15-Fluids & Properties of Matter

SECTION - I : STRAIGHT OBJECTIVE TYPE [k.M-I: lh/ks oLrqfu"B izdkj 15.1 A cylindrical vessel filled with water is released on an inclined surface of angle  as shown in figure. The friction coefficient of surface with vessel is  ( < tan ). Then the constant angle made by the surface of water with the incline will be: ty ls Hkjk ,d csyukdkj ik=k  dks.k okys ,d ur ry ij fp=kkuqlkj NksMk t k r k g S A i k = k d s l k F k l r g d k ? k " k Z . k x q . k k a d  ( < tan ) gSA ur ry ds lkFk ty dh lrg dk fu;r dks.k gksxkA (A*) tan1  (B)   tan1  (C)  + tan1  (D) cot1  Sol. (A) figure shows forces acting on a 'particle' on the surface, with respect to vessel.  (mg sin  &  mg cos  are pseudo forces). tan  =    = tan–1 .  is angle between normal to the inclined surface and the resultant force. The same angle will be formed between the surface of water & the inclined surface. { free surface is  to the resultant force acting on it.} gy (A)fp=klrgijik=kdslkis{kdk;Zjrcyksadksn'k ZrkgSA  (mg sin  o mg cos  Nn~e cy gS). tan  =    = tan–1 . urry dsvfHkyEc oifj.kkeh cyds e/;dks.k gSA;gh dks.kty dhlrg our ryds e/;gksxkA { eqDrlrg blij dk;Zjrifj.kkeh cyds yEcor~gS} 15.2 A cylindrical container of radius ' R ' and height ' h' is completely filled with a liquid. Two horizontal L shaped pipes of small cross-section area ' a' are connected to the cylinder as shown in the figure. Now the two pipes are opened and fluid starts coming out of the pipes horizontally in opposite directions. Then the torque due to ejected liquid on the system is: 'R'f=kT;ko'h'Å¡pkbZdk,dcsyukdkjik=knzolsiwjkHkjkgSAnks{kSfrtLvkd`frovYivuqizLFkdkV{ks=kQy'a' dsikbicsyulsfp=kuqlkjtqM+sgSAvcnksuksikbi[ksystkrsgS,oanzo{kSfrtikbiksalsfoijhrfn'k esackgjfudyuk izkjEHk djrk gSA rc ckgj fudy jgs nzo ds dkj.k fudk; ij cyk?kw.kZ gSA (A*) 4 aghR (B) 8 aghR (C) 2 aghR (D)noneofthesebuesalsdksbZugha Sol. (A) Velocity of efflux of water (v) = = gh force on ejected water = Rate of change of momentum of ejected water. =  (av) (v) =  av2 Torque of these forces about central line = (av2) 2R . 2 = 4av2 R = 4 agh R ckgj fudy jgs ty dk osx(v) = = gh ckgj fudyjgs ty ij cy=ckgj fudy jgs ty ds laosx esa ifjorZu dh nj =  (av) (v) =  av2 bucyksadkdsUnzh;js[k dslkis{kcyk?kw.kZ = (av2) 2R . 2 = 4av2 R = 4 agh R 15.3 A block of silver of mass 4 kg hanging from a string is immersed in a liquid of relative density 0.72. If relative density of silver is 10, then tension in the string will be:[ take g = 10 m/s2 ] 4kgnzO;ekudkpk¡nhdk,dxqVdk,dMksjhlsyVdkgS],oavkisf{kd?kuRo0.72dsnzoesaMwckgSA;fnpk¡nhdkvkisf{kd ?kuRo 10 gS, rks Mksjh esa ruko gksxk & :[ g = 10 m/s2 ] (A*) 37.12 N (B) 42 N (C) 73 N (D) 21 N Sol. (A) Let  ,  be the density