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**Hint**From the diagram, we can see that the battery, the galvanometer, and the resistance are in series with each other. The internal resistance of the galvanometer, $r$ is $25\Omega $. The initial resistance in series is $3000\Omega $. Thus we apply Kirchhoff’s Voltage Law in the above circuit to get the current flowing through the circuit. Next, we insert a new resistance $x$ in series with the previous resistance to get the new current flow.

**Complete Step by step solution**Applying Kirchhoff’s Voltage Law in the above circuit, we get

$2V = iR + ir$ .

The values of $R = 3000\Omega $, and $r = 25\Omega $ are given.

Substituting these values in the given equation, we get

$ \Rightarrow 2V = i(3000\Omega + 25\Omega )$

$ \Rightarrow i = \dfrac{2}{{3025}}A$

This value is described as $30units$ when the deflection of the galvanometer is at full scale.

Thus if $i = \dfrac{2}{{3025}}A = 30 units$ then

$1unit$ is $\dfrac{2}{{3025 \times 30}}A$.

Similarly, deflection $20units$ is equal to $\dfrac{{2 \times 20}}{{3025 \times 30}}A$current flow through the circuit which is the new current $I$ .

Thus keeping the numerator same, the denominator turns out to be

$\dfrac{2}{{3025 \times 1.5}}A = \dfrac{{2V}}{{4537.5\Omega }}$

; i.e. the denominator turns out to be $(4512.5 + 25)\Omega $.

We now get the required net resistance of the new circuit to be $4537.5\Omega $, which is including the internal resistance of the galvanometer.

This is because we can compare this equation with the K.V.L. equation of the new circuit, where $2V = I(r + R) = \dfrac{{2 \times 20}}{{3025 \times 30}}(25\Omega + R)$.

On further simplifying we get,

$ \Rightarrow 2 \times \dfrac{{3025 \times 30}}{{2 \times 20}} = 25 + R$

$ \therefore R + 25\Omega = (4512.5 + 25)\Omega $

Thus we get the required value of the new resistance $R$ as $4512.5\Omega $.

**Therefore the correct option is not present in the given question.**

**Note**A galvanometer measures the amount of current flowing through a given circuit. Here the galvanometer resistance is constant since we use the same galvanometer for both the readings. The value of the resistance $R$ changes as we wish to decrease the value of current passing through the galvanometer.

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