PHYSICS TARGET : JEE (MAIN + ADVANCED)
DPP No. : 5 to 8
Syllabus : XI class syllabus
DPP Sylla yllabu buss : Thermod ynamics , Circul ar mot ion, Soun d w ave, Project ile mo tio n, SHM, N ew ew t o n ’ s l a w , C en en t r e o f m a s s , G .O .O . , F l u i d m e c h a n i c s DPP No. : 05
ANSWER KEY OF DPP No. : 05 (A,C) 4. zero 3. 4 5. 10 . (A) p (B) q (C) p,q (D) s
1. 8.
(B) (B ) (D)
(C) ( B)
1.
Moment of inertia of a uniform quarter disc of radius R and mass M about an axis through its 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 ry
2. 9.
6.
1
7.
(A)
ds yEcor~ v{k ds lkis{k tM+Ro vk?kw.kZ gksxk& 2
An s.
MR 2 M 4R z (A) (A ) 2 3 2 MR 2 4R (C ) + M 2 3 MR 2 M.I. about ‘O’ is 2 By parallel-axis theorem :
MR 2 M 2 4R (B*) 2 3 2 MR 2 4R (D) +M 2 2 3
MR 2 2
2
= c m 2.
2
2
MR 2 4R 4R 2 c m = M 2. 1 + M 2 3 3
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°
A i
Sol.
i i
C
B
In the figure i + i = 90° 9 0° i = 45° 3.
Heat is supplied to a certain homogeneous sample of matter at a uniform rate. Its temperature is plotted against time as shown in the figure. Which of the following conclusions can be drawn? k Page No # 1
fdlh lekax inkFkZ ds uewus dks ,d leku nj ls Å"ek iznku dh xbZA uhps fn;s fp=k ds vuq:i mlds 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 Å"ek/kkfjrk Bksl voLFkk esa nzo voLFkk ds vis{kk vf/kd gSA (B) bldh fof'k"V Å"ek/kkfjrk nzo voLFkk esa Bksl dh vis{kk vf/kd gSA (C*) bldh ok"iu dh xqIr Å"ek xyu dh xqIr Å"ek ls vf/kd gSA (D) bldh ok"iu dh xqIr Å"ek xyu dh xqIr Å"ek ls de gSA S o l . Slope of graph is greater in the solid state i.e., temperature is rising faster, hence lower heat capacity. The transition from solid to liquid state takes lesser time, hence latent heat is smaller.
Bksl ls nzo ifjorZu esa de le; yxk vr% xqIr Å"ek de gSA
4.
An s. Sol.
5.
A block of weight W is dragged across the horizontal floor from A to B by the constant vertical force P acting acting at the the end end of the rope. Calcul Calculate ate the work done done on the block block by the the force force P = ( 3 + 1)N.Assu 1)N.Assume me 2 that block does not lift off the floor. (g = 10 m/s ) W Hkkj dk ,d CykWd] jLlh ds ,d fljs ij dk;Zjr fu;r Å/okZ/kj cy P }kjk {kSfrt Q'kZ ij A ls B rd ?klhVk tkrk gSA cy P = ( 3 + 1)N }kjk CykWd ij fd;k x;k dk;Z Kkr djksA ;g ekfu, fd CykWd] Q'kZ ls lEidZ ugha NksM+rk gSA (g = 10 m/s2)
4
W = Px h h = P sin 30 sin 60 1 2[ 3 1] Ph = 2Ph1 = 3 3 2( 3 1)( 3 1)( 3 ) = = 3
4J
Ans .
A Plane mirror revolves revolves as shown at constant angular velocity velocity making 2 rps about about its normal. With what velocity will the light spot spot move along a spherical screen of radius of 10 m if the mirror is at the centre of curvature of the screen and the light is incident from a fixed direction.
,d lery niZ.k vius vfHkyEc ds ifjr% 2
[email protected] ds fu;r dks.kh; osx ls ?kwerk gSA 10 m f=kT;k ds xksyh; xksyh; insZ ds vuqfn'k çdk'kh; /kCcs /kCcs dk osx D;k gksxk ;fn niZ.k insZ ds oØrk dsUnz ij gks rFkk çdk'k ,d fuf'pr fn'kk ls vkifrr gksrk gSA Sol.
Angular speed of reflected light = 0 rps Page No # 2
There is no change in angular of incidence due to rotation of mirror. An s . zero 6.
A water tank stands on the roof of a building as shown. Find the value of h (in m) for which the horizontal distance '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 ¼ehVj esa½ 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 = 2gh time of fall t =
(4 h)2 g
x = Vefflux t = 2 h( 4 h) 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 the maximum value of x in this interval.
Alit er Soluti on:
If the column of water itself were from ground upto a height of 4m, h = 2m would give the maximum range x. Farther the hole is from this midpoint, lower the range. Here the nearest point possible to this midpoint is the base of the container. Hence h = 1m.
Page No # 3
Hindi. Vefflux = 2gh
fxjus dk le; t = (4 h)2 g
x = Vefflux t = 2 h( 4 h) x ds oxZewy (0,4) gS rFkk x dk vf/kdre eku h = 2 ij gSA h dk vuqekfur eku 0 ls 1 gksxk vr% Li"V :i ls bl vUrjky esa h = 1, x dk vf/kdre eku nsxkA
oSdfYid gy :
ikuh ds LrEHk dh /kjkry ls ÅpkbZ 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. COMPREHENSION
A quantity of an ideal monoatomic gas consists of n moles initially at temperature T 1 . The pressure and volume are then slowly doubled i n such a manner so as to trace out a straight line on a P-V diagram. çkjEHk esa ,d vkn'kZ ,d ijek.kqd xSl ds T 1 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 ratio
W is equal to (where W is work done by the gas) : nRT1
bl izfØ;k ds fy,] W vuqikr cjkcj gS ( tgk¡ W xSl }kjk fd;k x;k dk;Z gS ) : nRT1
(A*) 1.5 Sol.
(B) 3
(C) 4.5
3 W = Area under the curve = P 1 V 1 2
W = oØ ls f?kjk {ks=kQy =
3 P V 2 1 1
(D) 6 V2
2 V1
P2
2P1
V2
2 V1
P2
2P1
and vkSj P 1 V 1 = nRT 1 3 .P1v1 w 2 Therefore vr% = nRT1 P1V1 8.
For the same process, the ratio
Q is equal to (where Q is heat supplied t o the gas) : nRT1
leku izfØ;k ds fy,] Q vuqikr cjkcj gS ( tgk¡ Q xSl dks nh xbZ Å"ek gS ) : (A) 1.5
nRT1 (B) 3
(C) 4.5
(D*) 6
Page No # 4
Sol.
Q = dU + W dU = nC v dT For final state
P2 V 2 = 2P 1 2V 1 vfUre voLFkk ds fy, P 2 V 2 = 2P 1 2V 1 = 4P 1 V 1 = nR(4T 1 ) Hence final temp. is 4T 1 vr% vfUre rki 4T 1 gS A 3 9 dU = n . R . 3T 1 = nR T 1 2 2 3 9 Q Q= . nRT 1 + nR T 1 = 6nRT 1 =6 nRT1 2 2 9.
Sol.
10 .
C has value R ;fn izfØ;k ds fy, vkSlr eksyj fof'k"V Å"ek C ls ifjHkkf"kr gks] rks C dk eku gS & R (A) 1.5 (B*) 2 (C) 3 (D) 6 nC T = 6n RT 1 nC T = Q dT = 4T 1 – T 1 = 3T 1 n . C . 3T 1 = 6nRT 1 C =2 R If C is defined as the average molar specific heat f or the process then
Consider a system of particles (it may be rigid or non rigid). In the column- some condition on force and torque is given. Column- contains the effects on the system. (Letters have usual meaning) d.kks dk fudk; ( ;g n`<+ ;k vn`<+ Hkh gks ldrk gS ) ysaA LrEHk - esa cy o cyk?kw.kZ dh dqN 'krsZ nh xbZ gSA LrEHk - esa fudk; ij izHkko fy[ks x;s gSA
¼inks dk lkekU; vFkZ gSA½
Column-I
C o l u m n - II (A) Fres 0
(p) L system will be constant
(B) res 0 (C) External force is absent ( D) N o n on co ns er va ti ve fo rc e a ct s.
(q ) L system will be constant (r) total work done by all forces will be zero ( s) t ot al m ec ha ni ca l en er gy wi ll b e c on st an t.
LrEHk - I
(p) Psystem fu;r jgsxkA
(A) Fifj.kkeh = 0
An s . Sol.
LrEHk - II
(B) ifj.kke = 0
(q) L system fu;r jgsxkA
(C) ckº; cy vuqifLFkr gS
(r) lHkh cyks }kjk fd;k x;k dqy dk;Z 'kwU;
gksxkA (D) dksbZ Hkh vlaj{kh cy dk;Zjr ugha gksrk gSA (s) gksxhA
dqy
;kaf=kd
ÅtkZ
fu;r
(A) p (B) q (C) p,q (D) s (A) If resultant force is zero, Psystem will be constant.
(B) If resultant torque is zero, L system will be constant.
