physics dpp 4 (jp & jr) advanced fc

8/9/2019 Physics DPP 4 (JP & JR) Advanced FC

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Page # 1

This DPP is to be discussed (22-04-2014)

CT-2 to be discussed (22-04-2014)

1. Course as per plan : WPE, Circular motion, COM

2. Course covered till previous week : Current Electicity, Capacitance, EMF.

3. DPP Syllabus : WPE, Circular motion, COM

DPP No. # 04 (JEE-ADVANCE)

Total Marks : 165 Max. Time : 180 min.Single choice Objective (1 negative marking) Q.1 to Q.15 (3 marks 3 min.) [45, 45]

Multiple choice objective ('1' negative marking) Q.16 to Q.21 (4 marks 4 min.) [24, 24]

Subjective Questions ('1' negative marking) Q.22 to 30 (4 marks 5 min.) [36, 45]

Comprehension ('1' negative marking) Q.31 to Q.39 (3 marks 3 min.) [27, 27]

Match the Following (no negative marking) (2 4 or 5)Q.40 to Q.42 (8 marks 10 min.) [24, 30]

Assertion and Reason (no negative marking) Q. 43 to 45 (3 marks 3 min.) [9 , 9]

ANSWER KEY OF DPP No. # 04 (JEE-ADVANCE)

1. (C) 2. (B) 3. (B) 4. (A) 5. (C) 6. (B) 7. (A)8. (D) 9. (C) 10. (4) 11. (B) 12. (D) 13. (C) 14. (C)

15. (D) 16. (A,C,D) 17. (B,C,D) 18. (A,B,D) 19. (A,B,C) 20. (A,B,C)

21. (A,B,C,D) 22. 1 23. 1 24. 8 25. 12 26. 5

27. 7 28. 2600 29. 20 30. 5 31. (A) 32. (D) 33. (B)

34. (C) 35. (A) 36. (D) 37. (A) 38. (A) 39. (B)

40. (A) (s) ; (B) (p,q,r,s,t) ; (C) (p,q,r,s,t) ; (D) (s,t) 41. (A) p,t ; (B) p,q ; (C) q,s ; (D) p,q,r

42. (A) (p, t) ; (B) (p, t) ; (C) (s) ; (D) (q, t) 43. (D) 44. (D) 45. (D)

1. Power of the only force acting on a particle of mass m = 1 kg moving in straight line depends on its velocity as

P = v2where v is in m/s and P is in watt. If initial velocity of the particle is 1 m/s, then the displacement of the

particle in n2 second will be :

ljy js[kk esa xfr dj jgs m = 1 kg nzO;eku ds d.k ij dk;Zjr dsoy cy dh kfDr blds osx ij P = v2ds vuqlkj fuHkZjdjrh gS tgka v, m/s esa gS rFkk P okWV esa gS A ;fn d.k dk izkjfEHkd osx 1 m/s gSA rc n2 lsd.M esa d.k dk foLFkkiu gSA(A) (n2 1)m (B) (n2)2m (C*) 1 m (D) 2 m

Sol. P = Fv

v2= Fv

F = v

Ma = v

1 vdx

dv= v

v

1

x

0

dxdv

v 1 = x

v = x + 1

TARGET : JEE (MAIN + ADVANCED) 2014

TEST SYLLABUS

TEST : PART TEST-2 (Test Date :23-04-2014)

Test Syllabus : Current Electricity, Capacitance, EMF, WPE, Circular Motion,

COM.

Course: JEE-ADVANCED

Date : 20-04-2014

DPP No. : 4

PHYSICS


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Page # 2

dt

dx= x + 1

x

01x

dx=

t

0

dt

n(x + 1) n(0 + 1) = t

x + 1 = et

x = et 1

x = et 1

at t = n2

x = 2 1 = 1 m.

2. Two blocks of mass m1and m

2(m

1< m

2) are connected with an ideal spring on a smooth horizontal surface as

shown in figure. At t = 0 m1is at rest and m

2is given a velocity v towards right. At this moment, spring is in its

natural length. Then choose the correct alternative :

m1rFkk m

2(m

1< m

2)nzO;eku ds nks CykWd fp=kkuqlkj ?k"kZ.kjfgr {kSfrt lrg ij ,d vkn kZ fLizax ls tqMs gq, gSA t = 0 ij

m1fojke voLFkk esa gS rFkk m

2osx v ls nka;h vksj xfr dj jgk gSA t = 0 ij fLizax viuh LokHkkfod yEckbZ esa gSA

m2

m1

v

(A) Block of mass m2will be finally at rest after some time.

(B*) Block of mass m2will never come to rest.

(C) Both the blocks will be finally at rest.

(D) None of these

(A) nzO;eku m2dqN le; ckn fojkekoLFkk esa vk tk,xkA

(B*) nzO;eku m2dHkh Hkh fojkekoLFkk ugha vk,xkA

(C) nksuks CykWd fojkekoLFkk esavk tk,xsA(D) mijksDr esa ls dksbZ ugha

Sol. If velocity of m2is zero then

;fn m2dk osx kwU; gS rksby momentum conservationlaosx laj{k.k lsm

1v= m

2v

v=1

2

m

vm

Now kinetic energy of m1

vc m1dh xfrt tkZ

=2

1m1 v2 =

2

1m

1

2

2

1

2 vm

m

=2

1

1

2

m

mm

2v2 =

1

2

m

m2

1m

2v2 =

1

2

m

m initial Kinetic energy izkjfEHkd xfrt tkZ

Kinetic energy of m1> initial mechanical energy of system

m1dh xfrt tkZ > fudk; dh ;kf=kd tkZ

Hence proved


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3. Consider the system shown in figure. Pulley, string and spring are ideal and m


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Sol. As string does no work on the ball, energy conservation can be applied.

pwfd jLlh xsan ij dksbZ Hkh dk;Z ugh djrh gS vr% mtkZ laj{k.k ykxw dj ldrs gSA

2

1mV2= mg (L L cos)

V = )cos1(Lg2

on putting values V = 10 m/s

eku j[kus ij V = 10m/s

5. Two blocks each of mass m are joined together using an ideal spring of force constant K and natural length 0.

The blocks are touching each other when the system is released from rest on a rough horizontal surface. Both

the blocks come to rest simultaneously when the extension in the spring is4

0. The coefficient of friction

between each block and the surface (assuming it to be same between any of the blocks and the surface) is :

m nzO;eku ds nks CykWd 0LokHkkfod yEckbZ rFkk cy fu;rkad K okyh vkn kZ fLizax ds}kjk ,d&nwljs ls tqMsgq, gS A CykWd ,d

nwljs dsk Li kZ dj jgs gStc fudk; [kqjnjh {kSfrt lrg ij fojke ls NksMk tkrk gS A nksuks CykWd ,d lkFk fojkekoLFkk es a vkrs

gS tc fLizax esa foLrkj 40

gSA izR;sd CykWd rFkk lrg ds e/; ?k"kZ.k xq.kkad gksxkA ;g ekfu, fd R;sd CykWd rFkk lrg dse/; ?k"kZ.k xq.kkad leku gSA

(A)mg40

K 0(B)

mg8

K 0(C*)

mg8

K3 0(D)

mg

K

20

17 0

Sol.

