Title 18. Current Electricity Med Ans.pdf ✅

1. (c): Here, number of electron, n = 10000000 = 7 10 Total charge on ten million electrons is, Q = ne [where e = 1.6 10 C −19  ] 10 1.6 10 C 7 −19 =   1.6 10 C −12 =  Time taken by ten million electrons to pass from point P to point Q is, t 1 s 10 s −6 =  = The current, 1.6 10 A 10 1.6 10 t Q I 6 6 12 − − − =   = = Since the direction of the current is always opposite to the direction of flow of electrons. Therefore due to flow of electrons from point P to point Q the current will flow from point Q to pint. P. 2. (a): Q = It Also Q = ne [e = 1.6 10 C −19  ]  ne = It or (;k) 19 1.6 10 1A 1s e It n −   = = 18 = 6.2510 electrons 1 s − 3. (b): As dt dQ I = dQ = Idt dQ = (2t 3t 1) 2 − + dt   = = = − + t 5 t 3 2 dQ (2t 3t 1)dt 3 5 3 2 t 2 3t 3 2t Q       = − +       = − − (5 − 3 ) + (5 − 3) 2 3 (5 3 ) 3 2 3 3 2 2 (25 9) 2 43.34C 2 3 (125 27) 3 2 =      = − − − + 4. (b): Let dq be the charge which passes in a small interval of time dt. Then dq = Idt or dq = (4+2t)dt On integrating, we get, q (4 2t)dt [4t t ] 48C 6 2 2 6 2 = + = + =  5. (b):Radius of electron orbit r 0.72A 0.72 10 m 10 o − = =  Frequency of revolution of electron in orbit of given atom) 18  = 9.410 rev/s (where T is time period of revolution of electron in orbit  Then equivalent current is, e 1.6 10 9.4 10 1.504A T e I 19 18 = =  =    = 6. (a): momentum will be conserve 7. (b): from high potential to low potential 8. (d): Let n be the number of turns in the coil. Then total length of wire used, l 2 r n 2 710 n 2 =   =   − Total resistance, A l R =  or , 3 2 7 2 (0.7 10 ) 2 10 2 10 n 4 − − −       =  n = 70 9. (d): The electrical resistance of a conductor is depend upon all factors size, temperature and geometry of conductor. 10. (b): The resistance of rod before reformation 2 1 1 1 r l R R   = =         =   = 2 r l A l R Now the rod is reformed such that, 2 l l 1 2 = 2 2 1 2 2 1 r l = r l (  Volume remains constant, or 1 2 2 2 2 1 l l r r = ...(i) Now the resistance of the rod after reformation 2 2 2 2 r l R   = 2 1 2 2 2 1 2 2 1 2 1 1 2 1 r r l l r l r l R R =       = or, 2 2 2 1 2 1 2 1 2 1 (2) l l l l l l R R =         =  = (using (i)) 4 R R 2 1 = 4 R R2 =
11. (d): If the wire is stretched by th (1/10) of its original length then the new length of wire become. 10 11 l 10 1 l l 2 = + = As the volume of wire remains constant then        =  =  10 11 l r l r l r 2 2 2 2 2 2 1 (using (i)) 2 1 2 2 r 11 10  r = Now the resistance of stretched wire., 2 1 2 2 1 2 2 2 r l 10 11 r 11 10 l 10 11 r l 10 11 R          =        =         =         =    = 15 r l R 2 1  1   =        = 15 18.15 10 11 R 2 2 12. (b): Since, 2 4 l dl R a l R  =   =  (where l is any radius and dl is small element). Total resistance,        −   =       −   =   = 2 1 1 2 r r 1 2 r r 2 r 1 r 1 l 4 1 l 4 dl 4 R         − = r r 4 r r R 1 2 2 1 13. (c): Potential of 20 V will be same across each resistance current. 10A 2 20 R V I 1 1 = = = 5A 4 20 R V I 2 2 = = = 4A 5 20 R V I 3 3 = = =  Total current drawn circuit, I = I 1 + I 2 + I 3 =10+ 5+ 4 =19A 14. (a): The slope of V − I graph gives the resistance of a conductor at a given temperature. From the graph, T1 is greater than at temperature T2 . As the resistance of a conductor is more at higher temperature and less at lower temperature, hence T1  T2 . 15. (b): Mobility of charged particle 10 4 d 3 10 7.5 10 E v − −    = = 6 2 1 1 2.5 10 m V s − − =  16. (a): E ne ne J nAe I vd        = = = 17. (b): (A) →(p) if the temperature is not very high. (B) → (r) (C) → (q) 18. (a): Ohm’s law V = IR is an equation of straight line. Hence I – V characteristics for ohmic conductors is also a straight line and its slope gives resistance of the conductor. 19. (b): The figure is showing I – V characteristics of non- ohmic or non linear conductors. 20. (d): Resistance of a wire in terms of conductivity () is given by A 1 l R  = where l is the length and A is the area of cross – section of wire respectively. As the wires are onnected in series, Rs = R1 +R2 A l A l A 2l s 1 2 +  =  A l A 2 1 1 s 1 2 2 +  +  =  where s is the effective conductivity 1 2 2 1 s 1 2 2 1 1    +  =  +  = 
