A potentiometer consists of a wire of length $$4\,m$$ and resistance $$10\,\Omega .$$ It is connected to a cell of emf $$2\,V.$$ The potential gradient of the wire is
A.
$$0.5\,V/m$$
B.
$$2\,V/m$$
C.
$$5\,V/m$$
D.
$$10\,V/m$$
Answer :
$$0.5\,V/m$$
Solution :
Potential gradient $$ = \frac{{{\text{Potential}}\,{\text{applied}}}}{{{\text{Length of wire}}}}$$
$$\therefore k = \frac{V}{l}$$
Given, $$V = 2\,V,\,l = 4\,m$$
$$\therefore k = \frac{2}{4} = 0.5\,V/m$$
Releted MCQ Question on Electrostatics and Magnetism >> Electric Current
Releted Question 1
The temperature coefficient of resistance of a wire is 0.00125 per $$^ \circ C$$ At $$300\,K,$$ its resistance is $$1\,ohm.$$ This resistance of the wire will be $$2\,ohm$$ at.
The electrostatic field due to a point charge depends on the distance $$r$$ as $$\frac{1}{{{r^2}}}.$$ Indicate which of the following quantities shows same dependence on $$r.$$
A.
Intensity of light from a point source.
B.
Electrostatic potential due to a point charge.
C.
Electrostatic potential at a distance r from the centre of a charged metallic sphere. Given $$r$$ < radius of the sphere.