Answer the question on basis of passage given below:

\(G = \frac{\kappa A}{l} = \kappa \) (both\(A\) and \(l\) are unity in their appropriate units in m or cm)

Molar conductivity of a solution at a given concentration is the conductance of the volume \(V\) of solution containing one mole of electrolyte kept between two electrodes with area of cross section \(A\) and distance of unit length \(l\). Therefore,

\(\Lambda_m = \frac{\kappa A}{l}\)

Since \(l = 1\) and \(A = V\) (volume containing 1 mole of electrolyte) \(\Lambda _m =\kappa V\)

If the length 'l' in the column of an electrolytic cell is halved and its area of cross-section is tripled then conductance (G) would become:

6 times

The correct answer is option 3. 6 times.

To determine how the conductance \(G\) would change if the length \(l\) is halved and the area of cross-section \(A\) is tripled, we can use the relationship between conductance, conductivity, area, and length:

The conductance \(G\) is given by the equation:

\(G = \frac{\kappa A}{l}\)

where:

\(\kappa\) is the conductivity of the electrolyte,

\(A\) is the area of cross-section of the electrodes,

\(l\) is the distance between the electrodes.

Given Changes,

The length \(l\) is halved, so \(l\) becomes \(\frac{l}{2}\).

The area \(A\) is tripled, so \(A\) becomes \(3A\).

New Conductance:

Substituting the new values into the conductance equation:

\(G_{\text{new}} = \frac{\kappa \cdot 3A}{\frac{l}{2}} = \frac{\kappa \cdot 3A \cdot 2}{l} = \frac{6\kappa A}{l}\)

Comparison with Original Conductance:

The original conductance \(G_{\text{original}}\) was:

\(G_{\text{original}} = \frac{\kappa A}{l}\)

The new conductance \(G_{\text{new}}\) is:

\(G_{\text{new}} = 6 \times \frac{\kappa A}{l} = 6 \times G_{\text{original}}\)

Conclusion:

The conductance \(G\) would become 6 times the original value.

So, the correct answer is option 3. 6 times.