)49.

mg

(d) none of these Equal volumes of two immiscible liquids ofdensities p and 2p are filled in a vessel ai shown in figure. Two small holes are p.rrrih"d at depth

(u) ,gl.r-P]

\

o./

(c) 6nqro53.

," ,r(r.*)

.3h :h and fa2

tuom the surface of lii;hter liquid.

If

t4and

are the velocities of efflux at these two

holes, then

aistJ2

Uranium ores on the earth at the present time have a composition consisting of 99.3oh of f.he isotope ,rIJ238 and.0.7o/o of the isotope ,rU235.The half lives of these isotopes are4.47 xl}eyears and 7.04 x 108years respectively. If these two isopopes were equally abundant when

'

the earth was formed, then the age of the earth is

lrake tn2 = o.7,rn( 99'3)= n.nut@) a.06 x 10e years (b) 2.03 '

\0'7 )

(c) 6.07'

10e

years

10e years

(d) none of thlse

(a)

J=2t21

b)+(d) none of these

54. If potential energy of electron revolving around nucleus in hydrogen atom is given by equation

"=-#:,

where k is.positive constant, e is

(c)50. The

T,

limbs of a manometer consist of uniform capillary tubes of radii 1.4 x 10-3 m and 7.2 x 1.0-a m. Find out the correct pressure difference if the level of the liquid (density 103 kg m-3, surface'tension 72 x '1,0a N m-1) in narrower tube stands 0.2 m above that in thebroader tube (assume angle of contact 0o).(a) 1386 Pa

the electronic charge and r is orbital radius of revolving electron. Application of Bohr,s theory to hydrogen atom in this shows that

(a) total energy of electron in nth orbit isproportional toton3.

(b) total energy of electron in nth orbit is 'proportionaln6.

(c) the velocity of electron inproportional to'n6.

nrhn

orbit is orbit is

(c)1

5170 Pa

(b) 1863 Pa (d) None of these,rr

(d) the velocity of electron inproportional toz-3.

th

2

pnvslcs FoR you rrsnunnv I

55.

One mole of

monoatomic ideal gas undergoes process AB in given P-V diagram. Then average specific heat for this Process isa

58. Statement-l : Heatingbystembased on circulation of steam are more efficient in warming a housethan those based on circulation of hot water. Statement-2 : The latent'heat of steam is high.

P

Linked ComPrehension TlPeThis section contains 3 paragraphs P5e-61, P62-6adnd Pur-ur' Based upon each paragraph,3 multiple choice questions have to be answered. Each question has 4 choices (a), (b), (c), and (d), out of which ONLY ONE is correct.

(a) (c)

2LR10

(b) 18R 10 _ 13R (d) _10

Paragraph for Question Nos. 59 to 61 A sphere of mass m and radius r is released from

9R10

rest while completely submerged in a river. The flow velocity is zrs and there is no turbulence. The specific

Assertion Reason IYPeThis section contains 3 questions numbered 56 to 58. Each question contains Statement-l (Assertion) and Statement-2 (Reason). Each question has 4 choices (a), (b), k) and (d) out

gravity of material of the sphere is 5. The force of buoyancy is equal to weight of the sphere.There is a cavity in the sphere.59. The fraction of the sphere which is empty willbe

of which ONLY 0NE is conect.

(a)

Statement -1 is True, Statement -2 is True; Statement -2 is a correct explanation for Statement -1. Statement -1 is True, Statement -2 is True; Statement -2 is not a correct explanation for Statement -1. Statement -1 is True, Statement -2 is False. Statement -'l is False, Statement -2 is True.

(a)

+ tu)9 (")?SJCJ

(d):

60. The acceleration of the sphere at the instant when velocity of the sphere is

(b)

(c) (d) A source of sound with frequency fsis fitted on a

(a) (c)61.

6r\raom 2turlrao

b)ry'@) nryo

P 2

will be

circular plank of radius 1 m which is moving with an angular velocity or' 51 and 52 are two stationary observers at 3 m and 4 m from the centre of circularplank.

3m

The graph of the velocity with time is bestrepresented byaag

(a) .. ,' UU'

(b) 56.

Statement-l :Statement-2: source.

51 and 52

will never hear

same

frequency at any instant.51

and S2are observing same moving

57. Statement-l : From a large metal sheet a small

circular piece of radius R is removed leaving a hole in the sheet now the sheet with hole and the piece are both heated to the same temperature. The piece can still exactly fit into the hole.Statement-2: The coefficient of surface expansion for both the piece and the hole is same.PHYslcS FoR You I rtanunnv 't

t

I3

Paragraph for Question Nos. 52 to

G4

The air column in a pipe closed at one end is made to vibrate in its second overtone in resonance with tuning

(a)

-+t43

,6' *fs^Po2

fork of frequency 440 Hz The speed of sound in dir is 330 m s-l and end corrections may be neglected.LetP6

denote the mean pressure at any poini in the

pipe, and APe tfe maximum amplituJe of pressure variation and AP = tMosinkr.

(d) None of these 42 Paragraph for Question Nos. 65 to 67 We have two radioactive nuclei A and B. A converts

(c)

into C after emitting two c,-particles and threeB-particles. Nucleus B converts into,C after emittingone o-particle and five B-particles. At time f = 0, nuclei of A are 4No and that of B are Ns. Half life of A (into the conversion of C) is 1 min and that of B is 2 min.

62. Find the length L of the air column.{a)7515

(b)

i**5

(c)

;-

(a) 9* (b)

tnitiatty number of nuclei of

C are zero.

63. Wavelength of air column is

(a)(c)

t'"

+"'

64, What is the amplitude of pressure variation atthe middle of the column?'

i*

(d) none of these

(c) both (a) ?nd (b) are correct (d) both (a) and (b) are wrong 66. \ /hat are number of nuclei of C when number of nudlei of A and B are equal?

65. If atomic numbers and mass numbers of A and B are 21, Zu A, and.A, respectively. Then.. (a) Z, - Zr= 6 (b) A, - Az= 4

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14

PHYSICS'FORYOU

I rranunnv,rr

(a)

2Ns

(b) 3No

69. Match the column'

(c)

9No2

(d)

5No2

Column

Ip)

Column

II

57. Ar what time rate of disintegration of are equal? (b) 5 min (a) 4 min (d) 2 min (c) 8 min

A and B

(A)

Heat current, in which heat transfer mechanism(s), is directly proportional to area of cross section

Conduction

(B)

For a small temPerature (q) Convectiondifference; heat current is directly proPortional to the temperature difference in case of

This

section contains 2 questions.

Each question contains

statements

A

B given in two columns which have (A,B,C,D) be matched.Statements c Column I have to be matched D statements (P,q,r,s) in Column ll. The answers to these questions have to be appropriately correct bubbled as illustrated in the following example'lf the A-s, B-q, B-r, C-p, C-q and D-s, then the matches are A-p, correctly bubbled 4 x 4 matrix should be as shown'

to in with

(c) The heat transfer

(r)

Radiation

mechanism(s) thatdensities

i'si

are due to difference of(D)

Fastest heat transfermechanism

(s)

KirchhoffLaw

's

58. A uniform solid cube is floating in a liquid

asCHEMISTRY

shown in the figure with part r inside the liquid some changes in the parameters are mentioned in column-I.

1. (a) 2. (b) 3.

(b)

4.18.

(d)

5'1s.

(b) (d) (b)

s,15.

(a)

s.16.

(d) 10' (d) 11' (a) 12' (a)17.

6. (a) 7. (d) 13. ($ 1a. (c)20. (c) 21. (c)

(b)

(b)

(c)

22. (A)-p, q; (B)-p; (c)-p, q,

r, s; (D)-q, r, s

23, (A)-p, q, s; (B)-p, q; (C)'r; (D)-F,qMATHEMATICS

Assuming no other changes, match thefollowing. Coltrmn(A)

24. (d) 25. (a) 26. (a) 27' (b) 28' (b) 2e. (c) 30' (a) 31. (b) 32, (b) 33. (b) 34. (d) 3s. (a) 36' (c) 37' 38. (a) 3e. (a) ao. (c) a1' (c) a2. 45. (A)-p; (B)-r;(C)-p; (D)-s 46. (A)-p, q; (B)-p, s; 1C)-p, q, r, s; (D)-FPHYSICS(b) (c)

I

Column

II

($

a3' (b) aa'

If

density of the liquid (p) increase

decreases, (B)

r will(q) decredse

If height of the cube isincreased keePing base area and densitY same,x

a7. (a) a8. (b) 4e. (c) 50. (b) 51. (c) 52' 54. (b) 55. (a) 56' (c) 57' (c) 58. (a) 5s'

(b) (a)

s3. (c) 60.(b)

willsYstery is (r)remalns same

61. (d) 62. (b) 63' (c) 6a' (c) 65. (b) 66' (c) 67, (b) 68. (A)-p; (B)-p; (C)t (D)-q 6e. (A)-p, q, r;,(B)-p, q, r;(C)-q; (D)-r

(c)

If the wholer will

accelerated upward, then

oo

(D)

If the cube is rePlacedbut lesser densitY,

bY

(s)

another cube of same size

may increase or decrease

For Paper I (Full Length); Refer to February Issue of Chemistry Today

r willPHYSICS FoR YoU I rrenunnv 't

t

1

5

S ffiss* PR

BLEMS

l.

