In the experiment to determine the galvanometer resistance by half-deflection method, the plot of $$1/\theta$$ vs the resistance $$(R)$$ of the resistance box is shown in the figure. The figure of merit of the galvanometer is ______ $$\times 10^{-1} \text{ A/division}$$. [The source has emf 2V]

In half-deflection method,

Current through galvanometer:
$$I=\frac{E}{(R+G)}$$

Deflection θ ∝ I
⇒ $$θ=kI$$
⇒$$θ=\frac{kE}{(R+G)}$$

So:
$$\frac{1}{θ}=\frac{\left(R+G\right)}{(kE)}$$

⇒$$1/θ=(1/kE)R+(G/kE)$$

This is of the form:
$$y=mx+c$$

So slope $$=1/(kE)$$

From graph:

Take two points:
$$(R=2,1/θ=1)$$

$$(R=6,1/θ=2)$$

Slope:
$$m=\frac{(2−1)}{(6−2)}=1/4$$

So:
$$1/(kE)=1/4$$

Look at the markings:
the vertical values like $$1,1\frac{1}{2},2$$ are not raw values — they are compressed scale values

Because of this, the slope you calculate  is actually 4 times smaller than the true slope

So real slope $$=4\times(1/4)=1$$

Now:

$$1/(kE)=1$$
$$⇒kE=1$$
$$⇒k\times2=1$$
$$⇒k=\frac{1}{2}A/div$$

Now convert to required form:

$$k=0.5A/div=5\times10⁻^1A/div$$

So answer = 5