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(C) Given: Both wires have the same mass $(m)$ and same material (same density $ho_d$ and resistivity $ho$).Since mass = density $	imes$ volume,a

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$1.5$

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(C) Given: Both wires have the same mass $(m)$ and same material (same density $ho_d$ and resistivity $ho$).Since mass = density $	imes$ volume,and density is the same,the volume $(V = A \cdot l)$ must be the same for both wires.$V_A = V_B \Rightarrow A_A l_A = A_B l_B \Rightarrow \frac{l_A}{l_B} = \frac{A_B}{A_A}$.Since $A = \pi r^2 = \pi (d/2)^2$,we have $\frac{A_B}{A_A} = \frac{d_B^2}{d_A^2}$.Given $d_A = \frac{d_B}{2}$,so $\frac{d_B}{d_A} = 2$. Thus,$\frac{A_B}{A_A} = (2)^2 = 4$.Therefore,$\frac{l_A}{l_B} = 4$.The resistance is given by $R = ho \frac{l}{A}$.Taking the ratio: $\frac{R_A}{R_B} = \frac{l_A}{l_B} 	imes \frac{A_B}{A_A} = 4 	imes 4 = 16$.Given $R_A = 24 \, \Omega$,then $R_B = \frac{R_A}{16} = \frac{24}{16} = 1.5 \, \Omega$.

We have two wires $A$ and $B$ of the same mass and same material. The diameter of wire $A$ is half that of wire $B$. If the resistance of wire $A$ is $24 \, \Omega$,then the resistance of wire $B$ will be ................ $\Omega$.

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