In a potentiometer arrangement, a cell of emf 1 . 20   V gives a balance point at 36   cm length of wire. This cell is now replaced by another cell of emf 1 . 80   V . The difference in balancing length of potentiometer wire in above conditions will be _____ cm .

Question

In a potentiometer arrangement, a cell of emf 1 . 20 &#160; V gives a balance point at 36 &#160; cm length of wire. This cell is now replaced by another cell of emf 1 . 80 &#160; V . The difference in balancing length of potentiometer wire in above conditions will be _____ cm .

Aicharya · Accepted Answer

In a potentiometer, null point or balance point is defined as the point where the deflection of the galvanometer is zero. If emf of the unknown battery is $ε$ then,
$ε =$ (Potential Gradient) $\times$ (Balancing length $l$ ). Therefore,
$\frac{ε_{1}}{ε_{2}} = \frac{l_{1}}{l_{2}}$
$\Rightarrow \frac{1.20}{1.80} = \frac{36}{l_{2}}$
$\Rightarrow l_{2} = 18 \times 3 = 54 cm$
Hence, difference in balancing length is $l_{2} - l_{1} = 54 - 36 = 18 cm$

In a potentiometer arrangement, a cell of emf 1 . 20   V gives a balance point at 36   cm length…

Solution

In a potentiometer arrangement, a cell of emf 1 . 20 &#160; V gives a balance point at 36 &#160; cm length…

Solution

In a potentiometer arrangement, a cell of emf 1 . 20 V gives a balance point at 36 cm length…