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113. Cell with exhernal ' i xna\ resistance U; (E,r),, Netemf =E Net aesisfonce | around loop = R+F curent IT? => V4 R Ve ment RAT Terminal voltage = PD acress ext resistance R ER Vee Nes ger % For max power in ext - resistance External resistance CR) = Internal resistance (7 \1l . Electzic Field befween the plates of a charged capacitor 3 . ae ig ie petenhal er PD air copacitor: E = ' a. separation = &: ar x. Fleld produced by individual plates = S. 2E0 Vv Dielechic capacitor : E= Coy Es = “ KEs Scanned with CamScanner \\l. Bullet sticks to pendulum block p conseavation (M+m)V = mu v= _mu M+ Then E conservalion 2 h= 9 | ‘ =x he = ( mu 23 \mem or us M+m [Zan m 2gh I$ u given, h will be asked [fh ou ,unu " Inelastic collision (don't use)clirect energy id 1 mu* = (Minjgh K WRONG APPRIACH 2 12. Propesties of EM waves phase clifference behween oscillations of =’ and B” =0 Angle behween planes of oscillations oF - B’ = 7 xad = 96° Band B = — Magnitude radto = =é Direction of propagation aleng E'xB” =? 9 Poynting vector 5° = E x8 He eee Scanned with CamScanner 106. Average Translatahona\ Kinehc energy For all molecules (meno,dia, poly) a CKE)irang = 2 *ZKT (3 translational deg of fred Total ke /ant for at) Total KE differs due te additional rotational modes CKE ) seta = 2kT (mong) 2ero rotahemal (KE) pra = okt (dia) 2 retatfonal CKE )ptar = £kT (pely) 3 retohonal der (A orf) oF Elechomagnehc Waves oa r( : useful 2 excellent Quy Ravi Mohan fs vony : ; ; , in Visible uv X-ray Y-7ay Rodi o Micro decreasing wavelongi(a) av) Va) ayn fren ee ee a jog. Excess pressure = —————$ $< Pinside - Poulstde = Paxcess DROP BUBBLE Cone cHedtie surjae) (Two efechve ia 4 = 2T smailert) Peness = 2 Pexcoss oo - ine Pee at Pin = Po + at Paneess T Pin = Po . (same for alrbubbie in water) Scanned with CamScanner 106. Average Translatatono\ Kinehc energy For all molecules (mens, dra, poly) i (KE) sans 4 3 KSkT (3 translatranal deg of freed Tota KEY /ant for ait) Total KE differs due te additional rotational medes 3 al f (KE )sotr = SAT — (mano) zero rotatrenal (KE) prot = okt (dia) 2 retatfona| (KE ) ptat = ET (pely) 3 retahonal ; i M rh rs veny wseful < excellent Quy ari 0 i r ' in Visible UV X-ray Y-ray Real o Micro ; decreasing wavelengin(a) fav) ava\ wy fem \o pmeee l a = tog. Excess pressure = FoR DROP BUBBLE Cone effechve suriace) (Two efechve surface) 2T smaller T] Penwess = 4ST Pinside - Poutside = Paxcess Pexass = ve oreater Pins P cat Pin = B+ 20 Paxcess T ee (same for alrbubble in waer) Scanned with CamScanner Int. Man ina lift CApparent weight) Wap = M(9+a) = M9 (1+2) = We (1+2) 4 @ upward, og = +a Wap = M(9-a) 7 Tsp =9-@ = Mg (-$) = Wa ('-) d Q downwards ss For a=o , lift may be moving soi allie a ¢ up ot doen but with vec) ¥Note: moving Upwards with decreasing speed means Qa doun, %B=9-a moving tlownwards with decreasing space] means @ UP, Gh = O+a : —— “i Block freed from the 102. Block celliding spring / Ber of compressed —p" spring : i Se ne z ‘ —_ u va S gs 2 bef] 3 Xs — -— 103. Emissive power of a pee bedy 4 E, _fTh Temp must ne a (=) Cahn 4% To make emissive power double , temp should be increased to 274 times 2"4 7” (Wz) = \iz = ray = 1-19 thnes Scanned with CamScanner NORMAL CONSTRAINT when two rigid suzfaces in contact they can have relaive motion along the surface but not along the normal ( They do not jump or penetrate wrt each ojher) At any instant for two points in contact yelative displacemen} (Xra) velocity (Vre!) and acceleration (Ari) along (hormai)to the surfaces of contact must be zero Approach: B, and Bz are two bodies in contact Pi and Pa be hve pojnts in contact Vv ayid Ve be the velocities of Pi and Pz resolve components along Nomal and e@ equate then (My = CVe)n same approach for TK! Body Bz @ displacement CX’) and ' eaueleration (a) Scanned with CamScanner Question: The avangement shown x “ «Be lying on a horizontal plane cae Unisorm transverse magnetic * c 7 Z field B 15 applied. The slider R rus i (m,&) is projected towards =, “t= rignt with velocity Ve. Find i's velocity as @ function of time and alse plot Yet graph. solulion : Vv be the Instantaneous velocity em} € = Bkv, curent I= om a *) opposing magnetic force on the a | = Thea I ower | re autv e— retardation = “pm 15 Ean ion b= - dv = Bur iv retardati = = dv _ . Bk at Vv mR v 4 gi t ry . at Vv - Integrating within limits {¢ = os fat Vs Vv a,* t [inviy. = 7 ot ce). _ ey in a= BE..t > V=Vo€ = Vo mR ea velocity ‘s exponentially My falling function from inl- iets _ tal value Ve to Scanned with CamScanner e@ oblique force to a block at rest over rough hoyi2ontal surtace PPO FGD: Finding Ne: Apply vertical equilibrium (equilibrium normal te surface) No+Fsiné =mg => No =(m9 -Fsiné) Finding limiting friction: fe = KsNo = Ms (mg -Fsine) check condition Jor sliding - Block will slide if (Net driving force along the surface) > limiting friction Heve: Feose 7 HS (mg - Fsiné) 7 Hsmg - MsFsin€ F (cas6 +Hs5in a) 7 Hsmg @ For sitding F 7( Msn ) Cos8 +HsSiné Scanned with CamScanner Bob at momentary rest (at the extreme position) “T=mgcos@o 4% Dont apply horizontal and vertical equilibrium Swinging /moving pendulum Net force towards centre (C) T-mgcose = centzipetal force (W's) L T= Mgcos6 + a Scanned with CamScanner Equilibrium of simple pendulum Bob (suspended object) attached to stzing Petmanently at rest % Apply equilibzium in any directions like Horizontal / veztical radial (along tength) FGD of the bob \ | Teese Fy = Tsine ‘ Fu 1 tone = Fu Fy Fy Squaring ond adding =p T= (Fith’ o@ Radial equilibrium + re! T= Fusine +Fycose Fu — — 6 Fv Fy cos8 +Fusing Foy swinging pendulum / momentary rest at extreme position ONLY RADIAL EQUILIBRIUM a ne dene ay Scanned with CamScanner LONGITUDINAL CONSTRAINT For (1) Linear rigid object all) String under tension Relative displacement (Xr) , velocity (Vre) and acceleration (rat) along the lenaih Clongitudinal) is 2eY0 @ Approach : resolve X,V ora of the end points along ine length and equate them @ Example: vi velocity Of end points of : Q@ rigid rod is as Snown == ™* FS, relate Vi and Vz . solution: Ma - 61 (Vs) y = V; cos6, Sa sat es bs ar “ors (Vid = Vi1Cos 6, Va equate longitudinal components Vi, cos: = Vi, COS AL @ Example: Find velocity of the bleck at the Instant shown b sclulion : fo st ne ot 58. ee —p Vv Ne Z ! ace AB, ty Equating velscities along Ve = Velocity of block the length : —— V,pCose = V Vp - Bh ne oat Scanned with CamScanner ~ Question : 2 —D — A and B are two vectors at an angle 120° such that resultant of these vectors ts per- -pendiculay to the smaller vector 2 and has magnitude 0 unit. Find the magnitudes of ® and B » Solution: The owangements of vectors is similar to RIVER- BOAT problem, crossing the river in or along shortest path (where VE and Va gives vesultant VR which mustbe absolutely transverse) Sa a BCos30* = R,---(4) BSIN30' =A ---(2) ! Given R=10 BCos30" Cresulfant) From (4): 120° 8.3 =10 BSin3o° | = (nullifies A’) = 5 = : inzo°=A - From (2) ° ce ee _ 20,1 2 - pA” W572 5 Scanned with CamScanner CASE-2- Vw >Ve- conceptual o,ueshan —? In such case Component of Vp can never nullify Vw —? Approach ty find direction of Va to have shortest route. ri myemember : Daift or disp ' along longitudina can not be zeros y > ra STEPS: ~. | Le Draw Vw along ~~}- dina\ / flow ; ri slg point of Va as centre construct circle of radius Vp w range : a pal the direction of Ve (bY Joining the centre to the point of tangent to the circle) 2 Scanned with CamScanner a 1 observer at rest ic: Cinertial frame) we: what do you observe ? ~~ Obsesver : The block is at rest -We: what are the forces ? there are two forces. Real forces pull of earth or weight, MQ vertrcally downwards and support of the table surface, norma) reaction Ne ver cally upwards “we: what Is your conclusion ? » observer . » observer: Find the block is at rest t% acceleration I$ 2e7r0; net force on the block Is 2ero as weight MQ downwards and normal reachon No upwards nullify €ach ofney —?p Fexr = 0 =D 420 Newton's law ts valid Scanned with CamScanner 