of silver and liquid. Also m and V be the mass and volume of silver block. S L  Tension in string = mg – bouyant force (A) ekukS,L pk¡nhonzods?kuRogSrFkkmoVpk¡nhdsxqVdsdsnzO;ekuovk;rugSA  Mksjh es ruko=mg–mRIykodcy T =  Vg –  Vg = ( –  ) Vg S L S L m Also o V = s  S  L  (10  0.72)  103  T =  s  mg = 3 × 4 × 10 = 37.12 N.   10 10 15.4. A vessel contains oil (density = 0.8 gm/cm3) over mercury (density = 13.6 gm/cm3). A uniform sphere floats with half its volume immersed in mercury and the other half in oil. The density of the material of sphere in gm/ cm3 is: ,dik=kesaikjs(?kuRo=13.6gm/cm3)dsÅijrsy(?kuRo=0.8gm/cm3)HkjkgSA,dlekaxxksykblizdkjrSjrk gSfdmldkvk/k vk;rursyesarFk vk/k ikjsesajgrkgSrksxksysdsinkFkZdk?kuRogm/cm3esagS (A) 3.3 (B) 6.4 (C*) 7.2 (D) 12.8 Sol (C) Weight = Buoyant force Hkkj =mRiykodcy V g = V  g + V  g m 2 Hg 2 oil m = Hg  oil 2 13.6  0.8 = 2 = 14.4 2 = 7.2 15.5 A small uniform tube is bent into a circular tube of radius R and kept in the vertical plane. Equal volumes of two liquids of densities  and  ( > ) fill half of the tube as shown in the figure.  is the angle which the radius passing through the interface makes with the vertical. ,dNksVhle:iuyhRf=kT;kdho`Ùk dkjuyhds:iesaeksM+htkrhgSrFk m/okZ/kjryesaj[khtkrhgSA?kuRoo (>)dsnksnzoksdslekuvk;run'k Z;svuqlkjvk/khuyhesaHkjsgSAbudslEidZrylsdsUnzdksfeykusokyhf=kT;k m/okZ/kj ls  dks.k cukrh gSA              (A*)  = tan–1   (B)  = tan–1     (C)  = tan–1     (D)  = tan–1     Sol. (A)           Pressure at 'A' from both side must balance. Figure is self–explanatory. Aijnkcnksuksavksjlarqfyrgksukpkfg,Afp=klsLi"VgSA  h g = h g  2 R sin (45º + ) = R [cos – sin]  [cos + sin] = [cos – sin] tan =       15.6 In the figure shown water is filled in a symmetrical container. Four pistons of equal area A are used at the four opening to keep the water in equilibrium. Now an additional force F is applied at each piston. The increase in the pressure at the centre of the container due to this addition is n'k Z;sx;sfp=kesa,dleferik=kesatyHkjkgSAleku{ks=kQyAdspkjfiLVupkj[kqysLFk uksijtydkslkE;koLFk esaj[krsgSvcizR;sdfiLVuijvfrfjDrcyFvkjksfirfd;ktkrkgSAbldsdkj.kik=kdsdsUnzijnkcesao`f)gSA (A*) F (B) 2F A A (C) 4F A (D) 0 Sol. (A) The four piston are initially in equilibrium. As additional force F is applied to each piston, the pressure in fluid at each point must be increased by F so that each piston retains state of equilibrium. A (A) pkjfiLVuizkjEHkesalkE;koLFk esagSAD;ksafdvfrfjDrcyFizR;sdfiLVuijvkjksfirgS]nzoesaizR;sdfcUnqijnkc F lsc<+tkukpkfg,rkfdizR;sdfiLVulkE;koLFk dhfLFkfresajgsA Thus the increment in pressure at each point is P = F (by Pascal’s law) A blizdkjizR;sdfcUnqijnkcesao`f)gSP= F (ikLdydsfu;els) 15.7 A block of