DPP_5 to 8

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TARGET : JEE (MAIN + ADVANCED)

Syllabus : XI class syllabus

PHYSICS

DPP No. : 5 to 8

DPP Syllabus : Thermod ynamics , Circul ar mot ion, Soun d w ave, Project i le mo tio n, SHM,Newton s law, Centre o f mass, G.O. , F lu id mechan ics

DPP No. : 05

ANSWER KEY OF DPP No. : 05

1. (B) 2. (C) 3. (A,C) 4. 4 5. zero 6. 1 7. (A)8. (D) 9. (B) 10 . (A) p (B) q (C) p,q (D) s

1. Moment of inertia of a uniform quarter disc of radius R and mass M about an axis throughits centre of mass and perpendicular to its plane is :,d leku ,d pkSFkkbZ pdrh] ftldh f=kT;k R, rFkk nzO;eku M gS] dk blds nzO;eku dsUnz o ryds yEcor~ v{k ds lkis{k tM+Ro vk?kw.kZ gksxk&

(A) 2MR

2

M2

3R4 z (B*) 2MR

2

M2

3R42

(C)2

MR2+ M

2

3R4

(D)2

MR2+ M

2

3R4

2

Ans. M.I. about O is2

MR2

By parallel-axis theorem :2

MR2

= cm +2

23R4

M

cm =22

3R4

.2M2

MR

1

2. Angle of incidence of the incident ray for which reflected ray intersect perpendiculaly the principal axis.vkifrr fdj.k ds fy, vkiru dks.k dk eku ftlds fy, ijkofrZr fdj.k eq[; v{k dks yEcor~izfrPNsn djrh gS] gksxk

i

C

(A) 0 (B) 30 (C*) 45 (D) 60

Sol.

i

C

i

i

A

B

In the figure i + i = 90 i = 45

3. Heat is supplied to a certain homogeneous sample of matter at a uniform rate. Itstemperature is plotted against time as shown in the figure. Which of the followingconclusions can be drawn? k

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fdlh lekax inkFkZ ds uewus dks ,d leku nj ls "ek iznku dh xbZA uhps fn;s fp=k ds vuq:imlds rki dk le; ds lkFk xzkQ [khapk x;kA uhps fn;s x;s fu"d"kks esa dkSulk lgh gS\

r ki

l e;

(A*) its specific heat capacity is greater in t he solid state than in the liquid state.(B) its specific heat capacity is greater in the l iquid state than in the solid state.(C*) its latent heat of vaporization is greater than its latent heat of fusi on.(D) its latent heat of vaporization i s smaller than its latent heat of fusi on.(A*) bldh fof'k"V "ek/kkfjrk Bksl voLFkk esa nzo voLFkk ds vis{kk vf/kd gSA(B) bldh fof'k"V "ek/kkfjrk nzo voLFkk esa Bksl dh vis{kk vf/kd gSA(C*) bldh ok"iu dh xqIr "ek xyu dh xqIr "ek ls vf/kd gSA(D) bldh ok"iu dh xqIr "ek xyu dh xqIr "ek ls de gSA

Sol . Slope of graph is greater in the solid state i.e., temperature is rising faster, hence lowerheat capacity.The transition from solid to liquid state takes lesser time, hence latent heat is smaller.

mkj xzkQ dk

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There is no change in angular of incidence due to rotation of mirror. Ans . zero

6. A water tank stands on the roof of a building as shown. Find the value of h (in m) for which the horizontaldistance 'x' covered by the water is greatest.fp=kkuqlkj ,d ikuh dh Vadh edku dh Nr ij j[kh gS] rks h dk og eku ehVj esa Kkr djks]ftlds fy, ikuh }kjk r; dh x;ha {kSfrt nwjh 'x' dk eku vf/kdre~ gksxk &

1m h

3m

x

Ans . 1

Sol. Vefflux = gh2

time of fall t =g

2)h4(

x = Vefflux t = )h4(h2

the roots of x are (0,4) and the maximum of x is at h = 2.

