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When a potentiometer is connected between the points $\mathrm{A}$ and $\mathrm{B}$ as shown in the circuit, balance point is obtained at $64 \mathrm{~cm}$. When it is cornected between $\mathrm{A}$ and $\mathrm{C}$, the balance point is $8 \mathrm{~cm}$. If the potentiometer is connected between $\mathrm{B}$ and $\mathrm{C}$ the balance point will be
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Verified Answer
The correct answer is:
$56cm$
Let the pot-ntial gradient of the potentiometer be $\mathrm{x}$.
$\begin{aligned}
& x=\frac{E_1}{64} \\
& \Rightarrow E_1=64 x ... (i)
\end{aligned}$
When it is connected between $\mathrm{A}$ and $\mathrm{C}$
$\mathrm{E}_1-\mathrm{E}_2=8 \mathrm{x}... (ii)$
Solving equation (i) and (ii)
$\begin{aligned}
& 64 x-E_2=8 x \\
& \Rightarrow E_2=56 x \\
& \Rightarrow x=\frac{E_2}{56}
\end{aligned}$
Therefore, when potentiometer is connected between B and $\mathrm{C}$ then the balance point will $56 \mathrm{~cm}$.
$\begin{aligned}
& x=\frac{E_1}{64} \\
& \Rightarrow E_1=64 x ... (i)
\end{aligned}$
When it is connected between $\mathrm{A}$ and $\mathrm{C}$
$\mathrm{E}_1-\mathrm{E}_2=8 \mathrm{x}... (ii)$
Solving equation (i) and (ii)
$\begin{aligned}
& 64 x-E_2=8 x \\
& \Rightarrow E_2=56 x \\
& \Rightarrow x=\frac{E_2}{56}
\end{aligned}$
Therefore, when potentiometer is connected between B and $\mathrm{C}$ then the balance point will $56 \mathrm{~cm}$.
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