A galvanometer of resistance $40\,\Omega $ gives a deflection of 5 divisions per mA. There are $50\,{\text{divisions}}$ on the scale. The maximum current that can pass through it when a shunt resistance of ${\text{2}}\Omega $ is connected is
A. $210\,{\text{mA}}$
B. $155\,{\text{mA}}$
C. $420\,{\text{mA}}$
D.$75\,{\text{mA}}$

Hint: First of all, we have to find the value of galvanometer current which can be found out by dividing the total number of divisions by sensitivity. After that we will then apply the formula for the maximum current for the galvanometer, followed by manipulation accordingly to obtain the result.

Complete step by step answer:
Given,
Deflection of galvanometer is $5\,{\text{div/mA}}$
Galvanometer resistance ${R_{\text{g}}} = 40\,\Omega $
Value of shunt resistance ${R_{\text{s}}} = 2\,\Omega $
Galvanometer current
$
{I_{\text{g}}} = \dfrac{{{\text{Total}}\,{\text{number}}\,{\text{of}}\,{\text{division}}}}{{{\text{Sensitivity}}}} \\
\Rightarrow {I_{\text{g}}} = \dfrac{{50}}{5} \\
\Rightarrow {I_{\text{g}}} = 10\,{\text{mA}} \\
$
Maximum current formula is given by
$I{\text{ = }}\dfrac{{{R_{\text{g}}}{\text{ + }}{R_{\text{s}}}}}{{{R_{\text{g}}}}}{\times }{I_{\text{g}}}$ …… (1)
Here, $I$is the maximum current pass through the galvanometer, ${R_{\text{g}}}$is the galvanometer resistance, ${R_{\text{s}}}$ is the shunt resistance and ${I_{\text{g}}}$ is the galvanometer current.
Substitute the value${R_{\text{g}}} = 40\Omega $, ${R_{\text{s}}} = 2\Omega $, ${I_{\text{g}}} = 10\,{\text{mA}}$ in the equation (1) and we get:
In the given equation
$
  I{\text{ = }}\dfrac{{{R_{\text{g}}}{\text{ + }}{R_{\text{s}}}}}{{{R_{\text{g}}}}}{ \times }{I_{\text{g}}} \\
\Rightarrow I = \dfrac{{40 + 2}}{{40}} \times 10 \times {10^{ - 3}} \\
\Rightarrow I = 210\,{\text{mA}} \\
$
Hence, the maximum current that can pass through it when a shunt resistance of ${\text{2}}\Omega $ is connected is $210\,{\text{mA}}$ .

The correct option is (A) $210\,{\text{mA}}$ .

Note:
A galvanometer is a device that is used to detect small electric current or measure its magnitude. The current and its intensity is usually indicated by a magnetic needle’s movement or that of a coil in a magnetic field that is an important part of a galvanometer. Most of the students tend to make the mistake while writing the formula. It is important to remember that the net resistance is present in the numerator. The current which passes through the shunt resistor is always greater than the galvanometer. This is because the shunt resistor is a low resistance device.