
A $10m$ long potentiometer wire is connected to a battery having a steady voltage. A Leclanche cell is balanced at $4m$ length of the wire. If the length is kept the same, but its cross section is doubled, the null point will be obtained at:
$A)\text{ }8m$
$B)\text{ 4}m$
$C)\text{ 2}m$
D) None of these
Answer
511.8k+ views
Hint: Students must recall that in a potentiometer, the balance point is obtained as a function of the voltage drop per unit length of the potentiometer wire. In this case, the total voltage drop and the total length remain the same. Therefore, the voltage drop per unit length also remains the same.
Complete step-by-step answer:
In a potentiometer, the balance point of a cell is obtained when the voltage of a secondary cell (whose voltage has to be measured) is equal to the voltage drop in the potentiometer wire for the length of the balance point. Therefore, it can be said that the voltage of the secondary cell can be written as a function of the voltage drop per unit length of the potentiometer wire.
Now, in this case, it is given that the primary battery of the potentiometer gives a constant voltage and also the length of the wire does not change. Hence, it can be inferred that the voltage drop per unit length of the potentiometer wire will also remain constant. This means that the balance point will also remain the same and not change.
Therefore, the new null point is the same as that of the old one, that is, $4m$.
So, the correct answer is “Option B”.
Note: Students must note and not get confused upon seeing that the cross sectional area of the wire has doubled. It means that the resistance of the wire decreases and hence, the current flowing through it increases by the same factor so that their product, that is, the voltage drop across the whole potentiometer wire remains the same. Therefore, changing the cross sectional area of the wire has no effect on the balance point of the secondary cell on the potentiometer wire.
Complete step-by-step answer:
In a potentiometer, the balance point of a cell is obtained when the voltage of a secondary cell (whose voltage has to be measured) is equal to the voltage drop in the potentiometer wire for the length of the balance point. Therefore, it can be said that the voltage of the secondary cell can be written as a function of the voltage drop per unit length of the potentiometer wire.
Now, in this case, it is given that the primary battery of the potentiometer gives a constant voltage and also the length of the wire does not change. Hence, it can be inferred that the voltage drop per unit length of the potentiometer wire will also remain constant. This means that the balance point will also remain the same and not change.
Therefore, the new null point is the same as that of the old one, that is, $4m$.
So, the correct answer is “Option B”.
Note: Students must note and not get confused upon seeing that the cross sectional area of the wire has doubled. It means that the resistance of the wire decreases and hence, the current flowing through it increases by the same factor so that their product, that is, the voltage drop across the whole potentiometer wire remains the same. Therefore, changing the cross sectional area of the wire has no effect on the balance point of the secondary cell on the potentiometer wire.
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