Question

A cylindrical conductor of length L and uniform area of cross – section A has resistance R. Another conductor of length 2L and resistance R of the same material has an area of cross-section,
$\text{A}\text{.)}\dfrac{\text{A}}{\text{2}}$
$\text{B}\text{.)}\dfrac{\text{3A}}{\text{2}}$
C.)2A
D.)3A

Answer 1

Hint: To solve this question, we will be using the resistance formula in which the length and cross sectional area is present with a constant, which is known as resistivity. We will put the given variables in the formula to get the answer, considering resistivity as a constant variable.

Formula used:
$\text{R= }\!\!\rho\!\!\text{ }\dfrac{\text{L}}{\text{A}}$

Complete step by step answer:
Let us first list the given data in the question,
So, in the question it is given that,
Length of a cylinder =L
Resistance of the cylinder =R
And the cross sectional area of the cylinder=A
Now, let’s put this in the formula of resistance, which is
$\text{R= }\!\!\rho\!\!\text{ }\dfrac{\text{L}}{\text{A}}$
In this formula, R=Resistance, L=Length and A= Cross sectional Area, and rho is the resistivity which is constant.
The cylinder’s values satisfies the formula,
So now, let’s write the data given in the question, about the another cylinder,
Another cylinder’s length=2L
Another cylinder’s Resistance =R
And here we need to find the area of cross section of another cylinder,
So let’s put the value in our formula,
$\text{R= }\!\!\rho\!\!\text{ }\dfrac{\text{2L}}{\text{?}}$
From our original formula and this value put on this formula, we can say that the area of the cross section must be 2A.
Hence, the area of cross section will be 2A of another cylinder,
So we conclude that the answer is option C.

Note:
The Resistance of the cylinder is directly proportional to the length of the cylinder, but it is inversely proportional to the area of cross section of the cylinder, and hence, the area of cross section of the another cylinder will be 2A, so that 2 will be common from denominator and numerator , and the resistance will be R.