Question

An aluminum rod and a copper rod are taken such that their lengths are same and their resistances are also same. The specific resistance of copper is half that of aluminum, but its density is three times that of aluminum. The ratio of the mass of aluminum rod and that of copper rod will be

• A. $\frac{1}{6}$
• B. $\frac{2}{3}$
• C. $\frac{1}{3}$
• D. $6$

Answer 1

The correct option is B $\frac{2}{3}$

The resistance of copper and aluminium wires is

$R_c=\frac{{\rho }_c{L}_c}{A_c}$

and $R_a=\frac{{\rho }_a{L}_a}{A_a}$

where, $L_c$ and $L_a$ are the lengths of of copper and aluminium

wires,

${\rho }_c$ and ${\rho }_a$ are specific resistances of copper and

aluminium wires,

$A_c$ and $A_a$ are area of cross sections of copper and aluminium wires.

Since, Resistance and length of both wires is same

$\frac{{\rho }_c}{A_c}=\frac{{\rho }_a}{A_a}$ .............................................$\left(1\right)$

Now, the density is the ratio of mass and volume hence for copper and aluminium wires

$d_c=\frac{m_c}{V_c}$

$d_c=\frac{m_c}{A_c{L}_c}$

$A_c=\frac{m_c}{d_c{L}_c}$

similarly

$A_a=\frac{m_a}{d_a{L}_a}$

Using these values in equation $\left(1\right),$ we get,

$\frac{{\rho }_c}{\frac{m_c}{d_c{L}_c}}=\frac{{\rho }_a}{\frac{m_a}{d_a{L}_a}}$

Given that, $L_c=L_a$, ${\rho }_c=\frac{1}{2}{\rho }_a$ and $d_c=3d_a$

Therefore

$\frac{3d_a{\rho }_a}{2m_c}=\frac{{\rho }_a{d}_a}{m_a}$

$\Rightarrow \frac{m_a}{m_c}=\frac{2}{3}$