A wire of resistance $$4\,\Omega $$ is stretched to twice its original length. The resistance of stretched wire would be
A.
$$2\,\Omega $$
B.
$$4\,\Omega $$
C.
$$8\,\Omega $$
D.
$$16\,\Omega $$
Answer :
$$16\,\Omega $$
Solution :
As the resistance of stretched wire to length $$n$$ times of original length is
$$R' = {n^2}R = {2^2} \times 4 = 4 \times 4 = 16\,\Omega $$
where, $$R =$$ original resistance
$${R'} =$$ final resistance
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.