When a wire is connected in the left gap of a | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let the resistance of the wire be $R$. In a metre bridge,the balancing condition is given by $\frac{R}{S} = \frac{l}{100-l}$,where $S$ is the resi

Vedclass · Accepted Answer

$\frac{800}{11} \ cm$

Vedclass · Answer

(B) Let the resistance of the wire be $R$. In a metre bridge,the balancing condition is given by $\frac{R}{S} = \frac{l}{100-l}$,where $S$ is the resistance in the right gap and $l$ is the balancing length from the left end.Initially,$l_1 = 40 \ cm$,so $\frac{R}{S} = \frac{40}{60} = \frac{2}{3}$,which implies $S = 1.5R$.When the wire is stretched to double its length,its new length $l' = 2l$ and its new area of cross-section $A' = \frac{A}{2}$ (since volume $V = Al$ remains constant).The new resistance $R'$ is given by $R' = \rho \frac{l'}{A'} = \rho \frac{2l}{A/2} = 4 \rho \frac{l}{A} = 4R$.Now,let the new balancing length be $l_2$. The new balancing condition is $\frac{R'}{S} = \frac{l_2}{100-l_2}$.Substituting $R' = 4R$ and $S = 1.5R$,we get $\frac{4R}{1.5R} = \frac{l_2}{100-l_2}$.$\frac{4}{1.5} = \frac{8}{3} = \frac{l_2}{100-l_2}$.$8(100 - l_2) = 3l_2 \implies 800 - 8l_2 = 3l_2 \implies 11l_2 = 800$.$l_2 = \frac{800}{11} \ cm$.

Similar Questions

$A$ balanced bridge is shown in the circuit diagram. The metre bridge wire has a resistance of $1 \Omega \text{ cm}^{-1}$. The current drawn from the battery is (Internal resistance of the battery is negligible). (in $\text{ A}$)

In a meter bridge experiment,initially the jockey is at the null point. Now,resistances $R_1$ and $R_2$ are interchanged. The shift in the position of the jockey is ................ $cm$.

In a meter bridge,the two gaps contain resistances of $10 \, \Omega$ and $30 \, \Omega$ respectively. If these resistances are interchanged,the null point shifts by ..... cm.

When a metal conductor connected to the left gap of a meter bridge is heated,the balancing point

In a meter-bridge,if the left and right gaps are connected with $2 \Omega$ and $3 \Omega$ resistances,respectively,then the bridge is balanced. What resistance should be connected with the $3 \Omega$ resistance to get the balancing point at the midpoint of the bridge wire?

Vedclass Test Series

Exam Paper Generator

Online Exam Module