Question

if a copper wire is stretched to decrease the radius by 0.1 %, what is the percentage change in its resistance ?

Answer 1

heyy dear ur answer.........

When wire is stretched, its radius decreases and its length increases.

We know R ~ length/cross section area

~ just assume this symbol for proportional to

Consider initial Resistance as R1Final resistance as R2Intitial length as L1Final length as L2

Below, I have Converted % value to numerical form. ie 0.1% = 1/1000

Initial Radius is R1

Final radius R2 = R1 - ( 1/1000)R1

So,R2 = (999/1000)R1

R2=0.999R1

Now, to find L2, we need to know that,

When wire is stretched its radius and cross sectional area changes. But volume of the wire always remains same.

V1 and V2 are volume of wire.

As i said volume remains same.

V1=V2

(Volume of wire is πr² * L)

πR1²*L1 = πR2² *L2

Put above values and solve foe L2

U get L2 = L1/0.99801

Take ratios of R1/R2

U get R1 as R1 = (0.9960)R2

So, change in resistance is R2-R1 = 0.004

convert to % change 0.004 * 100 = 0.4%

0.4% Increase in resistance.........hope this will help u.............@kundan

Answer 2

Nice question

Hope it helps you..