JEE Advanced 2021 · Paper 2 · Physics Q17

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PhysicsInteger+4Question 17 of 57

In order to measure the internal resistance $r_1$ of a cell of emf $E$ , a meter bridge of wire resistance $R_0 = 50\ \Omega$ , a resistance $R_0/2$ , another cell of emf $E/2$ (internal resistance $r$ ) and a galvanometer G are used in a circuit, as shown in the figure. If the null point is found at $l = 72$ cm, then the value of $r_1 =$ ____ $\Omega$ .

Meter bridge: a uniform wire (resistance R0) between two end points. An upper branch joins the same two ends through the cell E (internal resistance r1) followed by the resistance R0/2. From the junction between r1 and R0/2, a second cell E/2 (internal resistance r) and a galvanometer G lead down to a jockey on the wire at distance l from the left end.

Your answer

Answer and worked solution

Official answer: 3

Current in the main loop. At the null point no current flows through the $E/2$ cell. So $E$ drives a single current around $r_1$ , $R_0/2$ and the wire $R_0$ :
$I=\frac E{r_1+R_0/2+R_0}=\frac E{r_1+75}.$

Null condition. The potential difference between the junction (after $r_1$ ) and the jockey point equals $E/2$ . That difference is the drop across $r_1$ plus the drop along the 72 cm of wire:
$\frac E2=I\left(r_1+0.72\times50\right)=\frac{E(r_1+36)}{r_1+75}.$

Solve. $r_1+75=2r_1+72$ , so $r_1=3\ \Omega$ .

Topics:Potentiometer Kirchhoff's laws EMF and internal resistance

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JEE Advanced 2021 · Paper 2, Physics question 17