Question

A copper wire of unknown length has a resistance of 0.80 ohm. By successive passes through drawing dies, the length of the wire is increased to 2.5 times its original value. Assuming that resistivity remains unchanged during the drawing process, determine the new value of its resistance

Answer 1

Answer:

New Resistance is 5 ohm

Step-by-step explanation:

We are given,

Resistance of the copper wire is 0.8 ohm and then its length is increased by 2.5 times its original size.

So,

R = 0.8 ohm

Also,

R = ρL/A

Now,

Let L' be the new length,

but we also know that, when we stretch a wire its length increases but its Area of cross section decreases by the amount of stretch.

For eg:- When we stretch a rubber band, its length increases but at the same time, it gets thinner and thinner, in other words, Area of cross section decreases.

Thus,

At what ratio we stretch, at that ratio, we must decrease the area of cross section as well.

So,

Let A' be the new Area of cross section and the new Resistance be R'.

We know that,

L' = 2.5 × L

L' = 2.5L

A' = A ÷ 2.5

A' = A/2.5

Thus,

R' = ρL'/A'

R' = ρ(2.5L)/(A/2.5)

R' = ρ(2.5)(2.5) × L/A

R' = 6.25 × ρL/A

But R = ρL/A

So,

R' = 6.25R

Also, R = 0.8 ohm

So,

R' = 6.25 × 0.8

R' = 5 ohm

Thus, when we stretch the length to 2.5 times the original length, the new resistance will become 5 ohms.

Hope it helped and you understood it........All the best

Answer 2

Given: A copper wire of unknown length has a resistance of 0.80 ohm. By successive passes through drawing dies, the length of the wire is increased to 2.5 times its original value.

To find: Assuming that resistivity remains unchanged during the drawing process, new value of its resistance

Solution: When a wire is stretched to lengthen or shorten its length, the volume remains constant after the expansion or contraction.

Volume= Area × Length

Let length and area before stretching be L1 and A1 while after stretching be L2 and A2 respectively.

L2 = 2.5 L1

Therefore,

A1 × L1 = L2 × A2

=> A1 × L1 = 2.5× L1 × A2

=> A1 = 2.5 A2

=> A2 = A1/2.5

Now, resistance is given by the formula

= pL/A where p is the resistivity

Resistance before stretching= pL1/A1

Therefore, pL1/A1 = 0.80 ohm

Resistance after stretching

= pL2/A2

= p × 2.5 L1 / (A1/2.5)

= 2.5 × 2.5 × pL1/A1

= 2.5 × 2.5 × 0.80 (since pL1/A1= 0.80 ohm)

= 5 ohm

Therefore, new value of resistance is 5 ohm.