(C) If external forces are absent, both Psystem and L system will be constant. (D) If no non conservative force acts, total mechanical energ y of system will be constant. (A) ;fn ifj.kkeh cy 'kwU; gks rks] Pfudk; dk fu;r jgsxkA
(B) ;fn ifj.kkeh cyk?kw.kZ 'kwU; gks rks] L fudk; fu;r jgsxkA
(C) ;fn ckº; cy vuqifLFkr gS rks nksuks Pfudk; o L fudk; fu;r gksaxsA (D) ;fn dksbZ Hkh vlaj{kh cy dk;Z ugh djrk gS rks fudk; dh dqy ;kaf=kd ÅtkZ
fu;r jgsxhA
Page No # 5
PHYSICS TARGET : JEE (MAIN + ADVANCED)
DPP No. : 5 to 8
Syllabus : XI class syllabus
DPP Syllabus : Thermod ynamics , Circul ar mot ion, Soun d w ave, Project ile mo tio n, SHM, N ew t o n ’ s l a w , C en t r e o f m a s s , G .O . , Fl u i d m e c h a n i c s , F r i c t i o n , R o t a t i o n , DPP No. : 06
1. 8. 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 5ms – 1. The total time taken by the ball to come to rest finally is :
2. 9.
(C) (B)
ANSWER KEY OF DPP No. : 06 (C) (D) (A) 3. 4. 5. (A) s (B) q (C) r (D) q 10.
(A) (A)
6.
(D)
7.
(C)
;fn ,d xsan dks fLFkj voLFkk ls NksM+k tkrk gS rks ;g ckj & ckj ry ls Vdjkrh gSA izR;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 Sol.
(C) 0.5s (D) 0.25s v = 0 + gt t = 0.5 sec After first collision : Speed becomes 5 (0.5) = 2.5 m/s t1 = 2 (0.25) = 0.5 t2 = 2 (0.125) = 0.25 t3 = 0.125 and so on [where ti is 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 = 0.5 + (Since above is a G.P. with a = 0.5 and r = 0.5) 1 0.5 = 0.5 + 1 = 1.5 sec.
Page No # 6
2.
A section of fixed smooth circular track of radius R in vertical plane is shown in the figure. A block is released from position A and leaves the track at B. The radius of curvature of its trajectory when it just leaves the track at B is: fp=kkuqlkj Å/okZ/kj ry esa R f=kT;k dk fpduk fLFkj o`Ùkkdkj iFk çnf'kZr gSA ,d CykWd dks fLFkfr A ls NksM+us ij ;g iFk dks fcUnq B ij NksM+rk gS rks fcUnq B dks NksMus ds rqjUr ckn blds iFk dh oØrk f=kT;k gS :
(A) R Sol.
3.
Sol.
ugha
(B)
R 4
By energy conservation between A & B A rFkk B ij ÅtkZ laj{k.k ds fu;e ls 1 2R MgR Mg +0= + MV2 5 5 2 2gR V= 5 Now, radius of curvature r 2 V 2gR / 5 R a r g cos 37 2 2 oØrk f=kT;k r = V 2gR / 5 R ar g cos 37 2
(C*)
R 2
=
(D) none of these buesa ls dksbZ
R 53º
A R–R cos53 =2R/5
O 37º B
R–R cos37= R/5 Reference line 37º (funs Z ' kj s[ kk) g g cos37
A bob is attached to one end of a string other end of which is fixed at peg A. The bob is taken to a position where string makes an angle of 30 0 with the horizontal. On the circular path of the bob in vertical plane there is a peg ‘B’ at a symmetrical position with respect to the position of release as shown in the figure. If V c and V a be the minimum tangential velocity in clockwise and anticlockwise directions respectively, given to the bob in order to hit the peg ‘B’ then ratio V c : V a 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 Å/okZ/kj ckWc ds o`Ùkkdkj iFk ij [kwaVh (Peg) ‘B’ fLFkr gSA vc ckWc dks ;gka ls Li'kZ js[kh; osx nsdj NksM+k tkrk gSA nf{k.kkorZ rFkk okekorZ fn'kkvksa ls ckWc ds [kwaVh (Peg) B ij Vdjkus ds fy, U;wure osx Øe'k% V c rFkk V a gks rks V c : V a gS ::
(A) 1 : 1 (B) 1 : 2 (C*) 1 : 2 (D) 1 : 4 (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 ÅtkZ laj{k.k ls:
gR
gksuh pkfg,A (1) rFkk (2) ds e/;
1 2 R 1 mv a = mg m(gR) 2 2 2 Page No # 7
va2 = gR + gR = 2gR
va = 2gR
Page No # 8
For clock-wise motion, the bob must have atleast that much speed initially, so that the string must not become loose any where until it reaches the peg B.
nf{k.kkorZ xfr ds fy, ckWc ds ikl çkjEHk esa de ls de bruk osx gksuk pkfg, ftlls oks
4.
