
The resistance if a wire is 'R' ohm. If it is melted and stretched to 'n' times its original length, its new resistance will be:
(A) \[\dfrac{R}{{{n^2}}}\]
(B) \[nR\]
(C) \[\dfrac{R}{n}\]
(D) \[{n^2}R\]
Answer
146.4k+ views
Hint Given that the resistance of wire R has an initial length \[{l_1}\] and is stretched n times to new length \[{l_2}\]. It is observed that there is no change in radius. This means, assume volume is constant. Find a relation between areas of the wire before and after extension and substitute it in the resistance formula \[R = \dfrac{{\rho l}}{A}\].
Complete Step By Step Solution
It is given that there is a wire of radius r, length \[{l_1}\] having a resistance R. Now this wire is said to be extended to a new length \[{l_2}\], which is n times \[{l_1}\]. Since , there is no change in radius of the wire, let us assume that the volume before and after stretching are constant.
\[{V_1} = {V_2}\], where \[{V_1}\] is volume before stretching and \[{V_2}\]is volume after stretching.
Now we know that volume is a product of area and height. In our case, height is the length of the wire
\[{A_1} \times {l_1} = {A_2} \times {l_2}\]
We know that , \[{l_2} = n \times {l_1}\]. Substituting this , we get
\[ \Rightarrow {A_1} \times {l_1} = {A_2} \times n \times {l_1}\]
\[ \Rightarrow \dfrac{{{A_1}}}{n} = {A_2}\]
We know that resistance R of a material is calculated by using the formula\[R = \dfrac{{\rho l}}{A}\], where \[\rho \]is represented as resistivity of the material , \[A\] is area and \[l\]is length of the wire.
Now , before extension , resistivity is given as , \[{R_1} = \dfrac{{\rho {l_1}}}{{{A_1}}}\]
After extension, resistivity is given as, \[{R_2} = \dfrac{{\rho {l_2}}}{{{A_2}}}\]
Substituting for \[{A_2}\]in the equation for\[{R_2}\], we get
\[ \Rightarrow {R_2} = \dfrac{{\rho {l_2} \times n}}{{{A_1}}}\]
Now substituting\[{l_2} = n \times {l_1}\], we get
\[ \Rightarrow {R_2} = \dfrac{{\rho {l_1} \times {n^2}}}{{{A_1}}}\]
(The term \[\dfrac{{\rho {l_1}}}{{{A_1}}}\]is equal to\[{R_1}\] and \[{R_1}\] value given in the question is \[R\])
\[ \Rightarrow {R_2} = R \times {n^2}\]
Thus, Option (d) is the right answer for the given question.
Note Electrical resistivity is defined as the electrical property of a material which defines its strength to oppose electric current.
Complete Step By Step Solution
It is given that there is a wire of radius r, length \[{l_1}\] having a resistance R. Now this wire is said to be extended to a new length \[{l_2}\], which is n times \[{l_1}\]. Since , there is no change in radius of the wire, let us assume that the volume before and after stretching are constant.
\[{V_1} = {V_2}\], where \[{V_1}\] is volume before stretching and \[{V_2}\]is volume after stretching.
Now we know that volume is a product of area and height. In our case, height is the length of the wire
\[{A_1} \times {l_1} = {A_2} \times {l_2}\]
We know that , \[{l_2} = n \times {l_1}\]. Substituting this , we get
\[ \Rightarrow {A_1} \times {l_1} = {A_2} \times n \times {l_1}\]
\[ \Rightarrow \dfrac{{{A_1}}}{n} = {A_2}\]
We know that resistance R of a material is calculated by using the formula\[R = \dfrac{{\rho l}}{A}\], where \[\rho \]is represented as resistivity of the material , \[A\] is area and \[l\]is length of the wire.
Now , before extension , resistivity is given as , \[{R_1} = \dfrac{{\rho {l_1}}}{{{A_1}}}\]
After extension, resistivity is given as, \[{R_2} = \dfrac{{\rho {l_2}}}{{{A_2}}}\]
Substituting for \[{A_2}\]in the equation for\[{R_2}\], we get
\[ \Rightarrow {R_2} = \dfrac{{\rho {l_2} \times n}}{{{A_1}}}\]
Now substituting\[{l_2} = n \times {l_1}\], we get
\[ \Rightarrow {R_2} = \dfrac{{\rho {l_1} \times {n^2}}}{{{A_1}}}\]
(The term \[\dfrac{{\rho {l_1}}}{{{A_1}}}\]is equal to\[{R_1}\] and \[{R_1}\] value given in the question is \[R\])
\[ \Rightarrow {R_2} = R \times {n^2}\]
Thus, Option (d) is the right answer for the given question.
Note Electrical resistivity is defined as the electrical property of a material which defines its strength to oppose electric current.
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