The permitted value of h is 0 to 1 clearly h = 1 will give themaximum value of x in this interval.Alit er Solution:If the column of water itself were from ground upto a height of 4m, h = 2m would give the maximum rangex. Farther the hole is from this midpoint, lower the range. Here the nearest point possible to this midpoint isthe base of the container. Hence h = 1m.

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Hindi. Vefflux = gh2

fxjus dk le; t =g

2)h4(

x = Vefflux t = )h4(h2

x ds oxZewy (0,4) gS rFkk x dk vf/kdre eku h = 2 ij gSAh dk vuqekfur eku 0 ls 1 gksxk vr% Li"V :i ls bl vUrjky esa h = 1,x dk vf/kdre eku nsxkAoSdfYid gy :ikuh ds LrEHk dh /kjkry ls pkbZ 4m gS, h = 2m vf/kdre ijkl x nsxkA fNnz e/; fcUnq ij gS];gk blds e/; fcUnq dk lehiorhZ lEhko fcUnq ik=k dk vk/kkj gSA vr% h = 1m.

COMPREHENSIONA quantity of an ideal monoatomic gas consists of n moles initially at temperature T 1 . Thepressure and volume are then slowly doubled i n such a manner so as to trace out a straightline on a P-V diagram.kjEHk esa ,d vkn'kZ ,d ijek.kqd xSl ds T1 rki ij n eksy gSA nkc rFkk vk;ru dks/khjs&/khjs bl izdkj nqxquk djrs gSa fd P-V fp=k ij ,d ljy js[kk vkysf[kr gksrh gS&

7. For this process, the ratio1nRT

W is equal to (where W is work done by the gas) :

bl izf;k ds fy,]1nRT

W vuqikr cjkcj gS (tgk W xSl }kjk fd;k x;k dk;Z gS ) :

(A*) 1.5 (B) 3 (C) 4.5 (D) 6

Sol . W = Area under the curve =23

P1V1 12

12

P2P

V2V

W = o ls f?kjk {ks=kQy =23 P1V1 12

12

P2P

V2V

and vkSj P1V1 = nRT1

Therefore vr%1nRT

w=

11

11

VP

vP.23

8. For the same process, the ratio1nRT

Qis equal to (where Q is heat supplied t o the gas) :

leku izf;k ds fy,]1nRT

Q vuqikr cjkcj gS (tgk Q xSl dks nh xbZ "ek gS ) :

(A) 1.5 (B) 3 (C) 4.5 (D*) 6

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Sol . Q = dU + WdU = nCv dTFor final state P2V2 = 2P1 2V1vfUre voLFkk ds fy, P2V2 = 2P 1 2V 1 = 4P 1V1 = nR(4T1)Hence final temp. is 4T1vr% vfUre rki 4T 1 gSA

dU = n .

2

3 R . 3T1 =

2

9nRT1

Q =23

. nRT1 + 29

nRT1 = 6nRT11nRT

Q= 6

9. If C is defined as the average molar specific heat f or the process thenRC

has value

;fn izf;k ds fy, vkSlr eksyj fof'k"V "ek C ls ifjHkkf"kr gks] rksRC dk eku gS &

(A) 1.5 (B*) 2 (C) 3 (D) 6Sol . nC T = Q nC T = 6n RT1

dT = 4T1 T1 = 3T 1n . C . 3T1 = 6nRT1

R

C= 2

10 . Consider a system of particles (it may be rigid or non rigid). In the column- some conditionon force and torque is given. Column- contains the effects on the system. (Letters haveusual meaning)d.kks dk fudk; (;g n`

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TARGET : JEE (MAIN + ADVANCED)

Syllabus : XI class syllabus

PHYSICS

DPP No. : 5 to 8

DPP Syllabus : Thermod ynamics , Circul ar mot ion, Soun d w ave, Project i le mo tio n, SHM,Newton s l aw, Centre o f mass, G.O. , Flu id mechan ics , Fr ic t ion , Rota t ion ,

