A) \[1:3:5\]
B) \[5:3:1\]
C) \[1:15:125\]
D) \[125:15:1\]
Correct Answer: D
Solution :
: Mass of wire, M = Volume \[\times \]Density \[=Al\times d\] or \[A=\frac{M}{ld}\] Resistance of wire, \[R=\frac{\rho l}{A}=\frac{\rho l}{(M/ld)}\] or \[R=\frac{\rho {{l}^{2}}d}{m}\] Since all the three wires are made up of same material, therefore p and d remain the same for all the three wires. Hence, \[{{R}_{1}}:{{R}_{2}}:{{R}_{3}}=\frac{l_{1}^{2}}{{{M}_{1}}}:\frac{l_{2}^{2}}{{{M}_{2}}}:\frac{l_{3}^{3}}{{{M}_{3}}}=\frac{{{5}^{2}}}{1}:\frac{{{3}^{2}}}{3}:\frac{{{1}^{2}}}{5}\] \[=25:3:\frac{1}{5}=125:15:1\]
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