In a metre bridge experiment the balance point is obtained if the gaps are closed by 2$\Omega$ and 3$\Omega$. A shunt of X $\Omega$ is added to 3$\Omega$ resistor to shift the balancing point by 22.5 cm. The value of X is ___________.
Answer (integer)
2
Solution
$\frac{1}{100-1}=\frac{2}{3}$
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$\Rightarrow I=40 \mathrm{~cm}$
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as $3 \Omega$ is shunted the balance point will shift towards $3 \Omega$. So, new length $l^{\prime}=22.5+I=62.5$
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So, $\frac{62.5}{37.5}=\frac{2}{3 x}(3+x)$
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$\Rightarrow x=2 \Omega$
About this question
Subject: Physics · Chapter: Current Electricity · Topic: Wheatstone Bridge
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