a potentiometer wire of 10m length and 20 ohm resistance is connected in series with resistance r ohms and a battery of  emf 2v ,negligible internal resistance. potential gradient on the wire is 0.16 millivolt/centimeter  then r is.....ohms(atb)?(ans=230) (current electricity)

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R is the resistance of the wire = 20 ohmL is the potential meter length = 10 mE= 2 voltK =potential gradient = 0.16 milli volt/cm = 0.16×10-310-2=0.016 v/mcurrent across the wire is given by:I = ER+rpotential across the wire V = IR=ERR+rPotential gradient K = VL=ERR+rL=0.0162×2020+r=0.016×1020+r=40/0.16=250r=250-20=230 ohm

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