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Internal resistance of a cell is r = 5Omega & is connected in potentiometer (wire length = 1m) circuit arrangements as shown in figure, which shows two circular conducting rings R_(1) & R_(2)each having radii 20 cm cross each other at conducting joints A & B. Section AB subtends 120^(@) at the center of each ring. Resistance per unit length of R_(1) & R_(2) are (3)/(4)Omega//cm & (6)/(pi)Omega//cm respectively. Cell is connected across point C & D of rings lying on perpendicular bisector of AB. Now when switch Sw remains closed, balancing length at null point is 48 cm. Find the balancing length (in cm.) when switch Sw is opened. |
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Answer» SOLUTION :`R_(AC)=R_(CB)=(2pixx20)/3xx3/pi=40 Omega` & `R_(AD)=R_(BD)=(2pixx20)/3xx6/pi=80Omega` `rArr` Balanced W.S.B `rArr R_(CD)=120/2=60 Omega` `epsilonxx(60/(60+5))=48x` & `epsilon=x l` `rArr lxx60/55=48 rArr l=52 cm` |
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