A cell balances against a length of 150 cm o | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The internal resistance $r$ of a cell is given by the formula: $r = R \left( \frac{l_0 - l}{l} \right)$,where $l_0$ is the balancing length in an

Vedclass · Accepted Answer

$210$

Vedclass · Answer

(C) The internal resistance $r$ of a cell is given by the formula: $r = R \left( \frac{l_0 - l}{l} \right)$,where $l_0$ is the balancing length in an open circuit and $l$ is the balancing length when shunted by resistance $R$.Case $1$: $R_1 = 5 \ \Omega$,$l_1 = 150 \ cm$.$r = 5 \left( \frac{l_0 - 150}{150} \right) = \frac{l_0 - 150}{30} \quad \dots (1)$Case $2$: $R_2 = 10 \ \Omega$,$l_2 = 150 + 25 = 175 \ cm$.$r = 10 \left( \frac{l_0 - 175}{175} \right) = \frac{2(l_0 - 175)}{35} \quad \dots (2)$Equating $(1)$ and $(2)$:$\frac{l_0 - 150}{30} = \frac{2(l_0 - 175)}{35}$$\frac{l_0 - 150}{6} = \frac{2(l_0 - 175)}{7}$$7(l_0 - 150) = 12(l_0 - 175)$$7l_0 - 1050 = 12l_0 - 2100$$5l_0 = 1050$$l_0 = 210 \ cm$.

Similar Questions

The null point of a potentiometer with a cell of emf $\varepsilon$ is obtained at a distance $l$ on the wire,then

With a potentiometer,null points were obtained at $140 \, cm$ and $180 \, cm$ with cells of $emf$ $1.1 \, V$ and one unknown $X \, V$. The unknown $emf$ is .............. $V$.

For the measurement of potential difference,a potentiometer is preferred in comparison to a voltmeter because:

Vedclass Test Series

Exam Paper Generator

Online Exam Module