A conductivity cell has two platinum electrodes of `1.2cm^(2)` area, separated by a distance of `0.8cm`. The cell constant is

In a conductivity cell the two platinum electrodes each of area 10 sq. cm are fixed 1.5 cm apart. The cell. Contained 0.05N solution of a salt. If the two electrodes are just half dipped into the solution which has a resistance of 50 ohms, find equivalent conductance of the salt solution.

Each of the two platinum electrodes having area 64 mm^(2) of a conducitvity cell are separated by 8mm. The resistance of the cell containing 7.5xx10^(-3) M KCI solution at 298 K is 1250 ohm calculate (i) Cell constant (ii) Conductivity and (iii) Molar conductivity

A condenser has two conducting plates of radius 10cm separated by a distance of 5mm. It is charged with a constant cunent of 0.15 A. The magnetic field at a point 2cm from the axis in the gap is

0.1 mole per litre solution present in conductivity cell where electrode of 100 cm^(2) area placed at 1 cm and resistance observe is 5xx10^(3) ohm, what is molar conductivity of solution?

The conductivity of pure water in a conductivity cell with electrodes of cross-sectional area 4cm^(2) placed at a distance 2cm apart is 8 xx 10^(-7) S cm^(-1) . Calculate. (a) The resistance of water. (b) The current that would flow through the cell under the applied potential difference of 1 volt.

One normal salt solution surrounding two platinum electrodes, 2.1 cm apart and 6.3 cm^(2) in area was found to offer a resistance of 50 ohm. Calculate the equivalent conductivity solution.

The plates of a parallel plate capacitance 1.0 F are separated by a distance d =1 cm. Find the plate area .

Two parallel wires carry equal currents of 10A along the same direction and are separated by a distance of 2.0 cm. Find the magnetic field at a point which is 2.0 cm away from each of these wires.

(a). The resistance of a decinormal solution of an electrolyte in a conductivity cell was found to be 245 ohms. Calculate the equilvalet conductivity of the solution if the electrodes in the cell were 2 cm apart and each has an area of 3.5 sq. cm. (b). The conductivity of 0.001028M acetic acid is 4.95xx10^(-5)Scm^(-1) . Calculate its dissociation constant if ^^_(m)^(@) for acetic acid is 390.5Scm^(2_mol^(-1)