Question

A cell can be balanced against 110 cm and 100 cm of potentiometer wire, respectively with and without being short circuited through a resistance of $10\, \Omega.$ Its internal resistance is

• A. 2.0 ohm
• B. zero
• C. 1.0 ohm
• D. 0.5 ohm

Answer 1

Answer: (C)

1.0 ohm

Answer 2

[In the question, the length 110 cm & 100 cm are interchanged ] as $\varepsilon \frac{\varepsilon R}{R+r}$
Without being short circuited through R, only the battery $\varepsilon$ is balanced.
$\varepsilon =\frac{V}{L}\times l_1=\frac{V}{L} \times 110\,cm$ ....(i)
When R is connected across $\varepsilon ,$
$Ri =R \cdot\bigg(\frac{\varepsilon}{R+r}\bigg)=\frac{V}{L} \times l_2$
\, \, \, \Rightarrow \frac{R\varepsilon}{R+r}=\frac{V}{L} \times 100\hspace25mm.....(ii)
Dividing eqn. (i) and (ii), $\frac{R+r}{R}=\frac{110}{100}$
$\Rightarrow 1+ \frac{r}{R}=\frac{110}{100} \Rightarrow \times \frac{r}{R}=\frac{110}{100} -\frac{100}{100}$
$\Rightarrow r=R\cdot \frac{10}{100} =\frac{R}{10}$. As $R= 10\, \Omega;\,r=1\, \Omega$.