The balancing length for a cell is 460 cm in a potential experiment. When an external resistance of  $10\Omega$ is connected in parallel to the cell. The balancing length changes by 60 cm. If the internal resistance of the cell is $\frac{N}{10}$ , where N is an integer,  the value of N is 

$$\begin{gathered} Let the emf of cell is ‘E’ internal resis\tan ce ‘r’ and potential gradient is k. \\ When Only cell is connected : E=460k\to \left(1\right) \\ After connecting the resistor : \frac{\text{E}\times 10}{10+\text{r}}=400k\to \left(2\right) \\ From \left(1\right) and \left(2\right) , \\ \frac{460k\times 10}{10+\text{r}}=400k \\ \Rightarrow 46=4\left(10+\text{r}\right) \\ \Rightarrow 6=4\text{r} \\ \Rightarrow \text{r}=\frac{6}{4}=\frac{3}{2}=1.5\Omega \\ Given : \text{r}=\frac{\text{N}}{10} \\ \Rightarrow \text{N}=10\times \text{\hspace{0.17em}r\hspace{0.17em}=10\hspace{0.17em}}\times 1.5=15 \end{gathered}$$