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In a meter bridge two gaps in the metallic strip are connected by $3 \Omega$ and $9 \Omega$ resistors. What should be the value of shunt that needs to be added to $9 \Omega$ resistor to shift balancing point by $25 \mathrm{~cm}$ ?
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The correct answer is:
$4.5 \Omega$
The given situation is shown below. Suppose
bridge is balanced at a distance $l$ from one end.
Initially in balance condition,
When a shunt $S^{\prime} \Omega$ is connected to $9 \Omega$ resistor balance point shifts by $25 \mathrm{~cm}$.
So, new balance length $=25+25=50 \mathrm{~cm}$
$\frac{3}{l}=\frac{9}{100-l} \Rightarrow l=25 \mathrm{~cm}$
As meter bridge is balanced at centre point. This means resistances of left and right gap are equal.
$\Rightarrow \quad 3=\frac{9 S}{9+S} \Rightarrow S=4.5 \Omega$
bridge is balanced at a distance $l$ from one end.
Initially in balance condition,
When a shunt $S^{\prime} \Omega$ is connected to $9 \Omega$ resistor balance point shifts by $25 \mathrm{~cm}$.
So, new balance length $=25+25=50 \mathrm{~cm}$
$\frac{3}{l}=\frac{9}{100-l} \Rightarrow l=25 \mathrm{~cm}$
As meter bridge is balanced at centre point. This means resistances of left and right gap are equal.
$\Rightarrow \quad 3=\frac{9 S}{9+S} \Rightarrow S=4.5 \Omega$
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