17-Gravitation

SECTION - I : STRAIGHT OBJECTIVE TYPE [k.M-I: lh/ks oLrqfu"B izdkj 17.1 A tunnel is dug along the diameter of the earth (Radius R & mass M). There is a particle of mass 'm' at the centre of the tunnel. The minimum velocity given to the particle so that it just reaches to the surface of the earth is : i`Foh¼f=kT;kRonzO;ekuM)dsO;kldsvuqfn'k,dlqjaxcukbZtkrhgSAlqjaxdsdsUnzij'm' nzO;ekudk,d d.kgSAd.kdksfdrukU;wureosx fn;ktk,ftl ss;gi`Fohdhlrgrdrksigq¡pghtk,& 2GM (A*) (B) (C) R (D) it will reach with the help of negligible velocity. (;gux.;osxdhlgk;rklsigq¡ptk;sxkA) Sol. (A) Let the minimum speed imparted to the particle of mass m so that it just reaches surface of earth is v. Applying conservation of energy 1  3 GM  GM 2 Solving we get v = mv2 +    m =   R m + 0 gy ekukfdm nzO;ekudsd.kdksU;wurepkyv nhtkrhgSrkfd;gBhdi`FohdhlrgijigqWprkgSAÅtkZlaj{k.k yxkusij 1  3 GM  GM mv2 +   2  m =   R m + 0 gy djus ij ge izkIr djrs gSv = 17.2 A cavity of radius R/2 is made inside a solid sphere of radius R. The centre of the cavity is located at a distance R/2 from the centre of the sphere. The gravitational force on a particle of mass ' m ' at a distance R/2 from the centre of the sphere on the line joining both the centres of sphere and cavity is (opposite to the centre of cavity). [Here g = GM/R2, where M is the mass of the sphere ] R f=kT;kdsBkslxksysdsvUnjR/2f=kT;kdh,ddksVj(cavity)cuhgqbZgSAdksVjdkdsUnzxksysdsdsUnzls R/2nwjhijgSAdksVjoxksysdsdsUnzdkstksM+usokyhjs[k ij¼dksVjdsdsUnzdsfoijhr½xksysdsdsUnzlsR/2 nwjhijj[ksmnzO;ekudsd.kijxq:Rokd"kZ.kcygSA [;gkW g = GM/R2 gS] tgkW M xksys dk nzO;eku gS] (A) Sol. (B) mg 2 (B*) 3 mg 8 (C) mg 16 (D) noneof thesebuesalsdksbZugha Gravitation field at mass ‘m’ due to full solid sphere → E  r  R ....  1   1 30 60  0 4G  Gravitational field at mass ‘m’ due to cavity (–) → ()(R / 2)3  a3  E2  3 R2 .................  3 r 2  0 ()R3  R  0  = 240R2 = 240 Net gravitational field E  E1  E2 = R  R 60 240 R = 80 Net force on ‘m’  F = m E = mR 80 Here  = M 4 / 3R3 3mg & 0  1 4G then F = 8 gy (B) lEiw.kZBkslxksysdsdkj.k‘m’nzO;ekuijxq:Roh;{ks=k → E  r  R ....  1   1 30 60  0 4G  dksVj(–)dsdkj.km nzO;ekuijxq:Roh;{ks=k → ()(R / 2)3  a3  E2  3 R2 .................  3 r 2  0 ()R3  R  0  = 240R2 = 240 ifj.kehxq:Roh;{ks=k E  E1  E2 = R  R 60 240 R = 80 m ij ifj.kkeh cyF = m E = mR 80 ;gkW  = M 4 / 3R3 & 0  1 4G rc F = 3mg 8 17.3 A satellite is launched in the equatorial plane in such a way that it can transmit signals upto 60º latitude on the earth. The angular velocity of the satellite is : ,dmixzgdksHkwe/;js[k ryesablrjgiz{ksfirdjrsgSafd;gi`Fohij60ºva{k 'krdladsrlapkfjrdjldrk gSAmixzgdkdks.kh;osxgS& (A*) Sol. (A) (B) R (C) (D) In  AOB : - cos 60° = OB  OB = 2 R (where OB is orbital radius) Here gravitational force will provide the required centripetal force. Hence   = gy (A) GMm (OB)2 = m(OB) 2 =   =  AOB esa : - cos 60° = R  OB = 2 R (OBd{kh;f=kT;kgSA) ;gkWxq:Rokd"kZ.kcyvko';dvfHkdsUnzh;cyiznkudjrkgSA GMm vr% (OB)2   = = m(OB) 2 =   = 17.4 A satellite is seen after each 8 hours over equator at a place on the earth when its sense of rotation is opposite to the earth. The time interval after which it can be seen at the same place when the sense of rotation of earth & satellite is same will be : i`FohijHkwe/;js[k ijfLFkrfdlhLFk uij,dmixzgizR;sd8?k.Vscknfn[k bZnsrkgStcblds?kweusdhfn'k i`FohdsfoijhrgSAblhLFk uijmixzgdsns[kstkusdkle;vUrjkygksxktci`FohrFk mixzgds?kweusdhfn'k lekugks& (A) 8 hours (B) 12 hours (C*) 24 hours (D) 6 hours (A)8?k.Vs (B)12?k.Vs (C*)24?k.Vs (D)6?k.Vs Sol. (C) Given 8 = 2 1  2 2 = 2  2 , T1 = 24 hours for earth. T1 T2 (1 + 2 is the relative angular velocity for opposite direction)  T2 = 12 hours (T2 being the time period of satellite, it will remain same as the distance from the centre of the earth remains constant).  