A andBare connected eachotherby massless string and a spring, the string passes ovet a frictionless pulley as shown in figure.Twoblocks

Block B slides over the horizontal surface of

a

'

stationary block C and the block A slides along the vertical side of C, both with same uniform speed. The coefficient of friction between the surface of the blocks is 0.2. Force constant of the spring is 1960 N ml. If the mass of block A is 2 kg .ul*lut" the mass of the block B and the energy stored in

length L. A completely inelastic collision occurs in which the stone gets embedded in the object. Find (a) the speed of the combined,mass immediately after the collision with respectto an observeron the ground.

(b)

the length L of the string such that the tension

in the string becomes zero when the string becomes horizontal during the subsequent motion of the combined mass

the spring.

4.

What is the work done by gravitlz in equalising the,leygls when the vesse.ls are connected?S.'

Two identical cylindrical.vessels with their bases at the same level each contain a liquid of density p. The height of the liquid in one vessel is h1 and that in the other is h2.The area of Fither base is a.

A block is placed on the top of a inclined planeinclined at 37'with horizontal. The length of the

in tnir,,r.,ifofm circular disc of mass z and radius R is rotating in a horizontal plane about an axis passing through its centre and perpendicular to its plane with an angular. velocity or. ,Another disc of the same dimensions but of mass m/4 is placed gently on the first disc co-axially. Show that angular velocity of the system is 1ro.

plane is 5 m. The block slides down the planeand reaches the bottom. (a) Find the speed of the.block at the bottom the inclined plane is smooth. (b) Find the speed of the block at the bottom the coefficient of friction is 0.25. (Given, sin37" = 0.60 and cos37o = 0.80)3.

if if

6.

Abody is allowed to slide on an inclined frictionless track from rest position under earth's gravity. The track ends in a circular loop of radius R. Show that the minimum hbight h of the body so that it may successfully complete the loop is given by

A cart is moving along r-direction with a velocity of 4 m s-1. A person on the cart throws a stone with a velbcity 5 m 11 relative to himself. In the frame of reference of the cart the stone is thrown iny-z plane maklng an angle of 30' with vertical z-axis. At the highest point of its trajectory, the stone hits an object of equal mass hung vertically

r

= 1.9)n.

\2)

7.

Twoblocks of masses 5 kg and 2 kg are placed on a frictionless surface and connected by a spring. ;An external kick gives a velocity 14 m s-1 io the

heavier block in the direction of lighter one.Find

from branch of a tree by means of a string of

t5

PHYSTCS FOR YOU I rranunnv ',r r

(a) the velocity gained by the centre ofand

mass

while on block

A,

fn = PRa

:

0 = 0)'

(b) the separate velocities of the two blocks inthe centre of mass frame just after the kick.8.

For horizontat equitiurirlm

:f ;,**"o"n0'...(i)

T=fa=VmB{While for vertical equilibrium of A,

A thirr rpd.-o-f length L and mass M is held vertically

with one end on the floor and is allowed to fall. Find the velocity of the other end when it hitsthe floor, assuming that the end on the floor does

T=megFrom equations (i)rrmeg=

...(ii)

ind (ii), weget mAg ot *r=(*o)=2-=rc kg

not slip.9.

IuJm

0.2

For spring T = W, equation (ii), becomes

30 kg weight sitting'on his horse whips it. The horse speeds up at an average acceleration

Aboy of of 2 m

s-2.

(a) If the'b6y does not slide back, what is theforce of friction exerted by the horse on theboy?

W=*^s tn rQ 2x9.8 oT tt=43-=10-2 " k 1960=

The energy stored in the sPring,

(b) If the boy slides back during the acceleration,what is the coefficient of static friction between the horse and the boy? (Take I = 10 m s-2)

2.

Let L be the height of inclined plane as shown in figure,

"

(i)w,= (}) '

1e6o

"

(10-z;z

-

o.oe8l

lg. A particle thrown overof. a

a triangle from one end horizontal base falls on the other end of the

base after grazing the vertex. If 0 and Q are the base angle of projection, show that:

tand=141e+tanQ.

h=5sin37':5x0.5=3m(a) As the block slides down the inclined plane,

it loses PE and gains KE. Loss in PE: Gain in KEmgh=tma-

.1)

+

o= rlzgi

=,1z"zs*

=7.6T ms-l It gains

(b) As the block comes down, it loses PE.

KE and does work against friction. Loss in pB = gain in KE + work done against

friction

irririilj'iiili,ilIn this problem,

inl,Sii.iriiiiiil

or mgh=!*o'+ (prrgcos3T') x 5 1^ or 3mg = I mI)'+ 0.25 x zg x 0.8 x 5

(i)

Masses

A and B are moving with constant velocity, this is a problem of dynamic equilibrium, l.e. forces acting.on mass A (or

or 3.

a=J4g=6.26ms-1

(a) Let i,i unai be the unit vectors along x, y and z directions respectively. Given,

(ii)

B) balance each other. String and spring are weightless and no mass is involved between them, Trg.int = T.pri.g = T' (iii) Force of friction on block B, (': Rs = mBg) fn= VRn= lLmE

g.6 i) * t-l - doo* ""r, = (6 sin 30') i+ (6 cos 301AI = 1aNi + -A dstone = dstone, cart * D."rt = Qi + 3'j + e.6t; m r-1This is the absolute velocity of stone (with respectPHYslcS FORYou I

ocart=4tms'

1-r

rianulnv'tt 17

to ground). At highest point of its trajectory, the vertical component of its velocity (or) will becomezero, whereas the r and y components will rer4ain unchanged. Therefore, velocity of stone at highest

The height descended by this water is

AC=ht-n=(\:-b)So

point will

a=1+i+aj;ms-1

. be

work done by the force of gravity in equalising

the level is

or

Speed at highest point,r--=--

W=mgxAC=)pga(hr-hr)24s_1

1

=r a r=r/(+)t + (3)2 = 5 m Since, momenfum is conserved. mu = (2m)asa

The angular momentum of first disc

L=t(o=rmr-aIiy'hen another disc of

1r

Here,

oe

or ?o - a5 i= i^s-r = 2.5 m s-1 .'. Speed of combined mass just after collisionis 2.5 ms-1.

is velocity of combined mass,

*us A

and radius R is

placed co-axially, total momeXt of inertia of the combination is

in the string becomes zero at horizontal position. It implies that velocity of combined mass also becomes zero in horizontal position.

@) Tension

r,=!*,2

.I(t^)r =!*,,

As no extemal torque has been applied, angular momentum is conserved.

. r..-, , fto |*"' .'. IO=J(D Or 0)'=-::--a---tO

.

I' Z*"

4

s--

?0=z.5ms-' Applying conservation of energy, we have

6.

'

o=afi

-zgr2(e.8)

... L=4=gt=0.32m

29

Flence, length of the string is 0.32 m.4.

A

D.!

Suppose a be the velocity of the body at the highest point C. At point P, the potential energy of thebody = mgll At point C, the potential energy of the body

tTo,

tAs the total volume of water is constant, theheight lu in each vessel after interconnection be given by hp1 + h2a2= h(a1 + a2)

At point C, the kinetic energy of the body

= mg(2R)

will

2. According to law of conservation of energy Energy at point P = Energy at point C. mgh=mg(2R1+!mrP,1

=1 *oz

left vessel shown in figure drops from A to C and that in the right vessefrises from B to D. Effectively, the water in the part AC has

\z The level in the

or

h=(4:hr)

)

fas a1= a2 (given)l

:.

!*r'

= mg(h-

2Ri.o.

a2

=

zgltt

-

zn1 ...(i)

-Hence,

At point C downward force = mg + N for circular motion at point C :mg

dropped down to BD.

+N

=(+)

The mass of this volume of water is

*=rolr,-(r*\=-(!t+)1d pwslcs FoR you I reenunnv,r r

Since N cannot be negative, the velocity of the body at C must correspond to N = 0 if tire body is describe a circle i.e.

