An aluminium 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 aluminium, but its density is three timesthat of aluminium. The ratio of the mass of aluminium rod and that of copper rod will be :-

Options

A. `1//6`
B. `2//3`
C. `1//3`
D. 6

Explanation

To solve the problem, we need to find the ratio of the mass of the aluminium rod (m_a) to the mass of the copper rod (m_c) given that their lengths and resistances are the same, and we know the specific resistances and densities of both materials.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- Let the length of both rods be \( L \).
- The resistances of both rods are equal: \( R_a = R_c \).
- The specific resistance (resistivity) of copper \( \rho_c \) is half that of aluminium \( \rho_a \):
\[ \rho_c = \frac{1}{2} \rho_a \]
- The density of copper \( d_c \) is three times that of aluminium \( d_a \):
\[ d_c = 3 d_a \]

2. **Using the Resistance Formula**:
The resistance \( R \) of a rod is given by:
\[ R = \frac{\rho L}{A} \]
For both rods, since \( R_a = R_c \):
\[ \frac{\rho_a L}{A_a} = \frac{\rho_c L}{A_c} \]
Since the lengths \( L \) are the same, we can cancel them out:
\[ \frac{\rho_a}{A_a} = \frac{\rho_c}{A_c} \]

3. **Substituting the Specific Resistances**:
Substitute \( \rho_c = \frac{1}{2} \rho_a \) into the equation:
\[ \frac{\rho_a}{A_a} = \frac{\frac{1}{2} \rho_a}{A_c} \]
Cancel \( \rho_a \) (assuming \( \rho_a \neq 0 \)):
\[ \frac{1}{A_a} = \frac{1/2}{A_c} \]
Rearranging gives:
\[ A_c = \frac{1}{2} A_a \]

4. **Finding the Mass of Each Rod**:
The mass of a rod is given by:
\[ m = d \cdot V = d \cdot (A \cdot L) \]
For aluminium:
\[ m_a = d_a \cdot A_a \cdot L \]
For copper:
\[ m_c = d_c \cdot A_c \cdot L = 3 d_a \cdot \left(\frac{1}{2} A_a\right) \cdot L = \frac{3}{2} d_a \cdot A_a \cdot L \]

5. **Finding the Ratio of Masses**:
Now, we can find the ratio of the masses:
\[ \frac{m_a}{m_c} = \frac{d_a \cdot A_a \cdot L}{\frac{3}{2} d_a \cdot A_a \cdot L} \]
The \( d_a \), \( A_a \), and \( L \) terms cancel out:
\[ \frac{m_a}{m_c} = \frac{1}{\frac{3}{2}} = \frac{2}{3} \]

### Final Answer:
The ratio of the mass of the aluminium rod to that of the copper rod is:
\[ \frac{m_a}{m_c} = \frac{2}{3} \]