T = gy (C) 2 2  1 2 = 2  2 T2 T1 = 24 hours.(2 – 1 is the relative angular velocity for same direction) fn;k x;k gS8= 2 = 2 ,T =24?k.Vs]i`Foh dsfy, ( + mYVhfn'kkesalkis{k dks.kh;pky gSA) 1  2 2  2 1 1 2 T1 T2 T2=12?k.Vs(T2mixzgdkvkorZdkygS];gvifjofrZrjgsxkD;ksafdi`FohdsdsUnzlsnwjhvifjofrZrjgrh gS) 2 2  T = 2  1 = 2  2 T2 T1 = 24 ?k.Vs (2–1lekufn'k esalkis{kdks.kh;pkygSA) 17.5 Four similar particles of mass m are orbiting in a circle of radius r in the same angular direction because of their mutual gravitational attractive force. Velocity of a particle is given by m nzO;ekudspkj,dlekud.k,d&nwljsdsxq:Roh;çHk oesarf=kT;kdso`Ùkesa,dghdks.kh;fn'k esapDdjdkV jgs gSA d.k dk osx gksxk & Gm  1 2 1 2  2  1 Gm  1 1 2  2 (A*)    (B) (D)    Sol. (A)  r    2   Centripetal force = net gravitational force mv2 0 = 2F cos 45º + F r 1 Gm2 = + 4r 2 mv2 Gm2  Gm(2 2  1) 1/ 2 0 = [ 2 + 1]     gy (A) r 4r 2  4r  vfHkdsUnzh;cy=ifj.k ehxq:Rokd"kZ.kcy mv2 0 = 2F cos 45º + F r 1 Gm2 = + 4r 2 mv2 Gm2  Gm(2 2  1) 1/ 2 0 = [ 2 + 1]     r 4r 2  4r  17.6 Three particles P, Q and R are placed as per given figure. Masses of P, Q and R are m, respectively. The gravitational force on a fourth particle ‘S’ of mass m is equal to m and m rhud.k saP,QrFk Rdksfp=k uqlkjj[k tkrkgSAP,QrFk RdsnzO;ekuØe'k% m, 3 mrFk mgSApkSFksd.k‘S’ ftldk nzO;eku m gS] ij yxus okys xq:Rokd"kZ.k cy dk eku D;k gksxk ? (A) (B) (C*) (D) (A) (B) (C*) 3 GM2 2d2 3 Gm2 2d2 3 Gm2 2d2 3 Gm2 2d2 3 GM2 2d2 3 Gm2 2d2 3 Gm2 2d2 3 Gm2 in ST direction only in SQ direction and in SQ direction only in SQ direction and dsoySTfn'k esa ,SQdhfn'k esarFk dsoySQdhfn'k esa 3 Gm2 2d2 3 Gm2 2d2 3 Gm2 2d2 3 Gm2 in SU direction in ST direction SUdhfn'k esa (D) 2d2 SQdhfn'k esarFk 2d2 STdhfn'k esa Sol. (C) In horizontal direction Net force = G 3 mm 12 d2 cos30° – Gm2 4d2 cos60° Gm2 = 8d2 Gm2 – 8d2 = 0 in vertical direction G 3 m2 G 3 m2 Gm2 Net force = 12 d2 cos 60° + 3 d2 + 4 d2 cos30° 3 Gm2 = 24 d2 + 3Gm2 3 d2 + 3Gm2 8 d2 1 8  3  3Gm2 gy (C) = d2  24  = 2d2 along SQ {kSfrt fn'kk esa ifj.kkeh cy= G 3 mm 12 d2 cos30° – Gm2 4d2 cos60° = Å/okZ/kjfn'k esa Gm2 8d2 Gm2 – 8d2 = 0 G 3 m2 G 3 m2 Gm2 ifj.kkeh cy= 12 d2 cos60° + 3 d2 + 4 d2 cos30° 3 Gm2 = 24 d2 + 3Gm2 3 d2 + 3Gm2 8 d2 1 8  3  3Gm2 = d2  24  = 2d2 SQdhfn'kk es 17.7 The gravitational potential of two homogeneous spherical shells A and B of same surface density at their respective centres are in the ratio 3 : 4. If the two shells coalesce into single one such that surface density remains same, then the ratio of potential at an internal point of the new shell to shell A is equal to : ,dlekui`"Bh;?kuRookysnksle:idks'k sadsdsUnzijxq:Roh;foHko3:4dsvuqikresagSAvxjnksuksadks'kfeydj ,du;kdks'kcukrsgSftliji`"Bh;?kuRolekujgsrksu;sdks'kdsvkUrfjdfcUnqiju;sdks'krFk dks'kAdsfoHko dk vuqikr gS & (A) 3 : 2 (B) 4 : 3 (C*) 5 : 3 (D) 5 : 6 Sol. (C) MA = 4 R 2 , M = 4 R 2 where  is surface density. VA = GMA RA , VB = GMB RB VA MA RB 4R2 RB RA = = A  VB MB RA VA RA 4R2 RA RB 3 4 Given VB = RB = 4 , then RB = 3 RA for New shell of mass M and radius R - M = M + M = 4R 2 + 4R 2 4R2 = 4(R 2 + R 2) A B A V M R B A B 4[R2  R2 ] R 5 then VA = R A MA = A B (R2  R2 )1/ 2 A 4R2 = R = 3 A B A A gy (C) MA = 4 R 2 , M = 4 R 2 tgk¡lrg?kuRogS VA = GMA RA , VB = GMB RB VA MA RB 4R2 RB RA = = A  VB MB RA 4R2 RA RB VA RA 3 fn;k x;k gS = 4 rc R = R VB RB = 4 , B 3 A u;sdks'kdknzO;ekuMof=kT;kRgS- M = M + M = 4R 2 + 4R 2 4R2 = 4(R 2 + R 2) A B A B A B V M rc VA = R RA MA = 4[R2  R2 ] (R2  R2 )1/ 2 RA 5 4R2 = R = 3 A B A A 17.8 A point P lies on the axis of a fixed ring of mass M and radius R, at a distance 2R from its centre O. A small particle starts from P and reaches O under gravitational attraction only. Its