Narayana IIT Academy

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Darrell Walsh
6 years ago
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1 Narayana IIT cademy INDI Sec: LT_IIT JEE DVNED DTE :..8 TIME : 7: M to : M WT-5 MX MKS : 86 KEY SHEET 6-P PHYSIS D D HEMISTY 9 D D D 8 D 9 D MTHS D D D 46 D

2 --8_LT IIT_Jee-dv_6-P_WT_5_Key Sol s SLUTINS PHYSIS. The current density J, electric field E at the cross-section are related by E J multiplying both side by. E J= or I= Where is electric flu. ρ E J= or I= or I ρ. Let the equivalent resistance of one infinite ladder be. Then the complete network reduces to and Hence D. we have = 4( + ) so + = 5 so = 5 4. onceptual 5. onceptual 6. onceptual V V I I. I 9.9 I ampere g g urrent in will be zero in steady state I I V 6 F V + q - 6V + q - I I I V 4 V V V 6V q q q 44 F Sec: LT_IIT Page

3 --8_LT IIT_Jee-dv_6-P_WT_5_Key Sol s dq 9. urrent through branch containing capacitor is, I e t amp. dt t t=, I = in downward direction, i.e., towards P. Let V is the potential of point P then q V V 7 volt urrent through branch containing capacitor is, 8 7 I again towards P. pplying KL at P, current through is 4 and potential difference across is 7 4. harge on capacitor increases with time q potential V increases too urve I is for V and curve II is for V as current will decrease with time. lso V V V V V V V (when both were equal) V V V V V V V. urrent does not depend on area urrent will be same through the two sections ne v ne v 4 v v 4 ee v E v me v E lso. harge on plate ' + harge on plate Q ' S, d d I. Now if only S is closed then charge on plate will still be zero charge on plate total charge + Q is on plate ' Q amount of charge will flow through S So as to form a capacitor II. Similarly, if only S is closed Q amount of charges will flow through S in both of above cases harge of amount Q (sign is not taken) is flown Sec: LT_IIT Page

4 q Q +q _LT IIT_Jee-dv_6-P_WT_5_Key Sol s d d Now, if S and S both are closed let the distribution charges on the capacitors are q Q q Now q Q Q Q amount of charge in flown through S and Q i.e., Q amount of charge is flown through S.. If is less than ' then no neutral point will be found. Since potential drop across whole wire is which is less than '. If different terminals of and D are joined to either of and. Starting from if potential increase along then it will decrease in wire containing r by. No neutral point will be obtained. Q 4. F QE o 5. The circuit can be folded about and redrawn as 6. Let at any time t, the charge on the bigger sphere is Q q and on smaller sphere is Q q Then Q q Q q dq 4 r r dt t Q q dq dq. dt 4 r dt Q q 4 r q t Q 8 r Upon integration, q e t dq Q 8 r Q urrent i e e dt 8 r 8 r t 8 r. dq 5 Putting the values, at t s, dt e 7. pply Kirchhoff s rule and solve 8. Using KL and KVL we can find the potential difference across capacitor which comes out to be V. So energy stored in capacitor is, 6 4 u 8J Sec: LT_IIT Page 4

5 9. Glycoside formation also yields two anomers. II. ' l ' is less EN and cannot displace IV. Nal obalt hloride Nal. ased on half filled configuration of d-orbits --8_LT IIT_Jee-dv_6-P_WT_5_Key Sol s HEMISTY. Smaller the size with more number of charges on cation, the polarizing power is more. Na S Mn Ni..9 of rl6h. moles 4. The acid liberated.56 moles # moles of H ions formed Gives rl 6. 4 th elements 7. Tatutomerism 8. F F H F H H N 9. product formed is aldopentose. based on isoelectric point. polymerization. r HI Hl N Sec: LT_IIT Page 5 N More stableint ermediate

6 --8_LT IIT_Jee-dv_6-P_WT_5_Key Sol s. werner theory Postulates statements, 5, 6 are correct 4. onceptual 5. onceptual 4 6. H H Sn Sn H H H g H sample = = g pure H =.588 mol pure H and ML of M Sn + =. mol Sn + y reaction I:. mol Sn + reduces =. mol H Hence unreacted H =.588. =.88 mol y reaction II: obtained from.88 mol H =.88 =.94 mol nt Volume of gas = P =.98 L 7. Give of the digits to be omitted. The no. of 5 digit numbers formed from ( i),,, 4,5 4 4! 96 ( ii),,, 4,5 5! Total-6 MTHS 8. Let us consider two consecutive seats as a unit. To satisfy our required condition we need to select 6 such units no two units are consecutive, this can be done in 9 6 ways 84. The persons in these seats can be rearranged in! ways. Hence the number of required arrangements is 84! 6 K for some K N 6 9. Now K 5K 5 5K 5K = M, M N 4. The required number of ways is Let, y at, at,, y at, at Then y at t K y y m K n Number of diagonals 54 Number of triangles 8 Sec: LT_IIT Page 6

7 5! 44. equired rrangements!! 45. onceptual 46.. r r r 47. The required numbers can be calculated as follows ) / 456!! 6 ) / 56!! 4 ) 4 / 56!! 4 4) 4 / 56!! 4 5) 5 /46! 6) 5 / 46! 7) 5 / 46! 8) 45 / 6! 6 9) 45 /6! 6 ) 45 /6! 6 The total arrangements are 6. n n a b a b 48. n n n a b ab n ab n a b 49. onceptual 5. The required numbers are ntimes n n p K K 5. The number of ways the student can fail is n 6 n 6 n n n n n m m onceptual --8_LT IIT_Jee-dv_6-P_WT_5_Key Sol s Sec: LT_IIT Page 7

Chapter 28. Direct Current Circuits

Chapter 28. Direct Current Circuits Chapter 28 Direct Current Circuits Circuit Analysis Simple electric circuits may contain batteries, resistors, and capacitors in various combinations. For some circuits, analysis may consist of combining