Question

Two cylindrical rods $A$ and $B$ have same resistivities and same lengths. Diameter of rod $A$ is twice the diameter of the rod $B$. Ratio of voltage drop across rod $A$ to rod $B$ is:

• A. $\frac{1}{2}$
• B. $2$
• C. $4$
• D. $\frac{1}{4}$

Answer 1

Answer: (D)

$\frac{1}{4}$

Given,
two cylindrical rods $A$ and $B$ have same resistivities and same length.
We know that,
$R=\frac{\rho l}{A}=\frac{\rho l}{\pi r^2}$
For rod $A$;
Resistance of rod $A,R_A=\frac{\rho l}{\pi (2R{)}^2}....\left(i\right)$
For rod $B$;
Resistance of rod $B,R_B=\frac{\rho l}{\pi (R{)}^2}.....\left(ii\right)$
Dividing Eq. (i) by Eq. (ii), we get
$\frac{R_A}{R_B}=\frac{\rho l}{\pi (2R{)}^2}\times \frac{\pi (R{)}^2}{\rho l}$
$\frac{R_A}{R_B}=\frac{1}{4}$
$\frac{R_{A}I}{R_{B}I}=\frac{1}{4}$
$\Rightarrow \frac{V_A}{V_B}=\frac{1}{4}$.