*g=+or...h-i..

Maximum force of static friction

uhin=gN o1 ,,-i^=rfilR)gR = 29 (h*i,.

,[=p,n=p,(Mg)

p,(30x10) < 60 N

For minimum velocity, eq. (i) becomes

a2^6=2g(h^- - 2R) or\L,l

-

2R)

r'=#=o'20

=[+J. ,o= +ncM=-;r+;5x14+2x05+2

(a) Using, Hereo1

m.i. +m"i"ZkgFor a particle proiected with an initial velocity a at an angle o, the equation of trajectory is

= 14 m sa, az= 0, ffir= 5 kg and m2== 10 m s-l

;. ar,, Lrvr

(b) The centre of mass reference frame is one in which centre of mass is at rest' So the velocity of the heavier block in this frame just after the

!=xta^a

#ko-ztt2sinacosal

kick isa't =

'v = xtano,lt -{) \ R)

where R = range

r

at

-

a.,tt = 14- |0 = 4 m s-loCI'I = 0

l.

and that of lighter block is

L' "

I

I

o)=az-

-

10 =

-10 m s-1

The co-ordinates of A are (hcot0, h) and range-_

i.e., inthe centre of mass frame the blocks of 5 kg and 2 kg will approach each other with speeds 4 m s-1 and 10 m s-1 respectively.\A/hen rod is held vertically, its centre of mass is at a height (LIZ) fromthe floor, so tliat the potential energy of the rod is Mg(Ll2). On releasing, the rod

OB = hcotQ + hcotQ. Substituting the co-ordinates of A in the equation of trajectory, we get

lz=hcototanolr-

|

ftcot0 + hcotQl

r h,totrt,l

falls, i.e., it rotates about the end on the floor and

tane =

tana cotdcote + coto

the potential energy is converted into rotational

kinetic energy

] Iol2, *h"te I is the moment 2

1^

or

tan0 cotO + tane cotO = tanq' cot0

of

irtertia of the rod about the lower end and co the angular velocity when it hits the floor. Thus by conservation of mechanical energy,

l+-=-- tanQ or - tanO tancr tanQ or tanQ + tan0 = tanc,tfrffi!

OO

MnL =L 2

lr'

oror

Mgi=++o2 lFo,rod I= IrUt.=

{3s./

4[': r=L]

Ifuis thefloor, then

linear velocity of the end'hifting the

XPLO RE R10 ModelTest Papers 20 YRS. (1991-2010) Solved Papers with detailed solution

DUtl'HT

S*##JffiTEXPLSHGH

o=rr=,,{i{,9.

(a) The boy does not slide back, its acceleration = acceleration of the horse.

As friction is the only horizontal force, it mustact along the acceleration.

(b)

f,= IYla = 30 x 2.0 = 60 N. If the boy slides back, the horse could notPHYSIGS FOR YOU I rtanunRv',t

exert a friction of 60 N on the boY.

t

19

FastNUCLEUS

It exists at the centre of an atom, containing entirepositive charge and almost the whole of the mass' The electrons revolve around the nucleus to form an atom. The nucleus consists of protons (+ve charge)and neutrons (no charge). A proton has positive charge, equal in magnitude to that of an electron, (1.6 x 1Q-tr C) and a mass equal to 1836 times that of an electron. A neutron has no charge and its mass is approximately

ThetL Arz = Zmr -r (a

-

Z)m,

-

M(Z,A)

This mass defect is in form of energy and is responsible

for binding the nucleons together. From Einstein'smass-energy relation,E = mc2 (c speed of =a Binding energy = Lmc2

light, n ismass)

Generally,"An iimeasured in amu units. So let uscalculate the energy equivalent to L amu. It is calculated

in ev (electro"

equal to that of the proton (f.OfZOx10-27kg)(1837 times that of an electron).

E(=iamui=@er= 931

't xt.67 xto-27

"oit, "v

r

= 1.5 x 10-"

I)

r

(g

r rot)'

^

The number of protons in a nucleus of an atom is called as the atomic number (Z) of that atom. The number of protons and neutrons (together .called nucleons) in the nucleus of an atom is called the mass number (A) of the atom. A particular set of nucleons forming an atom is called a nuclide. It is represented as 7XA. The nuclides having same number of protons (Z),

x 106 eV = 931 MeV + B. E. = am (sgr) \4ev There is another quantity which is very useful inpredicting the stability of a nucleus calledenergy per nucleon.B.E. per nucleon =as

binding

o'f") t"u.

but different number of nucleons (A) are calledisotopes. The nuclides having the same number of nucleons (A), but different number of protons(Z) are called isobars. The nuclides having the same number of neutrons (A - Z) are called isotones.MASS DEFECT AND BINDING ENERGY

From the Plot of B.E./Nucleon Vs Mass Number (A),

we Observe that:0.)

:->YF

n lah.tboc

The nucleons are bound together in a nucleus and the energy has to be supplied in order to break apart the constituents into free nucleons. The energy with which nucleons are bound together in a nucleus is called binding energy (8.E.). In order to freenucleons from a bound nucleus, this much of energy ( = B.E.) has to be supplied. It is observed that the mass of a nucleus is always less than the mass of its constituent (free) nucleons. This difference in mass is called as mass defect and is denoted as L,m. lf m, = mass of neutron and m, = mass of a Proton

501 io(

, , ''o

,,,,.;.,.r.,11M4$S,hqmfug-r,(4),,':,

,

o o

M(Z'A) = mass of bound nucleusBy:

B.E./nucleon increases on an average and reaches a maximum of about 8.7 MeV for A = 50-80. For heavier nuclei, B.E./nucleon decreases slowly as A increases. For the.heaviestnafural element U238 it drops to about 2.5 VteV. From above observation, it follows that nuclei in the region of atomic masses50-80 are most

stable.

: -:::'

Akhil Tewari,

cRAViitY, 20 D, Ballygunge Tenace, KolkataPHYSlcs FORYOU I

rtenunnv'to 21

NUCLEAR FORCES The protons and neutrons are held together by the strong attractive forces inside the nucleus. These forces ale called as nuclear forces. Propertieri of the Nuclear Force

O

o o

separation more than 10 fm.

Nuclear force is short ranged. It exists in small region_(of diamete. i0-t5 m 1 ;). = The nuclear force between two nucleons decrases rapidly as the sepafation between them increases and becomes negligible at

Nuclear forcq is rnuch stronger than electromagnetic force and gravitationalforce.

be controlled and the chain reaction in such .;J;; known as controlled chain reaction. This forms the basis of a nuclear reactor.NUCLEAR FUSION

(like heavy water), then the number of fissions

absorbed by certain substances knowri as.moderators

explosion is created. In such cajes; the chain reaction is known as uhcontrolled chain reaction. This forms the basis of atomic bomb. In a chain reaction, the fast moving neutrons arecan

gf continuous fission by itself. If the number of fissions in a given interval of time goes on increasing continuously, then a condition of

a certain size called the critical size then

it is capable

Nuclear force is independent of charge. The nrrclear force between two protons is same aS that between two neutrons or between a neutron and proton. This is known as charge independent character of nuclear iorce.

The process in which two or'more light nuclei are combined into a single nucleus with the release of tremendous amount of energy is called as nuclear the fusion (1.e. of light nuclei) is more than the sum of masses after the fusion (i.e. of bigger nucleus) and this difference appears as the fusion energy. The most typical fusion reaction is the fusion of two deuteriumfusion. Like a fission reaction, the sum of masses before

NUCLEAR REACTION

Itr nuclear reaction, sum of masses before reacti6n is difference in masses-appears in the form of energy following the law of inter-conversion of mass and energy. The energy released i" u ,"".tio., rs called as Q valqe of a reaction "".tu", and is given asgreater than the sum of masses after the reaction. The

nuclei into

helium.

