Question

Using a particular conductivity cell, resistance offered is 60 $\Omega$. When the conductivity cell with area thrice the earlier and having 80% more distance between the electrodes is used, then resistance offered will be _______ $\Omega$.

Answer 1

36

Answer 2

To solve this problem, we use the formula for resistance:

$R = \rho \frac{l}{A}$

where:

• $R$ is the resistance
• $\rho$ is the resistivity of the solution (which remains constant as the solution is the same)
• $l$ is the distance between the electrodes
• $A$ is the area of the electrodes

Let's denote the initial conditions with subscript 1 and the new conditions with subscript 2.

Initial Conditions:

Given resistance $R_1 = 60 \, \Omega$.

Let the initial distance between electrodes be $l_1 = l$.

Let the initial area of the electrodes be $A_1 = A$.

From the resistance formula:

$R_1 = \rho \frac{l_1}{A_1}$

$60 = \rho \frac{l}{A}$ (Equation 1)

New Conditions:

The area of the conductivity cell is thrice the earlier:

$A_2 = 3 A_1 = 3A$

The distance between the electrodes is 80% more than the earlier:

$l_2 = l_1 + 0.80 l_1 = (1 + 0.80) l_1 = 1.80 l_1 = 1.8l$

Now, we write the resistance formula for the new conditions:

$R_2 = \rho \frac{l_2}{A_2}$

Substitute the expressions for $l_2$ and $A_2$:

$R_2 = \rho \frac{1.8l}{3A}$

$R_2 = \left( \frac{1.8}{3} \right) \times \left( \rho \frac{l}{A} \right)$

$R_2 = 0.6 \times \left( \rho \frac{l}{A} \right)$

From Equation 1, we know that $\rho \frac{l}{A} = 60 \, \Omega$. Substitute this value into the equation for $R_2$:

$R_2 = 0.6 \times 60$

$R_2 = 36 \, \Omega$

The resistance offered will be 36 $\Omega$.