Question

a conducting wire of resistance R ohm is stretched and in such a way that its area become 1/n th part of its original area ,now what will be the resistance of the wire? A. R/n² B.n²R C.R/n D. nR​

Answer 1

Explanation:

resistance is a characteristic of a wire that affects the flow of current.

the formula of resistance is

R = pl/A

where A is the area of that wire

in this question it is given that the area becomes 1/n

so now

R = pln/A

now the new resistance is nR

Answer 2

Answer:

The correct option is (B) $\rm n^2 R$.

Explanation:

Given:

• Original resistance of the wire = R.

Assumptions:

• Cross sectional area of the wire = A.
• Length of the wire = L.
• Resistivity of the wire = $\rho$.

According to the definition of the resistance, the original resistance of the wire is given by

$\rm R=\rho \dfrac LA.$

Now, when the wire is stretched such that its area becomes 1/n th part of its original area, A, its length will also change on stretching , such that, its volume remains same.

Let the new area and the new length of the wire be A' and L' respectively.

Therefore,

$\rm Volume= AL=A'L'\\AL=\dfrac An L'\\L'=nL.$

The resistance of the stretched wire is given by

$\rm R'=\rho\dfrac{L'}{A'}=\rho\dfrac{nL}{\dfrac An}=n^2\left (\rho\dfrac{L}{A} \right )=n^2R.$

Thus, the correct option is (B) $\rm n^2 R$.