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113. Cell with external resistan < V+ (E,7)) Netemf =E Net resistance around \oop = R+F nae, Le NTs Va 2 ~J RAF Termina\ voltage = PD acress ext resistance R V4 -V. = Vr s ER R+0 % For max power in ext . resisfance Extenal resistance CR) = Internal resistance (*) 14. Electxic $ield between the plates of a charged copacitor —E= , Ve petenhal or PD Air capacitor : a “} . separation = & Eo %. Field produced by individual plates = SZ 2Eo Vv Dielechic capacitor : Es Ear Oe E = air -. Kes K Scanned with CamScanner IIL. Bullet sticks to pendulum block Pp conseavation (M+m)V = mu * mu “Ss, Ve NA, Mm - “* f BS, M+ u as Then E conseyvalion —? h 2 22COCOte —) id pe A Ks Nv ” M ~ 29 2 As tt. mu ~ 23 (i or us M+m [zgn If u given, h will be asked CfA ou ,unu " Inelastic collision (don't use idivect energy lenin a + mut = (M4n)gh x WRONG APPROACH 2 112. Psoperties of EM waves phase clifference behween oscillations of —? a? E’ond B =0 Angle behween planes of oscillations oF EB’ and B° = T aad = 90° 2 Magnitude ratio = =¢ Direction of propagation along E xB. ~- Poynting vector S$’ = E XB Ke eee ee Scanned with CamScanner 106. Average Translatationa\ Kinehc energy For all molecules (meno, dra, poly) - (KE) ans = 3 id 3 translatrenal deg of freed for ay) Tétod KEY/Ant Total KE differs due t additional rotational modes (KE soto) = LAT — (rene) 22r0 rotatrenal CKE ) pra 2 kT (dia) 2 rotational (KE ) pat = ET (pely) 3 retahonal 10F. Order (A orf) of Elechomagnehc Waves s vony useful 2 excellent Quy . 4 Ravi Mohan i \ ‘ in Visible UV X-ray Y- ray o Micro decreasing wavelongin(a) nw we Mm om -—“— “NS log. Excess pressure = Pinside - Poutside = Parcess FoR DROP BUBBLE Cone effectve surJate) (Two effechve surface) = 2T smaller tT} Pens = ST Pexcss = eT = greater 0; o AT Pin = p+ 20 Pesess PT" SS (same for alr bubble in walker) Scanned with CamScanner 106. Average Tyanslatationa\ Kinetic energy For all molecules (meno, dia, poly) a (KE) vans = 3xX%zkT 3 translatyeno\ deg of fee Teta KE“ /Ant (3 translatteno\ deg of freed for a) Total KE differs due te additional rotational modes CKE )tota = 2kT (meno) 200 rotatrmal (KE) pra = KT (KE )ptar = S&T (dia) 2 rotational (pely) 3 retahonas \oF. order (A orf) ot Electomagnehe Waves guy j fs yeny useful 2 excellent Ravi Mohan t 7 : ‘ 3 X-ray Y-9ray Ri . . in Visible UV decreasing wavelangin(a) rr“ jog. Excess pressure = Pinside - Poulside = Paxcess FoR DROP BUBBLE (Two efechve surface) ar Cone effective suriae) = 2T smollert] Panes = Pexcess 7 Gecher ma 7 Pin = P+ 20 Penessf i" = as (same for alrbubbie in water) Scanned with CamScanner lol. Man ina lift CApparent weight) Wop = M(9+@) = Mg ('+2) = Wee (I+ 2) 4 a upward, 4 = 94a Wap = M(9-Q) , Qep = 9-2 = Mg (1-2) = Wa ('-2) d Q downwards = = Foy a=0 , lift may be moving ae Sen SE | up of down but with vec ) ¥Note: moving upwards with decreasing speed means Q down, %F=9-a moving tlownwards lth decreasing spc) means @ UP, Je = O+a ——— — ' = 4: j Block freed from the 102. Block colliding spring is of compressed nm , sparing vim} gints Fkxe ray n gm 2 belfry] — aX Xe | ennai 103. Emissive power of a black bedy Eat’ Es . Ley Temp must ue) E, Tt in kelvin 4 To moke emissive power double , temp should be increased to 2/4 times 2"4 - (\z)"* = Vv =dra = 1:19 thnes Scanned with CamScanner NORMAL CONSTRAINT when two rigid susfaces in contact they can have relaive métion along the surface but not along the normal ( They do not jump or penebate wrt each ojher) At any instant for two points in céntact yelative displacemen} (Xre) velocity CVret) and atceleralion (@rei) along (hormal)to the sursaces of contact must be zero Approach: B, and Bz Are two bodies in contact Pi and Pz be hwe pojnts in contact Ve and Vz be the velocities of Prand P: resolve components along Nowal and e@ equate then (Mi) ns = (Vi) same Opproach for @ displacement (x) and eaueleraion (a) Scanned with CamScanner Question : The arrangement shown x *% “Bx lying on a horizontal plane a= Unigorm transverse magnetic * a 7 Z field GB is Applied. the