
A thick wire is stretched so that its length becomes two times. Assuming that there is no change in its density, then what is the ratio of change in resistance of the wire to the initial resistance of the wire?
A. 2:1
B. 4:1
C. 3:1
D. 1:4
Answer
163.8k+ views
Hint: Before going to solve this problem let us understand the term resistance. It is defined as the property of a material that opposes the current flow in the conductor. Here, using the formula for the resistance we are proceeding further.
Formula Used:
The formula for the resistance of a wire is,
\[R = \rho \dfrac{l}{A}\]
where, \[\rho \] is the resistivity of a material, l is the length of wire and A is cross-sectional area.
Complete step by step solution:
Consider a thick wire that is stretched so that its length becomes two times. Assuming that there is no change in its density, then we need to find the ratio of change in resistance of the wire to the initial resistance of the wire. For that consider the formula to find the resistance of a wire,
\[R = \rho \dfrac{l}{A} \\ \]
\[\Rightarrow {R^1} = \rho \dfrac{{{l^1}}}{A}\]
If the length becomes two times, then \[{l^1} = 3l\]and the density and the area remain same, then,
\[\dfrac{{{R^1}}}{R} = \dfrac{{\rho \dfrac{{{l^1}}}{A}}}{{\rho \dfrac{l}{A}}} \\ \]
\[\Rightarrow \dfrac{{{R^1}}}{R} = \dfrac{{{l^1}}}{l} \\ \]
\[\Rightarrow \dfrac{{{R^1}}}{R} = \dfrac{{3l}}{l} \\ \]
\[\therefore \dfrac{{{R^1}}}{R} = \dfrac{3}{1}\]
Therefore, the ratio of change in resistance of the wire to the initial resistance of the wire is 3:1.
Hence, option C is the correct answer.
Note: Here, in this problem it is important to remember the formula to find the resistance of a given material and we found the ratio of the resistance as 3:1 as shown in this solution.
Formula Used:
The formula for the resistance of a wire is,
\[R = \rho \dfrac{l}{A}\]
where, \[\rho \] is the resistivity of a material, l is the length of wire and A is cross-sectional area.
Complete step by step solution:
Consider a thick wire that is stretched so that its length becomes two times. Assuming that there is no change in its density, then we need to find the ratio of change in resistance of the wire to the initial resistance of the wire. For that consider the formula to find the resistance of a wire,
\[R = \rho \dfrac{l}{A} \\ \]
\[\Rightarrow {R^1} = \rho \dfrac{{{l^1}}}{A}\]
If the length becomes two times, then \[{l^1} = 3l\]and the density and the area remain same, then,
\[\dfrac{{{R^1}}}{R} = \dfrac{{\rho \dfrac{{{l^1}}}{A}}}{{\rho \dfrac{l}{A}}} \\ \]
\[\Rightarrow \dfrac{{{R^1}}}{R} = \dfrac{{{l^1}}}{l} \\ \]
\[\Rightarrow \dfrac{{{R^1}}}{R} = \dfrac{{3l}}{l} \\ \]
\[\therefore \dfrac{{{R^1}}}{R} = \dfrac{3}{1}\]
Therefore, the ratio of change in resistance of the wire to the initial resistance of the wire is 3:1.
Hence, option C is the correct answer.
Note: Here, in this problem it is important to remember the formula to find the resistance of a given material and we found the ratio of the resistance as 3:1 as shown in this solution.
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