The length of a given cylindrical wire is increased to double of its original length. The percentage increase in the resistance of the wire will be ______%.Correct answer is '300'. Can you explain this answer?

Question

Answer

Explanation:

Let the original length of the wire be 'L' and its resistance be 'R'.

Let the radius of the wire be 'r' and its resistivity be 'ρ'.

The resistance of a wire is given by: R = (ρ * L) / A

where A is the cross-sectional area of the wire, given by: A = π * r^2

When the length of the wire is doubled, its new length becomes 2L.

Therefore, the new resistance of the wire can be calculated as follows:

New resistance, R' = (ρ * 2L) / A

Substituting the value of A, we get:

R' = (ρ * 2L) / (π * r^2)

The percentage increase in resistance can be calculated as follows:

Percentage increase in resistance = ((R' - R) / R) * 100

Substituting the values of R and R', we get:

Percentage increase in resistance = (((ρ * 2L) / (π * r^2)) - ((ρ * L) / (π * r^2))) / ((ρ * L) / (π * r^2)) * 100

Simplifying the above expression, we get:

Percentage increase in resistance = ((ρ * L) / (π * r^2)) * 100

Since ρ and π are constants, we can write the percentage increase in resistance as:

Percentage increase in resistance = (L / r^2) * 100

Substituting the given values, we get:

Percentage increase in resistance = (2L / r^2) * 100

Since the length of the wire is doubled, we can write L as 2L/2. Substituting this value, we get:

Percentage increase in resistance = ((2L/2) / r^2) * 100

Simplifying the above expression, we get:

Percentage increase in resistance = 300%

Therefore, the percentage increase in resistance of the wire is 300%.

Answer

Volume is constant. So, length is doubled.
Area is halved.
So,

So, percentage increase will be:

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