Solution - Career Point

∴17.[C]Potential of shell A is,1 ⎛ 2 2 2V A =4π∈⎟ ⎟ ⎞⎜ 4 πaσ − 4πb σ + 4πcσ⎜o ⎝a b c⎠σ= (a – b + c)∈ 0Potential of shell C is,1 ⎛ 22 2V C =4 π∈⎟ ⎟ ⎞⎜4πaσ − 4πbσ + 4πcσ⎜0 ⎝c c c⎠σ ⎛ 2 2 ⎞= ⎜ a b− + c⎟∈ 0⎜ ⎟⎝c c⎠As V A = V Cσ σ ⎛ 2 2 ⎞(a – b + c) = ⎜ a − b+ c⎟∈ 0∈ ⎜⎟0 ⎝c c⎠a − b a + bor a + b = ccor a – b = ( )( )≡A543211 23 2 5 43 4∈ AThe capacity of each capacitor is, C 0 = 0d5 5 ∈From fig. it is clear that C eq = C0 = o A3 3 d18.[B] Resistivity of conductors increases withincrease in temperature because rate ofcollisions between free electrones and ionsincreases with increase of temperature.However, the resistivity of semiconductorsdecreases with increase in temperature becausemore and more covalent bonds are broken athigher temperatures.19.[B] let L A and L B be length of parts A and BRThenA L=A[as cross-section is same]R B LBNow L c = 2 L A and (volume) c = (volume) Pi.e.L c × A c = 2 L A × A c = L A × A Awhere A c = A A are cross-sectional area of part Cand A.∴ A c = A A /2BAB∴R AR C/ρ LAA=A L=A×ρ Lc/A L c CL=A A×A/ 2 1=2LAAA4ACAA20.[A] Current flowing through potential wire is –EI =15r + r= E16rPotential drop across potential wire is,15EV = I × 15 r =1615EPotential gradient, K =1660021.[A]E E 15E∴ = Kl or = × l2 2 16×60016×600or l = = 320 cm.15×2AD i Cφ 1OPaφ 2φ 1φ 2Q bµ 0 iB AB = B CD = (sin φ4π( b / 2)1 + sin φ 1 )µ=0 4ia.4πb 2 2a + bB BC = B DA =Bµ 0 4 i b. .4πa 2 2a + b∴ B = B AB + B BC + B CD + B DAµ=0 4i ⎡ab.4π2 2⎢ + +a + b ⎣ba=µ4π0 .2 +28i a bab22.[A] In tan A position,µ 0 2 M= B4π3 H tan 30° =dabb ⎤+a⎥⎦B H…….(1)3Magnetic moment of second magnet,M' = (3m)(2 × 2l) = 6MIn tan B position,µ 0 6 M= B4π3 H tanθ …… (2)ddividing eq. (2) by (1) we getXtraEdge for IIT-JEE 103APRIL 2010