A wire stretched so that its length become 6/5 times of its original length .If its original resistance is 25 ohm meter Find the new resistance
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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 6/5 times its original length; hence there will definitely be some change in cross sectional area.
Now we know that, volume = Mass/Densityand 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 = 6/5 L1
Now, V2 = L2A2 & V1 = L1A1, we know that V1 = V2
i.e. L1A1 = L2A2
i.e. A2 = (L1/L2)A1
i.e. A2 = 5/6 A1
Putting the values of A2 & L2 into the following
R2/R1 = (L2/L1)*(A1/A2)
We get,
R2/R1 = 6/5*6/5
R2 =1.44 * R1
Thus, R2 = 1.44 *25 = 36 Ohm.
hope it helps u.... yarrr
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 6/5 times its original length; hence there will definitely be some change in cross sectional area.
Now we know that, volume = Mass/Densityand 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 = 6/5 L1
Now, V2 = L2A2 & V1 = L1A1, we know that V1 = V2
i.e. L1A1 = L2A2
i.e. A2 = (L1/L2)A1
i.e. A2 = 5/6 A1
Putting the values of A2 & L2 into the following
R2/R1 = (L2/L1)*(A1/A2)
We get,
R2/R1 = 6/5*6/5
R2 =1.44 * R1
Thus, R2 = 1.44 *25 = 36 Ohm.
hope it helps u.... yarrr
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sonam95:
so sorry
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