Question

In a meter bridge, the balance point is obtained at $40cm$. If a resistance equal to that in the left gap is shunted across itself, the new balance is:
A) $40$ $cm$
B) $15$ $cm$
C) $25$ $cm$
D) $32$ $cm$

Answer 1

Hint:A meter connect likewise called a slide wire connect is an instrument that chips away at the rule of a Wheatstone connect. A meter connect is utilized in finding the obscure opposition of a conductor as that of in a Wheatstone connect. The meter connection utilizes a similar rule as the Wheatstone bridge. It is utilized to locate the obscure Resistance of the Material. The meter is a wire of \[\;100\] cm, a scale, one obscures Resistor, one known Resistor or a Resistance Box, a Galvanometer and, a racer.

Formula used:
$\dfrac{{{R_1}}}{{{R_2}}} = \dfrac{l}{{100 - l}}$
${R_1}$ is the first resistance
${R_2}$ is the second resistance
$l = $ is the length

Complete step by step answer:
Let ${\rm I} = \dfrac{V}{R}$ are the resistance and $l$ be the length $\left( {40cm} \right)$, which is the balance point is obtained:
$\dfrac{{{R_1}}}{{{R_2}}} = \dfrac{l}{{100 - l}}$
Putting the values we get,
$\Rightarrow$ $\dfrac{{{R_1}}}{{{R_2}}} = \dfrac{{40}}{{100 - 40}}$
On subtracting the denominator term and we get,
$\Rightarrow$ $\dfrac{{40}}{{60}}$
On dividing we get,
$\Rightarrow$ $\dfrac{2}{3}$
If a left gap equal to that resistance and it is shunted across itself, so ${R_1}$with ${R_1}$
$ \Rightarrow $${R_{eff}} = \dfrac{{{R_1}}}{2}$
$ \Rightarrow $$\dfrac{{{R_1}}}{{2{R_2}}} = \dfrac{l}{{100 - l}}$
$\therefore $${R_1}$ is equate with ${R_1}$
$ \Rightarrow $$\dfrac{1}{2}\dfrac{{{R_1}}}{{{R_2}}} = \dfrac{l}{{100 - l}}$
Putting the values and we get,
$\dfrac{1}{2} \times \dfrac{2}{3} = \dfrac{l}{{100 - l}}$
Taking cross multiplication we get,
$\Rightarrow$ $3l = 100 - l$
Taking \[l\] as LHS and adding the term we get,
$\Rightarrow$ $3l + l = 100$
After doing adding we get,
$\Rightarrow$ $4l = 100$
On dividing \[4\] on both sides, we get
$\Rightarrow$ $l = \dfrac{{100}}{4}$
Let us divide the term we get,
$\Rightarrow$ $l = 25cm$
Hence the correct answer is $\left( C \right)$.

Note: As the racer slides over the wire AC, it shows zero avoidance at the adjusting point (invalid point). On the off chance that the length AB is at that point, the length BC is \[\left( {100 - l} \right)\].