DPP No. : 06

ANSWER KEY OF DPP No. : 061. (A) 2. (C) 3. (C) 4. (D) 5. (A) 6. (D) 7. (C)8. (A) 9. (B) 10. (A) s (B) q (C) r (D) q

1. If a ball is dropped from rest, it bounces from the floor repeatedly. The coefficient of restitution is 0.5 and

the speed just before the first bounce is 5ms1. The total time taken by the ball to come to rest finally is :;fn ,d xsan dks fLFkj voLFkk ls NksM+k tkrk gS rks ;g ckj & ckj ry ls Vdjkrh gSAizR;koLFkk xq.kkad dk eku 0 . 5 gS rFkk igyh VDdj ls Bhd igys xsan dh pky 5 eh-/ ls- gSA var esa xsan dks fLFkj gksus es yxk le; gS& (A*) 1.5s (B) 1s

(C) 0.5s (D) 0.25sSol. v = 0 + gt t = 0.5 sec

After first collision :Speed becomes 5 (0.5) = 2.5 m/st1 = 2 (0.25) = 0.5t2 = 2 (0.125) = 0.25t3 = 0.125 and so on[where t

iis the time taken to complete the i th to and fro motion after collision]

Total time = 0.5 + [0.5 + 0.25 + 0.125 + ...]= 0.5 +

5.015.0

(Since above is a G.P. with a = 0.5 and r = 0.5)

= 0.5 + 1 = 1.5 sec.

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2. A section of fixed smooth circular track of radius R in vertical plane is shown in the figure. A block isreleased from position A and leaves the track at B. The radius of curvature of its trajectory when it justleaves the track at B is:fp=kkuqlkj /okZ/kj ry esa R f=kT;k dk fpduk fLFkj o`kkdkj iFk nf'kZr gSA ,d CykWd dksfLFkfr A ls NksM+us ij ;g iFk dks fcUnq B ij NksM+rk gS rks fcUnq B dks NksMus ds rqjUrckn blds iFk dh ork f=kT;k gS :

(A) R (B)4R

(C*)2R

(D) none of these buesa ls dksbZ

ughaSol. By energy conservation between A & B

A rFkk B ij tkZ laj{k.k ds fu;e ls

Mg5R2

+ 0 =5

RMg +

21

MV2

V =5gR2

Now, radius of curvature r =

2R

37cosg5/gR2

aV

r

2

ork f=kT;k r =2R

37cosg5/gR2

aV

r

2

3. A bob is attached to one end of a string other end of which is fixed at peg A. The bob istaken to a position where string makes an angle of 30 0 with the horizontal. On the circularpath of the bob in vertical plane there is a peg B at a symmetrical position with respect tothe position of release as shown in the figure. If V c and Va be the minimum tangentialvelocity in clockwise and anticlockwise directions respectively, given to the bob in order tohit the peg B then ratio V c : Va is equal to :

,d jLlh ds ,d fljs ls ckWc tqM+k gS rFkk nwljk fljk [kwaVh (peg) A ls tqM+k gSA ckWc dks{kSfrt ls 30 0 fLFkfr rd ys tk;k tkrk gS rFkk ;gka ls NksM+k tkrk gSA /okZ/kj ckWc dso`kkdkj iFk ij [kwaVh (Peg) B fLFkr gSA vc ckWc dks ;gka ls Li'kZ js[kh; osx nsdjNksM+k tkrk gSA nf{k.kkorZ rFkk okekorZ fn'kkvksa ls ckWc ds [kwaVh (Peg) B ij Vdjkusds fy, U;wure osx e'k% Vc rFkk Va gks rks Vc : Va gS ::

(A) 1 : 1 (B) 1 : 2 (C*) 1 : 2 (D) 1 : 4

Sol. (C) For anti-clockwise motion, speed at the highest point should be . gR Conserving energy at (1) & (2) :

(C) okekorZ fn'kk esa xfr ds fy,] mPpre fcUnq ij pky gR gksuh pkfg,A (1) rFkk (2) ds e/;tkZ laj{k.k ls:

2amv2

1= )gR(m

21

2R

mg

53

37

R

ARR cos53=2R/5 B RR cos37=

R/5

g

37

g cos37

O

Reference line

( )funsZ' kj s[ kk

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va2 = gR + gR = 2gR va = gR2

