Question

In an experiment to measure the internal resistance of a cell by a potentiometer, it is found that the balance point is at a length of $2\text{m}$, when the cell is shunted by a $5\Omega$ resistance and at a length of $3\text{m}$, when the cell is shunted by a $10\Omega$ resistance. The internal resistance of the cell is

• A. $1.5\Omega$
• B. $10\Omega$
• C. $15\Omega$
• D. $1\Omega$

Answer 1

The correct option is B $10\Omega$

Let $x$ be potential gradient of the wire

When the cell is shunted by $5\Omega$ then.
$\frac{5}{5+r}E=x\left(2\right)$
When the cell is shunted by $10\Omega$ then,
$\frac{10}{10+r}E=x\left(3\right)$
By dividing both the equations, we get
$\frac{10+r}{2(5+r)}=\frac{2}{3}$
or $30+3r=20+4r\Rightarrow r=10\Omega$