tnerved that an object dropped fom ahegh fll towards Crm en obser ° falls aaa OTT oad Gade anid tar soioay aeaktacey erode wee oe ‘other object. This force of attraction betwee : 1 Nomis Laws of Planetary < pect. ion n two objects is called the force» Kenters Laws Regerteioon force. In this harecr, we allie about guriatiod tod" Me aw of gravitation. + Free Fall * Mass and Weight ~ Thrst and Pressure Ne enc iry eras cane tien sea | oe) for pulls) all objects lying on ot near its surface towards its centre. The force with. gychrredes’Prncile the earth pulls the objects towards its centre is called the gravitational force ofthe or gravity of the earth * Relative Densty al Law of Gravitation igiven by “Isaac Newton. According to this law, the attractive force between any two in the universe is directly proportional to the product of their masses and inversely onal to the square of distance between them. The direction of the force is along joining the centres of rwo objects. Consider two bodies A and B having masses ™, ‘whose centres are at a distance d from each other. 3D A 3 The gravitational force between two bodies is directed along the line joining their centres the force between rwo bodies is directly proportional to the product of Foemm, their masses, (i) ———— ‘eae Newton was born in Woolsthorpe near Grantham, Engiand in 1642. In 1685 he observed an {Be ag on ne ground wich prompted Newton 10 ore the possiblity of connecting gravity fe faoe thet kept ne moon in ts orb Thi led him to We universal law of gravationand the force berween two bodies is inversely proportional 0 the square ofthe distance berween them, ic. Fa (i, 7 ‘Combining Egg. () and (ji). we get aoe hers = 667% 10°'Nom"/kg? is called the universal ‘gravitational constant His value does nor depend on the n between the ew bodies and the masses of the bod he distance between them. Suppose the masses of two bodies a, ach and th distance d between them is 1 m, the Hence, the universal gravitational constan ined as the grvitational for dies of unit masses and separated by a unit ds m each other p in space. T Gis Nom’ hg? The value of G wa importance of Universal Law of Gravitation The universal law c 7 Phenomena given as b The force that binds us sch G9) The motion of the mo Gif) The motion of pla (0) The occurrence tides is 9) The flow of water in rivers is also duc to gravitation force of the earth Motion of Moon Around Earth Gnd Centripetal Force The force thas keeps a body moving along the circular pat puting towards the conse is called centripetal (centre se, force. The motion of the m feusPenl force. The cenuipetal force is provided by sh. Beatiational force of attraction of the earth. If there wee ni feb force, then the moon would pursiea uniform staighs ne notion v nod the orev (5 LO Sarin cette rem ae surface. (Given, mass ofthe earth = 6 mn jadi of bw ecrh=04 x10" i yo" oper Sol. Given, mas ofthe ath m, = 610% og ons Masot an obec. m, = 2g ¢ cath, R= 6410'm | ee by univer aw ofp termi ong ample 2. The mass of the mars is 639 4p and of fhe oper’ s0 10" ta rm 1090 ga fies : ‘elon Onn *tepler’s Seco 89 x e 1 a wast | Kepler's Thir t. Stes thar the cube 7 bis tly propor, 3 “expressed ng 4 Na 7 2thatthe path of any planet in an he shape ofa clipe with he nan oy raion An elipse showing Kepler's frst low point in the orbit of a planet nearest to the sun is call on and the point farthest from the san called “Fourstar an imaginary line from the sun tthe planet sweeps ‘etequal areas in equal intervals of time. Ths ifthe time of travelling ofa planet from A to Band from C ‘mDissame, then the areas AOB and COD ate equal ‘Areas showing Kepler's second law Kepler's Third Law Tesates that the cube of the mean distance of a planet from the ‘unis directly proportional to the squate of its orbital period 7 Kiverpresed as Pat, PakxT? ime period of the planet (around the sun), ius as mean distance of the planet from the sun wd b= Kepler constant. of autaction. Suppose the orbital loco wand che radian of he orbits. Then, he acting on an orbiting planet i given by where, is the mass of planet. TET denotes the ime period, then ae =; 2 br Substituting the value of» in Eq. (i), we get m= i a Bur aconling co Kepler’ third law of planetary motion, Peet? Putting this value into Eq, (i), we get AR mer _4nkm “Thus, gravitational force berween the sun and the planet is inversely proportional to the square of the distance between their centres, Free Fall ‘When objects fall towards the earth under the influence of earth's gravitational force alone, then these are called freely falling objects and such a motion is called free fll Acceleration due to Gravity (g) ‘Whenever an object falls towards the earth, an acceleration is involved. This acceleration is due to the carth’s gravitational pull and is called acceleration due to gravity, It is denoted by g.i.e fs? ‘The Slunit of gis the same a that of acceleration et mas of the earth Be Mandan objec aig ey towards ic be m. The disance beween centres of the earth and the objec is From Newton's law of gravitation, Gin = ‘Abo from second law of motion, force cxertcd on ano Fe Fame Since. «= ¢ (i. acceleration due to gravity) Fame Bguating RHS of Eqs. () a ee From the formula, iis lear that acceleration due to gravity does not depend on the mass ofa falling object. Itdeperch, ‘only on the mass ofthe earth or celestial bodies Calculation of the Value of g To alee