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For clock-wise motion, the bob must have atleast that much speed initially, so that the string must notbecome loose any where until it reaches the peg B.nf{k.kkorZ xfr ds fy, ckWc ds ikl kjEHk esa de ls de bruk osx gksuk pkfg, ftlls oks

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6. If F = 20 N, with how much force does block A presses the block B;fn F = 20 N, rks fi.M A, fi.M B dks fdrus cy ls nck;sxk &(A) 10 N (B) 20 N (C) 30 N (D*) Zero 'kwU;

Sol . If F = 20 N, 10 kg block will not move and it would not press 5 kg block So N = 0.

7. What should be the minimum value of F, so that block B can press the vertical wal lF dk U;wure eku D;k gksxk , ftlds dkj.k CykWd B /okZ/kj nhokj dks nck ldsaA(A) 20 N (B) 40 N (C*) 60 N (D) 80 N

8. If F = 50 N, the friction force (shear f orce) acting between block B and ground will be :;f n F = 50 N, rks fi.M B vkSj tehu ds chp yxus okyk ?k"kZ.k cy gksxk(A*) 10 N (B) 20 N (C) 30 N (D) None dksbZ ugha

Sol. If F = 50 N, force on 5 kg block = 10 N;f n F = 50 N, rks 5 kg ds CykWd ij cy = 10 N yxsxkA

So friction force = 10 N

vr% ?k"kZ.k cy = 10 N

9. The force of friction acting on B varies with the appl ied force F according to curve :B ij yxus okyk ?k"kZ.k cy] vkjksfir cy ds lkFk fdl o ds vuqlkj cnyrk gS %

(A) (B*) (C) (D)

Sol . Until the 10 kg block is sticked with ground (.. . F = 40 N),No force will be flet by 5 kg bl ock. After F = 40 N, the frictionforce on 5 kg increases, till F = 60 N,and after that, the k inet ic fr ict ion start act ing on 5 kg block,which will be constant (20N)

tc rd 10 kg dk CykWd lrg ds lkFk fpidk jgrk gS (... F = 40 N),rks 5 kg dk CykWd dksbZ cy eglwl ugh djsxkA F = 40 N, dsckn 5 kgokys CykWd ij ?k"kZ.k cy F = 60 N, rd c

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10 . A sample of gas goes from state A to state B in four different manners, as shown by thegraphs. Let W be the work done by the gas and U be change in internal energy along thepath AB. Correctly match the graphs with the statements provided.fp=k esa n'kkZ;s vuqlkj] ,d xSl dk uewuk pkj fofHkUu rjhdksa ls voLFkk A ls B rd igqprkgSA ;fn xSl }kjk ekxZ AB esa fd;k x;k dk;Z W rFkk vkUrfjd tkZ esa ifjorZu U gks] rksxzkQksa dks fn;s x;s dFkuksa ls feykbZ;s &

(A) V

P

A B (p) Both W and U are positive

(p) W rFkkU nksuksa /kukRed gSaA

(B)

P

T

A

B

(q) Both W and U are negative

(q) W rFkk U nksuksa _.kkRed gSaA

(C)

T

V

B

A

(r) W is positive whereas U is negative

(r) W rks /kukRed ijUrq U_. kkRed gSaA

(D)

V

P

A

B

(s) W is negative whereas U is positive

(s) W rks _.kkRed ijUrq U /kukRed gSAAn s . (A) s (B) q (C) r (D) qSol . in (A), V is on vertical axis.

(a) esa, V m/okZ/kj v{k ij gSA

Part-I(Hkkx -I)

As V is icreasing, W is positi ve. tSls V c

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/kukRed gS] fodYi (S), (a) ls tqM+k gSA leku rdZ vU; xzkQksa ij yxk;s tk ldrsgSA

TARGET : JEE (MAIN + ADVANCED)

Syllabus : XI class syllabus

PHYSICS

DPP No. : 5 to 8

DPP Sy l labus : Thermodynamics, Circular motion, Sound wave, Projectile motion, SHM,Newtons law, Centre of mass, G.O., Fluid mechanics, Friction, Rotation, sound wave