A virtual erect image in a concave mirror is represented, in th e above figures, by
,d vory niZ.k esa vkHkklh lh/kk izfrfcEc fuEu esa ls fdlds }kjk iznf'kZr gksrk gSA (A)
5.
(B)
(C)
(D*)
A driving mirror on a car is never concave because :
pyrh dkj ds fy;s ihNs ns[kus esa dHkh Hkh vory niZ.k mi;ksx ugh djrs D;ksfd (A*) its field of view is t oo small (B) the image would be inverted (C) the image would be virtual and therefore useless for t he driver (D) only a plane mirror forms true images. (A*) bldk n`'; {ks=k cgqr de gksrk gSA (B) izfrfcEc mYVk gksxk (C) izfrfcEc vkHkklh gksxk blfy;s pkyd ds fy;s cjkcj gS (D) flQZ lery niZ.k lgh izfrfcEc cukrk gS
COMPREHENSION vuqPNsn Two bodies A and B of masses 10 kg and 5 kg are placed very slightly separated as shown in figure. The coefficient of friction between the floor and the blocks is = 0.4. Block A is pushed by an external force F. The value of F can be changed. When the welding between block A and ground breaks, block A will start pressing block B and when welding of B also breaks, block B will start pressing the verti cal wall – nks fi.M A o B ftuds nzO;eku 10 kg o 5 kg gS cgqr gh de nwjh ij j[ks x;s gS tSlk fp=k esa çnf'kZr gSA fi.M vkSj ry ds chp ?k"kZ.k xq.kkad s = k = 0.4 gSA fi.M A dks ckº; cy F ls /kdsyk tkrk gSA F dk eku ifjorZu'khy gSA tc fi.M A vkSj tehu ds chp osfYMax VwV tkrh gSA rc fi.M A, fi.M B dks /kdsyuk izkjEHk dj nsrk gSA tc fi.M B dh osfYMax VwV tkrh gS rks fi.M B Å/okZ/kj nhokj dks nckuk
çkjEHk djrh gSA
Page No # 9
6.
Sol.
If F = 20 N, with how much force does block A presses the block B ;fn F = 2 0 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; 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 l F dk U;wure eku D;k gksxk , ftlds dkj.k CykWd B Å/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 ok yk ?k"kZ.k cy gksxk (A*) 10 N (B) 20 N (C) 30 N (D) None dksbZ ugha If F = 50 N, force on 5 kg block = 10 N ;f n F = 50 N, rk s 5 kg ds CykWd ij cy = 10 N yxsxkA
Sol.
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 oØ ds vuqlkj cnyrk gS % (A)
Sol.
(B*)
(C)
(D)
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 friction force on 5 kg increases, till F = 60 N, and after that, the kinetic friction start acting on 5 kg block, which will be constant (20N) tc rd 10 kg dk CykWd lrg ds lkFk fpidk jgrk gS (... F = 40 rks 5 kg dk CykWd dksbZ cy eglwl ugh djsxkA F = 40 N, ds ckn 5 kg okys CykWd ij ?k"kZ.k cy F = 60 N, rd c
N),
Page No # 10
10 .
A sample of gas goes from state A to state B in four different manners, as shown by the graphs. Let W be the work done by the gas and U be change in internal energy along the path 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 igq¡prk gSA ;fn xSl }kjk ekxZ AB esa fd;k x;k dk;Z W rFkk vkUrfjd ÅtkZ esa ifjorZu U gks] rks
xzkQksa dks fn;s x;s dFkuksa ls feykbZ;s & (A)
(p) Both W and U are positive
V A
B P
P
(p) W rFkk U nksuksa /kukRed gSaA
B
(B)
(q) Both W and U are negative A T
T
(q) W rFkk U nksuksa _.kkRed gSaA
A
(C)
(r) W is positive whereas U is negative B V
(r) W rks /kukRed ijUrq U _. kkRe d gS aA
V
(D)
A
(s) W is negative whereas U is positive
B P
An s . Sol.