.l

is An amu

follows : If difference in mass before and after the reaction

For the fusion reaction to occui the light nuclei are brought closer to each other (with a Jistance of10-14

,H2 + ,H2 -+ ,Hea + 24 MeV

followed. Total number.of protons and neutrons should also remain same on both sides of a nuclear reaction.NUCLEAR FISSION

Law of conservation of momentum is also

Am = mass of reactants minus mass of products, then Q value = Am(931) MeV

fusion reaction.

source of energy in sun and other stars is the nuclear

to counter the repulsive force between nuclei. Due to this reason, the fusion reaction is very difficult to perform. The inner core of sun is at very high temperature, and is suitable for fusion, in fact the

m). This is possible only at very high temperature

slow moving neutrons strikenuclear reaction takes place.

fission. The most typical fission reaction occurs when,rUzss. The

of tremendous amount of energy is called as nuclear

The breaking of a heavy nucleus into two or more fragments of comparable masses, with the release

following

uB^'n'* ,ukr, + 3ont +200 Mev more than one of the neutrons produced in the above fission reaction are capable of inducing a fission reaction (provided ll3i i, urruifuUiJ, tf,u., the number of fissions taking place at successi,re stages goes incre4sing at a very brisk rate and this generates a series of fission reactions. This is known as chain reaction, If mass of Ws sample greater than

lf

e2J}3s

+ont--1

reaction.

used with a25Yo efhciency in the reactor, how mairy grams of deuterium will be needed per day? (The masses of 1H2 and 2Hea are i.Ot+t umu und 4.0025 amu respectively.) Solution : Let us first calculate the value of nuclear e Q = Amc2 = An(931) MeV

Illustratiori 1 : It is proposed to use the nuclear fusion reaction, ,H, 1 ,H, -+ 2Hea in a nuclear reactor, of 200 MW rating. If the energy from above reaction is

+

e = (2 x 2.014't - 4.0026)x

931

MeV

==25

23.834 MeV = Zg.B3+

x

106

Now efficienry of reactor isSo effective energy used

eV

21o/o.

i6b-t

23,834x106 x 1.5 x 10-re ; = l.Se+ x 10-13;

22

pHyStcS FoRIOu I reenuenv,rr

Now 9.534 x2 deuterium.

1.0-13

J energy is released

by fusion of

y-Radiation

(#)

- '2

(9'534 x 10-13)

T/deuterium is released.

Requiremmt is 200 lvIW = 200 " 1ff lls

No. of deuterium nuclei required

'

8d[00 for 1 day.

These are electromagnetic waves of nuclear origih and of very short wavelength. They have no charge and no mass. They have maximum penetrating power and minimum ionising power. The energy released in a nuclear reaction is mainly emitted in the form

=-_ry*t,2

200 x 106 x 86400

= 3.624x7025

Number of deuterium nuclei =

ftx6xlOts

of 1 radiation Laws of Radioactive Decay O Rutherford-Soddy laws (Statistical Laws) O The disintegration of a radioactive substance is random and spontaneous.

3.624xt0* =Tx6x1023

O O

Radioactive decay.is puiely a nuclear phenomenon and is independent o{ anyphysical and chemical conditions.

+m=

2x3.624x1025 =t20.83 glday. 6x10ts

RADIOACTIVITY The phenomenon of spontaneous emission of radiation

or particles from the nucleus is called radioactivity. The substances which emit these radiations are called as radioactive substances. It was discovered by Henry Becquerel for atoms of radium, Later it wasdiscovered that many naturally occurring compounds of heavy elements like radium, thorium etc also emit

The radioactive decay follows first order kinetics, i.e., the rate of decay is proportional to the number of undecayed atoms in a radioactive substance at any time f . If dN bethe number of atoms (nuclei) disintegrating

in.

trme dt, the rate of decay is given as dN / dt

From first order of kinetic rate law dN :=_dt

l,N, where ), is called as decay or

radiations. At present, it is known that all the naturally occurring elements having atomic number greater than 82 are radioactive. For example some of them are; radium,

disintegration constant. Let N6 be the number of nuclei at time t = 0 and Nrbe the number of nuclei after time f, then according to integrated first order ratelaw, we have

polonium, thorium, actinium, uranium, radon etc. Later on Rutherford found that emission of radiationalways accompanied by transformation of one element

'The half

Nr = Noe-xt

(transmutation) into another' Actually radioactivityis the result of disintegration of'an unstable nucleus. Rutherford studied the nature of these radiations and found that these mainly consist of cr, p, Y rays.

period of a radioactive substance is defined as the time in which one-half of the radioactive substance is disintegrated. If N6 be the number of nuclei at f = 0, then in a half life Tyy the number of

life

'

+

)"f = ln

$ = ,'rgr log "N, N,

No

(Ty2)

a-Particles

(2Hea)

nuclei decayed will be No/2. Nt = Noe-lt

..

.(i)

These carry a charge of. +2e and mass equal to 4mr. These are nuclei of helium atoms. The energies of

a-particles vary from 5 MeV to 9 MeV and their velocities vary from 0.01-0.1 times of c (velocity of iighg. They can be deflected by electric and magnetic fields and have low penetrating Power but high ionizing power.p-Particles(-1eo)

= 2"= ry^r-lr,i,From (i) and (ii), we get

No

..

.(ii)

*=(1)"t"n = number of half lives

=(ll

These are fast moving electrons having charge equal

to -e and. mass me = 9.1 x 10-31 kg' Their velocities vary from 1"/" to 99oh of the velocity of light (c). They can also be deflected by electric and magnetic fields' They have low ionizing power but high penetrating power. p* particles are positrons.resnuenv'.ro 23

PHYSTCS FORYOU I

The mean life

(7.) of a radioactive

substance is equal

to the sum of life times of all atoms divided by the number of all atoms. It is given by

positron. p rays are electrons and p* arethe antielectrons or positrons.snr

_ _'l." -mIIllustration 2 : The mean lives of a radio active substance are'1,620 and 405 years for c-emission and p-emission respectively. Find out the time during which three fourth of a sample will decay if

+tpl

+ _reo +0(antineutrino)+ *, e0

,pt -ort

(positron) + u(neutrino)

Antineutrino and neutrino share the energyof electrons and positrons. That is the reason why the energy of p is continuous and B rays has a energy maximum.

p-emission simultaneously. Solution : When a substance decays by cr and p emission simultaneously, the average rate of disintegration l.uu is given by

it is decaying both the a-emission and

O

When a

y

particle is produced, both atomic

and mass number remain constant.

Activity of a Radioactive Isotope The activity of a radioactive substance (or radioisotope)means the rate of decay per second or the number of nuclei disintegrating per second. It is generally denoted by A.

l,ur=l,o+l"Uwhere l,o= disintegration constant for o-emission only. disintegration constant for p-emission only. Mean life is given byaB =1 t^=T

,dN A=If a time f = 0, the activity of a radioactive substance be As and after time f = f s, activity be A, thenrNo

+ .11.7=)

1,", = l,o + l,p

_=++_ T. Td=

^=l#],=,=TF

A,=|-4ry.l =- lrv. '*'f ' L dt

1'-'

h*#=3.08x10-3 Lout =2.go3toe I "25(e.oa

Unit of Activity

A' = A'e-Lt

The activity is measured in terms of Curie (Ci). 1 curie is the activity of 1 g of a freshly prepared sample of

'

ro*)r =2.303bs3.08 x 10-'

+t= 2.303x#"^1 log4= 450.17years. DSoddy Fajan Laws (Group-Displacement Laws) O When a nuclide emits one c-particle (2Hea), its mass number (A) decreases by 4 units and atomic number (Z) decreases by 2 units. ,X^ -,_r.yn-n + rHea + Energy

H

radium

Ra226

1Tr,, = 1602 years.)

1 curie = 1, C|= 3.7

x

1010

dps (disintegration per second)1010

1 dps is also known as 1 Bq (becquerel)

+

1Ci = 3.7 x

Bq

Illustration 3 : Radioisotopes of phosphorus p32 and P35 are mixed in the ratio of 2:L of atoms. Theactivity of the sample is 2 Ci. Find the activity of the sample after 30 days. 7112 of P32 = L4 days and Ty2 of P35 is 25 days. Solution: Let Ao = initial activity of sample. ,41e= initial activity of isotope 1 and ,426= initial activity of isotope 2.

O

When a nuclide emits a p-particle, its mass

number remains unchanged but atomicnumber increases by one unit.

,X^ - ,*ryA +_reo + where 6 is antineutrino.[rthe

d + Energy

4=At +4oSimilarly for final activity (Activity after time f)

nudeug due to conversion of neutron into.

protory antineutrino is produced. Ithas no ctrarge or mass/ but has momentum. When a proton is converted to a neutron, a neutron and a +ve

A,=Arr+AxNow in the given equatiory

-

A, =

Aroe-L't

*

Aroe-L't

p-particle is produced, which is calledrrenunnv,rr

as

4=2Ci +

Ao

= Aro+Aro=

2

..

..(i)

24

PHYSTCS FORYOU I

Initial ratio of atoms of isotopesFrom definition of activitY,

=2:

L

e. = ? x 0.2265 + L x 0.4354 = 0.5444 Ci.

