A) 125 : 15 : 1
B) 1 : 15 : 125
C) 5: 3: 15
D) 15 : 3 : 5
Correct Answer: A
Solution :
Here, ratio of masses of wires \[{{m}_{1}}:{{m}_{2}}:{{m}_{3}}=1:3:5\] and ratio of lengths of wires \[{{l}_{1}}:{{l}_{2}}:{{l}_{3}}=5:3:1\] The electrical resistance is given by as \[R=\rho \frac{l}{A}=\rho \times \frac{l}{A}\times \frac{l}{l}=\rho \frac{{{l}^{2}}}{V}=\rho \frac{{{l}^{2}}}{V}\times \frac{m}{m}\] \[R={{\rho }^{2}}\times \frac{{{l}^{2}}}{m}\propto \frac{{{l}^{2}}}{m}\] Hence, \[{{R}_{1}}:{{R}_{2}}:{{R}_{3}}=\frac{{{l}_{1}}^{2}}{{{m}_{1}}}:\frac{{{l}_{2}}^{2}}{{{m}_{2}}}:\frac{{{l}_{3}}^{2}}{{{m}_{3}}}\] \[=\frac{{{(5)}^{2}}}{1}:\frac{{{(3)}^{2}}}{3}:\frac{{{(1)}^{2}}}{5}\] \[=\frac{25}{1}:\frac{9}{3}:\frac{1}{5}\] Hence, \[{{R}_{1}}:{{R}_{2}}:{{R}_{3}}=125:15:1\]
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