Work energy theorem, dk;Z tkZ es;

4mg 00

=

20

2

0 K2

1

4K

2

1

=mg8

K3 0Ans.

6. A particle of mass m is connected to a block of mass M by an inextensible string of length . The block is free

to slide on the smooth horizontal plane. The particle is released from the horizontal position as shown in the

figure. If the coefficient of restitution for collision between block and particle is 1/3 , maximum height gained by

the particle from point of collision, after the f irst collision between block and particle will be (use M = m)

,d m nzO;eku dk d.k M nzO;eku ds Cykd ls YkEckbZ dh vforkfu; jLlh ls tqM+k gqvk gSA Cykd {kSfrt ry ij xfr djus

dsfy, LorU=k gS A d.k dksfp=kkuqlkj {kSfrt voLFkk ls NksM+k tkrk gSA ;fn d.k RkFkk Cykd dk izF;kLFkk xq.kkad3

1gks rks Cykd

rFkk d.k ds e/; izFke VDdj ds i pkr d.k dh vf/kdre pkbZ Kkr djksaA (M = m)

(A)3

(B*)

9

(C)

2

(D)


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Page # 5

Sol.2

mv1 mv

2= 0

u2

vv 21=

3

1

v1=

3

u, v

2=

3

u

(K.E.)in

=

2mu2

12 = mg (K.E.)

final=

9

um

2

12

2

=9

mu2

mu2= mg

let height attained by A after collision is mg =9

mu2 = /9

ekuk fd A }kjk VDdj ds ckn izkIRk pkbZ gks rks mg =9

mu2 = /9

7. Two spherical bodies of masses m and 5m and radii R and 2R respectively, are released in free space with initial

separation between their centres equal to 12 R. If they attract each other due to gravitational force only, the

distance covered by smaller sphere just before collision is

nks xksykdkj fi.M ftuds nzO;eku m rFkk 5m o f=kT;k,a R o 2R gSA tc buds dsUnzksa ds e/; dh nwjh 12R gS rcbUgsa eqDr:i ls NksM+k tkrk gSA ;fn fi.Mksa ds e/; dsoy xq:Rokd"kZ.k cy yxrk gkas rks NksV s fi.M )kjk VDdj ls iwoZ r; dh xbZ nwjh

Kkr dhft,A(A*)

2

R15(B)

2

R13(C) 10R (D)

2

R17

Sol.

Distance covered by the smaller sphere = 10R 2

R5=

2

R15


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8. In the figure shown a semicircular area is removed from a uniform square plate of side and mass m (before

removing). The x-coordinate of centre of mass of remaining portion is (The origin is at the centre of square)

iznf kZr fp=k esa nzO;eku m gVkus ls igys rFkk Hkqtk dh ,d leku oxkZdkj IysV ls ,d v)Zokkdkj {ks=kQy gVk gqvk

gSA ks"k Hkkx ds nzO;eku dsUnz ds xfunsZ kkad gksaxsA (ewy fcUnq oxZ ds dsUnz ij gS)

x

y

(A) )8(2

)2(

(B)

)8(2

)2(

(C)

8

)2( (D*)

)8(2

3

4

Sol. xcm=21

2211

mm

xmxm

x

y

2, 0

m1= mass of square plate oxkZdkj IysV dk nzO;eku = m

x1 = c. m. of square plate = 0

x1 = ox.kZdkj IysV dk nzO;eku dsUnz= 0m2= mass of removed part gVs gq, Hkkx dk nzO;eku

=

2

4m

2

2

= 8

m

x2= c.m. of removed part gVs gq, Hkkx dk nzO;eku dsUnz

=

23

4

2

=

2

3

41

xcm=m

8m

3

41

2.

8

m

xcm= )8(2

3

4

9. Sand is falling on a flat car being pulled with constant speed. The rate of mass falling on the cart is constant.Then the horizontal component of force exerted by the fall ing sand on the cart

fu;r osx ls [khaph tk jgh ,d dkj ij jsr fp=kkuqlkj fxj jgh gSA xkM+h ij fxj jgsa nz O;eku dh nj fu;r gSA xkM+h ij fxj jghjsr }kjk yxk;s x;s cy dk {kSfrt ?kVd

(A) increases (B) decreases (C*)remains constant (D) increases and then decreases

(A) c


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Sol. The horizontal component of velocity of sand just before falling on the cart is vs= 0.

The horizontal speed of cart = vC(constant).

The rate of mass falling on cart = (constant).

Horizontal force exerted by falling sand on cart = vrel

= (vc v

s) = v

c

and vcare constant, the horizontal force is constant.

gy% xkM+h ij fxjus ls Bhd iwo Z] jsr dk osx dk {k s frt ?kVdvs= 0.

xkM+h dk {kSfrt pky = vC(fu;r).

xkM+h ij nzO;eku fxjus dh nj = (fu;r)xkM+h ij fxjrh gqbZ jsr }kjk yxk;k x;k {kSfrt cy = vrel

= (vc v

s) = v

c

rFkk vcfu;r gS] vr% {kSfrt cy fu;r gSA

10. Two blocks A & B of masses m and 2m respectively, attached at opposite ends of a spring of spring constant K,

placed on smooth horizontal surface. Spring is initially at its natural length . A is given a velocity 2V0and B given

velocity V0as shown.

nks CYkkWdAoB ftudsnz O;eku e k% m rFkk 2m gS] KfLiazx fu;rkad dh fLiazx dsfoijhr fljks ls tqM+sgS ] fLiazx izkjEHk esaiz kdfkdyEckbZ esa gSA fp=kkuqlkj Adks 2V

0osx fn;k tkrk gS ,oa B dks V

0osx fn;k tkrk gSA

Maximum seperation between m and centre of mass of the system will be :

m rFkk fudk; ds nzO;eku dsUnz ds e/; vf/kdre~ nwjh gksxh :

(1)K3

mV8

3

20

(2)

K3

mV2

3

20

(3)

K3

mV2

3

220

(4*)

K3

mV8

3

220

Sol.

