Question

A uniform conductor of resistance R is cut into $10$ equal pieces. Half of them are joined in series and remaining half of them are connected in parallel, if these two combinations are joined in series, then the effective resistance of one piece is
A)$\dfrac{{13}}{{25}}R$
B)$\dfrac{R}{2}$
C)$\dfrac{R}{{15}}$
D)$\dfrac{{25}}{{11}}R$

Answer 1

Hint: The resistance of a conductor of some area and length is directly proportional to the length of the conductor and cross-sectional area of the conductor. Use the formula of series and parallel resistance after finding the resistance of one piece.

Complete step by step answer:
We know that,
The resistance of a conductor depends upon the cross-sectional area and and the length of the conductor as
$R = \rho \dfrac{l}{A}$
Where,
$\rho = $resistivity of the material
$l = $ length of the conductor
$A = $cross-sectional area of the conductor.
When a wire of resistance R is cut into $10$parts, the area of each part is the same but the length of each part is one tenth of the original length. So, the resistance of one part will also become one tenth of the original resistance.
$r = \dfrac{R}{{10}}$
Now, it is given in the question that five parts are connected in series,
Resistance of first five parts,${r_1} = 5 \times \dfrac{R}{{10}} = \dfrac{R}{2}$
The next five parts are connected in parallel,
Resistance of last five parts,${r_2} = \dfrac{1}{5} \times \dfrac{R}{{10}} = \dfrac{R}{{50}}$
Now,
Both ${r_1}$and ${r_2}$are connected in series,
So, the net resistance $ = {r_1} + {r_2}$
$ = \dfrac{R}{2} + \dfrac{R}{{50}}$
$ = \dfrac{{13}}{{25}}R$
A)is correct.

Note: It is given in the question that the conductor is uniform. If the conductor is non uniform then we cannot distribute the resistance in each part directly because it depends on both length and area.