Question

The circuit as shown in figure is used to compare the emf of two cells of 20v and30v.the null point is at point j when galvanometer s connected to 20 v battery .when the galvanometer is connected to 30v battery ,the null point is

Answer 1

Answer:

i hope it's help you to answer this question

Answer 2

The correct answer is option (2). Right of J.

Given:

The emf of the first cell, $E_1 = 20 V$.

The emf of the second cell, $E_2 = 30 V$.

The balancing length of the first cell, $l_1$ is at the point J.

To Find:

We have to find the balancing length of the second cell with respect to the point J.

Solution:

A potentiometer is the device that is used to compare the emf of two cells.

The balancing length of a potentiometer is the point on the slide wire where the galvanometer shows a null point or a zero deflection.

The equation for comparing the emf of 2 cells using a potentiometer is given by,

$\frac{E_1}{E_2} = \frac{l_1}{l_2}$

Substituting the values for $E_1$ and $E_2$ in the above equation, we get,

$\frac{20}{30} = \frac{l_1}{l_2}$

On rearranging, the above equation becomes,

$l_2= \frac{3}{2}.l_1 = 1.5 l_1$.

The value of $l_2$ is 1.5 times the value of $l_1$. That is, the null point is at the right of J.

Hence, when the galvanometer is connected to a 30 V battery ,the null point is at the right of J.

#SPJ2