Question

A potentiometer wire 10m long has a resistance of$40\Omega$. It is connected in series with a resistance box and a 2V storage cell. If the potential gradient along the wire is 0.1mV/cm, the resistance unplugged in the box is:
A. $260\Omega$
B. $760\Omega$
C. $960\Omega$
D. $1060\Omega$

Answer 1

Let us recall the definition of potential gradient and then substitute for potential difference in that expression using the ohm's law. Thus, you will get the value of current in that wire. Now you could get the current in terms of the unplugged resistance and then equate it to the above current and hence find the answer.

Complete step by step answer:
In the question, we are given a potentiometer wire of 10m long with $40\Omega$ and this is connected in series with a resistance box and a 2V cell. We are supposed to find the resistance unplugged in the box for the potential gradient along the wire as$0.1mV/cm$.
We know that potential gradient along the length of the wire by definition is given by $\dfrac{dV}{dl}$
And we are given its value to be,
$\dfrac{dV}{dl}=0.1mV/cm$
From Ohm’s law,
$V=IR$
$\Rightarrow \dfrac{I(40)}{10}=0.1mV/cm$
Where, I is the current in the wire.
$\Rightarrow I=\dfrac{1}{400}A$
If we take R to be the resistance that is unplugged in the box when it is connected in series, then, we could give the current as,
$I'=\dfrac{2}{R+40}$
As the connection between the potentiometer and resistance is in series, we could conclude that the same current flows through both. So,
$I=I'$
$\Rightarrow \dfrac{2}{R+40}=\dfrac{1}{400}$
$\Rightarrow R+40=800$
$\therefore R=800-40=760\Omega$

So, the correct answer is “Option b”.

Note: We should have a basic understanding that for a series connection, the current will be the same in every element in that connection and the voltage will be different. While for the parallel connection, the current is different in different elements while the voltage remains the same as it is measured between two points.