Reduction factor of tangent galvanometer is related to number of turns as:
A. \[n\]
B. \[{n^2}\]
C. \[{n^{\dfrac{{ - 1}}{2}}}\]
D. \[{n^{ - 1}}\]

Hint: To find the required relation between reduction factor and the number of turns of a tangent galvanometer we are using the formula for reduction factor (k) in the term of the radius of a coil (r), the intensity of magnetic field strength (H) and the number of turns (n): $k = \dfrac{{2rH}}{{{\mu _0}n}}$.
Complete step-by-step answer:
To calculate the required relation between reduction factor and the number of turns of a tangent galvanometer.
The formula used: $k = \dfrac{{2rH}}{{{\mu _0}n}}$…………(i), where r is the radius of the coil, H is the intensity of magnetic field ${\mu _0}$absolute permeability, and n be the number of turns in the coil.

Assuming all the terms except n in the eqn (i) is fixed then, the eqn (i) can also be written as :
$k = \dfrac{C}{n},$ [Where we take $C = \dfrac{{2r}}{{{\mu _0}}}$] So, C is the constant term having a fixed value for our consideration.
$ \Rightarrow k \propto \dfrac{1}{n}$
$ \Rightarrow k \propto {n^{ - 1}}$

Hence, the correct answer is (D).

Note:In order to find these kinds of formula-based questions the key is to remember the various fact-based short formulas and their implementation while solving the tricky numerical problems. One should also remember that the obtained relation is valid only until we do not change the value of H and the radius of the coil of a tangent galvanometer.