A) 1/6
B) 2/3
C) 1/3
D) 6
Correct Answer: B
Solution :
Resistance of wire is given by \[R=\rho \frac{l}{A}\] where, p is specific resistance of wire, A is area of cross-section of wire and\[l\]is length of wire. If Vis volume of wire, then if d is density and m is mass of wire, then \[R=\rho \frac{{{I}^{2}}d}{m}\] \[m\propto \rho d\] \[\frac{{{m}_{Al}}}{{{m}_{Cu}}}=\frac{{{\rho }_{Al}}}{{{\rho }_{Cu}}}\times \frac{{{d}_{Al}}}{{{d}_{Cu}}}\] \[=\frac{2}{1}\times \frac{1}{3}=\frac{2}{3}\]You need to login to perform this action.
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