A rod of length l having resistance R, is cut into two equal parts. These parts are connected in parallel then new resistance shall be

The correct option is R/4.

Step-by-Step Explanation:

1. Understanding the original resistance of the rod:
Let the initial resistance of the rod of length l be R. The resistance of a conductor is directly proportional to its length l when the cross-sectional area and material (resistivity) remain constant. This is represented by the formula:
$R=\rho \frac{l}{A}$
where $\rho$ is the resistivity of the material and $A$ is the cross-sectional area.

2. Finding the resistance of the cut parts:
When the rod is cut into two equal parts, the length of each part becomes:
$l^{\prime }=\frac{l}{2}$
Since the resistance is directly proportional to the length, the resistance of each half, $R^{\prime }$, will be:
$R^{\prime }=\rho \frac{l^{\prime }}{A}=\rho \frac{l}{2A}=\frac{R}{2}$
So, each part has a resistance of $\frac{R}{2}$.

3. Calculating the equivalent resistance in parallel:
When these two parts are connected in parallel, the equivalent resistance $R_{\text{eq}}$ is given by the parallel combination formula:
$\frac{1}{R_{\text{eq}}}=\frac{1}{R^{\prime }}+\frac{1}{R^{\prime }}$
Substituting the value of $R^{\prime }=\frac{R}{2}$:
$\frac{1}{R_{\text{eq}}}=\frac{1}{R/2}+\frac{1}{R/2}=\frac{2}{R}+\frac{2}{R}=\frac{4}{R}$
Taking the reciprocal to find $R_{\text{eq}}$:
$R_{\text{eq}}=\frac{R}{4}$

Thus, the new resistance of the parallel combination is R/4.