Q163 / 180 · COMEDK 2020 · Physics
To the potentiometer wire of length $L$ and $10 \Omega$ resistance, a battery of emf $2.5 \mathrm{~V}$ and a resistance $R$ are connected in series. If a potential difference of $1 \mathrm{~V}$ is balanced across $L$ / 2 length, the value of $R$ in $\Omega$ will be
- A1
- B$1.5$
- C2
- Correct answerD$2.5$
Solution
The given situation is shown below.
Resistance of potentioneter wire, $R_{p}=10 \Omega$ BmI, $E=25 \mathrm{~V}$ Total resistance, $R^{\prime}=R_{p}+R=(10+R) \Omega$ Current through potentiometer wire, $$ l=\frac{E}{R_{p}}=\frac{25}{10+R} $$ Since, potential difference across length $A L=\frac{L}{\mathrm{~g}}$, $$ \begin{aligned} & V &=l\left(\frac{R_{g}}{2}\right) \\ \Rightarrow & & 1 &=\frac{25}{10+R} \times \frac{10}{2} \\ \Rightarrow \quad & 10+n &=125 \\ \Rightarrow & & R &=25 \Omega \end{aligned} $$
Resistance of potentioneter wire, $R_{p}=10 \Omega$ BmI, $E=25 \mathrm{~V}$ Total resistance, $R^{\prime}=R_{p}+R=(10+R) \Omega$ Current through potentiometer wire, $$ l=\frac{E}{R_{p}}=\frac{25}{10+R} $$ Since, potential difference across length $A L=\frac{L}{\mathrm{~g}}$, $$ \begin{aligned} & V &=l\left(\frac{R_{g}}{2}\right) \\ \Rightarrow & & 1 &=\frac{25}{10+R} \times \frac{10}{2} \\ \Rightarrow \quad & 10+n &=125 \\ \Rightarrow & & R &=25 \Omega \end{aligned} $$
Hence, the correct answer is option: 'D'.