DPP No. : 07

ANSWER KEY TO DPP NO. # 7

1. (B) 2. (A) 3. (C) 4. (D) 5.4

125m 6. d = 4000 mm

7. (A) 8. (B) 9. (D)

1. A particle performs S.H.M. on xaxis with amplitude A and time period T. The time taken by the particle totravel a distance A/5 starting from rest is:,d d.k x- v{k ij ljy vkorZ xfr dj jgk gS ftldk vk;ke A rFkk vkorZ dky T gSA d.k }kjk fojke lspyrs gq, A/5 nwjh r; djus essa yxk le; gSA

(A)20T

(B*)2

Tcos1

54

(C)2

Tcos1

51

(D)2

Tsin1

51

Sol . Particle is starting from rest, i. e. from one of its extreme position.

As particle moves a distance5A

, we can represent it on a circle as shown.

d.k fLFkj voLFkk ls pyuk kjEHk djrk gS vFkkZr vius fdlh vUR; fcUnq esa pyuk kjEHkdjrk gSA

tSls fd d.k]5A nwjh r; djrk gS] bls ,d o`k ij n'kkZ;k tk ldrk gSA

cos =54

A5/A4

= cos1

54

t = cos 1

54

t = 1

co s 1

54

2

T

= cos 1

54

Method : As starts from rest i.e. fr om extreme position x = A sin (t + )

f}rh; fof/k% pqafd d.k fLFkjkoLFkk ls xfr djrk gS vFkkZr vUR; fcUnq ls xfr djrk gSA x= A sin (t + )

Px

y

A/54A/5

A

Q

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At t = 0 ij ; x = A =2

A 5A

= A cos t

54

= cos t t = cos154

t =

2

T cos 1

5

4

2. A body of mass 10 kg lies on a rough inclined plane of inclination = sin153

with the

horizontal. When a force of 30 N is applied on the block parallel to & upward the plane, thetotal reaction by the plane on the block is nearly along:

{kSfrt ds lkFk = sin153 urh dks.k okys [kqjnjs ur ry ij 10 kg nzO;eku dh ,d

oLrq j[kh gq;h gSA tc ur ry ds lekUrj ij dh vksj 30 N dk cy yxk;k tkrk gS rc ry}kjk oLrq ij dqy izfrf;k yxHkx fdlds vuqfn'k gSA

(A*) OA (B) OB (C) OC (D) OD

Sol . Frictional force along the in upwarddirection = 10 g sin 30 = 30 NtN = log cos = 80 Nt

Direction of R is along OA.

3. A large open tank has two small holes in its vertical wall as shown in figure. One is a square hole of side 'L'at a depth '4y' from the top and the other is a circular hole of radius 'R' at a depth y from the top. Whenthe tank is completely filled with water, the quantities of water flowing out per second from both holes arethe same. Then, 'R' is equal to :,d cM+h [kqyh Vadh dh /okZ/kj nhokj ij fp=kkuqlkj nks NksVs fNnz gSaA ,d 'L'Hkqtk dk oxkZdkj fNnz ijh lrg ls '4y' xgjkbZ ij o nwljk 'R' f=kT;k dk o`kkdkjfNnz ijh lrg ls y xgjkbZ ij gSA tc Vadh ty ls iwjh Hkjh gS]nksauks fNnzksa lsfr lSd.M ckgj fudy jgs ty dh ek=kk leku gSA rks 'R' cjkcj gS :

4yv1

v2L

2Ry

(A)2

L (B) 2L (C*)2 . L (D) 2L

Sol. Let v1 and v2 be the velocity of efflux from square and circular hole respectively. S 1 and S2 be cross-section areas of square and circular holes.

