A potentiometer wire has length $$4\,m$$ 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 $$1\,mV$$ per $$cm$$ on the wire is
A.
$$32\,\Omega $$
B.
$$40\,\Omega $$
C.
$$44\,\Omega $$
D.
$$48\,\Omega $$
Answer :
$$32\,\Omega $$
Solution :
Given, $$l = 4m,$$
$$R =$$ potentiometer wire resistance $$ = 8\,\Omega $$
Potential gradient $$ = \frac{{dV}}{{dr}} = 1\,mV/cm$$
So, for $$400\,cm,\,\Delta V = 400 \times 1 \times {10^{ - 3}} = 0.4\,V$$
Let $$a$$ resistor $${R_s}$$ connected in series, so as
$$\eqalign{
& \Delta V = \frac{V}{{R + {R_s}}} \times R \Rightarrow 0.4 = \frac{2}{{8 + R}} \times 8 \cr
& \Rightarrow 8 + R = \frac{{16}}{{0.4}} = 40 \Rightarrow R = 32\,\Omega \cr} $$
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.