Question

A potentiometer wire has length $4m$ and resistance $8\Omega$. The resistance that must be connected in series with the wire and an accumulator of emf 2V, so as to get a potential gradient $1mV$ per cm on the wire is
(A). $44\Omega$
(B). $48\Omega$
(C). $32\Omega$
(D). $40\Omega$

Answer 1

Hint- We will take the help from the formula for the potential gradient to find the unknown resistance. Formula for the potential gradient is voltage divided by the length of the wire. By putting the known values from the questions in the formula we will get the value of the resistance which is to be connected in series with the wire of the potentiometer and the accumulator.

Complete step-by-step answer:
Here in the question we have to know the value of the resistance that must be connected in series with the wire and an accumulator of emf 2V.
So, let the required value of the resistance is $R$.
Then after connection the equivalent resistance of the circuit
${R_{eq}} =$ $8\Omega + R\Omega$-----equation (1)
Then the current in the circuit,
$i = \dfrac{V}{{{R_{eq}}}}$
Here it is given that the accumulator is having the value of voltage=$V =$ $2Volt$
So, putting the value to get equation (2), as follows
$i = \dfrac{2}{{8 + R}}$------equation (2)
Now the potential drop across the given potentiometer,
${V_p} = i{R_p}$, where, ${V_p} =$Potential drop across potentiometer.
${R_p}$= Resistance of the potentiometer= $8\Omega$
So now,
${V_p} = \dfrac{2}{{8 + R}} \times 8$
We know the formula for the potential gradient, that is
Potential gradient =$\dfrac{{potential}}{{length}}$
So here potential gradient for the potentiometer= $\dfrac{{{V_p}}}{{{l_p}}}$
Where,${l_p} =$ length of the wire of potentiometer $= 4m$
So potential gradient =$\dfrac{{\dfrac{2}{{8 + R}} \times 8}}{4}$
$= \dfrac{{16}}{{4(8 + R)}} = \dfrac{4}{{(8 + R)}}$ volt per meter
And also there is given in the question the required potential gradient = $1mV$per $cm$=$0.1V$per $m$.
So, with the Above discussion finally we get the following relation to get the value of the $R$.
$\dfrac{4}{{(8 + R)}} = 0.1$
$\Rightarrow \dfrac{4}{{0.1}} = (8 + R)$
$\Rightarrow 40 = (8 + R)$
$\Rightarrow 40 - 8 = R$
$\Rightarrow R = 32\Omega$
Hence the required value of the resistance is $32\Omega$.
So, option (C) is the correct answer.

Note- Connecting the resistance in series, the current will be the same for the resistance as the potentiometer. Whereas the voltage across the resistor will be different from the voltage across the potentiometer.
The potential gradient is directly proportional to the voltage and inversely proportional to the length of the wire.