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Page No # 14

4yv1

v2L

2Ry

v1 = gy8 and v2 = )y(g8

The volume of water coming out of square and circular hole per second is

Q1 = v1S1 = gy8 L2 ; Q2 = v2S2 = gy2 R

2 Q1 = Q2

R = 2 . L

4. A ring of mass m and radius R rolls on a horizontal rough surface without slipping due to anapplied force F. The friction force acting on ri ng is : ,d m nzO;eku rFkk R f=kT;k dh oy; vkjksfir cy F ds izHkko esa fcuk fQlys {kSfrt[kqjnjs /kjkry ij yq

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Page No # 15

,d O;fDr 1m xgjs rj.k rky ds fdukjs ij [kM+k gS ftlesa 2/3 viorZukad dk nzo Hkjk gSAO;fDr dh vk[ks i`Foh ry ls 3 m pkbZ ij gS] rj.k rky ds isans ij ,d flDdk O;fDr dks 300

voueu dks.k ij fn[kkbZ nsrk gS] O;fDr dh vk[k ls flDds dh {kSfrt nwjh _________fp=k esax }kjk nf'kZr feeh esa D;k gksxh \

Ans. d = 4000 mm

Sol.

sin 60 =23

sinr r = 45

S = h = 1 my = H tan600 = 3m

x= S + y = 4m = 4000 mm

ComprehensionA source of sound, emitting frequency of 6000 Hz, moving towards a stationary reflecting wall with speed

50 m/sec. There are five observes A,B,C,D and E as shown in figure. Speed of sound is 350 m/sec./ofu dk lzksr] 6000 Hz vko`fk dh /ofu mRlftZr djrs gq, 50 m/sec. dh pky ls fLFkj ijkorZdnhokj dh vksj xfr'khy gSA ;gk fp=k esa n'kkZ, vuqlkj ikp s{kd A,B,C,D o E gSA /ofu dhpky 350 m/sec. gSA

7. The beat frequency appeared to observer A is(A*) zero (B) 2000 Hz (C) 1750 Hz (D) 400 Hzizs{kd A }kjk izsf{kr foLian vko`fr gSA(A*) 'kwU; (B) 2000 Hz (C) 1750 Hz (D) 400 Hz

8. The beat frequency appeared to observer B isizs{kd B }kjk izsf{kr foLian vko`fr gSA(A) 1750 Hz (B*) 400 Hz (C) 2000 Hz (D) zero 'kwU;

9. The beat frequency appeared to observer E is :izs{kd izs{kd E }kjk izsf{kr foLian vko`fr gSA(A) zero 'kwU; (B) 400 Hz (C) 1750 Hz (D*) 2000 Hz

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Page No # 16

Sol. B =s

00

vvfv2

=

300102

6000 = 400 Hz

A = 0

C = 2s

2s

vv

W2

=

3004006000503502

= 1750 Hz

D = E = f3f4

=3f

= 2000 Hz.

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Page No # 17

TARGET : JEE (MAIN + ADVANCED) 2015

Syllabus : XI class syllabus

PHYSICS

DPP No. : 5 to 8

DPP sy l labus : Thermodynamics, Circular motion, Sound wave, Projectile motion, SHM, Newtonslaw, Centre of mass, G.O., Fluid mechanics, Friction, Rotation, sound wave

DPP No. : 08

ANSWER KEY TO DPP NO. # 8

1. (D) 2. (A) 3. (C) 4. (A) 5. (A ) 6. 13 3

7. (A) 8. (B) 9. (A)

1. A particle is executing SHM according to the equation x = A cos t. Average speed of the particle during

the interval 0 t .

6

lehdj.k x = A cos t vuqlkj d.k ljy vkorZ xfr djrk gSA le; vUrjky 0 t 6 ds nkSjku d.k dh

vkSlr pky gS &

(A)2A3

(B)4A3

(C)

A3(D*)

A3

)32(

Sol . average speed =t

dt.dtdx

t

0

=t

dxt

0

=t

)0(x)t(x =

6/)16/(cosA

= )23(A3

since particle does not change it's direction in the given interval , average speed = V = )32(A3

2.A particle of mass 5 kg is moving on rough fixed inclined plane (making an angle 30 withhorizontal) with constant velocity of 5 m/s as shown in the figure. Find the friction forceacting on a body by the inclined plane. ( take g = 10m/s 2)5 kg nzO;eku dk ,d d.k ,d fLFkj [kqjnjs ur ry tks {kSfrt ls 30 dk dks.k cukrk gS ijfn[kk;s fp=kkuqlkj 5 m/s ds fu;r osx ls xfr djrk gSA ur ry }kjk d.k ij yxus okyk ?k"kZ.k cyKkr dhft,A ( g = 10m/s2)

(A*) 25 N (B) 20 N (C) 30 N (D) none of these buesa ls dksbZugh

Sol. Since the block slides down the incline with uniform velocity, net force on it must be zero. Hence mg sin must balance the frictional force f on the block.

