(C) In a meter bridge,the balance condition is given by $\frac{R_1}{R_2} = \frac{l}{100-l}$.Given $\frac{X}{Y} = \frac{40}{100-40} = \frac{40}{60} = \

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(C) In a meter bridge,the balance condition is given by $\frac{R_1}{R_2} = \frac{l}{100-l}$.Given $\frac{X}{Y} = \frac{40}{100-40} = \frac{40}{60} = \frac{2}{3}$.So,$X = \frac{2}{3}Y$.Now,we balance $3X$ against $Y$. Let the new null point be at $l' \ cm$.Then,$\frac{3X}{Y} = \frac{l'}{100-l'}$.Substituting $X = \frac{2}{3}Y$ into the equation:$\frac{3(\frac{2}{3}Y)}{Y} = \frac{l'}{100-l'}$$2 = \frac{l'}{100-l'}$$2(100 - l') = l'$$200 - 2l' = l'$$3l' = 200$$l' = \frac{200}{3} \approx 66.67 \ cm$.Rounding to the nearest integer,the null point is close to $67 \ cm$.

Class 12 Physics Current Electricity Meter Bridge

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In a meter bridge experiment,a null point is obtained at $40 \ cm$ from one end of the wire when resistance $X$ is balanced against another resistance $Y$. If $X < Y$,then the new position of the null point from the same end,if one decides to balance a resistance of $3X$ against $Y$,will be close to .............. $cm$.

JEE MAIN 2013 All JEE MAIN

A
$80$
B
$75$
C
$67$
D
$50$

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Current Electricity

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In a meter bridge,the point $D$ is the neutral point (null point) as shown in the figure.

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In the meter bridge experiment,the length $AB$ of the wire is $1 \ m$. The resistors $X$ and $Y$ have values $5 \ \Omega$ and $2 \ \Omega$ respectively. When a shunt resistance $S$ is connected in parallel to $X$,the balancing point is found to be $0.625 \ m$ from $A$. Then,the resistance of the shunt $S$ is (in $\Omega$)

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In a metre bridge experiment, the balance point is obtained if the gaps are closed by $2 \text{ } \Omega$ and $3 \text{ } \Omega$. $A$ shunt of $X \text{ } \Omega$ is added to $3 \text{ } \Omega$ resistor to shift the balancing point by $22.5 \text{ cm}$. The value of $X$ is

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