the valu of g, we should put the value of GM and Rin above formula, ic. ¢ =GM I Mass ofthe carh, M1=6%10™ ky Radius of the earth, 8 =6.4 10° m Universal gravitational constant, ( 7 X10" Nem? ig GM _ 667 x10 ® (64 10%) 6x10 Equations of Motion for Free Fall a pete equations of motion which we have dived cir Apes under uniform scceraion. In casrof san of bodies under free fall, there is a uniform acceleration, i.e acceleration due (0 gravity (g) acting downward, General equations Entre ty ng wher Hs the eight fom which the obey 8] oa i tne fils the inl velocity and vsthe agg, eal when the body accelerates a 5 eee tn solving numerical problems, we should meng fib z following points a OF an objet fall verily dovawan, Hei acceleration due to gravity is taken “ since its velocity increases while falling," P™ GE an object is thrown vertically upwand acceleration due ro gravity is taken’ a eaten Example 3. A car falls of a ledge and drops ground in OB s. The value of gis 10 mis? ton simplifying the calc ) What ih Whe Fe Example 4. An object is thrown vertically upwant \ And rises 1.4 height of 1307 Cateutate UW) the time taken by the highest paiT=Time of acent+Time of descent =2¢ “ime taken to return back, = 22.5" 55 [gate Kepler's est aw of pantary motion Tp iatie tne cterence between perinalion ane apaton? "a fammetiank. Fr detance ofa planet rm tho suns 40 times that of he Tein tre rate of ne tre periods of revoluton arcund the ae 1 Sippose 2 plane! exists whose mass and rads both are halt Bpatctine earth. The acoseration due to gravy onthe stace ‘iis planet willbe double sty $ Sate Te and Fae for the following siatoments Widcostraton due to gravity of the eats less at equator than al poles. Willen object fats vericaly downwards, then accleration fous to pray is taken a3 rogatve. Since, #s velooty ireeases whe ang I Abdy crops down a tower and reaches the ground nos GatOmet then tnd the nent ore tower ane. 18m) the toal content of the body which measures the ofa body Isa scalar quanciy and is SI wit Tn other words, mass is the quantity of matter in the universe, "The mass of the ‘Weight ofan object. = mg ‘where, m= mass and g = acceleration due vo peavey Here, M = mass ofthe earth and R = radius ofthe earth Important points rgaeding weight areas follow: (9 Weight ie a vector quantig, it ace in verily ownward direction and its SI unit is newton (N)- ‘Weight of tk mass is 9.8. (ie. 1kg-we=9-8N) Weight of an object is not constant, it changes From place to place. (it Ta the space, where g =0, weight of an abject is 2670. is) At the centre ofthe earth, weight becomes zero. This dducto the fc that on going doven tothe earth value of ‘deereses and at the cente ofthe earth, = 0. Note «From the above formula, tis ear tha weight of an bt Si change ona panet oer than te ea «Spring banc sued ta measure the weight of a boy and Pan thance tured to measure the mass of body. Weight of an Object on the Moon Let the mas of an object be mand its weight on che moon be a Suppose the mass of che moon is M, and it radius be Tar According co univer lw of ravitaion, che weight fn “objec on the moon will be Maxm Ry Ler the weight ofthe same object on the earth be w,.Let che meas ofthe earth be Mf, ad the radius of the earth be R, CMam @ R, [, =7.36X10™ kg 37 x 108 Now, My =598 x 107 gs R,=LTAx 10% R, tug 59810 (637x109)? 1 my 7396x107 (LAI)? 6 “Thus, the weight of an abject on the moon is one-sixth of is weight on the earth.Example 6. Mass ofan object is 12 ks. Cateuiate {is welght onthe earth. Gta weigton ne moon Sol Gia, mas ofan cet. m =k (0. Aaceraion du vi onc, ‘Weigh on theca, ~ (2) Accretion due gravity on mo00, fu 9 Weigh onthemoon tg = mpg =12%7 =98%2 =196N Example 7. A man weighs 600 N on the earth. What ‘is his mass, fg is 10 ms"? On the moon, his weight would be 100 N. What is the acceleration due to ‘gravity on the moon? Sol. Given, weighof man oa teeth, Aceon Sto prvi on te ca, Weigh fan oth moon, = 100 Aesheion doc vy on meen. Asweknow mas ofthe men, Simla, for the moon, g, = Thus, acceleration due ro gravity on the moon & 6 xample 8. A particle weighs 120 N on the surface {the earth. At what height above the earth's surface {Mlits weight be 30 Né Radius of the earth = 6400 kr. ML. Given, weight of particle on the surface of earth, = 120 N Weight of particle at height # above che eats sulace, 1% =30N ‘The weight ofa panicle onthe surface ofthe euch is be Ft ‘Let be the weight of a particle at eighth sbove che earths surface 1.66 mis, Le ge M 6 sy OR es wy poe a ® (ReoF 4 woman is wea Suing yen 10.4 "ine meant (Ee) =o) ale eet aad zt 7 ont og = 10m Figen ada on et a nce wile hey [Check point 03} Coston Tomat 4 haloes 70 Hono seven re eer ck of 5 Vrtolagin Mcei06N “olock is Thrust and Pressure SE ee ee ao a ‘of thrust is newton (N). I is a vector quanti Pressure isthe force acting perpendicularly on a unit an object ; Force(F) Thrust P ) ee Area (A Area The SI unit of pressure is Nm, which is also (Pa) named after the scientist Blaise Pascal. quantity (Gr=iNa?) From che formula of pressure, i is clear that che same an produce different presures depending on the a2 0% which it as force ating ona snl ara x2 pressure while the same force small presse. Example 9. Force of 200 N is applied to an abet acting on a large ar area 4 mi’. Find the pressure. Sol. Giveo, force, F= 200 N, area, A= 4 m Nom, Preaurs,p 200 _ 50 Nm Area(d) 4 fad the pr exerted by tht ble