Question

A 2V battery, a $990\Omega $ resistor, and a potentiometer of $2m$ length, all are connected in series of the resistance of the potentiometer wire is 10 ohm, then the potential gradient of the potentiometer wire is
A) $0.05V{m^{ - 1}}$
B) $0.5V{m^{ - 1}}$
C) $0.01V{m^{ - 1}}$
D) $0.1V{m^{ - 1}}$

Answer 1

Hint: A potentiometer is also called as a variable resistor. The potentiometer works on the principle that when a steady current flows through a wire of uniform cross-sectional area then the potential difference across two points of the wire is directly proportional to the length of the wire.

Complete step by step answer:
A potentiometer acts as an adjustable voltage divider. The potential gradient is defined as the potential drop per unit length of the wire.
when $l = 2m$
$R = 10\Omega $
Resistor ​$ = 990$,
$E = 2V$
Potential gradient of wire (k) is given by
$k = \dfrac{V}{l}$
Where k is the potential gradient
$V$ is the voltage
$l$ is the length
since the current in the potentiometer is given by
​​$I = \dfrac{E}{{R + {R_e}}}$
Also from Ohm’s Law, it is known that the voltage is directly proportional to the current and resistance of the circuit.
​​$V = IR = \dfrac{{ER}}{{R + {R_e}}}$
Substituting all the values and solving, the equation becomes,
​$\Rightarrow \dfrac{{2 \times 10}}{{10 + 990}} = \dfrac{{20}}{{1000}}$
$\Rightarrow V = 2 \times (10 - 2)$
Substitute the value of V in the formula of potential gradient,
$\Rightarrow k = \dfrac{{2 \times (10 - 2)}}{2}$
$\Rightarrow k = 0.01V{m^{ - 1}}$

Therefore, the potential gradient of the potentiometer wire is $0.01V{m^{ - 1}}$. Hence Option C is the right answer.

Note:
Sometimes there can be confusion between the voltmeter and potentiometer. This is because they both are used as the voltage measuring devices. But they are actually different. The potentiometer is used to measure the e.m.f of the circuit but the voltmeter is used to measure the voltage across the terminals in the circuit. Also, the voltage measured by the potentiometer has a high rate of accuracy, whereas the voltmeter provides an approximation of the voltage applied across the circuit.