1=

mm2

m2

=3

2

K1

1= KK

1

3

2= KK

1=

2

K3

|x1 max

| =

2

K3

)V2(m 20

=K3

mV8 20

|d1 max

| = 1+ x

1 max=

3

2+

K3

mV8 20


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11. A particle is moving along an ell iptical path with constant speed. As it moves from A to B, magnitude of its

acceleration :

,d d.k fu;r pky ls nh?kZokkdkj iFk ds vuqfn k xfr dj jgk gSA tSls gh ;gA ls B rd xfr djrk gSA rc blds Roj.kdk ifjek.k &

y

B

Ax

(A) continuously increases (B*) continuously decreases

(C) Remains constant (D) first increases and then decreases

(A) yxkrkj c


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Sol. (D) The maximum angular speed of the hoop corresponds to the situation when the bead is just about toslide upwards.The free body diagram of the bead is

For the bead not to slide upwards.

m2(r sin 45) cos 45 mg sin 45 < N .................... (1)where N = mg cos 45 + m2(r sin 45) sin 45 .................... (2)From 1 and 2 we get.

= 230 rad / s.

gy (D) tc chM ij dh vksj Bhd fQlyus dh fLFkfr esa gksrk gS rc ywi dk dks.kh; osx vf/kdre gSA chM dk

chM dks ij dh vksj ugha fQlyus ds fy,m2(r sin 45) cos 45 mg sin 45 < N .................... (1)

tgk N = mg cos 45 + m2(r sin 45) sin 45 .................... (2)

leh01 rFkk2 ls

= 230 rad / s.

13. Three point masses are attached by light inextensible strings of various lengths to a point O on the ceiling. All

of the masses swing round in horizontal circles of various radii with the same angular frequency (one suchcircle is drawn in the shown figure.) Then pick up the correct statement.

rhu fcUnqnz O;eku fHkUu&fHkUu yEckb;ksadh gYdh vforkU; Mksfj;ksa ls Nr ds ,d fcUnq Ols tq M+sgq,s gSaA lHkh nzO;eku fHkUUk&fHkUuf=kT;kvksads {kS frt orksaesaleku dks .kh; vkofk lsxfr djrsa gSa (,d ,slk ok fp=kkuqlkj fn[kk;k x;k gS)rks lgh dFku Nka fV;s&

(A) The vertical depth of each mass below point of suspension from ceiling is different.

(B) The radius of horizontal circular path of each mass is same.

(C*) All masses revolve in the same horizontal plane.

(D) All the particles must have same mass.

(A) izR;sd nzO;eku dh Nr ij fcUnq O ds uhps /okZ/kj nwjh fHkUu gSA(B) izR;sd nzO;eku ds {kSfrt okkdkj iFk dh f=kT;k leku gSA(C*) lHkh nzO;eku ,d gh {kSfrt ry ij ifje.k djrs gSaA(D) lHkh d.kksa dk nzO;eku leku gh gksxkA


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Sol. For conical pendulum of length , mass m moving

along horizontal circle as shown

T cos= mg .... (1)T sin= m2sin .... (2)

From equation (1) and equation (2), cos= 2g

cos is the vertical distance of bob below O point of suspension. Hence if of all three pendulums are same,they shall revolve in same horizontal plane.

Alternate :

If we remember that time period T of conical pendulum is

T = 2g

Lwhere L is the vertical depth of bob below point of suspension. If T is same for three

pendulums even L shall be also same. Hence all three particles shall revolve in same horizontal plane.

gy% yEckbZ ds dksfudy yksyd (conical pendulum) ds fy,]nzO;ekum {kSfrt ok ds vuqfn k djrk gqvk fn[kk;k x;k gS &T cos= mg .... (1)T sin= m2sin .... (2)

lehdj.k (1) rFkk lehdj.k (2) ls, cos= 2g

cos xksys dh fcUnqO (point of suspension) ds uhps /okZ/kj nwjh gSA blfy;s ;fn lHkh rhu yksydksa dk leku gS] rksos ,d gh {kSfrt ry esa xfr djsaxsAoSdfYid % dksfudy isaMqye dk vkorZdky T fn;k tkrk gS &

T =g

L2 tgk L fcUnq O ds uhps /okZ/kj nwjh gSA ;fn T rhuksa isaMqyeksa ds fy, leku gS rks L Hkh leku

gksxkA blfy;s lHkh rhu d.k ,d gh {kSfrt ry ij ifje.k djrs gSA

14. One end of a light rod of length 1 m is attached with a string of length 1m. Other end of the rod is attached

at point O such that rod can move in a vertical circle. Other end of the string is attached with a b lock of

mass 2kg. The minimum velocity that must be given to the block in horizontal direction so that it can

complete the vertical circle is (g = 10 m/s2).

1 eh- yEch gYdh NM+ ds ,d fljs ij 1m yEch Mksjh tqM+h gSA NM+ dk nwljk fljk fcUnq 'O' ls bl rjg tqM+k gS fd og /okZ/kj ok esa ?k we ldsA Mksjh ds nwljs fljs ls 2 kg nzO;eku dk CykWd tqM+k gSA CykWd dks fn;k x;k U;wure {kSfrt osx gksuk pkfg,tc fd og /okZ/kj ok dk pDdj iwjk dj lds &(g = 10 m/s2).

(A) 4 5 (B) 5 5 (C*) 10 (D) 3 5

Ans. Vmin

= gR5 = 2105 = 10 m/s


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15. The angle of deviation () vs angle of incidence (i) is plotted for a prism. Pick up the incorrect statement(s).

,d fTe ds fy, fopyu dks.k() v/s vkiru dks.k (i) ds chp xzkQ cuk;k x;k gSA xyr dFku pqfu;s &

(A) The angle of prism is 60

fTe dks.k60 gSA

(B) The refractive index of the prism is n = 3

fTe dk viorZukadn = 3 gSA

(C) For deviation to be 65 the angle of incidence i1= 55

65 fopyu ds fy, vkiru dks.k i1= 55 gSA

i

7060

60

65

i1(D*) The curve of '' vs 'i' is parabolic

'' vs 'i' dk o ijoy; gSA

Sol. (A) = i + e A (for minimum derivation i = e)minimum deviation = 2i A60 = 2 60 A60

(B) n =

2

Asin

2

Asin m

=

2

60sin

2

6060sin

= 3

(C) 1= i

1+ e A

65 = i1+ 70 60

55 = i1

(D*) Curve is not parabolicSol. (A) = i + e A (U;wure fopyu ds fy, i = e)

U;wure fopyu= 2i A60 = 2 60 A60

(B) n =

2

Asin

2

Asin m

=

2

60sin

2

6060sin

= 3

(C) 1= i

1+ e A

65 = i1+ 70 60 55 = i

1

(D) o ijoy; ugha gSA


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Page # 13

(C*) Angular velocity of the particle

(C*) d.k dk dks.kh; osx

(D*) x coordinate of the particle

(D*) d.k dk x funsZ kkad

Sol.

