Question

A 10m long potentiometer wire has a potential gradient of 0.0025 v/cm along its length.calculate length of wire at which the null point is obtained for a 1.025v standard cell. also find emf of another cell for which null point is at 860 cm.what maximum potential difference can be measured by the potentiometer?

Answer 1

As per the data given,

Length of the potentiometer wire = 10m =(10×100) cm =1000cm

Potential gradient of the wire(K) = 0.0025 V/cm

EMF of the standard cell (E) = 1.025 V

We have to find the length of the wire at which the null point is obtained

As per the formula

If a length L of the potentiometer wire balances a cell of emf E, then

$E = K* L$

∴$L = \frac{E}{K}$

⇒ $L = \frac{1.025}{0.0025}$

⇒ L = 410cm

Hence the length of the wire at which the null point is obtained for a 1.025V standard cell is 410 cm.

For the second part

Null point (L₁) = 860 cm

K = 0.0025V/cm

Emf of the cell (E) = $K * L_{1}$

∴$E = (0.0025 *860) V$

∴E = 2.15 V

Hence emf of the cell is 2.15 V

For the third part

Maximum potential difference that can be measured = Potential gradient (K) × Length of the wire(l)

Maximum potential difference that can be measured = $(0.0025 * 1000) V$

∴  Maximum potential difference that can be measured = 2.5 V

Hence the maximum potential difference that can be measured by the potentiometer is 2.5V