SAFE HANDS & IIT-ian's PACE EDT-12 (JEE) SOLUTIONS

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Hollie Montgomery
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1 Charge distribution will be as shown in figure: Q + Q + Q ( ) ( ) Answer : Option () Answer : Option () Field inside dielectric (E) = Answer : Option () As the radii of third sphere and A are same, charges will be equall distributed on both of them ie Q/ each Now when third sphere is touched with B, again both of them will have half of the total charge (Q + Q/) ie Q/ each Answer : Option () 5 Consider a spherical shell at distance r from the centre (a < r < b) Charge inside this shell will be q enclosed = Q + = ( ) B Gauss law, E(πr ) = Hence E = ( ) For E to be independent of r, numerator in the second term must be zero A = Q/πa Answer : Option () 6 Answer : Option () 7 Answer : Option () 8 Answer : Option () -(Q + Q ) -Q Q + Q Q EDT- (JEE) SOLUTIONS 9 E = = -ar Consider a sphere of radius r Then from Gauss law, E(πr ) = q = -8πϵ 0 ar ρ = = Answer : Option () 0 Answer : Option () Charge of -80μC will appear on the lower plate of μf capacitor This -80μC will be coming from upper plates of μf and μf capacitors So net charge of 80μC will appear on upper plates of μf and μf capacitors This charge will be divided in the inverse ratio of capacitances as these two are joined in parallel Answer : Option () When switch is turned to position, both the capacitors will be effectivel in parallel (Potential across them must be same) So effective capacitance will become 0μF ie 5 times the original capacitance So energ will becomes /5 times (We have to use as charge on the sstem remains same while potential will change) So 0% energ is left or 80% is lost Answer : Option () When the are connected (the will be in parallel), their effective capacitance will be C and total charge on them will be CV + CV Loss in energ = ( ) ( ) Answer : Option () Assume that there are si charges on the si vertices of the heagon Net force on the charge at the center will be zero as sstem is smmetric Now if we remove one particle, we are actuall removing amount of the force As initial net force was zero and we are removing ie then remaining force must be 0 magnitude wise Answer : Option ()

2 5 Consider a small element of length d at distance from the origin Charge on this small element = dq = d Force on this element = df = E dq = (E 0 + E ) d Total force on the rod = F = = ( ) = q (E 0 L/ + E ) Answer : Option () 6 As smaller sphere that is removed has half the radius of the larger sphere, its volume and hence charge on it are /8 times The same effect of removal of smaller sphere can be achieved if we place another sphere of radius R/ and charge Q/8 In that case, field at point P = E = ( ) ( ) ( ) = Answer : Option () 7 Note that net charge on capacitor is alwas zero and hence it stores onl energ and not charge Think about validit of this statement in case of isolated spherical capacitor Answer : Option () 8 One of them is shorted and other three are in parallel Answer : Option () 9 Work done b the agent pulling the plates will be stored as an energ inside capacitor Think about what will happen if batter is still connected! Where does work done b eternal agent go in this case? Answer : Option () 0 Answer : Option () Absolute potential can be defined to be zero anwhere ou want! Answer : Option () EDT- (JEE) SOLUTIONS Force on the charge will be zero no matter where it is placed insude the shell So even if we displace it slightl,force wil be zero and it will have no tendanc to move awa or towards the original position So it will be a neutral equilibrium Answer : Option () Because of electric field, positive charge will be induced inside the given Gaussian surface Answer : Option () Answer : Option () 5 Initiall the net charge on the assembl encircled in figure was zero as it is isolated V As switch S is closed, capacitor of 6μF is shorted Charge on the μf capacitor will be Q = CV = 6μC This charge must have flown through the switch as initial charge was zero Answer : Option () 6 We want potential to be same Suppose charge q is placed on A B will get Q - q charge = ( ) q = Answer : Option () 7 Weight must be balanced b the elctrostatic repulsive force Suppose charge Q is given to the plate ς = Field near the chraged sheet will be: E = mg = qe mg = Q = mgaϵ 0 /q Answer : Option () μf S 6μF

3 8 The given sstem is actuall a combination of two dipoles making angle 60 0 with each other as shown in the figure - q EDT- (JEE) SOLUTIONS 0 P q - q Both of them have magnitudes of ql Hence, net dipole moment will be vector sum of them ie ql Answer : Option () 9 Answer : Option () Remove this part Answer : Option () Q

4 EDT- (JEE) SOLUTIONS

5 EDT- (JEE) SOLUTIONS 5

6 (b) (a) log(log) ) log (log) log log(log) 0 f f e e '( ) '( ) (log) e sec cos 0, d sin sin EDT- (JEE) SOLUTIONS sin cos (c) ) f ( ) f ( ) (a) Taking log both sides, p log q log ( p q)log( ) 5(c) p q p q d d d to e e, 0, e Taking log to the both sides, log ( ) Differentiate both sides wrt, d d 6(a) Let tan and sin Differentiating wrt of and, we get Putting d d d tan tan, tan tan tan and sin sin tan d Again [ tan ] d d d and [ tan ] d d Hence (i) 7(b) From the given relation log log( a b) (ii) Differentiating we get d a b a b ( a b ) a (i) d a b Differentiating again wrt, we get d ( a b) a a b d d d d a ( a b) d ( a b) d a d ( a b) d