So X component of v elocity Vx= V sint

y component of force Fy= mv2/R sint = m2R sint

Angular veloci ty of part icle = constant.Xcoordinate of the particle x = Rcost. So B, C, D are correctly matched

osx X ?kVd Vx= V sint

cy dk y /kVd Fy= mv2/R sint = m2R sint

d.k dh dks.kh; pky= fu;rd.k dk X funsZ kkad x = Rcost. vr% B, C, D lgh gS

18. Two identical blocks A and B of mass m each are connected to each other by spring of spring constant k. BlockB is initially shifted to a small distance x

0to the left and then released. Choose the correct statements for this

problem, after the spring attains it's natural lenth.

nks le:i fi.MArFkk BnzO;eku m ijLij K fLizax fu;rkax dh fLizax lstq M+s gSA Cykd B dks izkjEHk esacka;h vkSj x0 foLFkkfir

djds NksM+k tkrk gSA fLizax ds izkdfrd yEckbZ esa vkus ds i pkr lgh fodYiks dk p;u dhft,A

mkA

m

xo

B


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Page # 14

(A*) Velocity of centre of mass of the system ism

k

2

1x

0

nzO;eku dsUnz dk osxm

k

2

1x

0gksxkA

(B*) Maximum elongation in spring during the subsequent motion is2

x0

xfr ds nkSjku fLizax esa vf/kdre izlkj2

x0gksxkA

(C) Maximum elongation in spring during the subsequent motion is x0

xfr ds nkSjku fLizax esa vf/kdre izlkj x0gksxkA

(D*) Maximum speed of block A during subsequent motion be 0xm

K

A dk vf/kdre osx 0xm

KgksxkA

Sol.

1/2 mv02= 1/2 kx

02

vcm

= v0/2

= 0x

m

k

2

1

1/2 kx2max

=

2

m

2

1v

02

1/2 kx2max=

20kx2

1

2

1

xmax

=2

x0

(VA)

max= (V

B)

max= v

0= 0x

m

k

19. Three identical particles A, B and C lie on a smooth horizontal table. Light inextensible strings which are just

taut connect AB and BC and ABC is 135 . An impulse J is imparted to the particle C in the direction BC. Massof each particle is m. Choose the correct options.

(A*) Speed of A just after the impulse imparted ism7

J2

(B*) Speed of B just after the impulse imparted ism7

J10

135

A

B

CJ

(C*) Speed of C just after the impulse imparted ism7

J3

(D) Speed of A just after the impulse imparted ism7

J2


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Page # 15

rhu le:i d.k A, B rFkkC ,d fpduh {kSfrt lrg ij j[ksa gSAAB rFkk BC nks vforkU; jfLl;k gS] rFkk ABC 135gSA d.kC ij ,d vkosx J, BCds vuqfn k yxrk gS] ;fn izR;sd d.k dk nzO;eku m gSA rc lfg fodYiksa dk p;u dhft,A

(A*) vkosx ds rqjUr i pkr~ A dh pkym7

J2gSA

(B*) vkosx ds rqjUr i pkr~ B dh pkym7

J10gSA

135

A

B

C

J

(C*) vkosx ds rqjUr i pkr~ C dh pkym7

J3gSA

(D) vkosx ds rqjUr i pkr~ A dh pkym7

J2gSA

Sol.

2

J2= 2mv

1

2

J2= mv

2 v

2= 2 v

1

J J2=

2

m(v

1+ v

2)

J 22 mv1= 2

mv3 1 J =

2

mv7 1v

1=

m7

J2

v2=

m7

J22

vA

= v1=

m7

J2v

B=

m7

J10

vc=

2

vv 21=

m7

J3.

20. Which of the following is/are conservative force(s) ?

fuEu esa dkSu lk@ls laj{kh cy gS

(A*) rr2F 3

(B*) rr

5F

(C*) 2/322 )yx(

)jyix(3F

(D) F

2/322 )yx(

)jxiy(3

Sol. (A, B,C) Since ;W = dr.FClearly for forces (A) and (B) the integration do not require any information of the path taken.

For (C) : Wc=

)jdyidx(.

)yx(

)jyix(32/322 =

2/322 )yx(

dyydxx3

Taking : x2+ y2= t

2xdx + 2y dy = dt

xdx + ydy =2

dt W

c= 2/32/3 t

dt

2

3

t

2/dt3

which is solvable.

Hence (A), (B) and (C) are conservative forces.But (D) requires some more information on path. Hence non-conservative.


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Page # 16

gy (A,B, C) pwafd ; W = dr.F

(A) rFkk (B)


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Page # 17

22. A ball thrown from position A against a smooth circular wall rebounds and hits position B at the other end of the

diameter through A. If the coefficient of restitution between ball & wall is e and = 30, f ind 3e. Do not considerany force on ball except contact force due to wall at time of collision.

,d xsan fLFkfrAls fpduh okkdkj nhokj ij Qsadh tkrh gS ,oa nwljs fljs ij fp=kkuqlkj f LFkfr B ij Vdjkrh gSA tksfdAlsxqtjus okys O;kl ds nwljs fljs ij gSA ;fn xsan o nhokj ds e/; R;koLFkku xq.kkad e gS ,oa = 30 gS] rks3edk eku KkrdjksA xsan ij VDdj ds le; lEidZ cy ds vfrfjDr dksbZ cy ugha gSA

Ans. 1

Sol.

v sin= e u cosv cos= u sin

tan=tan

e

e = tan2=3

1

3e = 1.

23. A rocket of total mass 1000kg initially is launched from ground. The gases are ejected at the rate 20kg/s with

velocity 1000 m/s relative to rocket vertically downwards. The initial acceleration of the rocket is a (in m/s2). Find

g

a. (Take g = 10m/s2)

1000kg dqy nzO;eku dk ,d jkWdsV izkjEHk esa /kjkry ls NksM+k tkrk gSA xSl jkWdsV ds lkis{k 1000 m/s ds osx ls 20kg/s dh

nj ls m/okZ/kj uhps dh vksj mRlftZr gksrh gSA jkWdsV dk izkjfEHkd Roj.k a (m/s2esa) gksxkA Kkr djks ga

(g = 10m/s2)

Sol. Thrust force iz.kksn cy

F =dtdm . u

rel = 20 1000 N

F

mgF

net= F mg

= 20000 10000

= 10000 N

a = 10 m/s2

g

a= 1

Ans. 1 [Answer Changed by GS Sir]


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Page # 18

24. AB is a long frictionless horizontal surface. One end of an ideal spring of spring constant K is attached to a block

of mass m, which is being moved left with constant velocity v, and the another end is free. Another block of mass

2m is given a velocity 3v towards the spring. Magnitude of work done by external agent in moving m with

constant velocity v in long time is times mv2 . Find the value of AB ,d yEch ?k"kZ.kghu {kSfrt lrg gSA K fu;rkad dh ,d vkn k Z fLiazx gS] bldk ,d fljk m nzO;eku ds CykWd ls tqM+k gS]tks ckW;h vksj fu;r osx vls xfr khy gS] rFkk vU; fljk eqDr gSA 2m nzO;eku ds vU; CykWd dks fLiazx dh vksj 3v osx fn;ktkrk gSA yEcs le; rd m nzO;eku dks fu;r osx v ls xfr khy j[kus ds fy, ckg~; dkjd }kjk fd;k x;s dk;Zdk ifjek.k

mv2 dk xquk gSA dk eku Kkr djs:

Ans . 8

Sol. Situation after long time

yEcs le; i pkr~ fLFkfr;kW

Work done fd;k x;k dk;Z= K =2

1(2m)(v)2

2

1(2m) (3v)2=

= mv2 9mv2= 8mv2.