"1516

A=l,N

Illustration

o

Am

Aro

lrNro

=&tL Nro

T1

^.2N20 where T represents half life

On solving equation (i) and (ii), we get

4o =?r25 -, Am '1. 1.4-so =25 1,4 7,

...(ii)

725 4o=G "ttd4o =tA, =Aros

count- ratq meter is used to measure the activity of a given sample. At one instant the rneter shows 4750 counts per minute. Five.minutes later it shows 2700 counts per minute. Find (a) decay constant (b) the half life of the sample. Solution: (a) Initial activity = 4 = ff at t = O dN Final activity = 4 =i at t = t4 :A

xJ

+

Arne-L't.

rq _o.b , go , + A.=2e la '-- +le '-t 76 75e-ALet y =4.7'4,Js

_o.9jls

E

,n

-

Consider the first exponential term:x _ 0.693 30

Usins )rf =2.303ton

-=-

4750 N" 2700 N,

&

=

e-1'485

-1.48s - Iogy=m +i.e.,

-

lo

Y

=

(-r.+ss'l Y= anurogl ,,m J

-

7'485

+ r(s) = 2.303 :-cm

So, frdm above calculations you can derive a general

=

)' =

2'303

, 5

lon "2700

4750

-0.1129 min-1

result

e_,

=antiloslffi)

/-r\

(b) T,,^ - 0'693 = 6.1.4 min. Lt z 0-1729.

oo

A Series of Quick Reference Books, Cover complete syllabus in

Easy

to Grasp

o

the form of po.,.n!s o A.!3ndy High Yield Facts Book Essential for all Competitive Examinations

Send DD/Mg in favour'of

MTG Book3.Available at all leading bookhops throughoutthe country.

PHYsrcs ron

v6u I rrenunnv'to

2,

'tlhoug it lProvallki tng lhlProlbllemns rtlmBy: Prof. Rajinder Singh Randhawa*

1. Ir the shown

figure

a vessel

density varies

with

filled with a liquid whose height /r from the bottom asps is a constant and h6 is

5. Find the natural frequenry of the semicircularshell of mass m and radius r which rolls without slipping as shown in figure.

, =rr(n-!)

*n*"

the height of the liquid in the container.

A solid block ofand mass

small'

dimensions of density

IL

5

po

n

is released from

5. A

the bottom of the tank. Show that the block will execute

SHM. Also find the frequency of oscillation.

If a mass M is suspended from the bottom of the rope, find the time for a transverse wave to travel the length of the rope.vertically.q

rope of mass m and length L is suspended

Apendulum clock is rnounted in an elevator whicha < g.

starts going up at constant acceleration 4, where At a height h the acceleration of the elevator

1.

Density of block is

Pbtock =

reverses

in

direction,

its

magnitude remains

constant. How soon after the start show the right time again?J.

will the

clock

(s

i

Oo,

to its volume is

\

IrPoJ

A plank of mass M is kept on rest on two identical spherical balls each of mass rn and radius R. The plank is connected to two springs which in turn

are attached to walls as shown in figure. Find the frequenry of small oscillation of the plank, assuming pure rolling of spheres on surface, andno slipping condition between plank and spheres.

^Ifrom the bottom,

I;+,,(n_ff\r_^r

The resultant force acting on the block at height h

\A/hen the block is in equilibrium position, then

'-=lffi]

4.

Find the natural freguency of vibration of thespring-mass-pulley systemas

shown in figure.

r-=o at n=b ,t2Let the block be displaced by r from mean position.

Now rsultant force on block at heightmean position.

r

from

. [+J''{-#lr--rRandhawa Institrite of Physics, S.C.O.208, First floor, Sector-36D, Chandigarh.PHYsrcs FoR You

I

rranumv'r

r

27

=-l%As ao=

(e*c\J"

{7o*=h =

E.r-

Resultant acceleratieo

(#)'

, - WIJT*"-l ''?-!;l:rotartime

LJt. "

-r)=hb-JI-"1

r

F

,z*

forSHM.

So,wehave,6-

t @-=rn or 1)=zn @ sho l \sho"=

=E*,r=ff1W111

l

The frequency of the pendulum clock, when theelevator is at rest, is given by

EIE,tr] -17lT-lConsider a small horizontal displacement r of the plank. Total energy of oscillation of the system,

The frequency when the elevator is accelerated up

+E

L

8l

'2n\I

tEi

Time gain in one oscillation

E=!kr2 *|*r*, *|mo, +zxlrro,Ic = M.I of spheres about contact pornt =

Time taken by elevator to travel a distance /r up,

L*pz

,_E "-!;'\a

No. of oscillations in time f = f x ui =rl-

' =ff =velocity of plank . = = rrr*lar velocity of spheres *The sphere is in pure rolling, therefore, aqM is half that of thg top most point. As there is no slipping between plank and spheres, the velocity of contact point is same. Thus, n=+ f;

fi1,

o

p-Type Semiconductor When a pure semiconductorof Sior Ge (tetravalent)

tr tr

The conductivity of the semiconductor is given by o = e(nelre+ nn\) ' where p, and p1 are the electron and hole mobilities and e is the electronic charge. r., ..' . The conductivity of an intrinsic semiconductor is o;= np(1t,+ 1to) The conductivity of n-type semiconductor is on= eNfl)", The conductivity of p-type semiconductor is

or=p-n JUNCTION

eNolt"y

is doped with a groirp III trivalent impurities like aluminium (Al), boron (B), indium (In) etc we obtain a p-type semiconductor. The trivalentimpurity atoms are calledas acceptor atoms.

When a p-type semiconductor crystal is brought into close contact with an n-type semiconductor

crystal, the resulting arrangementp-n junction or junction diode.

is called aa

o

The energyband structure of p-type semiconductor is as shown in the figure.

It is also called

semiconductor diode.'A p-n iunction cannot be made by simply pushing the two pieces together.

36

pxvslcs FoR You

I reanunRv 'r r

This would not lead to a single crystal strucfure. Special fabrication techniques are needed to form a p-n junction.

E Ir tr

Two important processes occur duringformation of ap-njunction: diffusion and drift. It is symbolically represented by

reverse biasing the width of the depletion region increases. The resistance of the p-n junction becomes high inreverse biasing.

the

O The p-n junction

in reverse bias can be considered to be equivalent to a capacitor with p and n regions act as the plates of the capacitor and the depletion region acts as the dielectric. Its capacitance is called depletion capacitance or transition capacitance.

The most important characteristic of a p-n junction

is its ability to conduct current in one direction only. In the other (reverse) direction it offers veryhigh resistance. The current in the junction diode is given by

J

where

k = Boltzmann

I = Io (dvrtr

-

1)

constant 10

=

reverse

saturation current, T = absolute temperature. Depletion Region In the vicinity of junction, the region containing the uncompensated acceptor and donor ions is called depletion region. There is a depletion of mobile charges (holes and free electrons) in this region. Since this region has immobile (fixed) ions which are electrically charged it is also called as the space charge region. The potential developed across the depletion layer is called barrier potential. The physicAl distance of the depletion region is called width of the depletion region. For silitortrdiode the barrier potential is about 0.7 V vFlieieai fot a germanium diode it is about0.3

B

Breakdown Voltage Ap-n junction allows a very small curient to flow when it is reverse biased. This current is due to the movement of minority charge carriers. It is almost independent of the voltage applied. However, if the reverse bias is made too higtr, the current through the p-n junction increases abruptly. The voltage at which this phenomenon occurs is called breakdown voltage. There can be two different process of breakdown. One is called zener breakdown and the other is called avalanche breakdown.

,-vCHARACTERISTICS OF A p-n JUNCTION The /-Vcharacteristics of. ap-n junctton do not obey Ohm's law. The experimental circuit arrangements

tr

for studying I-V characteristics of a p-n junctiondiode are shown in figure (i) and (ii).

v.

,i

tr The width of the depletion layer is about 10{ m. tr The width.ofthe depletion layer and the magnitude of potentiallbarrier depend on the nature of thematerial of semiconductor and the concentration of impurity,atoms. Forward Biasing ofa

p-n ]unction

When the positive terminal of extemal battery is connected to p-side and negative terminal ton-side of p-n junctiory then the p-n junction is said to be forr.vard biased.

In forward biasing, the width of the depletionregion decreases.The resistance of the p-zl junction becomes low in forward biasing. Reverse Biasing of ap-n |unction

When the positive terminal of the extemal battery is connected to n-side and the negative terminal to p-side of ap-n jvnctron, then the junction is said tobe reverse biased.

The I-V characteristics of a p-n junction areshown in figure below.PHYSICS FORYOU I

rranunRv'rr 37

t

tr In actual practice, no diode can act as an idealdiode.RECTIFIER

in reverse bias it acts as .u:r open switch.