speed at O will be ,dfcUnqP,MnzO;ekurFk Rf=kT;kdhoy;dsdsUnzls2RnwjhijgSA,dNksVkd.kPls'kq:gksdjdsoyxq:Rokd"kZ.k çHkko esa gh O rd igq¡prk gSA bldh O ij pky gksxh & (A) zero 'kwU; (B) Sol. (D) Gravitational potential at ‘P’  GM (C) (D*) VP = 5R Gravitational potential at 'O' GM V0 = – R work energy theorem W = K  m[V – V ] = 1/2 mv2 GM GM  1 gy (D) m    5R = 2 mv2 = v Pijxq:Roh;foHko VP = Oijxq:Roh;foHko GM V0 = – R dk;Z&ÅtkZ izes; W = K  m[VP – V ] = 1/2 mv2 GM GM  1 m    5R = mv2 = v 2 17.9 Gravitational field at the centre of a semicircle formed by a thin wire AB of mass m and length 𝑙 is : nzO;ekum ,oe~ 𝑙 yEckbZdsirysrkjAB lscusv)Zo`ÙkdsdsUnzijxq:Roh;{ks=kdkekuD;kgksxk: (A) Gm 𝑙2 along +x axis (B) Gm  𝑙2 along +y axis (C) 2 Gm 𝑙2 along + x axis (D*) 2 Gm 𝑙2 along + y axis (A) Gm 𝑙2 Gm +x v{kdhfn'k esa (B)  𝑙2 +y v{kdhfn'k esa (C) Sol. (D) 2 Gm 𝑙2 + x v{k dh fn'kk esa (D*) 2 Gm 𝑙2 + y v{k dh fn'kk esa Let mass per unit length of wire,  = mass of element, dm = rd m and r = 𝑙, r = 𝑙 𝑙  y then dE = Gdm r 2   dE = Grd ( ˆi cos  + ˆj sin ) x 0 0    G ˆ ˆ E =  i cos  d   j sin d r  0 = 2G ˆj = r 0  2Gm ˆj 𝑙 r gy (D) = 2Gm ˆj 𝑙2 (along y-axis) ekuk fd rkj dh izfr bdkbZyEckbZ dknzO;eku = m o r = 𝑙, r = 𝑙 𝑙 vYika'k dk nzO;eku , dm =rd y rc dE =  Gdm r 2 Grd dE =  2 ( ˆi cos  + ˆj sin ) x 0 0 r    G ˆ ˆ E =  i cos  d   j sin d r  0 0  = 2G ˆj = r 2Gm ˆj 𝑙 r = 2Gm ˆj 𝑙2 + y-v{k ds vuqfn'k 17.10 The percentage change in the acceleration of the earth towards the sun from a total eclipse of the sun to the point where the moon is on a side of earth directly opposite to the sun is lw;Zdsiw.kZxzg.kdhfLFkfrls]pUnzeki`Fohlslq;Zdsfoijhrfn'k esagS]bufLFkfr;ksaesai`Fohdklq;ZdhvksjRoj.k esa izfr'kr ifjorZu gksxk & M r M  r 2  r 2 M  r 2 M (A) s 2 100 Mm r1 (B) s  2  Mm r1  100 (C*) 2  1  r2 s 100 Mm (D)  1  r2 m  100 Ms       MS = mass of the sun, MM = mass of the moon, r1 = earth sun distance, r2 = earth moon distance. MS = lw;ZdknzO;eku,MM =pUnzekdk nzO;eku,r1 =lw;Zi`Fohdh nwjh,r2 = i`FohpUnzekdhnwjh Sol. (C) During total eclipse- F  GMsMe  GMmMe Total attraction due to sun and moon, 1 2 2 1 2 When moon goes on the opposite side of earth. F  GMsMe  GMmMe Effective force of attraction, 2 2 2 1 2 2GMmMe Change in force, F = F – F = 2 2 F Change in acceleration of earth a = Me 2GMm = 2 2 Average force on earth, Fav F1  F2 = 2 GMsMe 2 1 Average acceleration of earth, a = Fav  GMs av Me 2 a  2GMm  r 2  %age change in acceleration = 100 av 1 = 2 GMs 100  r 2 M = 2 1  r2 m 100 Ms   gy (C) iw.kZxzg.kdsnkSjku- F  GMsMe  GMmMe lw;ZopUnzekdsdkj.kdqyvkd"kZ.k, 1 2 2 1 2 tcpUnzeki`FohdsfoijhrfLFkfresatkrkgS F  GMsMe  GMmMe izHk ohvkd"kZ.kcy, 2 2 2 1 2 2GMmMe cyesaifjorZu, F = F1 – F2 = 2 2 F i`FohdsRoj.kesaifjorZua = Me 2GMm = 2 2 i`Foh ij vkSlr cy, Fav F1  F2 = 2 GMsMe 2 1 i`Fohdk vkSlrRoj.k,a = Fav  GMs av Me 2 a  2GMm  r 2  Roj.kesaizfr'krifjorZu = 100 = 2 av 2 1 GMs 100  r 2 M = 2 1  r2 m 100 Ms   17.11 A particle of mass M is at a distance 'a' from surface of a thin spherical shell of uniform equal mass and having radius a. ,d iryk [kks[kyk xksyk ftlesa leku nzO;eku M forfjr gS rFkk 'a' f=kT;k gS bldh a nwjh ij bldh lrg ls ,d M nzO;eku j[kk gS & (A) Gravitational field & potential both are zero at centre of the shell dks'kdsdsUnzijxq:Roh;{ks=krFk foHkonksuksadsUnzij'kwU;gSaA (B) Gravitational field is zero not only inside the shell but at a point outside the shell also xksysdsvUnjockgjHkhfdlhfcUnqijxq:Roh;{ks=k'kwU;gSA (C) Inside the shell, gravitational field alone is zero xksysdsvUnjdsoyxq:Roh;{ks=k'kwU;gSA (D*) Neither gravitational field nor gravitational potential is zero inside the shell. xksysdsvUnjurksxq:Roh;{ks=kvkSjughfoHko'kwU;gSA Sol. (D)AtcentredsUnzfcUnqij GM GM GM GM GM Vc = – a GM – 2a ; Ec = (2a)2 ; At