25. A ball of mass 'm' is suspended from a point with a massless string of length '' in form of a pendulum. This

ball is given a horizontal velocity g4 at bottom most point. When string makes an angle 60 from lower

vertical,tc

a

a= p. Write the value of p2. (g = 10 m/s2)

'm' nzO;eku dh ,d xsan '' yEckbZ dh nzO;ekughu Mksjh ls yksyd ds :i eas vkyfEcr gSA bl xsan dks fuEure fcUnq ls

{kSfrt osx g4 fn;k tkrk gSA tc Mksjh fupys v)Zok esa /okZ/ kj ls 60 dk dks.k cukrh gS] rc p =t

c

a

aA

p2 dk eku fyf[k,A (g = 10 m/s2)Ans. 12

Sol. at= g sin 60 =

2

g3

ac=

R

2

2

1m 2 mgR Cos60 =

2

1m (4gR) mgR.

2

1m 2 =

2

3mgR

2 = 3gR .

ac=

R

gR3= 3g

at=

2

g3, a

c= 3g

32g3

g32

a

aP

t

c

Ans.


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Page # 19

26. An ideal string ACB passes through a smooth ring of mass m = 5 kg as shown in the figure. The radius of circle

in which the ring moves is 2.5 m. Find the speed of the ring in m/s.

,d vkn kZ jLlh ACB, m = 5 kg nzO;eku dh fpduh oy; ls fp=kkuqlkj xqtjrh gSA ok dh f=kT;k ftlesa ok xfr djrk gS]2.5 m gSA oy; dh pky Kkr djksA

2.5 mC5 kg

53

37

A

B

Ans. 5Sol. T cos 37 + T cos53 = mg

T sin 53 + T sin 37 =R

mv2

v = Rg = 5 m/s

27. Two particles A and B are revolving with constant angular velocity on two concentric circles of radius 1m and 2m

respectively as shown in figure. The positions of the particles at t = 0 are shown in figure. If mA= 2kg,

mB= 1kg and

AP

andBP

are linear momentum of the particles then what is the maximum value of BA PP

in

kg-m/sec in subsequent motion of the two particles.nks d.k A rFkkB fu;r dks.kh; osx ls e k% 1m rFkk 2m f=kT;k ds nks ldsUnzh; okks ij fp=kkuqlkj pDdj yxk jgs gSA t =

0 ij d.kksa dh fLFkfr;k fp=k esa iznf kZr gSA ;fn mA= 2kg, m

B= 1kg rFkk

AP

oBP

d.kksa ds js[kh; laosx gS rks nksuksa d.kksa

dh vxyh xfr esa BA PP

dk vf/kdre eku kg-m/sec esa) Kkr djksA

2m/s

3m/s

BA


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Page # 20

Sol. Since angular velocities of the particles are different, after some time, two particles may move parallel. In such

case BA PP

is maximum.

pqafd d.kksa ds dks.kh; osx vyx&vyx gS dqN le; ckn nksuksa dk ,d nwljs ds lekUrj xfeku gksxsa A bl le; BA PP

vf/kdre gksxk

maxBA PP

= (2 2 + 1 3) kg m/s = 7 kg m/s

Ans. 7

28. A par ticle of mass m = 1 kg is ly ing at rest on x-axis, exper iences a net force given by law

F = x (3x 2) Newton, where x is the x-coordinate of the particle in meters. The magnitude of minimum

velocity in negative x-direction to be imparted to the particle placed at x = 4 meters such that it reaches

the origin is27

Pm/s. Find the value of P..

m = 1 kg nzO;eku dk d.k x-v{k ij fojkekoLFkk esa j[kk gqvk gS ;g d.k fu;e F = x (3x 2) }kjk dqy cy vuqHkodjrk gSA tgk x, d.k dk x-funsZ kkad ehVj esa gSaA x = 4 ij j[ks gq;s d.k dks ewy fcUnq ij igqpkus ds fy, _.kkRed

x-fn kk esa U;wure osx dk ifjek.k27

Pm/s gSA P dk eku Kkr djksA

Ans. 2600

Sol. The particle is at equilibrium at x = 0 and x =3

2.

The minimum speed imparted to the particle should be such that it just reaches x =3

2from there on it

shall automatically reach x = 0

2

1mv 2=

3/2

4

dxF = 3/2

4

dx)2x3(x =27

1300or v =

27

2600m/s

x = 0 rFkk x =3

2ij d.k lkE;oLFkk esa gksxk .

x = 0 ij d.k LFkkbZ lkE;oLFkk esa gksxk rFkk x =3

2ij vLFkkbZ lkE;oLFkk esa gksxkA

vr% d.k dks bruh U;wure pky nh tk;s fd ;g x =3

2ij igqap tk;s rFkk blds i pkr~ Lor% x = 0 ij igqap tk;s

2

1

mv2

=

3/2

4

dxF=

3/2

4

dx)2x3(x= 27

1300

or v = 27

2600

m/s


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Page # 21

29. Two blocks of masses m1= 10 kg and m

2= 20 kg are connected by a spring of stiffness k = 200 N/m. The

coefficient of friction between the blocks and the fixed horizontal surface is = 0.1. Find the minimum constant

horizontal force F (in Newton) to be applied to m1in order to slide the mass m

2. (Take g = 10 m/s2)

nks CykWd ftuds nzO;eku m1= 10 fdxzk- rFkk m

2= 20 fdxzk- gS k = 200 N/m fLizax fu;rkad dh fLax ls tqM+s gq, gSA CykWdksa

rFkk lrg ds chp ?k"kZ.k xq.kkad = 0.1 gSA m1ij vkjksfir U;wure fu;r {kSfrt cyF U;wVu esa Kkr djksftlls dh nzO;eku

m2fQly tk,A (g = 10 m/s2)

Ans. 20

Sol.