(A) s (B) q (C) r (D) q in (A), V is on vertical axis. (a) es a , V m/okZ/kj v{k ij gSA
(s) W rks _.kkRed ijUrq U /kukRed gSA
Part-I (Hkkx -I)
As V is icreasing, W is positi ve. tSls V c<+rk gS , W /kukRed
gSA V Part-II (Hkkx -II)
P
V is decreasing, W is negative.V ?kVus ij , W _. kk Re d gSA As negative work in part-II is greater than positive work in part-I, net work during the process is negative. Using PV = nRT and as V remains same for initial and final points of the process, it is obvious that final temp. is greater than initial temperature as pressure has increased. Therefore dU is positive. Hence option (S) is connected with (A). Similar arguments can be applied to other graphs. pwafd Hkkx -II esa _.kkRed dk;Z Hkkx -I, ls T;knk gS] vr% dqy dk;Z çfØ;k esa
_.kk Red gSA PV = nRT dk ç;ksx djus ij vkSj V çkjEHk ,oa vfUre fcUnqvksa ij leku jgrk gSA vr% Li"V gS fd vfUre rki] çkjfEHkd rki ls T;knk gS pwafd nkc c<+rk gSA vr% dU Page No # 11
/kukRed gS] fodYi (S), (a) ls tqM+k gSA leku rdZ vU; xzkQksa ij yxk;s tk ldrs gSA
PHYSICS TARGET : JEE (MAIN + ADVANCED)
DPP No. : 5 to 8 Syllabus : XI class syllabus
D P P Sy l l a b u s : Thermodynamics, Circular motion, Sound wave, Projectile motion, SHM,
Newton’s law, Centre of mass, G.O., Fluid mechanics, Friction, Rotation, sound wave DPP No. : 07
ANSWER K EY TO DPP NO. # 7 125 (C) (D) m 4. 5. 4 (D)
1.
(B)
2.
(A)
3.
7.
(A)
8.
(B)
9.
1.
A particle performs S.H.M. on xaxis with amplitude A and time period T. The time taken by the particle to travel 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 ls pyrs gq, A/5 nwjh r; djus essa yxk le; gSA T T T T 4 1 1 (A) (B*) cos1 (C) cos1 (D) sin1 20 2 2 2 5 5 5 Particle is starting from rest, i. e. from one of its extreme position. A As particle moves a distance , we can represent it on a circle as shown. 5
Sol.
6.
d = 4000 mm
d.k fLFkj voLFkk ls pyuk çkjEHk djrk gS vFkkZr vius fdlh vUR; fcUnq esa pyuk çkjEHk djrk gSA tSls fd d.k] A nwjh r; djrk gS] bls ,d o`Ùk ij n'kkZ;k tk ldrk gSA 5
4A / 5 4 cos = A 5 t = cos
– 1
T 2
M et h o d :
4 5
4 5
= cos
–1
4A/5 A/5
1
t=
y
co s
– 1
4 5
P
x
A
Q
4 = cos – 1 5
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 + ) Page No # 12
At t = 0 ij ; x = A
A –
=
t = cos –1
A = A cos t 5
4 = cos t 5 T t= co s – 1 2 2.
2
4 5
4 5
3 with the 5 horizontal. When a force of 30 N is applied on the block parallel to & upward the plane, the total reaction by the plane on the block is nearly along: {kSfrt ds lkFk = sin 1 3 urh dks.k okys [kqjnjs ur ry ij 10 kg nzO;eku dh ,d 5 oLrq j[kh gq;h gSA tc ur ry ds lekUrj Åij dh vksj 30 N dk cy yxk;k tkrk gS rc ry A body of mass 10 kg lies on a rough inclined plane of inclination = sin 1
}kjk oLrq ij dqy izfrfØ;k ¼yxHkx½ fdlds vuqfn'k gSA
(A*) OA Sol.
(B) OB
(C) OC
(D) OD
Frictional force along the in upward direction = 10 g sin – 30 = 30 Nt N = 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. When the tank is completely filled with water, the quantities of water flowing out per second from both holes are the same. Then, 'R' is equal to : ,d cM+h [kqyh Vadh dh Å/okZ/kj nhokj ij fp=kkuqlkj nks NksVs fNnz gSaA ,d 'L' Hkqtk dk oxkZdkj fNnz Åijh lrg ls '4y' xgjkbZ ij o nwljk 'R' f=kT;k dk o`Ùkkdkj fNnz Åijh lrg ls ‘y’ xgjkbZ ij gSA tc Vadh ty ls iwjh Hkjh gS]nksauks fNnzksa ls çfr lSd.M ckgj fudy jgs ty dh ek=kk l eku gSA rks 'R' cjkcj gS : y
4y
(A)
Sol.
L 2
(B) 2L
2R
v1
L
v2
(C*)
2 .L
(D)
L 2
Let v1 and v2 be the velocity of efflux from square and circular hole respectively. S 1 and S2 be crosssection areas of square and circular holes.