any point P inside fdlh vkUrfjd fcUnq P ij VP = – a – b EP = b2 {only duetooutsidemass M}(flQZ cká nzO;eku M ds dkj.k ) 17.12 A small area is removed from a uniform spherical shell of mass M and radius R. Then the gravitational field intensity near the hollow portion is nzO;eku M o f=kT;k R ds le:i xksyh; dks'k ls vYi {ks=kQy fudky fy;k tkrk gS rks [kks[kys Hkkx ds fudV xq:Roh; {ks=k dh rhozrk gS :- GM (A) R2 (B*) GM 2R2 (C) 3GM 2R2 (D) Zero Sol. Consider a small area (shaded strip) here Eself = Gravitational field due to this strip and Eext = Gravitational field due to the rest of spherical shell. Ein = Gravitational field just inside the strip Eout = Gravitational field just outside the strip Ein = Eext – Eself = 0  Eext = Eself GM Eout = Eext + Eself = R2  Eext = GM 2R2 After the shaded area has been removed there is no Eself and only Eext. GM hence, Enet = Eext = 2R2 Sol. vYi{ks=kQy¼Nk;kafdriV~Vh½ijfopkjdjrsgSaA ;gk¡Eself =bliV~Vhdsdkj.kxq:Roh;{ks=k vkSjEext=cpsgq,xksyh;dks'kdsdkj.kxq:Roh;{ks=k Ein = iV~VhdsBhdvUnjxq:Roh;{ks=k Eout = iV~VhdsBhdckgjxq:Roh;{ks=k Ein = Eext – Eself = 0  Eext = Eself E = E + E = GM  E = GM Eext out ext self R2 ext 2R2 Eself Nk;kafdr{ks=kQygVkusdscknogk¡dksbZEself ughagksxkdsoyEext. gSA Eext vr%, E net = Eext GM = 2R2 Eself 17.13 A uniform thin rod of mass m and length R is placed normally on surface of earth as shown. The mass of earth is M and its radius is R. Then the magnitude of gravitational force exerted by earth on the rod is nzO;ekum rFkkyEckbZRdh,dlekuNM+dksi`Fohdhlrgij fp=kkuqlkjvfHkyEcor~j[khtkrkgSAi`Fohdk nzO;eku M o f=kT;k R gSA i`Foh }kjk NM+ ij vkjksfir xq:Rokd"kZ.k cy dk ifjek.k & (A*) GMm 2R2 (B) GMm 4R2 (C) 4GMm 9 R2 (D) GMm 8R2 2R Sol. F =  R GM m  dx  R  x2 GMm = 2R2 17.14 Two particles of combined mass M, placed in space with certain separation, are released. Interaction between the particles is only of gravitational nature and there is no external force present. Acceleration of one particle with respect to the other when separation between them is R, has a magnitude : nksd.kftudkla;qDrnzO;ekuMgS]vkdk'kesafuf'prnwjhijj[ksgSa]mudksNksM+ktkrkgSAd.ksadse/;dsoyxq:Rokd"kZ.k cyyxrkgSrFk vU;dksbZckà;cyughyxrkgSAfdlh,dd.kdknwljsd.kdslkis{kRoj.kdkifjek.k]tcmuds e/;RnwjhgS]gksxk& (A) GM 2R2 2GM GM (B*) R2 (C) R2 (D) not possible to calculate due to lack of information (D)lwpukvksadhdehdsdkj.kx.kuklEHkoughgSA Sol. a1 = GM1M2 / M R2 1 a2 = GM1M2 / M R2 2 acceleration of M1 w.r.t. M2 (M1 dkM2 dslkis{kRoj.k) arel. = a1 + a2 G(M1  M2 ) = R2 GM = R2 . 17.15 Maximum height reached by a rocket fired with a speed equal to 50% of the escape velocity from earth's surface is : ,d jkWdsV dks i`Foh dh lrg ls iyk;u osx ds 50% eku ds cjkcj osx ls iz{ksfir fd;k tkrk gSA rks jkWdsV }kjk izkIr i`Foh dh lrg ls vf/kdre Å¡pkbZ gksxh & (A) R/2 (B) 16R/9 (C*) R/3 (D) R/8 Sol. v = 50 1 100 Ve = 2 2GM R Applying energy conservation ÅtkZlaj{k.kls   GMm  1 mv 2 =  GMm R 2GM 2 2GM (R  h) 1 2GM  1 1  1 h v2 =    . = 2GM R  R  h   4R R(R  h)  R + h = 4h  h = R/3 SECTION - II : MULTIPLE CORRECT ANSWER TYPE [k.M- II: cgqy lgh fodYi izdkj 17.16 A double star is a system of two stars of masses m and 2m, rotating about their centre of mass only under their mutual gravitational attraction. If r is the separation between these two stars then their time period of rotation about their centre of mass will be proportional to nksxzgksadsfudk;esam rFk 2m nzO;ekuokysnksrkjsmudsLoa;dsxq:Rokd"kZ.kçHk oesapDdjdkVjgsgSAvxjbu rkjksadschpdhnwjhrgSrksmudsnzO;ekudsUnzdslkis{kmudspDdjdkvkorZdky lekuqikrhgksxk\ (A*) r3/2 (B) r (C) m1/2 (D*) m–1/2 Sol. (A), (D) r2  2mr  2r m  2m 3 322r 3 2 3 2  2 2 Gm m 2m T2  27 Gm T  r3/2 T  m–1/2 17.17 A tunnel is dug along a chord of the earth at a perpendicular distance