O It is a device

which converts ac voltage to dc

voltage. Rectifiers are of two types: O Half wave rectifier O Full wave rectifier

Half Wave RectifierKnee Voltage

tr In forward

biasing, the voltage at which the

[ [r half wave rectifier one diode is used. tr The circuit diagram, input and output

current starts to increase rapidly is called cut-in qr knee voltage. For germaniurn diode it is about 0.3 V whereas for silicon diode it is 0.7 V.RESISTANCE OF A DIODE

voltage waveforms for a half wave rectifier are as shown in the following figure.

tr

Resistance of the diode is defined in two different

Output voltage

ways.

O

dc or static tesistance: It is defined as the ratio

of the voltage applied across the diode to the corresponding current flowing through it. It is given by

v rdc=7

o

It is defined as the ratio of small change in voltage across the diode to the corresponding small change inac or dynamic resistance :

={J

d. to

Time

current. It is given by

- _AV '""point.IDEAL DIODE

-T

Peak value of current is

It is equal to the reciprocal of the slope of thectraracteristic (1-V characteristics curve) at that

r-vo 'o-rf+&ry is the forward diode re'siiitance, R1 is the load resistance and Ve is the peak value of the input voltage. Average or dc value of current is '

where

tr

A diode permits only unidirectional conduction. It conducts well in the forward direction and poorly in the reverse direction. It would have 6eei ideal if a diode acts as a perfect conductor (with zero voltage across it) when it is forward biased, and as a perfect insulator (with no current flows through

'd.

, --Io tr

rms value of current is

, -T '^t -IoPeak inverse voltage is

tr

it) when it is reverse biased. The I-V characteristics qf an ideal diodein figure below.

as

shown

pIV = Vo Average or dc value of voltage is

Vd.=Id"R.=*R.It is also given

by rr Var=2IE

(Assuming diode

forward resistance is zero)

O O38

An ideal diode acts like an automatic switch,In forward bias, it acts as a closed switch whereasPHYSICS FOR YOU I rrenumv 'r r

Full Wave RectifierIn the fuII wave rectifier two diodes are used.

tr

The circuit diagram, input and output voltagewaveforms for a full wave rectifier are as shown in the figures.Centre tap

I

For half wave rectifiet

trmformer

Diode 1 (D1)

Rr I

I

outputvottage

=,ft.sry-toFor full wave rectifier,rrms

=t.zt

-+

, -Io rdc -2j : -f,, , - l

=0.482 +E" o;Peak value of current is

O

Rectif ication Ef f iciency The rectification efficiency tells us what percentage of total input ac power is converted into useful dc

output power. Rectification efficiency is definedas

,vo ^urac

Average or dc value of current is

rJU +R,2Io

n= ac input power fromn=Pd.x1oo% 'P^"

dc power delivered to load

the secondary of the transformer

,

=.:i-

rms value of current is

'*t=EPeak inverse voltage is

,10

For a half wave rectifier, dc power delivered to the load l- ;2Po"

is

:

=rl.R.

=[*J

o.

PIV rr2V3, Average ortdc Value of voltage is 2I^

Input

ac

power is

P". = r3,,,

(! +R.) =(21

n,+R1)

V1"=I6.R1

=fR1

Rectifi cation ef ficiency

It is also given by

,r. =ZVo 'cc ftRipple Frequency

(Assuming diode forwardtesistance is zero)= 50

n=Pdc ''

(h/n)2Ri =(Iol2)z (ry +R1) xloo% P". _i. l ,* R;

Hz tr For a half tr For a full wave rectifier, u, = 2u; = 100 Hzwave rectifier, D/ =D-

40.5 lr

o/

Ripple Factor The ripple factor is a measure of purity of the dcoutput of a rectifier, and is defined f=as

lf ry-'uThe Boolean expression for XNOR gate isY=

Y=A+B =A.B =A.Btr ORgate from NORgate

.-

A.B+A.B=ToBUniversal Gate

Y=A+B=A+BBoolean Identities

NAND Gate

as a

o NAND

gate is called as universal gate because

with

the repeated use of NAND gate we can constructany basic gate

A+B=B+AA+(B+C)=(A+B)+CA'(B+C)=A'B+A'C A+0= A

A'B =B'A A.(B.C)= (A'B).C

tr NOT gate from NAND gate

A+B.C=(A+B).(A+C),4.0=04.1.= A

Y=Ao ANDgate from NAND gate

A+L=1

A+A=AA+ A=1

A,A=A

A.A=0 A=AA.B= A+BA.(A + B)= A

Y=

:

A.B= A.B

o

OR gate from NAND gate

A=A A+B=A.E A+A'B=A a+A-n= A+B

A.(A+B)= A.B

44

PHYSTCS FOR YOU I resnurnv ',r

r

De Morgan's Theotems

tr A+B=A'E A.--{\. a.--> r, then

, _50x10-3x10=0.275x50x10-z

2y"

L(Using (i))

=Po'! (2m?,t)2y"

(b) : Here 2r = 0.1. nm/ r = 0.05 nm r = 0.05 x 10-e m = 5 x

10-11

m

B=747

Current due to motion of electron is3.

(a): Parallel currents attract and antiparallelcurrents repel.

_e.ea

T2r

(.,=?J

50

PHYSICS FOR YOU I reonuanv'r r

Magnetic field at a proton is

& -$o 2n (ero\= Fo ero 4Ttr 4nrlZn) 4nr or r=a[1n)ttB=2nI-

N,

For coil 2 (In vertical plane) = 400, Rz=2n cm = 2n x 10-2 m Iz=200 mA= 200' 10-3A

Magnetic field at the common centre O due to current in coil 1 is

IuoJ e 1 v_ 5 x 10-11 t4v--'' 10_7 1.6xt0_1eL019

& =Ig N{r2n ' 4n

Rl

=

FoNr/r2R1

(d): =

4.4x

rad

s-1.

c

1.6

n

ZRz Resultant magnetic field at O isB= JB?r*

Magnetic field at O due to current in coil 2 is Nzlz2n FoNzIz B, - [o

' 4n R2 -

q

('.' rrisr- to nr)

1m1m

'1mRefer figure,

tan 45'=

9lOF

OF=CF=

t*2

4nxl}r =11.

x200 x 10-3 x 10 x 25

The magnetic field at the centre O of the square frame isB=

Zx rcxl}-z10-3 Wb m-2 (a) : V/hen a charged particle of charge and 4 with velocity o perpendicular to magnetic field B, it follows a circular path whose radius is given bymass m entersma mBa

b, l-

[rir,+S. an\1.12)L + sin45"-lI x 4 =l&-16J

MJZ

\Alhen the frame is taken as circular coil of radius r,

then

2nr=4

Or

,=2

n

/

- -q

for the same value of o andrY-L- tmo-

B.

The magnetic field at the centre of the circular coil carrying same current l is

:,

ro i rd : ro

B,_lro

B .'. -= 1.6 B' Jz n210. (a):

_po 2nI _l!9nzl 4n r 4n (2ln) 4n"2nI

=flP 'qp4a?a

where subscripts p, d, and. q represent protoo deutron and a particle respectively. rp:rd

m2m4m ira=;,; tE='1.:2:2

12. (d) : Force acting on a charged particle moving with velocity d is subjected to magnetic field E

is given byF = q(Ax B) or, p

(ii)For coil 1. (In horizontal plane) N1 = 150, Rr = n cm = r x 10-2 m 1r = 200 mA = 200 10-3 A "

(i)

= qaBsin} When0 =0o,F=qoBsin0"=0When 0 = 90o, F = qoBsrn9}" = qaB=0

(iii) When 0 = 180o, F = 4aBsin180o

This implies, force acting on a charged particle is non-zero, when angle between d and B can have

any value other than zero and 180o.FORyOU I

PHYSTCS

rrenumv'rr 51

13. (d)

:

til = tzx

5o x 1o+

i

=5oo x

roal.^\

Torque acting on the loop is

i=titxB

= 6oox1o{ [x(tooo* 1,04i)

, A

/

=zeroNm17. (a) : Use 4oBsin0rConsider an element of thickness dr at a distance r from the centre of spiral coil. Number of turns in spiral = n Number of turns per unit thickness

(...i"i

=o)

*" =B

*4 sin01,

for the same values of m, a, q andsin3Oo

...18. (d)

rA _ sin9Oo _

ft=

rp

"1 t4T of oscillation of a magnet

Zrn

or

19

Number of turns in element dr is,

&-dndr

n

19. (d) : The time periodis given by

=b-a Magnetic field at its centre O due to current in6ln

,

element dr is

where,

T -2n.1

\ wn

ft'

dB =vo2ndnl

21 2r\b-a) 4n r =N-6=W(4) tr^I n dr =:-sX 2 @-a) rby

I = Moment of inertia of the magnet about the axis of rotation M = Magnetic moment of the magnetAs the

B: Uniform magnetic filed l and M remain the same

Total magnetic field at centre O due to current

-Xthrough the whole coil is . p pohar PoIn dr

-b1 Jz@-a)r

2(b-a)t

7

'lB T1 llBz According to given problem, h=24ttl

.'.

