
Two wires that are made up of different materials whose specific resistance are in the ratio of 2:3, length 3:4, area 4:5. Find the ratio of their resistance.
A. 6:5
B. 6:8
C. 5:8
D. 1:2
Answer
161.7k+ views
Hint: Resistance is defined as the property of a material that opposes the current flow in the conductor. Here, they have given the ratios of densities, length, and area. Using these data, we are going to find the ratio of the resistance.
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 two wires that are made up of different materials and their specific resistance is in the ratio of 2:3, length 3:4, area 4:5. We need to find the ratio of their resistance. For that consider the formula to find the resistance of a wire,
\[R = \rho \dfrac{l}{A}\]
Consider the formula for resistances of two wires are,
\[{R_1} = {\rho _1}\dfrac{{{l_1}}}{{{A_1}}}\] and \[{R_2} = {\rho _2}\dfrac{{{l_2}}}{{{A_2}}}\]
The ratio of resistance is,
\[\dfrac{{{R_1}}}{{{R_2}}} = \dfrac{{{\rho _1}}}{{{\rho _2}}}\dfrac{{{l_1}}}{{{l_2}}}\dfrac{{{A_2}}}{{{A_1}}}\]
Given that,
\[\dfrac{{{\rho _1}}}{{{\rho _2}}} = \dfrac{2}{3}\], \[\dfrac{{{l_1}}}{{{l_2}}} = \dfrac{3}{4}\] and \[\dfrac{{{A_1}}}{{{A_2}}} = \dfrac{4}{5}\]
Substitute the value in above equation, we get,
\[\dfrac{{{R_1}}}{{{R_2}}} = \dfrac{2}{3} \times \dfrac{3}{4} \times \dfrac{5}{4} \\ \]
\[\Rightarrow \dfrac{{{R_1}}}{{{R_2}}} = \dfrac{1}{4} \times \dfrac{5}{2} \\ \]
\[\therefore \dfrac{{{R_1}}}{{{R_2}}} = \dfrac{5}{8}\]
Therefore, the ratio of their resistance is 5:8.
Hence, option C is the correct answer.
Note:Here, in this problem, remember the formula to calculate the resistance of a wire and note down the values of given data, and thus we obtain the required solution. The resistance of a wire depends on its length, that is the resistance of a wire is directly proportional to its length. Hence, if the wire is longer, the resistance will be higher.
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 two wires that are made up of different materials and their specific resistance is in the ratio of 2:3, length 3:4, area 4:5. We need to find the ratio of their resistance. For that consider the formula to find the resistance of a wire,
\[R = \rho \dfrac{l}{A}\]
Consider the formula for resistances of two wires are,
\[{R_1} = {\rho _1}\dfrac{{{l_1}}}{{{A_1}}}\] and \[{R_2} = {\rho _2}\dfrac{{{l_2}}}{{{A_2}}}\]
The ratio of resistance is,
\[\dfrac{{{R_1}}}{{{R_2}}} = \dfrac{{{\rho _1}}}{{{\rho _2}}}\dfrac{{{l_1}}}{{{l_2}}}\dfrac{{{A_2}}}{{{A_1}}}\]
Given that,
\[\dfrac{{{\rho _1}}}{{{\rho _2}}} = \dfrac{2}{3}\], \[\dfrac{{{l_1}}}{{{l_2}}} = \dfrac{3}{4}\] and \[\dfrac{{{A_1}}}{{{A_2}}} = \dfrac{4}{5}\]
Substitute the value in above equation, we get,
\[\dfrac{{{R_1}}}{{{R_2}}} = \dfrac{2}{3} \times \dfrac{3}{4} \times \dfrac{5}{4} \\ \]
\[\Rightarrow \dfrac{{{R_1}}}{{{R_2}}} = \dfrac{1}{4} \times \dfrac{5}{2} \\ \]
\[\therefore \dfrac{{{R_1}}}{{{R_2}}} = \dfrac{5}{8}\]
Therefore, the ratio of their resistance is 5:8.
Hence, option C is the correct answer.
Note:Here, in this problem, remember the formula to calculate the resistance of a wire and note down the values of given data, and thus we obtain the required solution. The resistance of a wire depends on its length, that is the resistance of a wire is directly proportional to its length. Hence, if the wire is longer, the resistance will be higher.
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