Page No # 13
y
4y
2R
v1
L
v2
v1 = 8gy and v2 = 8g( y ) The volume of water coming out of square and circular hole per second is Q1 = v1S1 = 8gy L2 ; Q2 = v2S2 = 2gy R2 Q1 = Q2 R=
4.
2 .L
A ring of mass m and radius R rolls on a horizontal rough surface without slipping due to an applied 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<+drh gSA o y; ij yxus okyk ?k"kZ.k cy gS &
(A) So l .
(D)
F 3
(B) F + f = ma
Also ; FR – f R = F – f = ma From (1) & (2) f = 0. 5.
Sol.
6.
2F 3
(C)
F 4
(D*) Zero 'kwU;
.... (1) a R .... (2)
[ = mR 2 ]
A particle is projected from the ground level. It just passes through upper ends of vertical poles A, B, C of height 20 m, 30 m and 20 m respectively. The time taken by the particle to travel from B to C is double of the time taken from A to B. Find the maximum height attained by the particle from the ground level. ,d d.k dks tehu ls ç{ksfir fd;k tkrk gSA ;g Å/oZ [kEcs A, B, C ftudh Å¡pkbZ Øe'k% 20 m, 30 m rFkk 20 m gS] ds Åijh fljksa dks Bhd Nwrk gqvk fudyrk gSA B ls C rd tkus esa fy;k x;k le;]A ls B rd tkus esa fy, x, le; dk nqxuk gSA d.k }kjk tehu ls çkIr vf/kdre Å¡pkbZ dk eku gSA 125 [ Ans.: m] 4 tAB = t tBC = 2t So, for ABC part, Time of flight, 3 2u y tAC = 3t = uy = 2 gt g 1 Also, 10 = uyt – gt2 = gt2 t = 1s uy = 15 m/s 2 45 u2 y 225 h= = = m. 4 2g 20 45 125 Maximum height attained = 20 + = m. 4 4
A man is standing at the edge of a 1m deep swimming pool, completely filled with a liquid of refractive index. 3 / 2 The eyes of the man are 3 m above the ground. A coin located at the bottom of the pool appears to be at an angle of depression of 30 0 with reference to the eye of man. Then horizontal distance (represented by x in the figure) of the coin from the eye of the man is ____________ mm. Page No # 14
,d O;fDr 1m xgjs rj.k rky ds fdukjs ij [kM+k gS ftlesa 3 / 2 viorZukad dk nzo Hkjk gSA O;fDr dh vk¡[ks i`Foh ry ls 3 m Å¡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 _________¼fp=k esa x }kjk çnf'kZr½ feeh esa D;k gksxh \
An s.
d = 4000 mm
Sol.
3 sinr r = 45º 2 S = h = 1m y = H tan600 = 3m x= S + y = 4m = 4000 mm sin 60º =
Comprehension
A 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`fÙk dh /ofu mRlftZr djrs gq, 50 m/sec. dh pky ls fLFkj ijkorZd nhokj dh vksj xfr'khy gSA ;gk¡ fp=k esa n'kkZ, vuqlkj ik¡p çs{kd A,B,C,D o E gSA /ofu dh pky 350 m/sec. gSA
7.
8.
9.
The beat frequency appeared to observer A is (A*) zero (B) 2000 Hz izs{kd A }kjk izsf{kr foLian vko`fr gSA (A*) 'kwU; (B) 2000 Hz
(C) 1750 Hz
(D) 400 Hz
(C) 1750 Hz
(D) 400 Hz
The beat frequency appeared to observer B is izs{kd B }kjk izsf{kr foLian vko`fr gSA (A) 1750 Hz (B*) 400 Hz
(C) 2000 Hz
(D) zero 'kwU;
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 Page No # 15
Sol.
2v 0 f 0 2 10 = × 6000 = 400 Hz v vs 300 ƒA = 0 2W 2 350 50 6000 ƒC = 2 s 2 = = 1750 Hz 400 300 v vs 4f f ƒD = ƒE = f = = 2000 Hz. 3 3 ƒB =
Page No # 16
PHYSICS TARGET : JEE (MAIN + ADVANCED) 2015
DPP No. : 5 to 8
Syllabus : XI class syll abus
D P P s y l l a b u s : Thermodynamics, Circular motion, Sound wave, Projectile motion, SHM, Newton’s
law, Centre of mass, G.O., Fluid mechanics, Friction, Rotation, sound wave DPP No. : 08
1. 7.
(D) (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 ds nkSjku d.k dh 6
2. 8.
(A) (B)
ANSWER K EY TO DPP NO. # 8 (C ) (A) (A ) 4. 5. (A)
3. 9.
6.