R/2 from the earth’s centre. The wall of the tunnel may be assumed to be frictionless. A particle is released from one end of the tunnel. The pressing force by the particle on the wall, and the acceleration of the particle varies with x (distance of the particle from the centre) according to : i`FohdsdsUnzlsyEcor~nwjhR/2ijthokdsvuqfn'k,dlqjax[k snhtkrhgSAlqjaxdhnhokjksadks?k"kZ.kjfgrekuktk ldrkgSAlqjaxds,ddksusls,dd.kdksNksM+ktkrkgSAd.k}kjknhokjijyxk;kx;knckocyvkSjd.kdkRoj.k x nwjh ds lkFk fuEukuqlkj cnyrk gS (nwjh x dsUnz ls ekih xbZ gSA) (A) (B*) (C*) (D) (A) (B*) (C*) (D) Sol. (B) (C) Net force towards centre of earth = mg´ = Normal force N = mg´ sin  mgx R Thus pressing force N = mgx R R 2x Hence (B) N = Mg constant and independent of x. 2 tangential force F = ma = mg´ cos  R2  2 gx 4 Q = g´ cos  = R x a = gx R Curve is parabolic and at x = Hence (C) gy (B) (C) R , a = 0. 2 mgx i`Fohds dsUnzdh vkSjifj.kkeh cy=mg´ = R vfHkyEc cy N = mg´ sin  vr% ncko cy N = mgx R R 2x vr% B N = Mg vpj rFkk x ls LorU=k Li'kZ jsf[k; cy F = ma = mg´cos R2  2 gx 4 Q = g´ cos  = R x a = gx R oØ ijoy;dkj gS rFkk x = R ij, a = 0 gSA vr% (C) 17.18 A satellite revolves around a planet in circular orbit of radius R (much larger than the radius of the planet) with a time period of revolution T. If the satellite is stopped and then released in its orbit (Assume that the satellite experiences gravitational force due to the planet only). ,dmixzg]fdlhxzgdspkjksavksjRf=kT;k¼tksxzgdhf=kT;klscgqrT;knkgS½dso`Ùk dkjd{kesaTvkorZdky ls?kwerkgSA;fnmixzgdksviuhd{k esajksddjfLFkjvoLFk esayk;ktk;srFk fQjeqäfd;ktk;sarks¼;gekfu;s fdmixzgdsoyxzgdsdkj.kxq:Rokd"kZ.kcyvuqHko djrkgS½& (A*) It will fall on the planet rks;gxzgijfxjtk;sxkA (B) The time of fall of the satellite is nearly mixzg ds fxjus dk le;yxHkx gksxkA (C*) The time of fall of the satellite on the planet is nearly 2T 8 mixzg ds xzg ij fxjusdk le; yxHkx 2T gksxkA (D) It cannot fall on the planet so time of fall of the satellite is meaningless ;gxzgijughafxjsxk]vr%mixzgdsfxjusdkle;vFkZghugSA Sol. (A) It will fall because mg is acting on it towards the centre of planet and initial velocity is zero. It'll move in straight line. (C) Time of fall can be found by two methods : I Method : By energy conservation 1 mv2 – 2 GMm = 0 – r GMm – (1) R using this we get V = f(r). Now use dr V = – dt R' dr  f (r) = – dr dt t     f (r)  dt 0 ; R' = radius of the planet. In the final expression (or in the beginning itself) R'  0 { R > > R'} you will get t = Here GMm R 2  2 2 = m T  R   Note : This method is longer. If a student gets idea of solving the question only by this method then it is better to leave this question because it will consume more time. II Method : Kepler's Law : T2  r3 . Assume that the satellite moves in elliptical path with maximum and minimum distances from centre as R and R'.  R > > R'  velocity at R is very small ( ~ 0). When it reaches R' then it touches the surface of the planet. This motion (from R to R') is almost, same as given in the question. T2 r 3 Now 1  1 , T 2 2 R  R R r2 = 2 ~ 2 = T, r1 = R  T2 = Sol. (A);gfxjtk;sxkD;ksafdblijmgxzgdsdsUnz dhvksjdk;ZjrgSvkSjizkjfEHkdosx'kwU;gSA;g lh/khjs[k esaxfrdjsxkA (C)fxjusdkle;nksfof/k;ksa}kjkKkrfd;ktkldrkgSA izFkefof/k:ÅtkZlaj{k.kls 1 mv2 – 2 GMm = 0 – r GMm – (1) R bldkmi;ksxdjrsgq,geikrsgSaV=f(r).vcmi;ksxdjrsgSa dr V = – dt R' dr  f (r) = – dr dt t     f (r)  dt 0 ; R' = xzg dh f=kT;k vafreO;atdesa(;kizkjEHkesa;gLoa;)R'0 { R > > R'} vkiik;saxs t = ;gka GMm R 2  2 2 = m T  R   Note:;gfof/kyEchgSA;fnfdlhNk=kdksbliz'udksgydjukdsoyblfof/klsvkrkgSrksblsNksM+nsukcsgrj gSD;ksafdblesavf/kdle;yxsxkA f}rh; fof/k : dsiyj dk fu;e: T2  r3 . ekukfdmixzgdsUnzlsvf/kdreoU;wurenwjhØe'k%RoR'dslkFknh?kZo`Ùk dkjiFkesaxfrdjrkgSA R>> R' Rij