op, if it is made to Lie on 30cm x20 cm The dm 0 cm X30 ] sh Bread10, A woman is weating sharp, the mass of featetr pases, the mens woman weO kg ana‘ SoR@ Dally ite Bo ind out the pressure Applications of Pressure The as ota tn in ed a ce the oe il bh wih of tng psd oe enon. Reta nite ree a: | See eeee ees aes paaculecnmitaweee ‘6000 + Tre on grunt mre wen aman » wing tan whe ONES es Sah are ee . (weigh woman (ven, mas of woman. r= 60g] ad en Aste Fore() _ 600 10000 poate =e or 6000000) pen aly th worn) sadig on ely Biden ten" c00G00 Na M1, A block of wood is kept on a tabi top block is 6 ka potas of wooden Ds kg and is dimensions ~— Pragsure in Fluids. Pegpem x30 cm x20 cm e = Alig and gues ae vogethe called ids, Water and ‘rpendicur gy Midte te wo dr weight Fhe abo ave. wee be area on wi — ‘hte Buds ao exc pressure on the base and wal of pressure inal diecions. Buoyancy oe. the Sha tity, adhe pressure exerted by yon a unitamd slept is made to lie o tne table top with esol dimensions The tendency of aliquid to exer an upward force on a ined ts s20 cm ec immersed in itis called buoyancy. Gases alo exhibit oe (50cm x30.em this property of buoyancy Be I a en ofthe wooden bck = Ge Buoyant Force isan upward fore which acts on an object {Uso called pe Tedimesions = 50cm «30 cm x 2000 when iis immersed in aliquid. Ieis alo called upehrust. al Teis asa] Tiras, F=mg = 698-588 Itisthe buoyant force due ro which a heavy object seems t ) Ar ofa side = Lengeh x Breadth be lighter in water. As we lower an abject ino aliquid, the Tiquid underneath ie provides an upward force t che sameloesy PPMP 4 ae ya foe by our thumb the eatk immediatly cise dn he aes pun torte urdace. This is duet che fae chat every Tiquil exerts nom Gy When th block is on i side of dimesins an upward force on the objects immersed in it | ger a SKC Soc 0 cr iene the sets qud Ara= Length Breadth sr obec 4 spparene ‘45030 =1500 cm? =015 m aan objed? Prue, =f « 88 392 Nim fal262 a | em Factors Affecting Buoyant Force “The magnitude of buoyant force depends on the following I facrore {Density of the Fluid ‘The igi having higher density exer more upward buoyant ected ete deny. Ta ix the reson why it is easier to swim in sea water in ‘comparison ro normal water, Thesca water as higher density and bene exeresa greater buoyant fore on the swimmer than | the freshwater having lower dens {il Volume of Object immersed in the Liquid ‘As the volume of sold objec immersed inside the liquid increases, the upward buoyant force also increases, ‘The ‘magnitude of buoyant force acting ona soli object does not depend on the nature ofthe sold object. It depends only on is volume ‘2g, When wo balls made of different merals having different ‘weighs bu equal volume ate fully immersed in aliquid, they will experience an equal upward buoyant force as both the balls displace equal amount of che liquid due to their equal volumes, Floating or Sinking of Objects in Liquid When an object is immersed in a liquid, forces at oni: * Weight of the objece which acts in then following two downward direction, ic. ic tends +0 pee] all down the object. ie * Buoyant force (upthrust) which acsin. | 24-— | upward direction, i. it tends to push gid the object will float in the liquid, aad (i isbn oe mex a he wig ecient ese All 70H Seetea FE ain 2 Fainthe blank st perience 2 What is pascal? pote sd with broad foundation a 3 why are uldings responsible for this? a (° Weigh i - Density i The density of a substance is defined as mas pay sp paced ohne HL Soviet 5 Maas oF the sabecance > aise! So Volumes of Cie sobean cola ofA Volume ofthe mbrtanc spalicalions The Sunt of dense is hilogram per metre cube (gg | yesinele® PNT Ie'is a scalar quantity. The density of a substance ya wg ships anc specified conditions always remains same. Hens : : proper Tecan hep us to determine is purty. kes Gea How does a Bt ‘pth le 13 different substanes. The lighinss and the henge diferent substances can be described by using the wld Objects having less density than that of aliquid jen} : will oat onthe liquid. Ob cater densiytha| that of liquid, objects in the liquid. Iedereas} NSE with increase in temperatun bles ha id ilo oe gud ns ‘ ly incl Example 12. A sealed can of mass 700 g has , Volume of 500 cm. Will this can sink in water? (Density of water is 1 gem.) Sami Denti of can,p= Since, densy of 0 an wil sink in watemerged sian ‘ert + tow does a Boat Float i imedes’ Principle ta" When an objec sly of paral immened ane aech ran force or upthrs acting on an abjece Weight of liquid displaced by the object pecoring 0098 lec wi oat eve of obec eaua 1 he el of pu ets Peel scale law of floatation mcai ees pccln then inter toerer aps spplcations of Archimedes’ Principle pines principle is sein ‘ eiging ships and submarins. Be eter device used io devermine the pusiyof ml. ipdomecr (a device used for determining the densiy of “iui. Water? force calles buoyant (eat oats n water de 10 upwa force or opehrus) which is caused Posting wp the bosom of the boat. When Boats gradualy Preredinto water i cisplaces more and more water. Hence jan free on it also increases, When this buoyant oe ecames jut enough fo suppor the weigh of boat heb Tops sinng owe in water Now, according o Archimedes frp, “Bonyant force i equal tothe weight of Haut Iepaced byte boat” Hence, during the floating of he boat Ibe weignc of water displaced by the submerge par ofthe Beatie equal to the woh of the boot Erample 13. 