The sensitivity of a galvanometer is 60 divisions/amp. When a shunt is used, its sensitivity becomes 10 divisions/amp. If galvanometer is of resistance 20 Ω, the value of shunt used is :

Given information:

Original sensitivity of the galvanometer, $S_1=60divisions/amp$

New sensitivity of the galvanometer with shunt, $S_2=10divisions/amp$

Resistance of the galvanometer, $G = 20\Omega$

Let's denote the resistance of the shunt as S (in ohms).

Using the formula for sensitivity, we have:

$Sensitivity \left(S\right) = \frac{Number of divisions}{Current \left(I\right)}$

$S_1=60div/A$

$S_2=10div/A$

We can set up the equation:

$S_1 =\frac{I_g}{G+S}\_\_\_\_\_\_\_\_\left(1\right)$ 

$S_2 =\frac{I_g}{S} \_\_\_\_\_\_\_\_\_\left(2\right)$

where Ig is the current passing through the galvanometer.

Dividing equation (1) by equation (2), we get:

$\frac{S_1}{S_2} = \frac{G + S}{S}$

Substituting the given values, we get:

$\frac{60}{10} = \frac{20 + S }{S}$

$6 =\frac{20 + S}{S}$

$6S = 20 + S$

$5S = 20$

$S = 4\Omega$

Therefore, the value of the shunt resistance used is 4Ω.