iron is kept at the bottom of a bucket full of water at 2°C. The water exerts bouyant force on the block. If the temperature of water is increased by 1°C the temperature of iron block also increases by 1°C. The bouyant force on the block by water ,d yksgsdk xqVdk tylsHkjh,dckYVh dsisUns ij2°Cijj[kk gSAty CykWdij mRIykodcyvkjksfirdjrh gS;fntydkrki1°Clsc<+k;ktk;srFk yksgsdsxqVdsdkrkiHkh1°Clsc<+k;ktk;srksty}kjkxqVdsij mRIykod cy (A*) will increase (B) will decrease (C) will not change (D) may decrease or increase depending on the values of their coefficient of expansion (A)c<+sxk (B)?kVsxk (C)ifjofrZrughagksxk (D) mudsizlkjxq.k adksdsekuksdsvuqlkj?kV;kc<+ldrkgSA Sol. (A) Increasing the temperature of water from 20C to 30C increases its density while decreases the density of iron. Hence the bouyant force increases. Sol. (A) ty dk rki2ºCls3ºCrd c<+kus ij bldk ?kuRo c<+rk gS tcfd yksgs dk ?kuRo ?kVrk gSA bl izdkj mRIykod cy c<+rk gSA 15.8 An open tank 10m long and 2m deep is filled up to 1.5 m height with oil of specific gravity 0.82. The tank is uniformly accelerated along its length from rest to a speed of 20 m/sec horizontally. The shortest time in which the speed may be attained without spilling any oil is : [g = 10 m/sec2] 10m yEch o2m xgjh ,d [kqyh Vadh1.5m xgjkbZ rdfof'k"V ?kuRo0.82dsrsy lsHkjk gSA Vadh leku :i ls bldhyEckbZdsvuqfn'kfojkels20m/sec dh pkyrd {kSfrt:i lsRofjrgksrhgSArsydsckgjNyds(fudys) fcuk U;wure fdrus le; esa ;g pky izkIr dh tk ldrh gS: [g = 10 m/sec2] (A*) 20 sec. (B) 18 sec. (C) 10 sec. (D) 5 sec. Sol. (A) v ax v 0.5 v = u + axt , ax = t tan = g = tg = 5 (in triangle ABC) (f=kHkqt ABC esa)  t = 10  20 10 = 20 sec. 15.9 A U-tube of base length “𝑙” filled with same volume of two liquids of densities  and 2 is moving with an acceleration “a” on the horizontal plane. If the height difference between the two surfaces (open to atmosphere) becomes zero, then the height h is given by: vk/kkj yEckbZ“𝑙”dh,d Uuyh ?kuRoo2ds nksnzoksdsleku vk;ruls HkjhgS rFkk{kSfrt ryij“a” Roj.k lsxfr'khygSA;fnnkslrgksdse/;Å¡pkbZdkvUrj([kqysokrkoj.kesa)'kwU;gkstkrkgSrksÅ¡pkbZhgSA a (A) 2g 𝑙 3a (B*) 2g 𝑙 a (C) g 𝑙 2a (D) 3g 𝑙 Sol. (B) For the given situation, liquid of density 2  should be behind that of . From right limb : (B)nhxbZfLFkfrdsfy,]2?kuRodknzo?kuRodsnzodsihNsgksukpkfg, nkW;hHkqtkls PA = P atm +  gh PB = PA +  a 𝑙 = P atm +  gh + a 𝑙 PC = PB + (2) a 𝑙 = P atm +  gh + 3  a 𝑙 (1) Butfrom left limb:(ijUrq ckW;h Hkqtkls) PC = P atm + (2 ) gh (2) From (1) and (2) : (1) o (2) ls :) Patm +  gh + 3  a 𝑙 = P atm + 2  gh  h = 3a 2g 𝑙 Ans. 15.10 A narrow tube completely filled with a liquid is lying on a series of cylinders as shown in figure. Assuming no sliding between any surfaces, the