Question

A certain piece of silver of given mass is to be made like a wire. Which of the following combination of length (L) and the area of cross-sectional (A) will lead to the smallest resistance.
A. L and A
B. 2L and A/2
C. L/2 AND 2A
D. Any of the above, because volume of silver remains same

Answer 1

Hint:Here the length of the given piece of silver is L and the area of the same piece is A. Then find the relation between the resistance, area of the given silver piece and the value of length. Then use the method of elimination for each option one by one and check for the minimum value of resistance.

Formula used:
Resistance, $R = \rho \dfrac{l}{A}$
Where, $\rho $ is resistivity, A is the area of the given silver piece, and l is the length of the wire given silver piece.

Complete step by step solution:
We know that resistance is given as follows:
$R = \rho \dfrac{l}{A}$
Now, check for each option one by one:
For option A: $l = L,A = A$ (given)
Putting the given value in the known formula for resistance, we get;
$R = \rho \dfrac{L}{A}$

For option B: $l = 2L,A = \dfrac{A}{2}$ (given)
Putting the given value in the known formula for resistance, we get;
$R=\rho\,\dfrac{2L}{A/2} = \rho \dfrac{{4L}}{A} = 4R$

For option C: $l = \dfrac{L}{2},A = 2A$ (given)
Putting the given value in the known formula for resistance, we get;
$R=\rho\,\dfrac{L/2}{2A} = \rho \dfrac{L}{{4A}} = \dfrac{1}{4}R$
So, we can clearly see that the minimum value of R we get is in option C, that is R becomes one fourth times of its original value.

Hence, the correct answer is option C.

Note: The formula for the resistance also contain one quantity other than length and area of the silver piece given that is resistivity of the material but here in this case same material is used in all the three cases therefore the resistivity will be same for all the three cases and hence we only calculate value of R for l and A.