1/ an iron object is immersed in water. it iplaces 8 kg of water. How much is the buoyant force acing on the tron object in newton? (Gen, g=10ms*) $4. Acordingto Archimedes’ principle “The buoyant fore ating sigs sje lb equal othe weight of wae pase by this iron object. Wekaow that, weight.» Given, mass of water, m = Shy by 0 Acceleration duc to gravity. ¢ =10 >" Opting aus in Fa. (68 tse x10-40N Ste weigh of wae pd by ito be 80 therefore de buoyant force acting on the ion object (Ye cls be a0 N Relative Density ‘The relative demsicy of «substance isthe ratio ofits density to thar of water De aft bance Tin other words, we can say that Relative density of a substance ass ofthe substance, Volume of water Volume of the substance ~ Mass of water Mass Volus [: bemio = Since, the relative density ipa ratio of similar quanciies, soithas no unit, The relative density of a substance expresses the heaviness (or density) of the substance in comparison to water. By saying that relative density of iron is 87, we mean that ion 158.7 times heavy a8 an equal volume of water. The relative density of a substance is found accurately by using Archimedes’ principle. Example 14. Relative density of silver is 19.3 The density of water is 10" kg/m’. What is the density of silver in S! unit? Deny of wate Density of er Deny ofr = Relative densicyx Deo Example 15. A solid of mass 600 g has volume 450 em’, then i) what is the de i) what willbe the mass of wat (ii) tnd the relative density liv) wil it loat or sink in water? ‘displaced by this solid? Sol. Given, mas of said, m= 600 g Volume of slid = (0 ¢, Densinyofsolid = MB = 9 = 1.23 gem Volume * 450 Therefore density of rid = 1.33 g i ‘The sold will displace water equal 10 its 0 hich is 450 cm, So, ie will displace 45 Density of water is 1 gemen NCERT FOLDER —s | 4 What do you mean by acces= ae : SUR Biren mu i ete Deptt Ss The weight of che object, acting verily downwards, (Theos ofthe wats acting verily upwards “The objec wll oat on the surice of wace. ifthe phe greater than the weight of ee object. The thet wil sink, ifthe eight ofthe object is more than the upthras ofthe wate. D You ind your mass to be 42 kg on a weighing Machine. Is your mass more or less than AZkg? Pgis2 Mas is more than 42 hy. As the buoyant force duc 9 Airis actng on us, the reading of weighing machine eles than the actual mass of person ‘You have a bag of cotton and an iron bar, indicating a mass of 100 kg when don a weighing machine. In really. isheavier than other. Can you say which isheavier and why? Pale i the buoyant Force ating 0” ste sue aa ore the ron-piece cance is ede wohl? = 112. Then, new fee of gviation, Fe SRM Gm «gy mm gp ae le, The fre of vino becomes 4 times of che ‘original value. " an 2 Gravitational force acts on all objects in Proportion to their masses. Why, a heavy object. does not fall faster than alight object? Sol. Acceleration du ogra (independent of mass ofthe aling objec andi fll objects ata pine Sexton obj el ere err Ta objec ‘3 What is the magnitude of the gravitational force between the earh and a1 kg object ont surface? (Take, mass ofthe earths 6» 10% kg and radius of the earth is 6.4 x 10° m.) Sol. Grain fie brent ath nd an bj aa renby 2 Sv R wee, G =fravasonlconane (67 10°" N-mn?fg2, M = mas of thecath = 6X10" kg, radius ofthe exh 64108 ane msmasetanchjes =I 667 x10" x6 x10" x1 axe 97 -98N Thus, maginde of paviationa fre beowen the carthand | kgobjet 898. 4 The earth and the moon are attracted to each cther by gravitational force. Does the earth ‘tract the moon with a force thats greater than ‘or smaller than or equal to the force wih which the moon atracts the earth? Why? Sol. When ewo objets aract each other, then gravitational Foret of tation apple by fine objec on the second ‘Best issame asthe Force applied bythe second object Stk ie object So, both ext and moon act each Sther by the same raviaional force of aactionsha Porn of nai : 9 Whats the accelerator 10 What « gravitational force ban n Amit buys few grams of gold at the pies Hint 7 Colcutate the tor Sette tree ot tween the Sart = 6 50™ tg and of the san ; TL ANNAO® distance between ae greater at the pl 14 Aballis thro 'S A stone is height 19 Before tou,1 1on the moon = ofthe weight on the cath 1 MOBI ¢ apalis thrown vertically upwards with f49m/s. Caloulate men eee the maximum height to which irises. _(p thetotaltime ittakes to return tothe surface ofthe ‘Gives intial velocity, u = 49 mis { Atche maximum height velocity becomes ze. Final velociry, v =0 From the third equation of upward motion, aw —2gh 0= (49) -2x9.8%5 4 oe oe 2x98 Maximum height rained =122.5m 4 {Time taken by the ball co reach the maximum height From the frst equation of motion,» =~ & or 0249 -98%¢ 2.5m. same asthe time of ascent {ofl from maximum height is & 2 Total time taken by the ball co seeurn 100 % the earth =5 +5 = 105: ‘Astone is released from the t0P OF © height 196 m. Calculate its final veloc a ‘before touching the ground. 