A potentiometer having a wire 10 m long stretched on it is connected to a battery having a steady voltage. A cell gives a null deflection point at 700 cm. If the length of potentiometer wire is increased by 2 m, then the new position of null point will be

Solution:

Let the initial length of the potentiometer wire be $L_1 = 10$ m = 1000 cm.

Let the initial null point be $l_1 = 700$ cm.

Let the potential difference across the potentiometer wire be $V$. The potential gradient is $k_1 = V/L_1$.

The EMF of the cell is $E = k_1 l_1 = (V/L_1) l_1$.

When the length of the potentiometer wire is increased by 2 m, the new length is $L_2 = 10 + 2 = 12$ m = 1200 cm.

The battery voltage V is steady, so the new potential gradient is $k_2 = V/L_2$.

The EMF of the cell E remains the same.

The new null point is $l_2$ such that $E = k_2 l_2 = (V/L_2) l_2$.

Equating the expressions for E:

$(V/L_1) l_1 = (V/L_2) l_2$

$(1/L_1) l_1 = (1/L_2) l_2$

$l_2 = (L_2 / L_1) l_1$

$l_2 = (1200 ext{ cm} / 1000 ext{ cm}) imes 700 ext{ cm}$

$l_2 = (1.2) imes 700 ext{ cm}$

$l_2 = 840 ext{ cm}$

Thus, the new position of the null point will be 840 cm.