PHYSIS. If the acceleratin f wedge in the shwn arrangement is a twards left then at this instant acceleratin f the blck wuld be, (assume all surfaces t be frictinless) a () ( cs )a () a () cs a If the wedge mves leftward by x then the blck mves dwn the wedge by 4x hence acceleratin f blck ab a (4a) (a)(4a) cs( ) = (7 8 cs )a. In the arrangement shwn in the figure belw at a particular instant the blck is cming dwn with a speed f m/s and is mving up with 4 m/s. t the same instant it is als knwn that w.r.t. Pulley P, the blck is mving dwn with speed 3 m/s. etermine the velcity f blck w.r.t. grund. 4 m/s in dwnward directin () 3 m/s in upward directin () 7 m/s in dwnward directin () 7 m/s in upward directin Let Pulley P mves up with velcity V P then V P + = 4 V P = 4 m/s (up) S V = 4 + 3 = 7 m/s (up) 3. blck f mass 4 kg is pressed against the wall by a frce f 80 N as shwn in the figure. etermine the value f frictin frce and blck s acceleratin [take g = 0 m/s, s = 0., k = 0.5] 0 m/s () m/s () m/s () m/s The F f the blck is as shwn in the figure 80 N 4 kg 37 80 cs7 N = 80 cs 37º = 64 N 80 cs37 N S f L = 0. 64 =.8 N s 4g < 80 sin 37º, s frictin frce will act in dwnward 4g f www.edudigm.in 896066637 / 8585070735
directin. LSS XI SET Net applied frce in upward directin (excluding frictin frce) is 80 sin 37º 40 = 48 40 = 8 N as F applied in vertical directin is < F L S blck wn t mve in vertical directin, and value f static frictin frce is, f = 8 N. 4. Tw equal masses hang n either side f a pulley at the same height frm the grund. The mass n the right is given a hrizntal speed. fter sme time The mass n the left will be nearer t grund () The mass n the right will be hearer t grund () th the masses will be at equal distance frm the grund. () Nthing can be said regarding their psitins. v wnward frce n right blck is mre T T Grund mg mg 5. swimmer can swim in still water with a speed f 5 m/s. While crssing a river his average speed is 3 m/s. If he crsses the river in shrtest pssible time what is the speed f flw f water. m/s () 4 m/s () 6 m/s () 8 m/s (Vr t) Vmr t V avg = 3 = t V r = m/s 6. particle starts frm rest and mves with an acceleratin f a = { + t } m/s, the velcity f the particle at t = 4 sec is m/s () 4 m/s () zer () m/s a = t + t frm t, a = t + = 4 t dv = (4 t) dt www.edudigm.in 896066637 / 8585070735
V = 4t t / at t = sec, V = 6 m/s fr t > sec a = + t = t dv = tdt t V 6 = V = m/s 4 7. If cefficient f frictin between all surfaces is 0.4, then the minimum frce F t have the equilibrium f the system will be, (take g = 0 m/s ) 6.5 N () 50 N () 35 N () 50 N f Frm hrizntal equilibrium f blcks N = N = F hence limiting frictins are l f = l = F equatins are T + f = 5 g ( ) f = f ( ) and f + 5g = T + f ( ) Slving (), () and (3) f =.5 g F =.5 g.5 0 F = 0.4 = 6.5 N 8. Fr the situatin shwn in the figure, the blck is statinary wrt inclined fixed in an elevatr. The elevatr is having an acceleratin f 5 a 0 whse cmpnents are shwn in the figure. The surface is rugh and cefficient f static frictin between the incline and blck is s. etermine the magnitude f frce exerted by incline n the blck. g (take a 0, = 37º, s = 0.6) mg 3mg 9mg 3mg () 4 () () 0 5 5 Frces acting n blck wrt elevatr ther than that exerted by inclined plane are as shwn in figure. Hence frce exerted by incline n the blck www.edudigm.in 896066637 / 8585070735
R = LSS XI SET (ma ) [m(g a )] 0 0 = 3mg 9. particle mves with a cnstant speed u alng the curve y = sin x. The magnitude f acceleratin at the pint crrespnding t x = / is : u () Let at any instant the velcity makes an angle with the x-axis u u cs ˆ i sin ˆj u du dˆ d a u sin i cs ˆ j dt dt dt ( ) dy Nw Tan = dx = cs x d dx sin dt = sin x dt d dx cs sin x dt dt dx x, 0, u Nw at dt d u dt a u Putting this in () we get () u () u 0. In the situatin as shwn in figure if acceleratin f is a then find the acceleratin f a sin () a ct () a tan () a cs In a a a / a a sin sin( ) www.edudigm.in 896066637 / 8585070735
