If length of conductor whose volume /mass is constant increases by 3%,...
Solution:
Given:
Length of conductor increases by 3%
To find:
The percentage change in resistance
Concept Used:
The resistance of a conductor is directly proportional to its length.
Formula:
R = ρ(l/A)
where R is the resistance, ρ is the resistivity of the material, l is the length of the conductor, and A is the cross-sectional area of the conductor.
Calculation:
Let the initial length of the conductor be l1, and the initial resistance be R1.
Then, the initial volume of the conductor will be V1 = Al1, where A is the cross-sectional area of the conductor.
Also, let the final length of the conductor be l2, where l2 = 1.03l1 (since the length increases by 3%).
The final volume of the conductor will remain the same as the initial volume, so we have V1 = Al1 = Al2.
Therefore, the new cross-sectional area will be A2 = A(l1/l2).
Using the formula for resistance, we get:
R2 = ρ(l2/A2)
= ρ(l1/l2)(l2/A)
= ρ(l1/l2)(1.03l1)/A
= 1.03ρl1²/(A(l1/l2))
The percentage change in resistance can be calculated as follows:
% Change in Resistance = ((R2 - R1)/R1) x 100%
= ((1.03ρl1²/(A(l1/l2))) - R1)/R1) x 100%
Substituting A(l1/l2) for A2, we get:
% Change in Resistance = ((1.03ρl1²/A(l1/l2)) - R1)/R1) x 100%
= ((1.03ρl1²/A2) - R1)/R1) x 100%
Final Answer:
Therefore, the percentage change in resistance is ((1.03ρl1²/A2) - R1)/R1) x 100%.