WF + WSp + Wfric = K

Fx 2

1Kx2 m

1g x = 0 and Kx = m

2g

F 21 m

2g m

1g = 0 F = m

1g +

2gm2 = 0.1 10 10 +

210201.0 =20 N

30. Work done by force F to move block of mass 2kg from A to C very slowly is (76 )J. Force F is always actingtangential to path. Equation of path AB is x2= 8y and BC is straight line which is tangent on curve AB at point

B (between block and path ABC is 0.5). Then value of '' is [g = 10 m/s2]:2kg CykWd dksAls C rd /khjs&/khjs ystkus ij cy F }kjk fd;k x;k dk;Z (76 )J gSA cy F geskk iFk dh Li kZ js [kk ds vuqfn kdk;Zjr gSA AB iFk dh lehdj.k x2= 8y vkSj BC ljy js[kk gSA tks oAB ds fcUnq B ij Li kZ js[kk gSA iFk ABC o CykWdds e/; ?k"kZ.k xq.kkad = 0.5 gSA rc] '' dk eku gSS& [g = 10 m/s2]:

Ans. 5

Sol. Slope of line BC (js[kk BC dk


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Page # 22

If the mass m is taken from A to C slowly work done by friction will always be equal to the Wf= mgx

;fn O;eku m dks cgqr /kheh xfr ls AlsC ysdj tk;k tk;s rks ?k"kZ.k }kjk fd;k x;k dk;Z ges kk Wf= mgx gksxkA

Now, by vc Wnet

= KE = 0W

F mg(10 + 2) mg(10 + 4) = 0

WF= 380 = 76 5 = 5.

COMPREHENSION-1 (Q.34 to Q.36)

A particle of mass m collides elastically with the pan of mass (M = 2m) of a spring balance, as shown in figure. Pan

is in equillibrium before collision. Spring constant is k and speed of the particle before collision is v0. Answer the

following three questions regarding this collision.

,d m nzO;eku dk d.k fLizax rqyk ds iyMs ftldk nzO;eku M (M = 2m) ls fp=kkuqlkj VDdjkrs gSA iyMk VDdj dsiwoZ lkE;koLFkk esa gSA fLizax fu;rkad k gS] rFkk VDdj ds iwoZ d.k dk osxv

0gSA bl VDdj ds QyLo:i fue iz uksa ds

mkj nhft,A

31. Maximum compression in the spring after the collision is

VDdj ds i pkr~ fLizax esa vf/kde~ laihMu gksxk &

(A*)k

m2

3

v0(B)

k3

m2v

0(C)

k3

mv

0(D)

k

mv

0

32. Maximum height attained by the particle from the point of collision after collision is

VDdj ds fcUnq ls d.k }kjk izkIr vf/kdre~ pkbZ Kkr dhft,A

(A)g16

v20(B)

g8

v 20(C)

g36

v20(D*)

g72

v20

33. Minimum kinetic energy of the particle after collision is

VDdj ds i pkr~ d.k dh U;wure xfrt tkZ gksxhA

(A)8

mv 20(B*)

8

mv3 20(C)

4

mv3 20 (D)2

mv 20


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Page # 23

Sol. JBC JAC

v1+ v2= V0/2

2mv2 mv

1= m

2

v0 v1=

6

v0v

2=

3

v0

vf/kdre lEihM+u Maximum compression =m2

K

v2

=k

m2

3

v0

vf/kdre pkbZ Maximum height =g2

v21=

g72

v20

U;wUkre xfrt tkZ minimum kinetic energy =

2

0

2v3m

21

=8

mv3 20

COMPREHENSION -2 vuqPNsnTwo particles are moving in diff erent circles in same plane with different angular velocities as shown in

figure. At t = 0, initial positions of particles A and B are shown by dots on the respective circles. Initial

distance between particles is 1m. Particle A move anticlockwise in the first circle whereas B moves

clockwise in the second circle. Angle described (rotated) by A and B in time 't ' are A=

t2 and B=

t) respectively. Here is in radian and t is in second. Radius of each circle is shown in diagram.

nks d.k] fHkUu okks aesa (leku ry esa) fp=kkuqlkj fHkUu dks.kh; osx esa xfr dj jgs gSA t = 0 ij d.kA rFkk B dh kjfEHkd fLFkfr

laxr okksa esafcUnq }kjk n kkZbZ xbZ gS A d.kksads e/; kjfEHkd nwjh 1m gSA d.kA Fke ok esa okekorZ rFkkB nwljs ok esanf{k.kkorZ

?kwerk gSAA rFkk B }kjk le; 't'esa fu:fir (?kwf.kZr)dks.k e k% A=

t

2rFkk

B=

t) gSA ;gk jsfM;u esarFkk t lSd.M

ek=kd esa gSA R;sd ok dh f=kT;k fp=k esa n kkZ,uqlkj gS&

34. Find the magnitude of acceleration of A at = 1 sec

(A)3

2m/s (B)

7

2(C*)

4

2(D) None of these


24/30

Page # 24

Sol. At t ime t = 1 sec positions of A and B are

t = 1 sec ijA rFkk B dh fLFkfr;k gS

acceleration of A )j(ra 121A

= )j()1(

2

2

A dk Roj.k )j(ra 121A

= )j()1(

2

2

35. At time t = 1 sec, the magnitude of accelerati on of A with respect to B is

le; t = 1 sec, ij A dk B ds lkis{k Roj.k dk ifjek.k gksxk :

(A*) 2

2

sec

m65

4

(B) 2

2

sec

m7

2

(C) 2

2

sec

m15

3

(D) 2

2

sec

m7

4

Sol. At time t = 1 sec

t = 1 sec ij

)i(ra 222B

= )i(2 2

)i(2)j(4

aa 22

BA

arel

=

2/12 4

16

1

= 65

4

2m/sec2

36. At time t = 2 second, the angular velocity of the particle A with respect to the par ticle B is

le; t = 2 lSd.M ij A dk B ds lkis{k dks.kh; osx gksxk :

(A) 5rad/sec (B)2

3rad/sec (C)

3

2rad/sec (D*)

6

5rad/sec

Sol. At t ime t = 2 sec, position of A and B are

le; t = 2 lSd.M ij, A rFkk B dh fLFkfr gS

vA=

1r

1=

2

)1(

2

m/sec.

vB=

2r

2= 2m/sec.

distance nwjh AB = 3m

=AB

vv BA =

3

22/ =

6

5rad/sec.


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Page # 25

COMPREHENSION-3

Figure shows an irregular wedge of mass m placed on a smooth horizontal surface. Part BC is rough.

fp=kkuqlkj m nzO;eku dk ,d vle:i ost,d fpduh {kSfrt lrg ij fLFkr gSA Hkkx BC [kqjnjk gSA

h

H

m B Cm ost

D

E

37. What minimum velocity should be imparted to a small block of same mass m so that it may reach point B :

leku nzO;eku m ds ,d NksVs CykWd dks fdruk U;wure osx iznku djsa] ftlls ;g B fcUnq ij igqp lds &

(A*) gH2 (B) gH2 (C) )hH(g2 (D) gh

Ans. Let u be the required minimum velocity. By momentum conservation :

ekuk u vko ;d U;wure osx gSA laosx laj{k.k ls]mu = (m + m)v v = u/2.

m

h

v

v

mB

C

( )mPpre fcUnqij

Energy equation : (tkZ lehdj.k ls)

2

1mu2=

2

1(2m)v2+ mgH.