13 3
vkSlr pky gS & (A)
3 A 2
(B)
3 A 4
t
Sol.
average speed =
dx . dt dt 0
t
(C)
3 A
(D*)
3 A (2 3 )
t
=
dx 0
t
=
x( t ) x(0) A(cos / 6 1) 3 A = = ( 3 2) / 6 t
since particle does not change it's direction in the given interval , average speed = V = 2.
A particle of mass 5 kg is moving on rough fixed inclined plane (making an angle 30° with horizontal) with constant velocity of 5 m/s as shown in the figure. Find the friction force acting 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 ¼tks {kSfrt ls 30° dk dks.k cukrk gS½ ij fn[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 cy Kkr dhft,A ( g = 10m/s 2 )
(A*) 25 N
ugh
Sol.
3 A (2 3 )
(B) 20 N
(C) 30 N
(D) none of these buesa ls dksbZ
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 ½ = 25 N.
pw¡fd CykWd ur ry ij uhps dh vksj ,dleku osx ls fQlyrk gS] rks bl ij usV cy 'kwU; gksxkA blfy;s mg sin CykWd ij ?k"kZ.k cy ‘f’ dks larqfyr djrk gSA blfy;s f = mg sin = 5 10 ½ = 25 N. Page No # 17
3.
A sphere rolls without sliding on a rough inclined plane (only mg and constant forces are acting on the body). The angular momentum of the body:
,d xksyk [kqjnjs ur ry ij fcuk f Qlys yq<+drk gSA oLrq dk dks.kh; laosx & dsUnz ds lkis{k lajf{kr gSA (A) about centre is conserved laidZ fcUnq ds lkis{k lajf{kr gSA (B) is conserved about the point of contact
(C*) is conserved about a point whose distance from the inclined plane is greater than the radius of the sphere
So l .
ml fcUnq ds lkis{k lajf{kr gksxk ftldh ur ry ls nwjh xksys dh f=kT;k ls vf/kd gksxh fdlh Hkh fcUnq ds lkis{k lajf{kr ugha (D) is not conserved about any point. gS (C) Angular momentum will be conserved if the net torque is zero .
dks.kh; laosx lajf{kr jgsxk ;fn dqy vk?kw.kZ 'kwU; gSA Now for the sphere to move down: vc xksys dks uhps vkus ds fy, mg sin > mg cos
Let x be the perpendicular distance of the point (as shown in figure) about which torque remains zero. x ml fcUnq dh yEcor~ nwjh ¼fp=kkuqlkj½ gS ftlds ifjr% cy vk?kw.kZ 'kwU; gSA for = 0 ; x > R as shown ds fy, = 0 gksxk tSlk çnf'kZr gS x > R
Note: As mgsin > mgcos , the point should be inside the sphere. uksV : pwafd mgsin > mgcos , og fcUnq xksys ds vUnj gksxkA 4.
Sol.
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 points having amplitude 2 mm is: 1.5 eh0 yEch Mksjh tks nksuksa fljksa ij c¡/kh gS] ewyfo/kk esa dEié dj jgh gSA Mksjh ds e/; fcUnq ¼dsUæ½ ij vk;ke 4 eh0eh0 gSA mu nks fcUnqvksa ds chp dh nwjh ftudk vk;ke 2 eh0eh0 gS] gksxh (A*) 1 m (B) 75 cm (C) 60 cm (D) 50 cm = 2 = 3m Equation of standing wave y = 2A sin kx cos t y = A as amplitude is 2A. A = 2A sin kx 1 2 x= x1 = m 4 6 2 5 and .x= x 2 = 1.25 m x 2 – x 1 = 1m 6
Page No # 18
g y% = 2 = 3m
vizxkeh rjax dk lehdj.k (x = 0 ij fuLiUn ysrs gq, ) y = 2A sin kx cos t y = A pqfd vk;ke 2A gS a A = 2A sin kx 2 x= 6 vkSj . 2 x = 5 6 5.
Sol.
x1 =
1 m 4
x 2 = 1.25 m x 2 – x 1 = 1m
The source (S) of sound is moving constant velocity v 0 as shown in diagram. An obserever O listens to the sound emmited by the source. The observed frequency of the sound : fp=kkuqlkj /ofu dk L=kksr (S) fu;r osx v 0 ls lh/kh js[kk esa xfr'khy gSA ,d izs{kd O, S }kjk mRiUu dh x;h /ofu lqurk gSA izs{kd }kjk lquh x;h /ofu dh vko`fÙk
(A*) continuously decreases (B) continuously increases (C) first decreases then increases (D) first increases then decreases. yxkrkj c<+sxh (A*) yxkrkj ?kVsxh (B) igys ?kVsxh fQj c<+sxh igys c<+sxh fQj ?kVsxh (C) (D) From figure , the velocity of approach (Vcos ) decrease as the source comes closer (as increases).And the velocity of separation also increases as will decrease. Hence the frequency of sound as heared b y the observed decreases continuously
6.