LSS XI SET a = a cs. man rwing acrss a km wide river, trying t reach the ppsite bank in the shrtest pssible time, reaches a jetty km dwnstream in 30 min. t a certain time f the year, when the river flw has halved, he embarks frm the same pint and rws t reach the same jetty. The shrtest pssible time in which he can reach his destinatin is very nearly 60 min () 5 min () 33 min ().5 min. raw the resultant velcity f the man w.r.t. the grund in the first case as well as the secnd; a 7 cmparisn gives the speed in the secnd case t be 4 times the previus ne. The time shuld be divided by this factr.. m ice gram f ice (at 0º) is kept in an insulated calrimeter f negligible heat capacity and m s gram f steam (at 00º) is passed int it, while n steam is allwed t escape. The final temperature is. Given : L ice = 80 cal/g, L water = 540 cal/g and s water = cal/g/º. Which f the fllwing statements, is/are true? 0º 00º () If m ice > m s, = 0ºs () If 80 m ice > 540 m s, = 0º () If 80 m ice > 540 m s, 00º. 3. simple pendulum attached t the ceiling f a statinary left has a time perid T. The distance y cvered by the lift mving upward varies with time t as y t where y is in meter and t in secnd. If g 0 m/ s, the time perid f the pendulum will be 4 5 T () 5 6 T () 5 4 T () 6 5 T Hint : Given y t dy t dt d a m/ s cceleratin = dt directed upwards. l T g a Hence l T g and T g 0 5 T g a 0 6 www.edudigm.in 896066637 / 8585070735
4. particle is executing linear simple harmnic mtin abut the rigin x 0, which f the graphs shwn in figure represents the variatin f the ptential energy functin U( x) versus x U (x) U (x) U (x) U (x) x x x x () () () Hint : In simple harmnic mtin, the frce acting in the particle (restring frce) is given by F Kx du Kx dx x U( x) Kxdm kx 0 5 The bb f a simple pendulum executes simple harmnic mtin in water with a perid t, while the perid f scillatin f the bb in air is t 0. If the density f the material f the bb is 3 (4/ 3) 000 Kg / m and the viscsity f water is neglected, the relatinship between t and t is 0 Hint : t t0 () t t0/ () t t0 () t 4t0 3g g geff g g 4 4 t 0 l g l t g and t g g t 0 g g/4 Nw t t 0. 6. Tw particles are executing simple harmnic mtins f the same amplitude and the same frequency alng the same straight line and abut the same mean psitin. If the maximum separatin between them is times the amplitude, the phase difference between them is () / () /3 () /4 Hint : Let x sint x sin( t ) Separatin at time '' t x x x x sin( t ) sint dx 0 Fr maximum separatin, dt www.edudigm.in 896066637 / 8585070735
cs( t ) cst 0 t t Plus sign is nt pssible because 0than x 0 t t t xmax sin sin sin Nw sin 4 7. Tw bdies and f equal masses are suspend frm tw separate springs f frce cnstants K and K respectively. If the tw bdies scillate such that their maximum velcities are equal. The rati f the amplitudes f scillatin f and will be K K K () K Hint : Kinetic energy at equilibrium psitin = Ptential energy at extreme psitin Kx Kx x K x K K () K K () K 8. The P V diagram fr mles f an ideal gas underging a prcess is shwn in figure. The maximum temperature f the gas during the prcess is 9PV 4nR 0 0 () 3PV 0 0 nr () PV 0 0 nr () 3PV nr 0 0 Hint : Equatin f Straight line is P mv c ( ) Where m slpe c Intercept P0 mv0 c and P0 m( V0) c P0 m there V 0 and 3P0 nrt nrt PV P q V www.edudigm.in 896066637 / 8585070735
Putting P in equatin () ( ) T mv cv nr ( ) dt d T 0 & 0 T will be maximum is dv dv dt 0 Putting dv V m (3 P ) 9PV Tmax 4nRm 4 nr P / V 4nR 0 0 0 0 0 Paragraph fr Questins Ns. 9 t system f tw blcks is placed n a rugh hrizntal surface as shwn in the figure. frce F is hrizntally applied n the upper blck. Let f and f represent the frictinal frces between upper and lwer surfaces f cntact, respectively and a, a represent the acceleratin f 3 kg and kg blck respectively. (F is maximum value f F at which f is limiting frictin). F 3 kg kg s = 0.5 k = 0.3 s = 0. k = 0. 9. If F is gradually increasing frce then which f the fllwing statements wuld be true? Fr a particular value f F (< F ) there is n mtin at any f the cntact surface. () The value f F is 0 N. () s F increases beynd F, f increases and cntinues t increase until it acquires its limiting value. () all f the abve. 0. Fr F = N, mark the crrect ptin. f = 7.8 N, f = 7.8 N, a =.4 m/s, a = 0 m/s () f = 7.8 N, f = 0 N, a = a =.4 m/s () f = 7.8 N, f = 5 N, a = a =.4 m/s () f = 7.4 N, f = 5 N, a = a =. m/s. Fr relative mtin t be there between tw blcks, the minimum value f F shuld be 5 N () 30 N () 5 N () 3 N Frm the given data we can find the limiting frce fr the tw surfaces. f L = 0.5 3 0 = 5 N f L = 0. 