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\[\begin{align}

& \text{A}\text{.}\dfrac{s}{3} \\

& \text{B}\text{.}\dfrac{s}{6} \\

& C.\dfrac{s}{9} \\

& D.\dfrac{s}{12} \\

\end{align}\]

Answer

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A galvanometer is said to be sensitive, if it can show large deflections even when a small current flows through the circuit and thus sensitivity of the galvanometer is the ratio of change in deflection $d\theta$ to change in the current \[dI\]. This can be represented mathematically as $S=\dfrac{d\theta}{dI}$ .

We also know that, if $s$ is the sensitivity of the galvanometer, then $s=\dfrac{G}{n-1}$ where $G$ is a constant, which is the character of galvanometer and $n$ is the range of the galvanometer, which is the number of divisions the pointer moves to detect the small current.

Then, we can write, $S=\dfrac{G}{n-1}$, also given that $s=\dfrac{G}{8}$

Then, $\dfrac{G}{8}=\dfrac{G}{n-1}$

Or, $n-1=8$

Or $n=9$

Clearly, the range of the galvanometer increased to $9$.i.e. for small current, it points away $9$ divisions, then the sensitivity is said to become $S=\dfrac{s}{9}$