This question aims to calculate the resistivity of leaf tissue. Resistivity is a characteristic property of a material that refers to the capability or the resistive power of a material to resist the flow of electric current. This property of any material opposes the flow of electric current and protects the material from an electric shock. The greater the resistivity of a substance, the greater the resistance in the electric current’s flow.

The material mentioned in this question is leaf tissue. Leaf tissues are composed of groups of plant cells. In the given question, all properties of the leaf tissue are mentioned, which are required to calculate the resistivity. The formula for calculating resistivity is discussed in the solution.

**Expert Answer**

The resistivity of a material is its ability to restrict the flow of electric current. Several factors are required to calculate the resistivity of the material, such as the material’s area, length, resistance, etc. The formula for calculating resistivity can be obtained from the formula of resistance:

\[ R = \frac{\rho L}{A} \]

Rearranging the above equation:

\[ \rho = \frac{RA}{L} \]

The data stated in the question is given below:

Resistance of leaf = $R$ = $2.4 M$ $\Omega$

Electrode’s Distance = $L$ = $23 cm$ = $0.23 m$

Width of leaf = $w$ = $2.7 cm$

Thickness of leaf = $t$ = $0.20 mm$

For calculating the resistivity, the first thing needed is the area.

Calculating the area of the leaf:

\[ Area = A = w \times t \]

\[ A = (2.7) \times (0.02) \]

\[ A = 0.054 cm^{2} \]

Converting this area into meters:

\[ A = 0.054 x 10^{-4}m^{2} \]

Inserting the values into the equation:

\[ \rho = \frac{RA}{L} \]

\[ \rho = \frac{(2.4 x 10^{6}) \times (0.054 x 10^{-4})}{0.23} \]

\[ \rho = \frac{12.96}{0.23} \]

\[ \rho = 56.34 \Omega m \]

**Example**

The resistance of a material is $0.0625$ $\Omega$ and its area is $3.14 x 10^{-6}$ $m^{2}$. The length of this material is $3.5 m$. Determine its resistivity.

For calculating the resistivity, the following formula is used:

\[ \rho = \frac{RA}{L} \]

As the question has provided all the necessary information, simply insert the values into the formula.

Inserting the values:

\[ \rho = \frac{(0.0625) \times (3.14 x10^{-6})}{3.5} \]

\[ \rho = \frac{1.962 x 10^{-7}}{3.5} \]

\[ \rho = 5.607 \Omega m \]