5 0 = 0 N Fr F < f L th the blcks remain at rest and f = F, f = f and a = a = O Fr F > f L and F < a certain value say F the mtin starts at lwer surface but bth the blcks cntinue t mve with sme acceleratin. The frictin n lwer surface becmes kinetic in nature. F f K m/s Here, a = a = a = 5 F f = 3a and f f k = a www.edudigm.in 896066637 / 8585070735 3 kg kg F
F 3 f k ll these equatins give f = 5 Fr relative mtin t start between tw bdies, f > f L F 30 N S minimum value f F t cause relative mtin between blcks is 30 N Paragraph fr Questins Ns. t 4 balln starts rising frm the surface f the earth. The vertical cmpnent f the balln s velcity is cnstant and it is equal t V. ue t wind the balln gathers the hrizntal velcity v x = ay when a is a cnstant and y is the height frm the surface at the earth.. The tangential acceleratin f the balln as a functin f y is a vy a y v () 3a vy a y v () av ay v 3. The nrmal acceleratin f the balln as a functin f y is a y av v () av a y v () 4. Ttal acceleratin f the balln as a functin f y is av a y () () a v y a y v 4 av ay av () av () av () 3aV dvx dy a x a avy dt dt = av dvy ay 0 dt attal a ˆ ˆ ˆ xi a y j avi dy v tan dx a y we have, ay a n a ttal cs = t ttal (ay) a a cs v nrmal acceleratin = a vy (ay) v a N = attal sin = av (ay) v Paragraph fr Questins Ns. 5 t 7 student perfrms tw experiments t determine the cefficient f static and kinetic frictin between a blck f mass 00 kg and the hrizntal flr. www.edudigm.in 896066637 / 8585070735
Ist Experiment : He applies a gradual increasing frce n the blck and is just able t slide the blck when frce is 450 N. IInd Experiment : He applies cnstant frce f different magnitudes fr a duratin f s and determine the distance traveled by the blck in this duratin SET FORE ISTNE.. 3. 300 N 600 N 750 N 0.5 m.0 m 3.0 m ssume all frces are applying hrizntally (Take g = 0 m/s ) 5. The cefficient f static frictin between blck and flr is 0.45 () 0.3 () 0.5 ().45 6. Which set f readings f experiment II is abslutely wrng? () () 3 () Nne f these 7. The speed f the blck after 3S (beginning frm starting f applicatin f frce) in set fr nd experiment is 6 m/s () m/s () 3 m/s () Infrmatin is in sufficient. 5. Frm exp., f = 450 N sn = 450 s = 0.45 [N = mg = 000 N] 6. In st set the frce applied is less than f L s the blck can t mve. 7. Frm nd and 3 rd set f experiment. 600 f k t m 750 f k 3 t m f k = 300 N k = 0.3 600 300 Fr nd set, a = 00 = m3 m/s Velcity f the bject after S, V = 3 = 6 m/s Fr t 3 S, frictin frce (kinetic in nature) retards the mtin V required = 6 3 = 3 m/s Paragraph fr Questin Ns. 8 t 30 Tw mles f an ideal mn-atmic gas is taken thrugh a cycle as shwn in the P-T diagram. uring the prcess, pressure and temperature f the gas vary such that PT = K, where K is a cnstant. www.edudigm.in 896066637 / 8585070735
8. nstant K is given by PT () PT () PT () PT 9. The wrk dne in prcess (R is the gas cnstant) RT () RT () 3RT () 4RT 30. The heat energy absrbed in prcess is RT () 3RT () 5RT () 7RT 8. () Hint : Using the ideal gas equatin PV = nrt, the vlume f the gas in states, and are nrt nr( T) nrt V P P P ( ) nrt nr( T) nrt V P P P ( ) nrt nr( T) nrt V P P P ( ) It is given that in the prcess, the pressure and temperature f the gas vary such that PT K Where K is a cnstant. Thus fr pint, we have K PT a P ( T ) PT. 9. () Hint : Fr the prcess, we have and PT = K eliminating T u PV nr P PV nrt r P V nrk / nrk P V The wrk dne in prcess is given by V V nrk V V V V dv nrkv V / W PdV dv nrk ( V V ) www.edudigm.in 896066637 / 8585070735
nrt nrt nr( PT ) P P nrt 4RT 30. () Hint : The prcess takes place at a cnstant pressure P P. Therefre, the wrk dne in this prcess is W P. V P( V V ) nrt nrt P P P nrt RT The change in the internal energy y is prcess is V n T n T T ( ) n ( T T ) V V 3R n V T T 3RT ( V) W 3RT RT 5RT V www.edudigm.in 896066637 / 8585070735