slider v, { (m,&) is projected towards , - —*---- rignt with velocity Ve. Find i's velocity as @ funcHon of time and alse plof solubon : v be the Instantaneous velocity tv em - phy, cuvent I = otv f £ 4 ra ® Opposing magnetic force on the a |v Slider = TSB ; Tis . = BYY q retardation = =m = “mE Fm yetardation b= - IV, BL v a mR dv _ _ Br at Vv mR Vv : t > . - imits dv _- pte at Integrating within limi J wy Lea Vv a,* t . = -S4 [t). [invIy. — ced _ox's Inv -- s8.t pVewe ™ Vo mR ea velocity 's exponentially i failing function f7om inl- _ tal value Ve — t— i Scanned with CamScanner @ oblique force to a block at rest over rough horyi2ontal surface POO FGD: Finding No: Apply vertical equilibrium (equilibrium normal fo surface) No+Fsiné =mg => No =(mg -Fsiné) Finding limiting friction: fe = BsNo =}s (m9 -Fsin@) check condition for sliding - Block will slide if (Net driving force along the surface) > limiting friction Have: cepse > Hs (M9- Fsiné) > Hsmg - HsF sine F (Cas6 +Hs5in 6) 7 #smg @ For siding F 7 ( Msg ) Cos8 +HsSin& Scanned with CamScanner Bob at momentary rest (at the extreme position) [ T= mgcos Bo | + Dont apply horizontal and vertical equilibrium = Net force towards \ centre (Cc) T-mgcose = centzipetal force (mv) : T= Mgcos6 + os Scanned with CamScanner Equilibrium of simple pendulum Bob (Suspended object) attached to stzing Petmanently at rest % Apply equilibzjium in any directions like Horizontal / veztical radial (along tength) FGD of the bob % | Tceose Fy = Tsing ‘So Fu 1 tone = Fu Fy Fy squaring and adding =p T= (Firth o@ Radial equilibrium + no! T=Fusino+Fycose | get 6 FY YY eycose + Fusing For swinging pendulum / momentary rest at extreme position ONLY RADIAL EQUILIBRIUM Scanned with CamScanner LONGITUDINAL CONSTRAINT For (1) Linear rigid object ail) string under tension Relative displacement (Xr) 5 Velocity (Vre) and acceleration (Arai) along the lengin (longitudinal) is 2eYo @ Approach : resolve X,V ora of the end points along tne length and equate them @ Example : vi Velocity Of end points of ' Q@ rigid rod is as Snown = = Te relate Vi and Vz V2 Solution: M (Va)u = Vv: cos, --— 6 li Ly, =<" (Vida = V1 C056) Vz equate longitudinal components V, COSO1 = V2 COS Oz @ Exomple: Find velocity of the block at the Instant shown & solution : iM p 8 @ 7 $ co? - Vv Ae, Wg quating velocities along k the length Vw aveloclty of loc v Vpcose = V Vo = case Scanned with CamScanner ~ Question : 2 a] —P A and 8 aye two vectors at an angle |20° such that resultant of these vectors ts per- -pendicular to the smaller vector Z and has magnitude 'O unit. Find the magnitudes of R and B » Solution: The owangements of vectors is similar to RIVER- BOAT problem, crossing the river in or along shortest path (where Vp and Vs gives vesultant Vp which mustbe absolutely transverse) —— BC0s30* =R ---(1) BSIN3BO' =A ---(2) BC0s30° 1 Given R=10 Cresultant) From (4) : ; 8.3 =10 BSIn3o° | A 20 (hullifies A’) ° iNZ0°=A From (2) ° ape = 2 ak (2 =D A= 2 a 2 Scanned with CamScanner CASE-2 .- ' Vw >Ve_ conceptual question —? In such case component Of Vp can never nullify Vw —o Approach +o find direction of Va to have shortest route. # Z. 4 P 4 - oe oi 2 = % yemembe? : Vw ! Daift or disp | } along longitudina | ; ero can not be 2 \ : J » ” ae | . Draw Vw along —~t}- inal / flow 4 or seagoe point of Yu as centre construct radius circle @ t to ine circle from the tail of Dyaw fangen : eer —? wet the direction of Ve (by J NY the centre to the point of tangent to the cl Scanned with CamScanner No observer at rest Cinertial frame) We: what do you observe ? ~» obsexver : The block is at rest We: what are the forces ? there are two fortes. Real forces pull of earth or weight, MQ verhcally downwards and support of the table surface , normal reacHon Ne vertically upwards -wWe: what Is your conclusion 2 » observer . obserwer: 1 find the block is at rest its acceleration Is 2e7r0> net force on the block IS 2e7ro as weight MQ downwards and normal reachon No upwards nullify €ach other ——D Fexrr =O =D A420 Newton's law Is valid Scanned with CamScanner 