value of acceleration of the cylinders for which liquid will not come out of the tube from anywhere is given by ,dnzolsiw.kZr;kHkjh,diryhuyhcsyuksadhiafDrijfp=k uqlkjj[khgSAfdlhHkhlrgijfQlyuughagS ;gekursgq,csyuksdsRoj.kdkeku]ftldsfy,nzodghalsHkhuyhlsckgjughfudysxk]fn;ktkrkgSA open to atmosphere a (A*) gH 2L (B) gH L (C) 2gH L (D) (D) Sol. (A) No sliding  pure rolling Therefore, acceleration of the tube = 2a (since COM of cylinders are moving at 'a') (A) dksbZfQlyuugha'kq)yq<+du blfy,]uyhdkRoj.k=2a(pwafdcsyuksadknzO;ekudsUnz'a'lsxfr'khygS) PA = Patm +  (2a) L (From horizontal limb) ({kSfrt Hkqtk ls) Also ; PA = Patm +  g H (From vertical limb) (Å/okZ/kjHkqtkls) gH  a = 2L Ans. 15.11 The velocity of the liquid coming out of a small hole of a vessel containing two different liquids of densities 2 and  as shown in figure is ,dNksVsik=kds ,dNksVs fNnzls ckgjfudy jgsnzo dkosx D;kgS ftlesa2o?kuRods nksvyx vyxnzo fp=kkuqlkj Hkjs gSA (A) (B*) 2 (C) 2 (D) Sol. (B) Pressure at (1) : (B) (1) ij nkc P1 = P atm +  g (2h) Applying Bernoulli's theorum between points (1) and (2) fcUnqvksa(1)o(2)dse/;cjuksyhizes;vkjksfirdjusij 1 [Patm + 2  g h] + 2g(h) + 2 (2 ) (0)2 = Patm + (2 ) g (0) + 1 (2 ) v2 2  v = 2 Ans. 15.12 A non uniform cylinder of mass m, length 𝑙 and radius r is having its cetnre of mass at a distance 𝑙/4 from the centre and lying on the axis of the cylinder. The cylinder is kept in a liquid of uniform density  The moment of inertia of the rod about the centre of mass is . The angular acceleration of point A relative to point B just after the rod is released from the position shown in figure is nzO;eku m,yEckbZ𝑙o f=kT;krds ,dvle:i csyudk nzO;ekudsUnz bldsdsUnz ls𝑙/4nwjh ijcsyu dhv{k ij gSAcsyule:i?kuRodsnzoesaj[k gSAnzO;ekudsUnzdslkis{kNM+(csyu)dktM+Rovk?kw.kZgSAfp=kesan'k Z;h fLFkfrlsNksM+usdsrqjUri'pkr~fcUnqAdhfcUnqBdslkis{kNM+dkdks.kh;Roj.kgSA g𝑙2r 2 (A)  (B*) g𝑙2r2 4 (C) g𝑙2r2 2 (D) 3g𝑙2r2 4 Sol. (B) Torque about CM : (B) nzO;ekudsUnzdslkis{kcyk?kw.kZ Fb . 𝑙 =     = 1 (r2) (𝑙) () (g). 𝑙 I 4 r 2𝑙2g  = 4 I '' will be same for all points. Hence (B). lHkhfcUnqvksdsfy, lekujgsxk bl izdkj (B) 15.13 The coefficient of viscosity  of a liquid is defined as the tangential force on a layer in that liquid per unit area per unit velocity gradient across it. Then a sphere of radius ‘a’, moving through it under a constant force F attains a constant velocity ‘V’ given by : (where K is a numerical constant) ,dnzodk';kurkxq.kkaddksmlnzodh,dijrdsizfrbdkbZ{ks=kQyrFk izfrbdkbZosxço.krkdsfy,Li'kZ js[kh;cyds:iesaifjHk f"krfd;ktkrkgSArksbleasxfr'khy‘a’f=kT;kdkxksykfu;rcyFdsizHk oesafu;r osx‘V’izkIrdjrkgStksfn;ktkrkgS: (tgkW K ,d la[;kRed fu;rkad gS) F (A) K Fa  (B) K a Sol. (C) F  (C*) K a (D) K a F by