1% Given, height, b = 19.6 om tid ution noi! = +296 00+ 298196 =196 196 8 DEA =9.mh endl i56 a a ee on A stone is thrown sae inal vloty ot 40 is Taking gs 1015 find the maximum height reached by the ‘stone. What isthe net displacement and the total distance covered by the stone? | Given, inital velocity = 40 ms Final velociy becomes eo, ie.» [sc maximum height) From third equation of upward mation, Paw -2gh (07 = (40)? -2«10 <4 "Maximum height reached by the stone = [Afer reaching the maximum height, the sone will {allcowands the earch and wil each the earths ouriace covering thesame distance So, distance covered by the stone = 80+ 80 =160 Displacement ofthe stone =0. Because che stone starts fom the earths surface and Bally caches the earths susface again i.e. the initia and final posions ofthe stone are same. A stone is allowed to fall from the top of a tower 100m high and at the same time, ‘another stone Is projected vertically upwards from the ground with a velocity of 25 mis. Calculate when and where the two stones will meet. Lecafer ier both stones meet and sbe she distance tall by the stone dropped from che top of rower At which the stones will meet Distance travelled by the stone dropped = + «Distance teavelled bythe stone projected upwards Lj somoe 22942 ¢0)0° sams dgt=0+ 200) eae tae asset 5 el 2 Focthesone projected upwards =~ gr [due co upward motion, negative sign i taken] 00-1) = 25! 1002 Substiuting value of from Eq, (i), we get So, the stones will mer afer 4s, 22= 5x (4)! = 80m So, the scones willbe aa distance of 80 m fiom the top ‘of tower or 20 m (100 m — 80 m) from the base ofthe 18 Aball thrown up vertically returns tothe thrower after 6s, find (i) the velocity with which it was thrown up, (i) the maximum height it reach (i) and its position ater 4 5 Sol. Total time taken Time keno each he mati hight = S = 3 ~~ time of ascent = time of descent} (@) From the fst equation of motion, st negative sign is taken due to upward motion) ~98 x3 (¢*4t maximum height, » = 0] = u=D4ms (2) From the third equation of motion, it 3 Unegsive sign ip taken due to upward motion) 0= (294) By pu 294? 2x98 Maximum height attained bythe ball is 44.1 m, i) Ini 3s he bal willie hen a toward the earth, ¢ mo ‘The position aier 4 = Distance covered in 1s in the downward t “7 axa lg Teball wl 49m dow fag thewower or the eight ofall rom fe 19 In what rection. do0s the buoyant an object immersed in aliquid act? Sol. ‘The buoyant fre onan objet immenadigg thon wine rely prone a 20 Wry doas a beck of Pate releases water, come up to the surface of water Sol. The upehust or uoyane force acting on he be at(441-49) lc faa i ns gear a fe plod oe ge 21 The volume of 50 g of a substanceie2o—a) If the density of water is 1g cm wl substance float or sink? Sol. Given, mass of substance, m =50 g lume of substance, V=20 em’ - Density of substance Mas _50_, 0= te = 0-25 gem i.e. The density of che substance is greater tan de density of water, soit will sink in water 22 The volume of a 500g sealed packet is 360 cm*. Will the packet float or sink als if the density of water is 1 g cm *? What wl be the mass of the water displaced by tis packet? Sol. Given, mass of packet;m =500 g Volume of packet, V = 350 cm? Mass Densiyof packet, p ts Volume 500 g 350m? 1.43 gem i.e. The density of packer is greater chan densi +0 itwill sine in water Mass of water displaced by the packet Volume of packet x Density of watet =350g uneimasine et Seoanhss ine atthe surface The SI unit of + Equations of» here, is th offal, wis the the body acce * The total amo Mass. The Si ‘wantity * Tho weight o towards the « The St unit g ‘Wantityproportional tothe square of its orbital object falls towards the earth under the force alone, then such e motion is called swith which an object falls towards the cath ‘ull is called acceleration due to ce of earth. § = re [Shunit of is ms" and its value is 8 mis" ‘of motion for freely falling bodies, veurgt heute hat viau' +2ah ‘his the height from wh ‘vis the inital velocity 2 accelerates at g - otal amount of matter containe sealant of mass i Kilogram (kg) and tis. aacaler Jof an object is the force the earth, i.e. weight of an object. = 8 Lunt of weight is Newton (Nand sts 8 vector + Presrae ithe fore acting perpndicaary om unit area. of ich the object falls. i the ime nd vis the final velocity, wher din an object is called its with which it attracted For Mind MaP Vi ns/090 (OR. Scan the Code usHias ‘surface The Su vecorguantty mobject ean be caleultodas. Proscure Thrust (PO pc A) Ae “The Sl unit of pressure is Na or paseal(Po)- Freee aaa qua “+All the liquids and gases are called fluids. 1 Rigjatlrer an opard fo which ace nn te Seamed iq eo ald oe + acre cng tw Bayan Force 1 Volume of obec mere in tbe Haid «the Boopent fre exerted bythe ia se than tbe trou ofthe bj the objet wil snk te gui 1 Tra baoent fore oul tothe weight the bec the ‘bj wl loa i the ins. «eins peop force emore han te weight ofthe obec, he Ue ls in te ii and then lat «The dennty ofa substance is defined as mass pe unit volume A: Donaty = Mas the sata Volume of the substance ‘The St unit of density is kilogram per metre cube (ks) and {tsa sealar quantity. 1 essing to Archimedes’ peinciple, When an objects ally acca tmmeroed na iquid,kexpriance sbuoyast aoaarseahrast which equa tothe weight of ied dlsplaced by the cbc’ «+ Thereatve density of thatof Trpaltve density ofa substance ‘Danaty ofthe substance = Density of water substance is the ratio ofits density tO «+ Relative dansity has no unitMultiple Choice Questions 1 Law of gravitation gives the gravitational force between (a) the earth and a point mass only (6) the earth and the sun only (oj any two bodies having some mass {a} two charged bodies only NCERT Exemplar Sol. (¢) Law af raviation is applicable tall bodies having some mass and is given by F = where, F = Foreeofanraction beeen the wo bods, im and my, = Masses of wo bods, ravieational constant and r=Distance beowsen the rwo badics, 2 The value of quantity G in the law of gravitation (a) depends on mess of the earth only (b) depends on radius of earth only (€) depends on both mass and radius ofthe arth (@) is independent of mass and radius ofthe earth NCERT Exemplar Sol. (d) G isthe constan of proportionality and is called the univesal gravicational constant, Its independent of mass and radius ofthe earch 3 The weakest force in the following is (a) magnetic force (€) gravitational force Sol. (c) Gravitational force is che weakest force among sven force (b) nuclear force (a) electric force 4A planet is moving around the sun with mean distance rand une period T, then, @r u=15.33 mis § On veu-ge SE 0 £98 2% bn hypothetical case, it he dame the earth bec ‘omes half of its present va and mass becomes four times its Pf value, then how would the weight im object on Surface of the earth be affect NCERT! 25 Apa he ¢ Sol. L 26 sng oy geSo ‘Mass bosoms four times its present valu, On bringing modificatior ia 25 A particle weighs 120 N on the surface of the earth. At what height above the eanhve surface will ts weight be 30 Nt Radios ot the earth = 6400 km Sol Let the weight ofthe particle on the nts of the Gm Mn am athe height above the carts surface, where its weigh willbe 30. Gum ae ey b+R) ‘On dividing Eq, () by Eq, (i), we get 120_ GMm , (b+ RY 30 * Gin (aR Re b+R Hence, 120 Given) o 26 Shashank placed an iron cuboid of imensions 4 cmx 7 cmx 10 cm on a tray Containing fine sand. He placed the cuboid insuch a way that it was made to lie on the sand with its faces of dimensions ©) 4emx7em, (i) Tem x10em (ii) and-4em «100m. the density of iron is nearly 8 gcm™* and Gstoms) tnd the, minima end Taximum pressure as calculated by Shashank. = Wem 7 em x10 em) x Bem” Le Volume oF cuboid 1% 6 =22405=24 kg Force applied bythe cuboid on the snd =mg= 224 x10=224N fe g=10m5) yy Pressure wil be minimum when aes of he fice af cuboid kept on snd is maximum, Le inthe ase of face with Fem 10 em. ‘Areaofthefice= 7m 10cm Force _ 224 Minimum presure= FO" = 224 _ 3290 Pe Aree O07 0 Pressure will be maximam when area of fice of cuboid kep on the sand i minimum, ie. the case office with 4 cm 7m, ‘Aes ofthe face = 4 cm = 7 em 4 ex 7m =0.0028m? 100 “100. Force _ 224 ea 00028 000 Nmn* 0 27 A cubical tub of side 2 m is full of water Calculate the total thrust and pressure at the bottom of tank due to water, (Take, density of water =1000 kgm~and g=10ms*) Sol. :. Volume of cubical cub Length > Breadth x Heigh eee length = breadth Maximum presuce Volume = 2x2 Mas of water= Volume x Density of water % 1000 = 8000 Kg 0 & Weight of water Mass x Acceleration due to gravity = 8000 «10 8x10" N. ‘Total crust = Weight of water x10°N Area of square shape of tub 0 Area of boro Pressure of water atthe bottomfilled with airhas a volume of Calculate the minimum force put it completely p) 28 Avail es child t applied by « child 16 : ide the water. Take, g = 108 Sot. Gna . ‘Volume, V= 500.cm* = 500 x 10° m?, g-lome Force euired to put the bl inside she water = Buoyant force Weigh water diplaced = me) Now, we know that o Mas of water= Deny of wate x Volume m=pV ‘Oa sbsiing this value in Ea (0), we get Force pV 1000 kgm) x (500 10° m") (10 ms) N o =1000 x 500 x10 x10 N & Minimum force applied by child ro pu the ball comply inside the water i 5N. o 29 A boat of mass 50 kg is oating in the river with (J) ofits volume inside tne water 2 Calculate the buoyant force acting on the boat. (Take, g = 10 ms-*) Sol Given that, mas of the bout, =50 ky As we know that, when a body ost in river. Its apparent weigh is zero ny Therefore, buoyant force = weight of bly me=0x10 «ny it. Buoyant foree= 500 N 0”) Long Answer (\4) Type Questions 1G) Write the formuta to find the magnitude of gravitational force between the earth 4and an object on the earth's surface. ii) Derive how does the value of Gravitational force F between two objects change when (a) distance between them is re ) distance between them is reduced (©) and mass of an object is and mass ject is increased Sol. () Formula to find the ns ula to find the magnitude of gravitat : gravitational 30 Sol 31 ‘Acubleal object of side 4 cm hag all It mass of the cube is2 whether this object will toa Water (Take, density of water og ays Given that, Mass ofthe object, m= 2 kg Sideofcube=4cm=0.04m fst me ‘Since , volume of cube = (side Maa! Volume ofthe object, V = (0.04 mip (0000064 : \s we know that, 2 ‘Mass of the object Deity = tame ofthe ober = Densicy= 2 been 2000000 sy 700 b= 3250p Here, density of che object (.. 31250 gg arene than the density of water Gi Dig Thee he objet winking The dimensions of wooden Ball 2mx 0.25 mx 0.10 m-fvelatve daa wood is 0.4, calculate mass offal ig. (ake, donsity of waler= 10! tga 0.6.= Density of wood _ Denisyfad Density of water” 10 ga | 20.25 0.10 m? =005a! wien bloc As we know, mass of w Density x Volume 0.6 x 10° 0.05 kg = 0.030210 | [5 Marks el where, M= mass ofthe earth, ‘m= mass ofthe object, R = radius of che earth 6=66; (@) Lec gravitational force be F when te | between them is Now, when the distance reduces to al _GMm _4GMm @) 3 one so— Gravit ondividit ‘when the F a 2 F ii) Why ) Let Ib 3 wa ju athat, ifthe earth attracts two placed at the same distance the centre of the earth with equal force, then their masses will be the ‘same. ‘Mathematically express the ‘cceleration due to gravity in terms of ‘mass of the earth and radius of the ‘earth. ia) Why is G called a universal constant? 9 Lette two bodies have masses m, and m, and they are placed at che same distance R from the entre of the earth, According tothe question, if the same force acts on both of them, then @ “ Rr R Soom =m, theie masses willbesame, (G= universal gravitational constant ‘Mz=mass of the earth and R = radius of the earth. G iis known as the univer wn sal gravitational as same al the Sol. (3 (2) Gravizational force acting onthe 50 Kg, Gravitational constant, G=657 x10 N-m* kg? ‘Acceleration due to gravity on the earth, =98m/s Gi) A bag of sugar weighs w at a certain place on the equator. If this bag is taken To Antarctica, then will it weigh the samme for more or less, Give a reason for your answer ig = 50% 9.8 = 490 aa (6) Gravitational force acting on the 50 kg mass du t jupiter, : GX Mig ® My ot Geeance of jupiter from the earth ay 667 x10" x2 x10" x50 fame = G3 x10" x6.3%10" Fue = 1.604109 . (@ Comin fc acingon he 50g mas pean Mga ‘amen (distance of saturn from the earth)? 5, sor xoxtxso 2 Param = 1 28x 10 1.28% 107 657 x6x50 en 18x 128 fag = 02x10 “ avon reduc the pier and 68x 107 + 0.12% 10°) N 8x 10°N “Than, the conbined free due 1, he planes jupiter and sana (1.8 x10) Nis nelgbe as eae 10 the gravitational force due wo the x0 thesarurn a Gig We know thas, at euator i es than gat poles Antarctica). Thus, weight ax equaor i es than (rei at pole Antartica). A bag of sugar weighs Toes certain place onthe equator. this bag i ken vo Antarctica, then twill weigh more due to greater vale of oa ne ml he eth om 18 slat Bones (Weigh of person onthe ath = mg 8x98 = 6664 N-m Ihe” (Val of go cath i rated ch earth ond iad, Derivation From Newon’ iw of ration, # Ha Mira ea jes ing mas fling tomar ana bere cents of eat and the objet Fresco lof mosin, fre exer on obec Sines = ge, acceleration dc to grav Euatng RHS: ae ou me - 5 “wo objects of masses mm and m, having the same size are dropped simultaneousiy fom heights hy and h,, respectively Find fut the ratio of time they would take in aching the ground. Wil this ratio remain the same, it {one ofthe objects is hollow and the other one is solid Gi) and both of them are NCERT Exe 6A. (i) Howl (i), vig the fie fl she acceleration prod ody emsins conan Acceleration afer 25= 98, 7 W) Astoct needte sinks in waterbst ast | ship oats. Explain, how i) Why do you preter a broad and ™ | handle of your suitcase? depend on volume of o he volume of o pus ect, more Up presure act 0 the bo Peal vo dhe uct at al on oper 1 chat more he ate cop the body. As the a een andl of pes ‘es Pom one place oanotber (0 Det matcal orm un The Mength 13 em tui) THe cue cm. Find the asubaane Relative density Density of the subst nother words ‘Mass ofthe subsance pe ofthe substance Masa en Sep Ena uc at town a FE mn een a Seece Beer tthe cs ese ‘take from one place to another. a ae em density. Give its i The mass of ano cube hang an joe length 13 con is 50 g. Find ie density. a 7 IB Tam vole of = 250 9 seded tne 1%00 cubic cm. Find the density of te tn ing (cc), State, ifthe object would sink or loat in water. Sol () The reaive density of ubsance ithe ro fe emg otha of ete Reaive densi of ubwance Density ofthe obance Desay afrae In other words Reative density ofaubance Mas ofthe sbsance Volume of wate “Volume ofthe sabrance” Mas of wae ee [mersied a Given hat, mas of he be, M =508 Side of abe = 1.5 cm ear Molumeofeabe, V 257 om? = 3375 Mas M50 + Demity= Yoiume 3575 21481 gem? i) Given ha mass = 250 Volume, V=400 ce Man 250 ad = Damiy= Ma = 250g) = 0.625 p(y ‘As we how that, dese of water = 1 6 (0) Sov density of tin lear shun that of water and hence in wil ont ” 9 ‘The radius ofthe earth at the poles is 6357 kim and the radius at the equator is. 6378 km. Calculate the percentage change in the weight of a body when itis taken from the equator tothe poles. Let acceleration duet gravity at equator, GM, B “ and acceleration du to gait ples GM, & “The variation of celertion duct gavin na % Peceange rion ag = ——eptg 100 om(4-}) RR . 100% 8 RR (ors? 577 a ane 100 «0.79% ‘svar nthe weight of body hange in g = 0.7%‘Muttiple Type Questions 1 Newton's aw of gravitation is universal lav, because it (a) acts om all bodies and particles in the universe j (©) acts on all the masses at all distances and not affected by the medium (6) is always attractive (4) None of the above 2 “Gravtatonal fore beeen to objets 10 Nf | Imac ofboth objects ae dood eeu ning ih aisance berween them hen he ene ty would tenons BN CN (jw (4) 108 | 3 the vale of acceleration due to pot (a)tstamconcquser nd ys t (b) is leaston poles | eel | (4) increases fom poe to cquatr NCERT Exemplar H 4 Tree spheres have radi! 1 cm. 2 cm and Sem respects Which sphere tena makina ea onan? He) (e) second (c) ind (4) Al equ 5 When a body is dipped complty ot partly tee liquid, then weight (w) of the liquid displaced by the body and upthrust force on the body inte the quid are related as, (a)w=F (b)wF (d)Noneof these Fillin the Blanks $ The aceleration duc o gravity on the moon is about ofthat onthe earth, 7 ‘The weight of an object on the earth s about of ts weight onthe moon 8 A heavy ship floats in water because its density les than that of water 9 Archimedes’ principle applied to liquids and 10 eis the x force which