osx cgqr de gS( ~ 0)tc ;g R'ij igqaprk gS rks ;g xzg dh lrg dks Nwrk gSA ;g xfr (Rls R'rd)yxHkxiz'uesanhxbZtSlhghgSA T2 r 3 vc 1  1 , T = T, r = R 2 3 1 1 2 2 R  R R r2 = 2 ~ 2  T2 = SECTION - III : ASSERTION AND REASON TYPE [k.M- III: dFku vkSj dkj.k izdkj 17.19 STATEMENT-1 : In free space a uniform spherical planet of mass M has a smooth narrow tunnel along its diameter. This planet and another superdense small particle of mass M start approaching towards each other from rest under action of their gravitational forces. When the particle passes through the centre of the planet, sum of kinetic energies of both the bodies is maximum. oDrO;-1:eqävkdk'kesa,dlekuxksykdkjxzgftldknzO;ekuMgSAbldsO;kldsvuqfn'k,diryhfpduhlqjax [k snhxbZgSA;gxzgonwljklekunzO;ekuMdkvfrl?kuNksVkd.k,dnwljsdhrjQfojkelsmudsxq:Rokd"kZ.k cydsizHk oesaiklvkuk'kq:djrsgSAtcd.kxzgdsdsUnzlsxqtjrkgSrcnksuksaoLrqvksadhxfrtÅtkZvksadk;ksx vf/kdregksxkA STATEMENT-2 : When the resultant of all forces acting on a particle or a particle like object (initially at rest) is constant in direction, the kinetic energy of the particle keeps on increasing. oDrO;-2:tcd.k;kd.kdhrjg,doLrq(izkjEHkesafojkeesaFkh)ijdk;ZjrlHkhcyksadkifj.k ehcyfn'k esafu;r gksrksd.kdhxfrtÅtkZc<+rhtk;sxhA (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 (A*) oDrO;-1lR; gS] oDrO;-2lR; gS ;oDrO;-2, oDrO;-1 dk lgh Li"Vhdj.k gSA (B)oDrO;-1lR; gS] oDrO;-2lR; gS ; oDrO;-2,oDrO;-1 dk lgh Li"Vhdj.k ugha gSA (C) oDrO;-1lR; gS] oDrO;-2vlR; gS ; (D) oDrO;-1vlR; gS] oDrO;-2lR; gS Sol. (A) Till the particle reaches the centre of planet, force on both bodies are in direction of their respective velocities, hence kinetic energies of both keep on increasing . After the particle crosses the centre of planet, forces on both are retarding in nature. Hence as the particle passes through the centre of the planet, sum of kinetic energies of both the bodies is maximum. Therefore statement-1 is True, Statement-2 is True; Statement- 2 is a correct explanation for Statement-1. (A)tcrdd.kxzgdsdsUnzrdigq¡prkgSnksuksaoLrqvksaijcymudslaxrosxksadhfn'kvksaesagSa]blfy;snksuksadhxfrt ÅtkZ;sac<+rhtkrhgSAd.k}kjkxzgdsdsUnzdksikjdjusdscknnksuksaijcyeandizd`frdsgSaAblfy;stcd.kxzg dsdsUnzlsxqtjrkgS]rksnksuksaoLrqvksadhxfrtÅtkZvksadk;ksxvf/kdregksrkgSAblfy;soDrO;-1lR;gS]oDrO;-2 lR;gS( oDrO;-2oDrO;-1dklghLi"Vhdj.kgSA SECTION - IV : COMPREHENSION TYPE [k.M- IV:cks/ku izdkj Changing from a circular to An elliptical orbit Let us identify the system as the spacecraft and the Earth but not the portion of the fuel in the spacecraft that we use to change the orbit. In a given orbit, the mechanical energy of the spacecraft – Earth GMm system is given by E = – 2r . This energy includes the kinetic energy of the spacecraft and the potential energy associated with the gravitational force between the spacecraft and the Earth. If the rocket engines are fired, the thrust force moves the spacecraft through a displacement. As a result, the mechanical energy of the spacecraft – Earth system increases. The spacecraft has a new higher energy but is constrained to be in an orbit that includes the original starting point. It can not be in a higher energy circular orbit having a larger radius because this orbit would not contain the starting point. The only possibility is that the orbit is elliptical as shown in the figure. o`Rr kd kj d {kk Circular orbit Elliptical orbit nh?kZo`Rr kd kj d {kk i `Foh Earth Rocket engine is fired here j kWd sV bat u ; gkWa i zTofy r fd ; k t kr k g vuqPNsn # 1 E = – GMm 2a o`Ùkh; d{kk ls nh?kZ o`Ùkh; d{kk esa ifjorZu ;fn ge varfj{k ;ku ,oa i`Foh dks ,d fudk; ekusa ijUrq mlds bZa/ku Hkkx dks NksM+rs gq,] tks d{k cnyus ds dke esa vkrk gSA nh xbZ d{kk esa] varfj{k ;ku&i`Foh fudk; dh ;kaf=kd ÅtkZ fuEu