A wire with a circular cross section and a resistance R is lengthened to 9.66 times its original length by pulling it through a small hole. The total volume of the wire is unchanged. Find the resistance of the wire after it is stretched. Answer in units of R

The resistance of the wire after it is stretched is 93.31R.

Explanation:

Resistance is the property of the material to oppose the current flow through it. It is given by the relation :

R = (ρl)/A

Here ρ is resistivity, l is length of wire and A is the area of the wire.

Let l₀, and A₀ are the original length and original circular cross section area of the wire. while l₁ and A₁ are the new length and new circular cross section area of the wire.

Volume of the original wire, V₀ = A₀ x l₀

Volume of the new wire, V₁ = A₁ x l₁

According to the problem. volume remain same. So,

V₀ = V₁

A₀ x l₀ = A₁ x l₁

It is given that l₁ = 9.66 x l₀. Substitute this value in the above equation;

A₀ x l₀ = A₁ x 9.66 x l₀

A₁ = A₀/9.66

Resistance of the original wire, R = (ρl₀)/A₀

Resistance of the new wire, R₁ = (ρl₁)/A₁

Substitute the value of l₁ and A₁ in the above equation.

R₁ = (ρ x l₀ x 9.66)/(A₀/9.66) = 93.31 x (ρl₀)/A₀

But (ρl₀)/A₀ = R. hence,

R₁ = 93.31 R

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2

false

explanation: