Saved Bookmarks
| 1. |
A uniform wire is cut into two pieces such that one piece is twice as long as the other. The two pieces are connected in parallel in the left gap of a metre bridge. When a resistance of 20 Omega is connected in the right gap, the null pooint is obtained at 60 cm from the right end of the bridge wire. Find the resistance of the wire before it was cut into two pieces. |
|
Answer» Solution :Let `R_(w)` be the resistance of the wire before it was CUT in two. Let `L_(1), L_(2)` and `X_(1),X_(2)` be the lenghts and resistance of the two PIECES. `therefore R_(w) = X_(1) + X_(2)` and `(X_(1)/X_(2))= (L_(1)/L_(2))` (`rho` and A being the same) Data: `L_(1)=2L_(2), R=20 Omega` (in the right gap), `L_(R) = 60 cm` `therefore L_(x) = 100-L_(R) = 40 cm` and `X_(1)/X_(2) = L_(1)/L_(2)=2` Since, the pieces are CONNECTED in parallel, their EQUIVALENT resistance is `X_(p) = (X_(1)X_(2))/(X_(1)+ X_(2)) = X/(X_(1)/X_(2) + 1) = X_(1)/(2+1) = 1/3X_(1)`.................(1) And, with the bridge balanced, `X_(p)/R = L_(X)/L_(R)` `therefore X_(p) = R(L_(X)/L_(R)) = 20 xx 40/60, 40/3 Omega`..................(2) From Eqs. (1) and (2), `1/3X_(1)= 40/3 therefore X_(1) = 40 Omega` `therefore X_(2) = 1/2X_(1) = 20 Omega` `therefore` The ORIGINAL resistance of the wire is `R_(w) = X_(1) + X_(2)= 40 + 20 = 60 Omega`. |
|