
A wire of $9\Omega $ resistance having $30cm$ length is tripled on itself. What is its new resistance?
a. $81\Omega $
b. $9\Omega $
c. $3\Omega $
d. $18\Omega $
Answer
465.3k+ views
Hint: 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 $\dfrac{1}{3}$ times the initial area of the $30cm$ wire. For the conservation of the volume, area decreases at the same rate as length increases.
Formula used:
\[R = \dfrac{{\rho l}}{A}\]
$\rho $ is the resistivity of the wire,
$R$ is the Resistance of the wire,
$l$ is the length of the wire and
$A$ is the area of cross-section of wire.
Complete answer:
When the wire is stretched to three times its original length the area becomes $\dfrac{1}{3}$ 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.
\[R = \dfrac{{\rho l}}{A}\]
\[\Rightarrow 9 = \dfrac{{30\rho }}{A}\]
\[\Rightarrow \rho = \dfrac{{9A}}{{30}}\]
\[\Rightarrow \rho = \dfrac{{3A}}{{10}}\]
Now we have a value of $\rho $ so, we can now calculate the value of the area using the value of $\rho $ since the length is increased to three times of its original length.
Let, new Resistance be $R'$ ,
By conservation of volume
$A \times l = {A^ {'} } \times {l ^ {'} }$
$A \times l = {A^ {'} } \times 3l$
New Area , ${A^ {'} } = \dfrac{1}{3}A$
and new length is ${l ^ {'} } = 3l$
Since,
\[R' = \dfrac{{\rho {l ^ {'} }}}{{{A^ {'} } }}\]
So, new length is $3l$
Therefor $\rho = \dfrac{{R'A}}{l}$
$\Rightarrow \rho = \dfrac{{3A}}{{10}} = \dfrac{{R' \times {A^ {'} } }}{{{l ^ {'} }}} $
$\Rightarrow \dfrac{{3A}}{{10}} = \dfrac{{R' \times \dfrac{1}{3}A}}{{3l}} $
$\Rightarrow \dfrac{{3A}}{{10}} = \dfrac{{R' \times \dfrac{1}{3}A}}{{90}} $
$\Rightarrow R' = 81\Omega$
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.
Formula used:
\[R = \dfrac{{\rho l}}{A}\]
$\rho $ is the resistivity of the wire,
$R$ is the Resistance of the wire,
$l$ is the length of the wire and
$A$ is the area of cross-section of wire.
Complete answer:
When the wire is stretched to three times its original length the area becomes $\dfrac{1}{3}$ 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.
\[R = \dfrac{{\rho l}}{A}\]
\[\Rightarrow 9 = \dfrac{{30\rho }}{A}\]
\[\Rightarrow \rho = \dfrac{{9A}}{{30}}\]
\[\Rightarrow \rho = \dfrac{{3A}}{{10}}\]
Now we have a value of $\rho $ so, we can now calculate the value of the area using the value of $\rho $ since the length is increased to three times of its original length.
Let, new Resistance be $R'$ ,
By conservation of volume
$A \times l = {A^ {'} } \times {l ^ {'} }$
$A \times l = {A^ {'} } \times 3l$
New Area , ${A^ {'} } = \dfrac{1}{3}A$
and new length is ${l ^ {'} } = 3l$
Since,
\[R' = \dfrac{{\rho {l ^ {'} }}}{{{A^ {'} } }}\]
So, new length is $3l$
Therefor $\rho = \dfrac{{R'A}}{l}$
$\Rightarrow \rho = \dfrac{{3A}}{{10}} = \dfrac{{R' \times {A^ {'} } }}{{{l ^ {'} }}} $
$\Rightarrow \dfrac{{3A}}{{10}} = \dfrac{{R' \times \dfrac{1}{3}A}}{{3l}} $
$\Rightarrow \dfrac{{3A}}{{10}} = \dfrac{{R' \times \dfrac{1}{3}A}}{{90}} $
$\Rightarrow R' = 81\Omega$
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.
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