makes objects appear lighter in water True and Folse Whe acceleration duc to gravity acting on tee falling bodys ditetly proportional othe masse body CHAPTER EXERCISE Objective Type Questions (M1COs) (5 sake») 12. the weight of an object 6 20 conta Mth the change in acceleration duet 13 The relative density of @ substance eq pete ce ay 15. when a body ally do acceleration due to gravity 9 is taken a a * oa ‘Match the Columns 16 Match the fol Column | Cotmna a Pascatsiaw 3 22 Assertion-Reason Direetion (Nos. 17- -_ 23 statement of Reason. Of te statem ¢s, mark he coma (8) Itboth Assertion and Reason are te, but Bama (0) I Assertion is true, but Reason ttt () 1 Assertion i lalse but Reneen ta V7 “Assertion A planet is heavently bad Reason A planet revolving ar nd the sun ie Kepler aw 18 Assertion at th ant #ravitational acceleration is ze1 Very Sho Reason On going below he earth surface, ee of gravitational accelerati increases. e. w Assertion To whose weig Sloat «tay must deel? 25 Ms han actual we z Reason During floating, the b et downw: ard force in that case 27pede en tac vn noeye ae ‘contact area of your whole ec que on different areas are differen, The pends on the aca on which i acts, The effet is larger while standing than while lying. BG che our a ep ipcording 0 given passage when wil the effect ks ‘on sand is larger? ‘hen you lie down on Toose sand, force acts on an ‘equal 10 (apeontact arca of whole body “[p)area of your feet only {e) weight of sand {@) None of the above you stand on loose sand, then the direction of force acing on an. area is Jalhorizontal to the surface of contact {b) perpendicular to the surface of contact {e)parallel to the surface of contact {d) Both (a) and () Answers py 2) >to 6 One-sixth 7. Six times: 9. Gases "10. Buoyant WeFalse 12-True 19 True 14. False 15+ True Bs (ayes (a). (2) 719) (0 24 3 ae yh By i 4a) 50) 8. average ‘Answer (vSAl Type Question cn) the once of gravitation exist a all Be places of the force ‘them is reduced to (1/4)1h? ot Bg cgan tye ana sven crn mass. Comment. Long Answer (LA) Type Questions 31. A sphere of mass 40 kg is attracted bya second sphere ‘of mass 15 kg when their centres 320.cm apart with 2 fare ot llgram weight caeuate the val of ‘gravitational constant. [ANs-L7_ c10°#N- my kg? 1 Does the acceleration of ely aling ob Pees A oy extent on te ocation, whether he depend vo arg of Mount Everest or in Death Valley California? Expain Find the ato of weights af «body on the earth and Find the eng on lupe is 25. tmes that 8 he crt th mnan is asked to run with a ag containing 20 KE A an Sl be easier for him tous with 206 steel loc gt block? Bxpain with reason. secre floats in ater with a smal oi at He Aes ae Heat of test be ena mass of seven coins and external volume is 16 em), It just serene ed coun s aed, Calculate he mass of each coin, (Density of water is 1g cm?) (Ams. tres! [5 Marks each) 36 A stone is thrown vertically upward with an initial elocty of 50 ms. Take, 9 as 10 mi: Find ihe weerinlam height reached by the stone, What is the maenfplacement and the total stance covered by the stone? Ans. 125 m, zero, 250m] fa planet existed whose mass was twice that of the eee peal whose radius is 3 tes greater How much, Catlin 1 kg mass weigh in the planet? [Ans. 2.17 NI What is upthrust? What are the quantities that can, Wetmupnrust? How does it acount fr the Hating of what when a patially immersed body is pressed a pea tle, what will happen to the uplbrust? 4) How does a boat float in water? {uy Stae wo factors on which magnitude of buoyant Sue Feng on a body, immersed in a fluid depends. uly A block of wood can Moat whale a simular sized thock of concrete can be used as an anchor: Why? How does the force of atiraction between the Hoodies depen! upon their masses and distance between them? vermafent thought that twobricks ted together would Ajitanerthan.a single one under the action of gravity. bo you agrce with his hypothesis oF not? Comment.1 tra sates revolving around a planet of mass An an lpia orbit of semi-major axis fin the orbtal speed of the satelite when tis ata distance rom the i + auf? 1] wm cud 1] cotaouf?-!] —— eyvtacuf3 2] ovtsouf3 oreo2-3] 2 Distance between the / The masses ofthese stars are Mand 16 Mand thet radii areaand 2s respectively. A body of mass misfired Straight from the surface ofthe lager sar towards the 6M i. 26 salle star. The miniqwum initial speed forthe body 3 satelite is launched in a circular 0 fs Aboter satelite sano launched in an obi ofradus 1.1 R The period of the second satelite i larger than the frst by approximate 4 trearth comes closer sun by (/4) thf the present distance, th th consists of how many =) 5 Communication steites move in obits of ad 44400 km around the earth. Assume that the ony fo acting on itis that due tothe earth, then what's the Take, mass of earth = 6 x 10" carth= 6x10" kg) Challengers* 6 emean rads ofthe cat's ott und 15x10" The mean rads of he ott at a the sun e610", The merry wl a 7 Aank Sms high filled with wate and they the top with ol of density 0.85 bottom ofthe tank, duc to these liquids (@) 138 seer ) 88.25 nee 8 Awonien locket volume 100.1" pene spingtlanee regs 12 in aie ts apg statr sch ha halo the block seo hee reading ofthe spring balance is water 1 9 bodys ating in mater wth (2) rots pata kom hen eparna |grigr Answer Key 1 2 3 4 se 6. 7 8 9.) wll Work OR Scan the Code | Wet tosis soy OR. Scan the Code