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Question: A wire of \(9\Omega \) resistance having \(30cm\) length is tripled on itself. What is its new resis...

A wire of 9Ω9\Omega resistance having 30cm30cm length is tripled on itself. What is its new resistance?
a. 81Ω81\Omega
b. 9Ω9\Omega
c. 3Ω3\Omega
d. 18Ω18\Omega

Explanation

Solution

Here we need not convert the length to its S.I. unit but we would consider simple unitary methods to solve this problem. Here the length of the wire is increased three times of its original length. Similarly, the area of the cross-section becomes 13\dfrac{1}{3} times the initial area of the 30cm30cm wire. For the conservation of the volume, area decreases at the same rate as length increases.
Formula used:
R=ρlAR = \dfrac{{\rho l}}{A}
ρ\rho is the resistivity of the wire,
RR is the Resistance of the wire,
ll is the length of the wire and
AA is the area of cross-section of wire.

Complete answer:
When the wire is stretched to three times its original length the area becomes 13\dfrac{1}{3} times the original area. According to the formula above we know that length is directly proportional to the resistance and area is inversely proportional to the resistance.
R=ρlAR = \dfrac{{\rho l}}{A}
9=30ρA\Rightarrow 9 = \dfrac{{30\rho }}{A}
ρ=9A30\Rightarrow \rho = \dfrac{{9A}}{{30}}
ρ=3A10\Rightarrow \rho = \dfrac{{3A}}{{10}}
Now we have a value of ρ\rho so, we can now calculate the value of the area using the value of ρ\rho since the length is increased to three times of its original length.
Let, new Resistance be RR' ,
By conservation of volume
A×l=A×lA \times l = {A^ {'} } \times {l ^ {'} }
A×l=A×3lA \times l = {A^ {'} } \times 3l
New Area , A=13A{A^ {'} } = \dfrac{1}{3}A
and new length is l=3l{l ^ {'} } = 3l
Since,
R=ρlAR' = \dfrac{{\rho {l ^ {'} }}}{{{A^ {'} } }}
So, new length is 3l3l
Therefor ρ=RAl\rho = \dfrac{{R'A}}{l}
ρ=3A10=R×Al\Rightarrow \rho = \dfrac{{3A}}{{10}} = \dfrac{{R' \times {A^ {'} } }}{{{l ^ {'} }}}
3A10=R×13A3l\Rightarrow \dfrac{{3A}}{{10}} = \dfrac{{R' \times \dfrac{1}{3}A}}{{3l}}
3A10=R×13A90\Rightarrow \dfrac{{3A}}{{10}} = \dfrac{{R' \times \dfrac{1}{3}A}}{{90}}
R=81Ω\Rightarrow R' = 81\Omega

Hence, option A is correct.

Note:
Resistance of a wire depends on the length of the wire, the area of the cross-section, and the resistivity of the material which itself depends on the material type and finally the temperature of the conductor.
So, any kind of change in these values may lead to the change in the value of resistance of the material.