Substituting v = u/2 : (v = u/2 j[kus ij)

u = 2 gH

38. The velocity of wedge when the block comes to rest (w.r.t. wedge) on part BC is :ost dk osx D;k gSa] tc xqVdk ost ds BC Hkkx ij ost ds lkis{k :d tkrk gS &

(A*) gH (B) hH(g (C) gH2 (D) none of these buesa ls dksbZ ugha

Ans. When the block comes to rest, the wedge continues to move at V =2

u= gH on the smooth surface. (since,

momentum of wedge-block system remains conserved).

tc CykWd fojke ij vkrk gS] rks ost fpduh lrg ij V =2

u= gH ls xfr djrk gSA(pwafd] ost CykWd fudk; dk laosx

lajf{kr jgrk gS).

39. If the coefficient of friction between the block and wedge is, and the block comes to rest with respect to wedgeat a point E on the rough surface then BE will be

;fn xqVds rFkk ost ds chp ?k"kZ.k xq.kkad gSa] rFkk xqVdk [kqjnjh lrg ij fcUnq E ij ost ds lkis{k xqVdk fLFkj voLFkk esavk tkrk gS] rc BE gksxh &

(A)H

(B*) hH

(C)h

(D) none of these buesa ls dksbZ ughaAns. By work-energy theorem on the system : fudk; ij dk;Z&tkZ izes; ls]

(mg) (BE) mgh =

4

1mu2

= 41 m (4 gH) BE =

hH


26/30

Page # 26

MATCH THE COLOUMN

40. In columnII different situations are shown in which one object collides with the another object. In each case

friction is absent and neglect effect of nonimpulsive forces. In columnI different direction are given.

You have to match the directions for each case in which momentum conservation can be applied on

object A or object B or system A & B.(Assume that objects do not bounce off the ground)

LrEHkII esa fofHkUu fLFkfr;k n kkZbZ xbZgS ,ftlesa ,d oLrq nwljh oLrq ls Vdjkrh gSA R;s d fLFkfr esa ?k"kZ.k vuqifLFkr gS,oa vuvkosxh cyksa ds izHkko dks ux.; ekusaA LrEHkI eas fofHkUu fn kk,sa nh xbZ gSA

vkidks fn kkvksa dks R;sd ml fLFkfr ls lqesfyr djuk gS ftlesa oLrq A ;k oLrqB ;kA o B ds fudk; ij laosx laj{k.kyxk;k tk ldrk gSA (ekurs gq;s fd oLrq;s Hkw fe ls Vdjkdj mNyrh ugh gSA)

Column-I LrEHkI Column-IILrEHkII

(A) Along the line of impact (p)

VDdj dh js[kk ds vuqfn k

(B) Perpendicular to line of impact (q)

VDdj dh js[kk ds yEcor~

(C) In horizontal direction (r)

{kSfrt fn kk esa

(D) In vertical direction (s) RA= R

B

/okZ/kj fn kk esa

(t)

Ans. (A) (s) ; (B) (p,q,r,s,t) ; (C) (p,q,r,s,t) ; (D) (s,t)

Sol. (p)

(A) Normal force from ground lies along line of impact. Hence (A) is not answer.

/kjkry ls vfHkyEc cy VDdj dh js[kk ds vuqfn k yxrk gS vr% (A) lgh ugh gSA(B) Since no external force act perpendicular to the line of impact. (B) is an answer.

D;ksfd VDdj dh js[kk ds yEcor~ dksbZ cy ugh yxrk vr% (B) lgh gSA(C) Horizontal direction is same as direction perpendicular to the line of impact. (C) is an answer.

{kSfrt fn kk] VDdj dh js[kk ds yEcor~ fn kk ds leku gS vr% (C) lgh gSA(D) Normal impulse from ground lies in vertical direction. (D) is not an answer.

/kjkry ls vfHkyEc vkosx yEcor~ fn kk esa gS vr% (D) lgh ugh gSA

(q)


27/30

Page # 27

(A) The component of normal force from ground lies along the line of impact. Hence not an answer.

/kjkry ls vfHkyEc cy dk ?kVd] VDdj dh js[kk ds vuqfn k gS vr%lgh mkj ugh gSA(B) No external force perpendicular to the line of impact for A.

A ds fy;s VDdj dh js[kk ds yEcor~ dksbZ ckg~; cy ugh gSA(C) For the system A + B there is no external froce along horizontal direction. Hence an answer.

A + B fudk; ds fy;s {kSfrt fn kk esa dksbZ ckg~; cy ugh gSA vr% lgh mkj gSA(D) For B the normal force from ground is balanced by the impulsive force by A. Hence momentum will not

remain conserved in the vertical direction. B ds fy; /kjkry dk vfHkyEc cy] Ads vkosxh cy ls larqfyr gksxkA vr% /okZ/kj fn kk ds vuqfn k laosx lajf{kr ughajgsxkA

(r)

(A) The component of tension force of thread lies along the line of impact. Hence not an answer.

/kkxs ds ruko cy dk ?kVd VDdj dh js[kk ds vuqfn k gksxk vr% ;g lgh ugh gSA(B) No external force perpendicular to the line of impact for A.

A ds fy;s VDdj dh js[kk ds yEcor~ dksbZ ckg~; cy ugh gSA

(C) For system A + B there is no external f roce along horizontal direction. Hence an answer.fudk; A + B ds fy;s {kSfrt fn kk esa dksbZ ckg~; cy ugh gSA vr% lgh gSA(D) For B the tension force from thread is balanced by the impulsive force by A. Hence momentum will not

remain conserved in the vertical direction.

B ds fy;s /kkxs dk ruko cy A ds vkosxh cy ls larqfyr gksxkA vr% /okZ/kj fn kk ds vuqfn k laosx lajf{kr ugha jgsxkA

(s)

(A) & (C) are the same direction and there is no external force for the system A + B. Hence answer.

fudk; A + B ds fy;s dksbZ ckg~; cy ugh gS rFkk (A) & (C) leku fn kk esa gS vr% lgh gSA(B) & (D) are the same direction and there is no net force for the system A + B. Hence answer.

B vkSj D leku fn kk esa gS rFkk fudk; A + B ds fy;s dqy cy kwU; gSA vr% lgh gSA

(t)

(A) The component of normal force from ground lies along the line of impact. Hence not an answer.

