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**Hint:**To solve this problem, use the formula for internal resistance. Substitute the values in the formula of internal resistance for $5 \Omega$ resistor and calculate the internal resistance for it. Then, use the same formula to find the internal resistance for $4 \Omega$ resistor. Substitute the internal resistance obtained above in this equation and find the unknown variable which is the new position of the balance point.

**Formula used:**

$r= \dfrac {{l}_{1}-{l}_{2}}{{l}_{2}}R$

**Complete answer:**

Given: ${l}_{1}= 84 cm$

${l}_{2}= 70 cm$

${R}_{1}= 5 \Omega$

${R}_{2}= 4 \Omega$

The formula for internal resistance is given by,

$r= \dfrac {{l}_{1}-{l}_{2}}{{l}_{2}}R$

Internal resistance when $5 \Omega$ resistor is connected is given by,

$r= \dfrac {{l}_{1}-{l}_{2}}{{l}_{2}}{R}_{1}$

Substituting values in above equation we get,

$r= \dfrac {84-70}{70}\times 5$

$\Rightarrow r= \dfrac {14}{70}\times 5$

$\Rightarrow r= 1 \Omega$

Internal resistance when $4 \Omega$ resistor is connected is given by,

$r= \dfrac {{l}_{1}-{l}_{3}}{{l}_{3}}{R}_{2}$

Substituting values in above equation we get,

$1= \dfrac {84-{l}_{3}}{{l}_{3}} \times 4$

$\Rightarrow {l}_{3}=(84-{l}_{3}) \times 4$

$\Rightarrow {l}_{3}= 336 – 4{l}_{3}$

$\Rightarrow 5{l}_{3}= 336$

$\Rightarrow {l}_{3}= \dfrac {336}{5}$

$\Rightarrow {l}_{3}= 67.2 cm$

Hence, when $5 \Omega$ resistor is changed by $4 \Omega$ resistor, the new position of the balance point is 67.2 cm.

**So, the correct answer is option C i.e. 67.2 cm.**

**Note:**

Internal resistance is the opposition offered by the cells and batteries to the flow of current flowing in the generation of heat. Internal resistance depends upon the nature of the material of the wire. The potentiometer is an arrangement that can be used to find the unknown value of resistances and the cell’s internal resistance. Potentiometer is also used to determine the unknown values of potential differences.

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