Ic-OI-=

1T"E

i-

14.

voln r^, fl) =2(b d--"'\a)' (a): B forsolenoi4= '|u '|ux19x3 ltnnl=Pn 0.4Magnetic morrient of the coil = I x A x N

Bz=24pT-18pT = 6pT Tt=2s

t.

Tz=(2t)il6uT=4s

En:tm-\ H=150

M=0.4xpxr = BMsin90'

(0.01)2x10

20. (b) : Here, B

:

1 Wb

Am-l

B = Volt,H

:. r=4nx10-715. (c)

x500 x 3x0.4xnx(o.ot)20.4

xto21. (b)

u--- Bn'=

FoH

= 6n2 x10-7 N m

: A galvanometer can be converted into an ammeter of given range by connecting a shunt of suitable resistance S (in parallel) across thegalvanometer.

: The amount of work done by the external torque is given by0"

Zn,

roTtso = ?7

1

105

W -15

f r.*rd}01

Then s = !-t "=16.

I-Ir =ry= 5-1

a in parallel

0.

(a): Here, E=roooic=roooxrorA = 10 cm x 5 cm = 50 cmz= 50

lr

'01

= f mn sinodo = MB(cosor -

cos02)

t

10a m2

I=t2 AMagnetic moment of the loop is

Here, 01 = 0o; 0z = 180o; M = 1'.5J f1; B = 1 T Substituting the given values, we get 1,lg = (1.5 I Tr) (i T) [coso. - cos180.]

=3I

62

PHYsrcs FoRYou I rranunRv'rr

(a) : Here

Magnetic moment of the coil,

' \ffi Asu=nl7or or "MB D'=.7 4it'I .MB l=-

i=lAiu=2s-1 'M=Ninrz

N = 16; r = 10 cm = 0'1 m; B = 5 x 10-27...(i)

Here, r = 0.1 m;N = 10, 0 = 50o, Bn= 4 x 10-s T In case of a tangent galvanometer

(!)

;

1=2'8, t*reNFo

Substituting the given values, we get

where / is the moment of inertia of the coil about its axis of rotation

l- 2x0.1x4x10-s10

x 4nx10a

tan 60o

.l=7.14(Using (i))

4iE'1)'

Ninr2B =w

(c): Here, Xq=0.0075; Tt r - 73"C = (- 73 + 273) K= 200 K Tz = - L73"C = (-173 + 273) K=

=R

Ni12B

As x*, =!= # T2 1ocxmi.

=' (. r* * +)..'(i)

100 K

Substituting the given values, we get

L6xLx(0.1)2x5x1.0-2

26. (b)

:

xm2=27*r= 2 x 0'0075 = 0'015 Magnetic moment of a wire of length I is

4xnx2'=9x1oakgm2 n23. (c):

M--mIa-

where m is the strength of each pole.

When the wire is bentintoa

semicircle of radius

r, then l" l:nr ot r=TC

t+-Zr#l

Now, the distance between two Poles

-Zr-!TE

Its new magnetic moment isHere, 0 = 60"; Bt = L.2 x 10-2 T 0r = 15"; 0z = 60o - L$o = 45o. In equilibrium, torques due to two fields mustbalance

M'27.

=

mxvr

=

*rI IE=2;{^rD =2!TE

(Using(i))

xt=12MB1 sin01 = MBzsin$z.

(d): tan6 =+=pr=t (': B, =Bu(Given)

.'.

6 = 45ocase,

-

D_.:

B"

sin02sin45o1.2 x 10-2

sin0'

1.2 x L0-2

sinl'S' sin45'

2g. (b): In first

o

_ t.z x to-2 sin(+s'-go')

1, @; \=E{f

...(i)

-

(sin45'cos30'- cos45'sin30')sin 45q

where Bs is the horizontal component of earth's magnetic field

In second1

case,

=1.2x10-2 [cos30'

-

sin30"]

=72x70_2tf=6(J5- r)xro-3

r

;)

'2nmagnetPHYSICS FoRYOu I

...(ii)

where B is the magnetic field due to external

rranunnv'tt 53

In the thirdt)a=1

case,

"2nIBH+B

...(iii)

or (r^,or32. (d) : Here,r

+)'

= (+oo)r- (2oo)2=2oo x ooo

Divide (ii) and (i), we get

4.\6u = 2rE x 100 or u = 50 Hz.= ssinfroor

_ [-{x2

t2 !41 25,g=

,s=f,Ba

B"

1') \2) - o

V = 200sin 1ioo4 v .'. Phase difference between V and i is d=-

[-zBaB BH

.n=

2

Power consumed, P =

V*" lrzero W

-=l+16

V*, i*,

cos9Oo =

"

cosQ

! = BH t6u3_

2

33. (b)...(iv)

:

Here,

Divide (iii) by (i), we get

/E[-r .t -{ B"(Using iv)

field B = 0.025 T Radius of the loop, r = 2 cm= 2 x L0-2 m Constant rate at which radius of the loopMagnetic

shrinks, 4=t*10-3ms-l 'dtMagnetic flux linked with the loop is0 = Bz{cos0 =.B(n'2)cosl = Bnr2 The magnitude of the induced emf is

DaO -=124xs=g,lV vibs/min =J63 vibs/min29. (a) : The maximum curent is obtained at resonance where the net impedance is only resistive which is the resistance of the coil only. This gives the resistance of the coil as 10 Q. Now, this coil along

H=4 =!Gn\=Bnzr!! dt dt dt= 0.025

xnx2x2xI0-2 x 1x 10-3

34. (c) : Power factor becomes equal to one at theresonant frequency.

=rx10{V=npV-

with the internal resistance of the cell gives acurrent of 0.5 A.30.

35. (a) : Here, 0 = t$t'The induced

50t + 250

(d)

:

The phase diagram is as shown

in

the

figure.

r-

vc=vr

= _(20f_50) At t = 3 r = - (20 x 3-5b) :- 10 V. (a) : Let the current I be flowing in the larger 35.loop.

46 -dr ir =: Il0t2 _ 50f + 250) dt dt'

emf is

F-L_-_+ R2difference. 16. To measure the current passing through a particular part of the circuit, the ammter is connected in series. To convert a moving coil galvanometer into an ammeter, resistance is connected so that the

(a)(b)

(c)

the potential difference across resistance R1 is greater than of R2. potential difference across R2 is greater as this resistance is smaller. the potential differences are the same acrossR2 as

full current passing through the ammeter is notdissipated. How is the galvanometer converted to an ammeter?PHYslcsFOR

well

as R1.

(d) One must know the value of the emf applied.

You I rrsnumY't

t

69

17. How is the voltmeter connected in a circuit tomeasure the potential difference across the wire?

22.

Choose the right statement.

18. The essential principle of the Wheatstone bridge is inthe given circuit, no current will be flowing through the

galvanometer.Rz

If Rl = 1 C), ='2 Q, and one has also

3C),5(land5f2"R3andRaare replaced by (a) 3 C),5 (b) 5 cr,6 Cl (c) 3 C),5 O

fi

(a)

The current passing through the branch

(d)

6 cr,3 cl

ABC > the current throudh ADC. (b) The current throughADC > the current through ABC. (c) The currents through ABC, ADC and AC are the same. (d) There is current only through AC.The equivalent resistance between A and C is

What is the total resistance connected to thebattery? 20.

All resistances are equal. The total resistance betweenA and C is(a)R

(o)*

t.r

Jn

(d)

2R

(a)(c)

T8R

2R

(b)

6R

Give the equivalent diagram.