1 2 1 2 s 2  +     = 21. (a): For a good conductor the graph between voltage and current does not obey exactly ohms law it shows some deviation from straight line. 22. (a) Materials Resistivity ( )( mat 0 C) o   , ( )(0 C o  ij m) Copper (Cu) 8 1.7 10−  Nichrome (alloy of Ni, Fe, Cr), 8 100 10−  Germanium 0.46 Silicon 2300 Hence, the option (a) is correct 23. (d): Wire bound resistors are made by winding the wires of an alloy, viz. manganin, constantan, nichrome etc. the choice of these materials is dictated mostly by the fact that their resistivities are relatively ijsensitive to temperature. These resistances are typically in range of a fraction of an ohm to a few hundred ohms. 24. (c): The typical range of resistivity for metals is from 10 m 8 − to 10 m 6 − 25. (c): The specific resistance is independent of the dimensions of the wire but depends upon the nature of the material of the wire. 26. (a): Nearly independent of temperature 27. (c): The graph (a) is true for conductors whose resistivity increases with temperature. But for semiconductors, resistivity decreases as the number of conducting electrons increase with rise of temperature. 28. (b): 10k 29. (d): (1 10 5%) 6   30. (a): 0.2A 10 2 3 7 9 7 r r I 1 2 = = + − = +  = Potential difference across 8.4V 9 0.2 3 9 0.6 1 =  = −  = − Potential difference across 2 AB 2 2  ;V =  + 0.2r = 7 + 0.27 7 +1.4 = 8.4V 31. (b): These materials exhibit a very weak dependence of resistivity on temperature. Their resistance values would be changed very little with temperature as shown in figure. hence these materials are widely used as heating element. 32. (b): Semiconductors having negative temperature coefficient of resistivity whereas metals are having positive temperature coefficient of resistivity with increase in temperature the resistivity of metal increases whereas resistivity of semiconductor decreases. 33. (d): Here, R 99 , 0 =  T 27 C o 0 = 4 o 1 T 1.710 C R 116 − −  = =  R R [(1 (T T )]  T = 0 + − 0 1 (T T ) 99 116 1 (T T ) R R 0 0 0 T  − =  −  − =  − 99 17 1.7 10 1 99o 17 99 1 116 99 T T0 4   = =       −  − = − 1010.10 C 99 10 T T o 5  − 0 = = T 1010.1 T 1010.1 27 1037.1 C o  = + 0 = + = 34. (b): Here, R0 = 5,R100 = 5.25,Rt = 5.5 As, R R (1 t) R R (1 100) t = 0 +  100 = 0 + R 100 R R 0 100 0  −  = ...(i) Let the temperature of hot bath be t C o R R (1 t) t = 0 +  R t R R 0 t 0  −  = Equating equations (i) and (ii), we get R t R R R 100 R R 0 t 0 0 100 0  − =  − 100 5.25 5 5.5 5 100 R R R R t 100 0 t 0  − −  = − − = 100 200 C 0.25 0.5 o =  =
35. (c): Here, R 2.5 ,T 28 C o 1 =  1 = R2 = 2.9 and T 100 C o 2 = As R R [1 (T T )] 2 = 1 +  2 − 1 2.9 = 2.5[1+ (100− 28)] 1 [72] 2.5 2.9 − =  or, 3 o 1 2.22 10 C 2.5 0.4 72 1 2.5 2.9 2.5 72 1 − − =  =  −  =  36. (d): Resistance of heater coil, =   = = 400 100 200 200 P V R' 2 Resistance of either half part = 200. Equivalent resistance when both parts are connected in parallel, 100 . 200 200 200 200 R' =  +  = Energy liberated per second when combination is connected to a source of 200 V, 400J. 100 200 200 R' V 2 =  = = 37. (c): A- r, B- p, C-s, D –q 38. (a): When the heater is connected to the supply its initial currecnt will be slightly higher than its steady value but due to heating effect of the current the temperature will rise. This causes in resistance and a slight decrease in current to steady current. 39. (a): In series combination, current across its circuit components is always constant and in parallel combination the voltage across the circuit components is constant. 40. (d): Wire of length 20.1 m of 1 12 m −  is bent to a circle. Resistance of each part =120.1 =1.2  Total resistance = 0.6. 41. (a): For maximum equivalent or total resistance of the resistors must be combined in series. Req = R1 + R2 = R3 = 5 + 4.5 + 3 =12.5 42. (a): Since the voltage across the circuit is constant. Then currect through 3 resistor, 4A 3 12 R V I 1 1 = = = The current through 4 resistor 3A 4 12 R V I 2 2 = = = and the current through 5 resistor, 5 12 R V I 3 3 = = = 2.4A 43. (b): The equivalent circuit is, eq eq eq eq eq R R R R R + + = +  = + 2 2 3 (2 ) 2 1 i.e. 2 0 2 Req − Req − = 1 1 8 2 . 2 1  Req =  + =  44. (d): In the parallel combination of three resistances, the equivalent resistance is., 1 2 3 1 1 1 1 Req R R R = + + or, 315 143 315 35 45 63 5 1 7 1 9 1 1 = + + = + + = Req = = 2.02 143 315 Req 45. (c): If one had considered a solid cylinder of radius b, one can suppose that it is made of two concentraic cylinders of radius a and the outer part, joined along the length concentrically one inside the other. If q I and x I are the currents flowing through the inner and outer cylinders. b a x total b a x R V R V R V I = I = I + I  = + where Rb is the total resistance and R x is the resistance of the tubular part. x b Ra 1 R 1 R 1  = − But, a 2 a l R   = and b 2 b l R   =