JEE MAIN - Physics (2023 - 29th January Morning Shift - No. 20)
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
2
Explanation
$\frac{1}{100-1}=\frac{2}{3}$
$\Rightarrow I=40 \mathrm{~cm}$
as $3 \Omega$ is shunted the balance point will shift towards $3 \Omega$. So, new length $l^{\prime}=22.5+I=62.5$
So, $\frac{62.5}{37.5}=\frac{2}{3 x}(3+x)$
$\Rightarrow x=2 \Omega$
$\Rightarrow I=40 \mathrm{~cm}$
as $3 \Omega$ is shunted the balance point will shift towards $3 \Omega$. So, new length $l^{\prime}=22.5+I=62.5$
So, $\frac{62.5}{37.5}=\frac{2}{3 x}(3+x)$
$\Rightarrow x=2 \Omega$
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