(d) none of these Lerrgth of AB and CD are the same but ABthicker than CD.

ts

(d) : ?r*. for the electron =

ff

='1..17x10sms-1

This will be much higher than that of the gas. This should be (b). However this theory is notapplicable; therefore (d). (d) : The drift velocity = 4E 1(a) The'ctirrent density in AB is the same as that

As initiat velocity = O afte%very collisiory

inCD.(b) The field inAB> the field in CD. (c) The current in the thicker wire AB > the current in the'thinner wire CD. (d) The current is the same inAB and CD.

o=accelerationxtime

(a,c):

y=R[ =+ v=9!*I A v=Field E,I=MI

70

PHYsrcs FoRYou I rrenurnv'rr

+ El=p-AI - I-:E = I=oE Ap(b) :AV= IR; LV =(1 A) x (8.13' 10-3 Q)

t1

+ 4I = a.rs x 1o-3 v m-l LThe velocity of the electric field is the velocity of light. It is the field that causes current to flow:

Applying the second law of Kirchhoff, in the loop ACBDA, -J1Rr + izRz = 0. It follows I1R1 = I2R2. The potential difference along the path ACB as well as BDA are the same. However as R1 is greater, the current is smaller than that through R2.

$)

12. (c):

and not the electrons that travel from one end tothe other.6.

(a) : The resistivity and hence the resistance increases with temperature. This is the principleof the platinum resistance thermometer.

G) : They are superconductors. The theory of superconductivity will be studied by you later, asquantum mechanics is applied for this study. (b) : The work done per second is the power consumed. Work done is Vq arrd per second, it is

100

Q

r,

r is the intemal resistance. The resistance of 100 Q and 100 Q in parallel

112

=+ The total resistance in parallelThe current in the circuit13. (c, d):

100 ---=-L00

100 = 50 Q

t Unlike the situation in charging a capacitor, the moment the potential difference is applied, the maximum current flows in the'resistor, unlike acapacitor where work done is capacitor.as

V

!=VI

The total resistance in the circuit = 50 + 10 = 60 Q

-

1'08 60

-

0.018

A

!qV. 2'

The current is

not connected to an extemal resistance, the potential difference across the cell is emf, 1.08 V.

\

y'hen the cell is

also not a constant when charging or discharginga

(!)

;

The intemal resistance of this battery is rather

high (10 Q) as given in Q. No. 12. Therefore thecurrent flowing will be(a, d)

(b) : When power is taken through long wires,

P:IV + I= L. vThe power loss along the transmission lines is

iry 10f,)

= 0.108

A

=I2R=[#)^For a given resistance, as power loss is inversely proportional to lZ, higher the transmission voltage, lower is the power loss, when transmifting a given power and given transmission wires.

The current passing through which the circuitshould not be disturbed. Therefore the ammeter which is connected in series with the circuit shouldpass all the curcent i.e. the ammeter must have low resistance. This is achieved by connecting a thick

10'

fi*ffi?-ilN^trr.t$i*.r- D rc rB.

Rl

At B, when I, is entering from A, flows out of

Therefore the current entering B, I is the same through R1. Similarly by taking points C, Dsuccessively, one finds, the same current is flowing

wire of low resistance in parallel (shunting) to theammeter. 17. The voltmeter is connected in parallel so that the current passing through the voltmeter is negligible. A high resistance is connected in series with the galvanometer. A high resistance is in series with the galvanometer so that its effective resistance is very high. The dissipation of the current through the voltmeter will be negligible.I

through

R2

and R3.

PHYSTCS FORYOU I

neanunnv',rr 71

18. (c):

B, O, D are identical.

Three resistances areIf galvanometer gives null position, A is common for AB andCD.

inR

parallel connection

between A and B and three more between B and C. B, O, D have the same potential.Vn = Vo.

The total resistance

i, 4*

:'

Va-Vs=Ve-Vo.Va

33

=

2R3

Similarly,common.

-

Vc

should be equal to Vp

-

Vs. C is

21. @t:

ven

vnc vncRt=R,

_veoRn

rrRz

IlRl _ 12R3lzRq The current is the same as R1 and R2 are in series. As R1 < R2 (because R1 is thicker), Vae is less than V6p. Therefore E1 < E2 where E1 and E2 are the fields. (a) is not correct as the areas of cross-section of AB and CD are different.

RzAs

Rt=1 ... Rs=q=1 R22Rt62

The other combinations for R3 and Ra do not fit the

Wheastone's bridge. 19. One can find various problems of this type. The method of solution is the same. Let the total resistance to right of CD be X. The circuit is now

22. (dl: AC is only a connecting wire between A and C which has negligible resistance. The currentpasses from A to C.

23. (a)

: This is a Wheatstone's bridge. There is no current along BD or DB. AB and BC are in series and CD and DA are in series. The equivalent value is_t_-_

111 2R2RRone has to mark the points.

24. First,2 Q is connected

in parallel with X which gives

2X

2+XAs this is ari infinite series, X.+

2X

2X

-

2+X

ls also eoual to

2+X--4

-v

X+1=TotalresistanceL Q is

where

the resistance of AC

i.r. 2X

" X'-X-2=O = X=1+.h+8 2Y

2+X

+ 1=

X. the total resistance

Total resistance, X not valid here.

-

2 Q as the negative value isi:i:";!::ti:ii

20. (a): This is an extended Wheatstone's Bridge. This is equivalent to72PHYsics FoR You I irsnunnv 'rr

!

:tt:":,i.,,i

:

This is a lrVheatstone's bridge. There is no current through BD.

oo

1{1usrr$*rN

lf you have any difficult / unsolved problem or you are unable to understand one, then write to us. Our team of experts will diagnose your problems. The diagnosed problems willbe published in the subsequent issues. u[hey.,often,eh oose,,,g

,

Wgestio,ns u.ppertor,Lgernr,rn,uch

'oi* -,1:\\$',.\iN+

iht'**it*'*ii}i

1.

We know according to the law of conservation of energy, P.E. + K.E. = constant. If a car running on a highway is accelerated, its kinetic energy increases

P.E.

of the electron + K.E. of the electron = Total energy

which is a constant for a particular level.

but the gravitational potential energy remains the same. Kindly explain this.[Mr. Sumant Kalra,Ferozepur City (Puniab)]

III. Aconstant force is acting on abody continuously to move it through a particular distance x.

Soln.: Let us study this question part by part.

I.

Abody is falling through

a

height

h.

As it is falling,

Work done = constant force F x r, the displacement.This is ubo 1 ma2

the potential energy decreases and the kinetic energy increases. The total remains the same.

22 -!

muz,

the final

- initial kinetic

energy. The potential energy remains the samebut the

u{*,

increase in kinetic energy is due to the work done.

.ihi

,

8t

i

'."r

,,

The mechanical energy and eiectrostatic energy is normally used in classical physics. In modem physics, according to Einstein, the total energy E = mc2.wherci the energy may be any type of energy.

There was no external force acting on the body'as

it

For a photorr, E = hu (Einstein's photoelectricequation).The classical theory is applicable to a closed system. Here the P.F.+

is falling from A to B. We can assume that the body had been already falling and it passes A at time zeroof our experiment. In that case it has also some kinetic

K.E.

energy at A. Initial K.E + Initial P.E. is equal to final K.E. + Final potentiil energy..

=

constant because we are studying

only the kinematics chapter of Newtonian mechanics where things are ideal and calculations are simple. Ybur question is good because it ls orrly by searching for contradictions and resolving them that one makesprogress.

This is

a

ponservative system.

II. This is the same in the case of the H atom model(classical theory).

74

pHyslcsFoRyou I rrsRunny'rr

2.

What is the difference between emf and supply

The potential difference across

voltage?

[Mr.Soln.: Suppose one takes a 2 V battery.5sr

Tas GurungJ

33 Total potential difference =Vr+Vr=: - + =: JJJ:+V -- 2V.

1

e=L

x

1=I y

The emf of the battery is the maximum potential difference in the circuit. The potential difference is not for the whole circuit but across the variousresistances.

In the first circuit, the resistance connected is R1, whereas in the second, it is R1 + R2 in series. Letus assume that the battery is new and its internal resistance is negligible. If Rl = 5 O and R, = I 9,The total resistance is 6 Q.

The voltage or potential difference across various parts is supply voltage to that part.The voltage across a battery is called its emf. This isthe maximum voltage one can obtain from the battery.

This is the voltage obtained only due to chemicalaction in the battery.

)\/ 1 The current flowing through the circuit = 1= i e " 6Q 3The potential difference across 5Q= 5 x

*= * JJ

U

Whatever yoltage is supplied from any source - battery or dlmamo to a particular circuit is calledthe supply voltage.

oo,1

Amazing Life Lessons You Can LearT f,ro.m AlbertFollow your curiosity.

Ein*t

"l

have no special talent. I am only passionately curious."

Perseverance is priceless.

"lt's not that l'm so smart; it's just that lstay with T_-- -"'- longer." " -'- problems - J---J , -- - ---

The imagination is powerful.'f lmaginatio