A) 125 : 15 : 1
B) 1 : 15 : 125
C) 5 : 3 : 1
D) 1 : 3 : 5
Correct Answer: A
Solution :
The electrical resistance of a wire would be expected to be greater for a longer wire, less for a wire of larger cross-sectional area, and would be expected to depend upon the material out of which the wires are made. Experimentally, the resistance of a wire can be expressed as \[R=\frac{\rho l}{A}\] where, \[\rho =\]resistivity, \[l=length\] A = cross-sectional area Also, \[area=\frac{volume}{length}=\frac{V}{l}\] \[\therefore \] \[R=\rho \frac{l}{A}=\rho \frac{l}{A}\times \frac{l}{l}=\rho \frac{{{l}^{2}}}{V}\times \frac{\rho }{\rho }\] \[R={{\rho }^{2}}\frac{{{l}^{2}}}{m}\] \[(\because \,\,m=V\rho )\] \[\therefore \] \[{{R}_{1}}:{{R}_{2}}:R{{ & }_{3}}=\frac{l_{1}^{2}}{{{m}_{1}}}:\frac{l_{2}^{2}}{{{m}_{2}}}:\frac{l_{3}^{2}}{{{m}_{3}}}\] Given, \[{{m}_{1}}:{{m}_{2}}:{{m}_{3}}=1:3:5\] \[{{l}_{1}}:{{l}_{2}}:{{l}_{3}}=5:3:1\] \[\therefore \] \[{{R}_{1}}:{{R}_{2}}:{{R}_{3}}:=\frac{{{(5)}^{2}}}{1}:\frac{{{(3)}^{3}}}{(3)}:\frac{{{(1)}^{2}}}{5}\] \[=\frac{25}{1}:\frac{9}{3}:\frac{1}{5}\] \[\therefore \] \[{{R}_{1}}:{{R}_{2}}:{{R}_{3}}:\,=125:15:1\]
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