7 EDT- (JEE) SOLUTIONS 8(d) ) p 6 p p ) ( 6 p p ) ( p p) ( p 6p) d d ) 6 p p p p p 8p ) 8p d and d p p ) 6p p p and 8 p 6p 6p 8 p 6p d ) 6p p = a constant d 6p 9(b) We have loge log log tan tan loge log 6 log tan tan tan log tan log tan tan ( log ) tan d 0(d) ()! ()! (b) e e () d log 6 log tan d 0 0 n d n ( log ) d and ( ){ e ( )} e d 0 d 0 0 cos cos cos cos 0 d (b) Curve e Differentiating with respect to e or d d d d e e e 0 ( ) 0 This hold for, 0 (c) a( t sint), a( cos t) / dt a(sint) = tan t d d / dt a( cos t) Length of the normal = d t = a( cos t) tan ( / ) = a( cos t)sec( t / ) = a sin ( t / )sec( t / ) = a sin( t / )tan( t / ) (b) Let diameter of sphere AE r Let radius of cone is and height is AD since BD AD DE A B D C

8 EDT- (JEE) SOLUTIONS or (r ) (i) Volume of cone V (r ) (r ) dv ( dv r ) 0 (r ) 0 ( r ) 0 r, 0 d V Now (r 6), put r d V r 6 r = negative value So, volume of cone is maimum at r Height = Radius r log log 5(d) Let log d log Put 0 0 d log 0 e and d log d d At e, 0 d e log In [, ) the function will be maimum and minimum value does not eist ( ) ( ) 6(a) ) e e ( ) e ( ) { ( )} e ( ) Now b the sign-scheme for + ( ) / f ( ) 0, if,, because e ( ) is alwas positive So, f () is increasing on, 7(b) f ( ) 6 9, For decreasing f ( ) ( ) ( ) 0, (, ) 8(d) ) 6 a b f ( ) a f ( c) 0 f 0

9 0 a 0 a a 0 a 9(c) From mean value theorem a 0, a) 0 b b) a) c) b a, b) EDT- (JEE) SOLUTIONS 8 f ( ) ( )( ) ( ) ( ) f ( c) ( c )( c ) c( c ) c( c ) f ( c) c = c c c c c c 6c According to mean value theorem, b) a) c) b a ( / 8) 0 5 c 6c c 6c 0 ( / ) 0 c (b) Let the point be, ) Therefore ( (i) ( ) Now slope of the tangent at (, ) is ( ), but it is equal to Therefore, 7 ( ) (a) 7 7 Hence the point is, ( log) loglog log loglog[log loglog] d d (b) Let tan log log Putting tan, we get tan d ( Putting (log loglog) loglog (log loglog) loglog, z tan sec tan tan sin, we get ) sin cos z tan tan cos tan tan (tan ) log log

10 z sin Thus dz dz d EDT- (JEE) SOLUTIONS / d dz / d ( ) d (c) Let ep( cos sin) 0 ep( cos sin)( sin cos ) Now 0 sin cos 0 sin 0 6 Now is ve at Maimum value of 6 6 ep = ep( ) (c) f ( ) 9a a 5 (d) f ( ) 6 8a a f ( ) 8a For maimum and minimum, 6 8a a 0 a a 0 a or a, at a maimum and at a minimum p q, a a a or a 0 But a 0, therefore a 5 ) ( ) f' ( ) 5 75 (75)( ) For maima and minima, 75 5 ( ) ( ) 5 ( ) 5 ( ) [( ) ] 0 Either 0 or or 75 ( ) 0 At, f' h 0 and f' h 0 ) is maimum at 6(d) Let t, then d f'( t) dt d d ( ) sint sin d ( ) 75 5

11 EDT- (JEE) SOLUTIONS 7(c) u(, ) log log u log ; u log Now, u u u log u log log log log log log log log log log log log log 8(a) Given curve 9(a) p q Differentiate with respect to, d d p, p p For given line, slope of tangent p p From equation (i), p d 9 8 q q 7 (i) (i) d Now tangent makes equal angle with ais o 5 or o 5 o o tan( 5 ) tan(5 ) d From equation (i), (taking +ve sign) 0 ( )( ) 0, From the given curve, when 6 Therefore, required points are, 0(d) Slope of the normal = tan / d / d (, ), f ( ) d (, ), 5 5 and when, ( ) = 8 8 and, 6 6

Physics 2135 Exam 1 September 20, 2016

Physics 2135 Exam 1 September 20, 2016 Eam Total / 200 Phsics 2135 Eam 1 September 20, 2016 Printed Name: Rec. Sec. Letter: Five multiple choice questions, 8 points each. Choose the best or most nearl correct answer. 1. Two positive charges