A potentiometer has been connected between A and B and the balance point has been obtained at $203.6cm$. When the end of the potentiometer which has been connected to B is shifted to C, then the balance point has been found to be at $24.6cm$. Let us assume that the potentiometer is connected between B and C, then where will be the balance point?
$\begin{align}
& A.179.0cm \\
& B.197.2cm \\
& C.212.0cm \\
& D.228.0cm \\
\end{align}$
Answer
640.5k+ views
Hint: The potential gradient of the potentiometer can be found by taking the ratio of the potential to the length of the potentiometer wire. Using this find the potential gradient of the potentiometer wire. Similarly find the potential gradient at the second case also. Substitute and rearrange the equation. This will help you in answering this question.
Complete step by step answer:
Let the potential gradient of the potentiometer be $x$.
Therefore this value can be found by the equation,
$x=\dfrac{V}{L}$
Where $V$be the potential and $L$be the length of the potentiometer wire.
When the potentiometer has been connected between AB,
$\begin{align}
& v={{E}_{1}} \\
& L=203.6cm \\
\end{align}$
Substituting this value in the equation obtained above can be given as,
$\therefore x=\dfrac{{{E}_{1}}}{203.6}$
Therefore the potential will be,
$\therefore {{E}_{1}}=203.6\times x$……… (1)
When it is connected between AC then we can write that,
\[\begin{align}
& V={{E}_{1}}-{{E}_{2}} \\
& x=\dfrac{{{E}_{1}}-{{E}_{2}}}{L} \\
& \Rightarrow {{E}_{1}}-{{E}_{2}}=24.6x \\
\end{align}\]………… (2)
Substituting equation (1) from equation (2) can be written as,
\[\begin{align}
& V={{E}_{1}}-{{E}_{2}} \\
& 203.6x-{{E}_{2}}=24.6x \\
& {{E}_{2}}=\left( 203.6-24.6 \right)x \\
& \Rightarrow {{E}_{2}}=179x \\
\end{align}\]
Rearranging this equation can be shown as,
\[x=\dfrac{{{E}_{2}}}{179}\]
Therefore when the potentiometer is connected between B and C then the balance point will be found to be at \[179cm\].
So, the correct answer is “Option A”.
Note: A potentiometer is a device which is defined as a three-terminal resistor having a sliding or rotating contact that creates an adjustable voltage divider. When only two terminals are used, like one end and the other end as the wiper, it will be performing as a variable resistor known as rheostat.
Complete step by step answer:
Let the potential gradient of the potentiometer be $x$.
Therefore this value can be found by the equation,
$x=\dfrac{V}{L}$
Where $V$be the potential and $L$be the length of the potentiometer wire.
When the potentiometer has been connected between AB,
$\begin{align}
& v={{E}_{1}} \\
& L=203.6cm \\
\end{align}$
Substituting this value in the equation obtained above can be given as,
$\therefore x=\dfrac{{{E}_{1}}}{203.6}$
Therefore the potential will be,
$\therefore {{E}_{1}}=203.6\times x$……… (1)
When it is connected between AC then we can write that,
\[\begin{align}
& V={{E}_{1}}-{{E}_{2}} \\
& x=\dfrac{{{E}_{1}}-{{E}_{2}}}{L} \\
& \Rightarrow {{E}_{1}}-{{E}_{2}}=24.6x \\
\end{align}\]………… (2)
Substituting equation (1) from equation (2) can be written as,
\[\begin{align}
& V={{E}_{1}}-{{E}_{2}} \\
& 203.6x-{{E}_{2}}=24.6x \\
& {{E}_{2}}=\left( 203.6-24.6 \right)x \\
& \Rightarrow {{E}_{2}}=179x \\
\end{align}\]
Rearranging this equation can be shown as,
\[x=\dfrac{{{E}_{2}}}{179}\]
Therefore when the potentiometer is connected between B and C then the balance point will be found to be at \[179cm\].
So, the correct answer is “Option A”.
Note: A potentiometer is a device which is defined as a three-terminal resistor having a sliding or rotating contact that creates an adjustable voltage divider. When only two terminals are used, like one end and the other end as the wiper, it will be performing as a variable resistor known as rheostat.
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