Find out the moment of inertia of the following structure (writt en as ) about axis AB made of thin uniform rods of mass per unit length . fuEu
Sol.
The moment of inertia of all seven rods parallel to AB and not l ying on AB is lHkh 7 NM+ dk tM+Ro vk?kw.kZ AB ds lekUrj ij AB ij ugha gS = 7 × ( ) 2 = 7 3 the moment of inertia of all five r ods lying on AB = 0 lHkh 5 NM+ tks AB ds ifjr gS dk tMRo vk?kw.kZ 'kwU; gS The moment of inertia of all 18 rods perpendicular to AB is = 18 ( )
2
3
= 6 3
2
lHkh 18 NM+ dk AB ds yEcor~ tM+Ro vk?kw.kZ AB is = 18 ( ) = 6 3 3
Hence net MI of rod about AB
=7
+6 = 13 Ans. vr% ifj.kkeh tM+Ro vk?kw.kZ AB ds ifjr% = 7 3 + 6 3 = 13 3 A n s . 3
3
3
Page No # 19
COMPREHENSION A concave mirror of radius of curvature 20 cm is shown in the figure. A circular disc of diameter 1 cm is placed on the principle axis of mirror with its plane perpendicular to the principal axis at a distance 15 cm from the pole of the mirror. The radius of disc starts increasing according to the law r = (0.5 + 0.1 t) cm/sec where t is time is second. ,d vory niZ.k ftldh oØrk 20 cm fp=kkuqlkj gSA ,d o`Ùkkdkj pdrh ftldk O;kl 1 cm gS rFkk bldks niZ.k ds eq[; v{k ij eq[; v{k ds yEcor~ niZ.k ds /kzqo (pole) ls 15 cm dh nwjh 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<+uk izkjEHk djrh gS tgk¡ t le; lsd.M esa gSA
7.
The image formed by the mirror will be in the shape of a
niZ.k }kjk cu;s x;s izfrfcEc dh vkd`fr gksxh &
Sol.
(A*) circular disc (B) elliptical disc with major axis horizontal (C) elliptical disc with major axis vertical (D) distorted disc (A*) o`Ùkkdkj pdrh (B) ,d nh?kZo`Ùkkdkj pdrh ftldh nh?kZ v{k {kSfrt gS (C) ,d nh?kZo`Ùkkdkj pdrh ftldh nh?kZ v{k Å/okZ/kj gSA (D) vfu;fer vkd`fr okyh pdrhA All dimensions of the disc are perpendicular to the principal axis. Hence all dimensions are equally magnified, resulting in an image in the shape of a cir cular disc.
lHkh pdrh dh foek, eq[; v{k ds yEcor~ gSA vr% lHkh foek, cjkcj :i ls vkoZf/kr gksxhA rFkk izfrfcEc o`Ùkkdkj pdrh ds vkd`fr dk gksxkA 8.
Sol.
9.
In the above question, the area of image of the disc at t = 1 second is : pdrh ds izfrfcEc dk t = 1 lsd.M ij mijksDr iz'u d s fy, {ks=kQy gksxk & (A) 1.2 cm2 (B*) 1.44 cm2 (C) 1.52 cm2 (D) none of these buesa ls
dksbZ ugha
At t = 1 sec. r = 0.5 t + 0.1 t = 0.6 cm f 10 m= = =–2 f u 10 15 Radius of image = 2r = 1.2 cm Area of image = (1.2)2 = 1.44 cm2 . izfrfcEc dh f=kT;k = 2r = 1.2 cm izfrfcEc dk {ks=kQy = (1.2)2 = 1.44 cm2 . What will be the rate at which t he radius of image will be changing (A*) 0.2 cm/sec increasing (B) 0.2 cm/sec decreasing (C) 0.4 cm/sec increasing (D) 0.4 cm/sec decreasing
izfrfcEc dh f=kT;k ds ifjorZu dh nj D;k gksxh & (A*) 0.2 cm/sec c<+rh gqbZ gqbZ (C) 0.4 cm/sec c<+rh gqbZ gqbZ Sol.
(B)
0.2
cm/sec
?kVrh
(D)
0.4
cm/sec
?kVrh
dr = 0.1 dt r image = |m|r object = 2r object dr image = 2. dr = 0.2 = 0.2 cm/sec. dt dt Page No # 20
