Question

A wire of resistance 3 ohm is drawn out so that its length is increased by twice its initial length.Find the new resistance.

Answer 1

Answer:

Resistivity is the property of the material and it will not change with changing the length of the wire. Now, we will focus on the effect of resistance due to change in length of the wire.

We know that,

R = (rho)*length/Cross Sectional Area

Since, the length of the wire becomes 3 times its original length; hence there will definitely be some change in cross sectional area.

Now we know that, volume = Mass/Density and since neither mass nor density has changed due to change in length, thus volume must remain constant.

We further know that,

Volume = Length * Cross Sectional Area

Assuming, Initial Volume = V1 & Final Volume = V2

Initial Cross Section Area = A1 & Final Cross Sectional Area = A2

Initial Length of Wire = L1 & Final Length of Wire = L2

Thus we have, L2 = 3 L1

Now, V2 = L2A2 & V1 = L1A1, we know that V1 = V2

i.e. L1A1 = L2A2

i.e. A2 = (L1/L2)A1

i.e. A2 = 1/3 A1

Putting the values of A2 & L2 into the following

R2/R1 = (L2/L1)*(A1/A2)

We get,

R2/R1 = 3*3

R2 =9 * R1

Thus, R2 = 9 *10= 90 Ohm.......

Explanation:

...........m...............

Answer 2

Given:

Initial resistance, R = 3 ohms

New length, L' = 2L

To Find:

New resistance, R'

Solution:

$R=p\frac{L}{A} ........... (1)\\\\R=p\frac{L'}{A'}=p\frac{2L}{A/2}=4p\frac{L}{A} =4R (from (1))\\R= 4 \times3=12ohms$

(A' = A/2 because the length of the wire depends indirectly on the area of the cross-section)

Hence, the new resistance will be 12 ohms

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