A cell balances against a length of 150 cm on a potentiometer wire when it is shunted by a resistance of $5…

A cell balances against a length of 150 cm on a potentiometer wire when it is shunted by a resistance of $5 \Omega$. But when it is shunted by a resistance of $10 \Omega$, then balancing length increases by 25 cm . The balancing length when the cell is in an open circuit is
  1. 200 cm
  2. 225 cm
  3. 210 cm
  4. 250 cm

Solution

$\begin{array}{ll} & \mathrm{r}=\mathrm{s}\left(\frac{l-l_{\mathrm{b}}}{l_{\mathrm{b}}}\right) \\ \therefore \quad & 5\left(\frac{l-150}{150}\right)=10\left(\frac{l-175}{175}\right) \\ \therefore & \frac{l-150}{6}=2\left(\frac{l-175}{7}\right) \\ \therefore & 7 l-1050=12 l-2100 \\ \therefore & 5 l=1050 \\ \therefore \quad & l=210 \mathrm{~cm}\end{array}$

Asked in: MHT CET 2024 (15 May Shift 1)

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