A galvanometer is connected in series with a 280 \ resistance and a...

Let $G$ be the resistance of the galvanometer and $E$ be the EMF of the battery.

Initial current through the galvanometer is $I_1 = \dfrac{E}{280 + G}$.

Since deflection is proportional to current, $I_1 = 60k$, where $k$ is a constant of proportionality.

When a shunt of $28 \ \Omega$ is connected, the equivalent resistance of the galvanometer and shunt is $R_{eq} = \dfrac{28G}{28 + G}$.

Total resistance of the circuit becomes $R_{total} = 280 + \dfrac{28G}{28 + G}$.

The new main current is $I_2 = \dfrac{E}{280 + \dfrac{28G}{28 + G}}$.

Current through the galvanometer is $I_g = I_2 \times \dfrac{28}{28 + G} = \dfrac{28E}{280(28 + G) + 28G}$.

Given that the new deflection is $20$ divisions, $I_g = 20k$.

Dividing the two equations, we get:

$\dfrac{I_1}{I_g} = \dfrac{\dfrac{E}{280 + G}}{\dfrac{28E}{280(28 + G) + 28G}} = \dfrac{60k}{20k} = 3$

$\dfrac{280(28 + G) + 28G}{28(280 + G)} = 3$

$\dfrac{7840 + 308G}{7840 + 28G} = 3$

$7840 + 308G = 23520 + 84G$

$224G = 15680$

$G = 70 \ \Omega$

Answer: $70 \ \Omega$