Therefore f = mg sin = 5 10 = 25 N.pwfd CykWd ur ry ij uhps dh vksj ,dleku osx ls fQlyrk gS] rks bl ij usV cy 'kwU; gksxkAblfy;s mg sin CykWd ij ?k"kZ.k cy f dks larqfyr djrk gSA

blfy;s f = mg sin = 5 10 = 25 N.

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Page No # 18

3. A sphere rolls without sliding on a rough inclined plane (only mg and constant forces areacting on the body). The angular momentum of the body:,d xksyk [kqjnjs ur ry ij fcuk fQlys yq R as shownx > R ds fy, = 0 gksxk tSlk nf'kZr gS

Note: As mgsin > mgcos , the point should be inside the sphere.

uksV : pwafd mgsin > mgcos , og fcUnq xksys ds vUnj gksxkA

4. A string of length 1.5 m with its two ends clamped is vibrating in fundamental mode.Amplitude at the centre of the string is 4 mm. Minimum distance between the two pointshaving amplitude 2 mm is:1.5 eh0 yEch Mksjh tks nksuksa fljksa ij c/kh gS] ewyfo/kk esa dEi dj jghgSA Mksjh ds e/; fcUnq dsU ij vk;ke 4 eh0eh0 gSA mu nks fcUnqvksa dschp dh nwjh ftudk vk;ke 2 eh0eh0 gS] gksxh(A*) 1 m (B) 75 cm (C) 60 cm (D) 50 cm

Sol . = 2 = 3m

Equation of standing wavey = 2A sin kx cos ty = A as amplitude is 2A.A = 2A sin kx

2x =

6

x1 = 41

m

and2 . x =

65 x2 = 1.25 m x2 x1 = 1m

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Page No # 19

gy% = 2 = 3m

vizxkeh rjax dk lehdj.k (x = 0 ij fuLiUn ysrs gq,)y = 2A sin kx cos ty = A pqfd vk;ke 2A gSaA = 2A sin kx

2x =

6

x1 = 41

m

vkSj . 2

x = 65

x2 = 1.25 m x2 x1 = 1m

5. The source (S) of sound is moving constant velocity v 0 as shown in diagram. An obsereverO listens to the sound emmited by the source. The observed frequency of the sound :fp=kkuqlkj /ofu dk L=kksr (S) fu;r osx v0 ls lh/kh js[kk esa xfr'khy gSA ,d izs{kdO, S }kjk mRiUu dh x;h /ofu lqurk gSA izs{kd }kjk lquh x;h /ofu dh vko`fk

(A*) continuously decreases (B) continuously increases(C) f irst decreases then increases (D) f irst increases then decreases.(A*) yxkrkj ?kVsxh (B) yxkrkj c

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Page No # 20

COMPREHENSIONA concave mirror of radius of curvature 20 cm is shown in the figure. A circular disc of diameter 1 cm isplaced on the principle axis of mirror with its plane perpendicular to the principal axis at a distance 15 cmfrom the pole of the mirror. The radius of disc starts increasing according to the law r = (0.5 + 0.1 t) cm/secwhere t is time is second.,d vory niZ.k ftldh ork 20 cm fp=kkuqlkj gSA ,d o`kkdkj pdrh ftldk O;kl 1 cm gSrFkk bldks niZ.k ds eq[; v{k ij eq[; v{k ds yEcor~ niZ.k ds /kzqo (pole) ls 15 cm dhnwjh ij fp=kkuqlkj j[kk tkrk gSA vc pdrh dh f=kT;k fu;e r = (0.5 + 0.1 t) cm/sec ds

vuqlkj c