dimensional analysis, (c) is the only correct answer. Sol. (C) foeh;fo'ys"k.klsdsoy(c)lghmÙkjgSA 15.14 A square box of water has a small hole located in one of the bottom corner. When the box is full and placed on a level surface, complete opening of the hole results in a flow of water with a speed v 0, as shown in figure (1). When the box is still half empty, it is tilted by 45º so that the hole is at the lowest point. Now the water will flow out with a speed of tylsHkjsgq;soxkZdkjckWDlesafupysdksusij,dNksVkfNnzfLFkrgSAckWDliw.kZ:ilsHkjkgSolerylrg ij j[kk gSA fNnz dks iwjk [kksyus ij ty ds izokg dh pky v0 gS, fp=k (1) ds vuqlkjA tc ckWDl vk/kk [kkyh gksrk gS] ;g 45º dks.k ij >qdk;k tkrk gS rkfd fNnz lcls fupys fcUnq ij jgsA vc ty fdl pky ls cgsxk & V0 Fig. (1) Initial (fp=k(1) çkj fEHkd ) (A) V0 (B) V0/2 (C) V0/ Sol. (D) (D*) V0/ V0 = V2 = = 15.15 A steady stream of water falls straight down from a pipe as shown. Assume the flow is incompressible, then tydhLFk ;h/k jkikbilsn'k Z;svuqlkjlh/khfxjrhgSekuksfdizokgvlEihM~;gSrc (A) the pressure in the water is higher at lower points in the stream. (B) the pressure in the water is lower at lower points in the stream. (C*) the pressure in the water is the same at all points in the stream. (D) pressure variation will depend upon density and exit speed of the water. (A) /k jkesafupysfcUnqvksaijtydknkcmPpgSA (B) /k jkesafupysfcUnqvksaijtydknkcdegSA (C*) /k jkesalHkhfcUnqvksaijtydknkclekugSA (D) nkcesaifjorZutyds?kuRoockgjfudyus dhpky ijfuHkZj djsxkA Sol. (C) Pressure at all points in stream will be atmospheric. /kjkesalHkhfcUnqvksaijnkcok;qe.Myh;gksxkA 15.16 Water filled in an empty tank of area 10 A through a tap of cross sectional area A. The speed of water flowing  out of tap is given by  (m/s) = 10 (1 – sin 30 t) where ‘t’ is in seconds. The height of water level from the bottom of the tank at t = 15 sec will be {ks=kQy10Adh,d[kkyhVadhesavuqizLFkdkV{ks=kQyAds,duylstyHkjktkrkgSAuylsckgjfudyjgsty  dh pky (m/s)=10(1–sin 30 t) ls nh tkrh gS tgkW‘t’lSd.M esa gSA Vadh ds isUns ls ty ds ry dh Å¡pkbZ t = 15 sec ij gksxh 30 5 30 (A) 10 m (B) 15 +  m (C) 4 m (D*) 15 –  m Sol. (D) Volume of water filled in tank in t = 15 sec. 15  cos  / 30t 15 V = 0 A  10[1 sin t]dt 30 V = 10A[t    / 30 0 V = 10[15  Sol. (D) 30 ] A h =  V 10A  = 15  30    m t= 15secesaVadhesaHkjsikuhdkvk;rugS 15  cos  / 30t 15 V = 0 A  10[1 sin t]dt 30 V = 10A[t    / 30 0 V = 10[15  30 ] A h = V 10A  = 15  30    m 15.17 A mosquito with 8 legs stands on water surface and each leg makes depression of radius ' a'. If the surface tension and angle of contact are ' T ' and zero respectively , then the weight of mosquito is : 8VkWxksokyk,dePNjtydhlrgij[kM+kgS]izR;sdVkaxf=kT;k'a'dkxM~