
The length of the resistance wire is increased by $10\% $ . What is the corresponding change in the resistance of wire?
(A) $10\% $
(B) $25\% $
(C) $21\% $
(D) $9\% $
Answer
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Hint:Derive the relation between the resistance and the wire, substitute the known resistance value and find the relation of the initial and the final resistance. Substitute this relation in the formula of the change in resistance, to find the answer for it.
Useful formula:
(1) The resistance of the wire when the density and the volume of the wire remains same is given by
$R\alpha {l^2}$
Where $R$ is the initial resistance of the wire and $l$ is the length of the wire.
(2) The formula of the change in the resistance is given by
$c = \dfrac{{R' - R}}{R} \times 100$
Where $R'$ is the final resistance of the wire and $c$ is the change in the resistance.
Complete step by step solution:
It is given that the
length of the resistance wire is increased by $10\% $
From the formula of the resistance,
$R = {l^2}$ --------------(1)
If the length of the wire increased by $10\% $ , then the length changes to $1 + 10\% l$
Substituting this in the above equation, we get
$R' = {\left( {1 + 10\% l} \right)^2}$
$R' = {\left( {1 + \dfrac{{10}}{{100}}l} \right)^2}$
By performing the various arithmetic operations, we get
\[R' = {\left( {\dfrac{{100 + 10}}{{100}}l} \right)^2}\]
By simplification of the above equation, we get
$R' = {\left( {\dfrac{{110l}}{{100}}} \right)^2}$
Squaring the right hand side of the equation,
$R' = \dfrac{{12100{l^2}}}{{10000}}$
By further simplification and also substituting the equation (1) in the above equation, we get
$R' = 1.21\,R$ --------------(2)
Using the formula of the change in the resistance,
$c = \dfrac{{R' - R}}{R} \times 100$
Substituting the equation (2) in above formula
$c = \dfrac{{1.21R - R}}{R} \times 100$
By simplifying the above equation, we get
$c = 21\% $
Thus the option (C) is correct.
Note: The density of the material remains the same even after the length of the wire changes and it depends only on the material in which the wire is made. The cross sectional area changes only when the diameter changes but remains the same when the length of the wire changes.
Useful formula:
(1) The resistance of the wire when the density and the volume of the wire remains same is given by
$R\alpha {l^2}$
Where $R$ is the initial resistance of the wire and $l$ is the length of the wire.
(2) The formula of the change in the resistance is given by
$c = \dfrac{{R' - R}}{R} \times 100$
Where $R'$ is the final resistance of the wire and $c$ is the change in the resistance.
Complete step by step solution:
It is given that the
length of the resistance wire is increased by $10\% $
From the formula of the resistance,
$R = {l^2}$ --------------(1)
If the length of the wire increased by $10\% $ , then the length changes to $1 + 10\% l$
Substituting this in the above equation, we get
$R' = {\left( {1 + 10\% l} \right)^2}$
$R' = {\left( {1 + \dfrac{{10}}{{100}}l} \right)^2}$
By performing the various arithmetic operations, we get
\[R' = {\left( {\dfrac{{100 + 10}}{{100}}l} \right)^2}\]
By simplification of the above equation, we get
$R' = {\left( {\dfrac{{110l}}{{100}}} \right)^2}$
Squaring the right hand side of the equation,
$R' = \dfrac{{12100{l^2}}}{{10000}}$
By further simplification and also substituting the equation (1) in the above equation, we get
$R' = 1.21\,R$ --------------(2)
Using the formula of the change in the resistance,
$c = \dfrac{{R' - R}}{R} \times 100$
Substituting the equation (2) in above formula
$c = \dfrac{{1.21R - R}}{R} \times 100$
By simplifying the above equation, we get
$c = 21\% $
Thus the option (C) is correct.
Note: The density of the material remains the same even after the length of the wire changes and it depends only on the material in which the wire is made. The cross sectional area changes only when the diameter changes but remains the same when the length of the wire changes.
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