/kjkry ls vfHkyEc cy dk ?kVd] VDdj dh js[kk ds vuqfn k yxrk gS vr% ;g lgh ugh gSA(B) Is answer because the normal force from the ground is balanced for A. Hence an answer.

lgh gS D;ksfd /kjkry ls vfHkyEc cy A ds fy;s larqfyr gSA(C) For the system A + B there is no external froce along horizontal direction. Hence an answer.

fudk; A + B ds fy;s dksbZ ckg~; cy {kSfrt fn kk esa ugh gSA vr% lgh gSA

(D) For A the normal force from ground is balanced by the impulsive force by B. Initial and final momentum is zero. Hence an answer.

A ds fy;s /kjkry dk vfHkyEc cy] B ds vkosxh cy ls larqfyr gSA vr% izkjfEHkd vkSj vfUre laosx kwU; gSA vr% ;g lgh gSA


28/30

Page # 28

41. A small spherical ball of mass m is projected from lowest point (point P) in the space between two f ixed,

concentric spheres A and B (see figure). The smaller sphere A has a radius R and the space between the two

spheres has a width d. The ball has a diameter very slightly less than d. All surfaces are frictionless. Speed of

ball at lowest point is v. NAand N

Brepresent magnitudes of the normal reaction force on the ball exerted by the

spheres A and B respectively. Match the value of v given in columnI with corresponding results in columnII.

fp=kkuqlkj nks ldsUnzh; fLFkj xksyksaA o B ds e/; ds {ks=k ds fuEure fcUnq Pls m nzO;eku dh NksVh xksykdkj xsan dks {ksfirfd;k tkrk gSA NksVs xksysAdh f=kT;k R gS o nksauks xksyksa ds e/; {ks=k dh pkSM+kbZ d gsA xsan dk O;kl d ls gYdk lk de gSA lHkh

lrg ?k"kZ.kghu gSA fuEure fcUnq ij pky v gSA NAoNBxsan ij e k% xksyAo B }kjk vkjksfir vfHkyEc frf;k cy dsifjek.kgSA dkWyeI esa fn;s x;s v ds eku ds fy, dkWyeII esa lEcfU/kr ifjek.k ls feyku dhft,A

ColumnI ColumnII

(A) gRv (p) maximum value of NA= 0

(B) gR2v (q) minimum value of NB= 0

(C) gR3v (r) maximum value of NB= 6 mg

(D) gR5v (s) maximum value of NB= 4 mg

(t) maximum value of NB= 2 mg

dkWyeI dkWyeII

(A) gRv (p) NAdk vf/kdre eku = 0

(B) gR2v (q) NBdk U;wure eku = 0

(C) gR3v (r) NBdk vf/kdre eku = 6 mg

(D) gR5v (s) NBdk vf/kdre eku = 4 mg

(t) NBdk vf/kdre eku = 2 mg

Ans. (A) p,t ; (B) p,q ; (C) q,s ; (D) p,q,r

Sol. Ball only loose contact with surface B when v is in range Rg5vRg2 so for A,B,D maximum value of NA

is zero for option C ball lose contact with surface B at some point.

xsan Blrg ls lEidZrHkh Nks M+rh gS tc v dh ijkl Rg5vRg2 dse/; gS ] vr%A,B,D dsfy, NAdk vf/kdre eku

kwU; gSA fodYi Cds fy, leku fcUnq ij Bls ls lEidZ NksM+ nsrh gSAmaximum value of N

Bis lowest point and given N

Bdk vf/kdre eku fuEure fcUnq ij gS

N = mg +R

mv2.


29/30

Page # 29

42. Figure shows four situations in which a small block of mass 'm' is released from rest (with respect to smooth

fixed wedge) as shown in figure. Column-II shows work done by normal reaction with respect to an observer who

is stationary with respect to ground till block reaches at the bottom of inclined wedge, match the appropriate

column (Assume that there is infinite friction between block and floor of cabin) :

fp=k esa pkj fLFkfr;ka nf kZr gSA ftuesa 'm' nzO;eku dk NksVk CykWd fLFkjkoLFkk ls NksM+k tkrk gS (fpdus tM+or~ ost ds

lkis{k)A dkWye-II esa tehu ij fLFkr fLFkj s{kd ds lkis{k CykWd }kjk urry ds ry rd igqpus esa vfHkyEc cy }kjk

fd;k x;k dk;Z nf kZr gSA laxr dkWye dk feyku dhft,A (ekuk CykWd rFkk dsfcu ds ry ds e/; vuUr ?k"kZ.k cy

mifLFkr gS) :ColumnI ColumnII

(A)

45

h v = 2gh

(p) Positive /kukRed

(B)

45

h

v = 2gh

(q) Negative _.kkRed

(C) 45

h

45

v = 2gh

(r) equal to mgh in magnitude ifjek.k esa mgh ds cjkcj

(D)

45

h

v = 2gh (s) equal to zero kwU; ds cjkcj

(t) equal to 2 mgh in magnitude

ifjek.k esa 2 mgh ds cjkcjAns. (A) (p, t) ; (B) (p, t) ; (C) (s) ; (D) (q, t)


30/30

43. STATEMENT-1 : In a perfectly inelastic coll ision between two bodies, velocity of each body after collision

is same.

STATEMENT-2 : In a perfectly inelastic coll ision between two bodies, velocity of seperation is zero.

(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.

(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1

(C) Statement-1 is True, Statement-2 is False

(D*) Statement-1 is False, Statement-2 is True

oDrO;-1 : nksa fi.M+ksa ds e/; iw.kZr% vR;kLFk VDdj gksus ij] VDdj ds i pkr~ izR;sd fi.M+ dk osx leku gksrk gSAoDrO;-2 :nksa fi.M+ksa ds e/; iw.kZr% vR;kLFk VDdj gksus ij vyxko osx kwU; gksrk gSA

(1) oO;-1 lR; gS, oO;-2 lR; gS ; oO;-2, oO;-1 dk lgh Li"Vhdj.k gSA(2) oO;-1 lR; gS, oO;-2 lR; gS ; oO;-2, oO;-1 dk lgh Li"Vhdj.k ugha gSA(3) oO;-1 lR; gS, oO;-2 vlR; gS(4*) oO;-1 vlR; gS] oO;-2 lR; gS

44. STATEMENT-1 : For a stone projected horizontally, the magnitude of the tangential acceleration keeps on

decreasing even through the speed of particle keeps on increasing.

STATEMENT-2 :A given particle speeds up under action of constant acceleration. The tangential acceleration of

any particle is the component of its acceleration in the direction of its velocity. The tangential acceleration of this

given particle increases if angle between velocity and acceleration keeps on decreasing.

(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.

(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1

(C) Statement-1 is True, Statement-2 is False

(D*) Statement-1 is False, Statement-2 is True

oDrO;-1 :{kSfrtr% iz{ksfir ,d iRFkj ds fy,] Li khZ Roj.k dk ifjek.k ?kVrk tkrk gS ;|fi d.k dh pky c

physics dpp 4 (jp & jr) advanced fc

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