izdkj iznf'kZr GMm djrs gSaA E = – 2r . blÅtkZesavarfj{k;kudhxfrtÅtkZ,oai`FohvkSjvarfj{k;kudsxq:Rokd"kZ.kcylslEcfU/krxq:Roh;ÅtkZ 'kkfeygSA;fnjkWdsV&batudksTofyrdjrsgS]rksvarfj{k;ku/kDdsdscy}kjkfoLFk firgksdjpyrkgSA ifj.k eLo:i]varfj{k;ku&i`Fohfudk;dh;kaf=kdÅtkZc<+rhgSA varfj{k;kudhu;hÅtkZvf/kdgksrhgSysfduogmld{k esajgrkgS]ftlesaewyizkjfEHkdfcUnq'k feygksA;g mPpÅtkZokyho`Ùkh;d{k esaughajgldrk]ftldhf=kT;kT;knkgksaD;ksafdbld{k esaizkjfEHkdfcUnqughagksrkA rcdsoynh?ko`Ùk dkjd{k dhlEHk oukgksrhgS]tSlkfp=kesan'k Z;kx;kgSA o`Rr kd kj d {kk Circular orbit Elliptical orbit nh?kZo`Rr kd kj d {kk i `Foh Earth Rocket engine is fired here j kWd sV bat u ; gkWa i zTofy r fd ; k t kr k g E = – GMm 2a 17.20 If the spacecraft-earth system had initial energy ( – E0), then the total mechanical energy of the system after firing the rocket will be : ;fnvarfj{k;ku&i`Fohfudk;dhizkjfEHkdÅtkZ(–E0),Fkh]rksjkWdsVdksTofyrdjusdsi'pkr~fudk;dhdqy ;kaf=kdÅtkZgksxh& (A) – 1.1 E0 (B*) – 0.9 E0 (C) – E0 (D) Noneof these mijksä esa ls dksbZugha Sol. (B) The total mechanical energy of the system after firing the rocket will increase by 10%. Hence (B) Note : – 0.9 E0 > – E0 fudk;dhdqy;kaf=kdÅtkZjkWdsVdsizToyudsi'pkr~10%lsc<+tk;sxhA vr%(B) uksV : – 0.9 E0 > – E0 17.21 Semimajor axis of the new elliptical orbit is u;hnh?kZo`Ùk dkjd{k dhv)Znh?kZv{kgS& 6.7  104 (A*) 9 Sol. (A) 6.4  104 Km (B) 9 7.1104 Km (C) 9 6.1 104 km (D) 9 Km Because the mechanical energy is negative, a decrease in magnitude is increase in energy.  GMm Eell  2a  r (6400  300) 6.7  104 Ecir.  GMm a 2r  0.9 = a  a = 9 Km. Sol. (A) D;ksafd;kaf=kdÅtkZ_.k RedgS]vr%ifjek.kesadehgksusijÅtkZesao`f)gksxh  GMm Eell  2a  r (6400  300) 6.7 104 Ecir.  GMm a 2r  0.9 = a  a = 9 Km 17.22 Maximum height of the spacecraft above the surface of the Earth will be : varfj{k;kudhi`FohlrglsegÙkeÅ¡pkbZgS& 1.06  104 (A) 9 0.61104 km. (B) 9 1.61104 km (C) 300 km (D*) 9 km Sol. (D) Maximum distance from the centre of the Earth will occur when the spacecraft is at apogee thus 6.7  104 rmax = 2a – h – r = 2 × 9 7.37 104 7.37 104 – 6700 = 9 1.61104 km. hmax = rmax – RE = 9 – 6400 = 9 km. gy tc;kuviHkwijgS]rcbldhi`FohdsUnzlsegRrenwjhgksxh& 6.7  104 7.37 104 rmax = 2a – h – r = 2 × 9 7.37 104 – 6700 = 9 1.61104 km. hmax = rmax – RE = 9 – 6400 = 9 km. SECTION - V : MATRIX - MATCH TYPE [k.M- V:eSfVªDl&lqesy izdkj Z 17.23 A satellite is revolving around the earth in a circular orbit of radius ‘a’ with velocity v 0. A particle is  projected from the satellite in forward direction with relative velocity v =     1 v . During subsequent  0 motion of particle, match the following : (p) – 3GMem a 5 (B) Minimum distance of particle from the earth (q) 8 GMem a (C) Maximum distance of particle from the earth (r) 5a/3 (s) 2a/3 (t) a Column-I Column-II (A) d.k dh dqy ÅtkZ (p) – 3GMem a 5 (B) d.kdhi`FohlsU;wurenwjh (q) 8 GMem a (C) d.kdhi`Fohlsvf/kdrenwjh (r) 5a/3 (s) 2a/3 (t) a Ans. (A) - p ; (B) - (t) ; (C) - (r) Sol. (B) Angular momentum of particle = m(v0 + v) a = mv0a v0 = Total energy of particle = 1 m(v 2 0 + v)2 – GMem a = 1 × 5 mv 2 – GM em 2 4 0 a = 5 GM em 8 a – GM em a = – 3GMem a At any distance ‘r’ T.E. = 1 mu2 – 2 GMem r but angular momentum conservation mur = m a u = T.E. at any distance ‘r’ = 1 m 5 2 4 GMe a r 2 – GMe m r but through conservation of total energy = 1 m 5 2 4 GMe a r 2 – GMe m r = – 3GMe m a on solving 3r2 – 8ar + 5a2 = 0 (r – a) (3r – 5a) = 0 r = a, r = 5a/3 minimum distance = a maximum distance = 5a/3 gy (B) d.kdkdks.kh;laosx = m(v0 + v) a = mv0a v0 = d.k dh dqy ÅtkZ = 1 m(v 2 0 + v)2 – GM em a = 1 × 5 mv 2 – GM em 2 4 0 = 5 GM em – 8 a a GM em a = – 3GMem 8a fdlh nwjh‘r’dhT.E. = 1 mu2 – 2 GM em r ysfdudks.kh;laosxlajf{krjgrkgS mur = m a u = fdlh nwjh‘r’ij dqy ÅtkZ = ysfdudqyÅtkZlaj{k.kls = 1 5 2 m 4 1 5 2 m 4 GMe a r 2 GMe a r 2 – GMe m r – GMe m = – r 3GMe m 8a gy djus ij 3r2 – 8ar + 5a2 = 0 (r – a) (3r – 5a) = 0 r = a, r = 5a/3 U;wure nwjh=a vf/kdre nwjh= 5a/3 17.24 Let V and E denote the gravitational potential and gravitational field respectively at a point due to certain uniform mass distribution described in four different situations of column-I. Assume the gravitational potential at infinity to be zero. The value of E and V are given in column-II. Match the statement in column-I with results in column-II. ekukfdlhlekax(uniform)nzO;ekuforj.kdsdkj.kVrFk EfdlhfcUnqijØe'k%xq:Roh;foHkooxq:Roh;{ks=k dksn'k ZrsgSA;gnzO;ekuforj.kvyx&vyxfLFkfr;ksaesaLrEHk-Iesafn;kx;kgSAvuUrijxq:Roh;foHkodks'kwU; ekusAErFk VdsekuLrEHk-IIesafn;sx;sgSALrEHk-Iesafn;sx;sdFkuksadksLrEHk-IIdsifj.k edslkFklqesfyr dfj,A Column-I Column-II (A) At centre of thin spherical shell (p) E = 0 (B) At centre of solid sphere (q) E  0 (C) A solid sphere has a non-concentric spherical cavity. At the centre of the spherical cavity (r) V  0 (D) At centre of line joining two point masses of equal magnitude (s) V = 0 LrEHk-I LrEHk-II (A) irysxksykdkjdks'k(shell) dsdsUnzij (p) E = 0 (B)BkslxksysdsdsUnzij (q) E  0 (C) ,d Bkslxksyk ftlesa ,d xksykdkjxqfgdk (cavity) gS;g xqfgdkladsUnzh;ughgSrksblxksykdkjxqfgdkdsdsUnzij (r) V  0 (D) nkslekunzO;ekudsfcUnqnzO;ekuksadkstksM+usokyhjs[k dse/;fcUnqij (s) V = 0 Ans. (A) p,r (B) p,r (C) q,r (D) p,r Sol. (A) At centre of thin spherical shell V  0, E = 0. (B) At centre of solid sphere V  0 , E = 0. (C) At centre of spherical cavity inside solid sphere V  0, E  0. (D) At centre of two point masses V  0, E=0. Sol. (A)iryh xksykdkj dks'k ds dsUnz ij V0,E=0 (B) Bksl xksys ds dsUnz ij V 0 , E=0 (C) Bkslxksys dsvanj xksykdkjxqfgdkdsdsUnzijV0rFk E0 (D)nksfcUnqnzO;ekuksadkstksM+usokyhjs[k dse/;fcUnqijE=0 , V 0 SECTION - VI : INTEGER TYPE [k.M-VI:iw.kkZadizdkj SHORT SUBJECTIVE y?kqÙkjkRediz'u 17.25 Ravi can throw a ball at a speed on earth which can cross a river of width 10 m. Ravi reaches on an imaginary planet whose mean density is twice of the earth. if maximum possible radius of planet so that if Ravi throws the ball at same speed it may escape from planet.is x km. then x is. (Given radius of earth = 6.4 × 106 m.) jfo i`Foh ij ,d xsUn dks ,d pky ls QSad ldrk gS ftlls ;g 10 m pkSMh unh dks ikj dj ldrh gSA jfo ,d dkYifud xzg ij vk tkrk gS ftldk ek/ ; ?kuRo i`Foh dk nqxuk gSA xzg dh vf/kdre lEHko f=kT;k x km. gSA rkfd ;fn jfo xsUn dks leku pky ls QSads rks ;g xzg ls iyk;u dj lds rkss x gksxkA i`Foh dh f=kT;k = 6.4 × 106m. Ans. 4 Sol. Speed of the ball which can cross 10 m wide river is V2 sin(2  45º ) R = g = 10, v = Let the radius of planet is ‘R’, then 4 4 2  Me 2MeR3 Mass of planet M = 3 R3 × 2 = 3 R3 × 4 / 3R3 = 3 Escape velocity on planet V = = g  GMe  2R = 8 103 = 2R = .......   2  R = = 4 × 103 = 4 Km. 2 gy xsUn dh pky ftlls ;g10m pkSMh unh dks ikj dj lds] gS V2 sin(2  45º ) R = g = 10, v = ekukfdxzgdhf=kT;kRgS]rc 4 4 2  Me 2MeR3 xzgdknzO;eku M = 3 R3 × 2 = 3 R3 × 4 / 3R3 = 3 xzg dkiyk;u osx V = = g  GMe  2R = 8 103 = 2R = .......   2  R = = 4 × 103 = 4 Km. 2 LONG SUBJECTIVE fucU/k Rediz'u 17.26 The gravitational field in a region is given by →  (3ˆi  4ˆj ) N/kg. Find out the work done (in joule) in displacing a particle by 1 m along the line 4y = 3x + 9. fdlh {ks=k esa xq:Roh; {ks=k →  (3ˆi  4ˆj ) N/kg }kjk O;Dr fd;k tkrk gSA ,d d.k dks js[kk 4y = 3x + 9 ds v u q f n ' k 1ehVjdhnwjhrdfoLFkkfirdjusesafd,x,dk;Z(twyesa)dhx.kukdhft,A Sol. Slope of displacement vector m = 3 1 4 Slope of force vector m =  4 2 3 y 90º displacement direction x force direction m × m = 3 ×  4 = – 1 1 2 4 3 i.e. force and displacement directions are perpendicular. The work done is zero gy foLFk iulfn'kdk