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**Hint:**The Metre Bridge is an arrangement derived from the Wheatstone’s network. The Wheatstone’s network is an arrangement of 4 resistors as shown:

The Wheatstone’s network is said to be balanced if the reading in the galvanometer becomes zero. This condition is possible when –

$\dfrac{P}{Q} = \dfrac{R}{S}$

**Complete step by step answer:**

Step 1: Calculating the unknown resistance.

In a meter bridge, the ratio of the known resistance to the unknown resistance is equal to the ratio of the corresponding lengths.

If the galvanometer reading turns 0 when the jockey is pressed at a length $l$, the formula for finding the unknown resistance R –

$\dfrac{R}{S} = \dfrac{l}{{1 - l}}$

where S – known resistance.

In this problem,

$

\Rightarrow S = 90\Omega \\

\Rightarrow l = 40cm = 0.4m \\

$

Hence, substitute these values to obtain the unknown resistance R –

$

\Rightarrow \dfrac{R}{S} = \dfrac{l}{{1 - l}} \\

\Rightarrow R = S\dfrac{l}{{1 - l}} \\

Solving, \\

\Rightarrow R = 90\dfrac{{0.4}}{{1 - 0.4}} \\

\Rightarrow R = 90\dfrac{{0.4}}{{0.6}} = 60\Omega \\

$

Step 2: Finding the change in resistance

Let us consider the equation for unknown resistance in the meter bridge

$R = S\dfrac{l}{{1 - l}}$

To find the incremental value of change in resistance, we need to differentiate the equation with respect to length

$

\Rightarrow \dfrac{{dR}}{{dl}} = R\left( {\dfrac{1}{l} + \dfrac{1}{{1 - l}}} \right) \\

rearranging, \\

\Rightarrow \dfrac{{dR}}{R} = \dfrac{{dl}}{l} + \dfrac{{dl}}{{1 - l}} \\

$

Least count,$dl = 1mm = 0.001m$

Substituting in the formula, we get:

$

\Rightarrow \dfrac{{dR}}{R} = \dfrac{{dl}}{l} + \dfrac{{dl}}{{1 - l}} \\

\Rightarrow \dfrac{{dR}}{R} = \dfrac{{0.001}}{{0.4}} + \dfrac{{0.001}}{{1 - 0.4}} \\

\Rightarrow \dfrac{{dR}}{R} = \dfrac{{0.001}}{{0.4}} + \dfrac{{0.001}}{{0.6}} = \dfrac{{0.01}}{4} + \dfrac{{0.01}}{6} \\

\Rightarrow \dfrac{{dR}}{R} = 0.25\Omega \\

$

Hence, incremental change = $ \pm 0.25\Omega $

**Hence, the correct option is Option C, $60 \pm 0.25\Omega.$**

**Note:**Students can, often, get confused between R and S in the formula. Always, note that the known resistance will be in the denominator as it is connected on the left. So, the known resistance S will be equivalent to (1-l).

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