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  • A student is conducting an experiment to determine the resistivity (ρ) of a given wire using a meter bridge. The student sets up the meter bridge experiment as shown below: The follow

A student is conducting an experiment to determine the resistivity (ρ) of a given wire using a meter bridge. The student sets up the meter bridge experiment as shown below:

The following information is provided:

Known resistance (R1) = 150.0 ohms

Length of wire on one side (l1) = 60.0 cm

Length of wire on the other side (l2) = 120.0 cm

Balance point length (l) = 90.0 cm

Battery voltage (V ) = 5.0 volts

Current through the circuit (I) = 0.3 amperes

Given the formula for the resistivity of a wire:\rho=\frac{R_1 \cdot l \cdot A}{l_1 \cdot l_2}, where A is the cross-sectional area of the wire, calculate the resistivity (ρ) of the given wire.

 

 

Option: 1

9.5736 ohm · cm


Option: 2

170 ohm · cm


Option: 3

200 ohm · cm


Option: 4

20 ohm · cm


Answers (1)

best_answer

Using Ohm’s law, we can find the resistance of the wire (R) using the equation

\begin{aligned} &R=\frac{V}{I}\\ &R=\frac{5.0 \text { volts }}{0.3 \text { amperes }}=16.67 \mathrm{ohms} \end{aligned}

Now, we can substitute the given values into the formula for resistivity (ρ) to solve for it:

\rho=\frac{R_1 \cdot l \cdot A}{l_1 \cdot l_2}

Plugging in the known values:

\rho=\frac{150.0 \mathrm{ohms} \cdot 90.0 \mathrm{~cm} \cdot A}{60.0 \mathrm{~cm} \cdot 120.0 \mathrm{~cm}}

To isolate A, we can rearrange the equation:

A=\frac{\rho \cdot l_1 \cdot l_2}{R_1 \cdot l}

Substituting the given values:

A=\frac{\rho \cdot 60.0 \mathrm{~cm} \cdot 120.0 \mathrm{~cm}}{150.0 \mathrm{ohms} \cdot 90.0 \mathrm{~cm}}

Now, we can substitute the calculated value of R back into the equation for A:

A=\frac{\rho \cdot 60.0 \mathrm{~cm} \cdot 120.0 \mathrm{~cm}}{16.67 \mathrm{ohms} \cdot 90.0 \mathrm{~cm}}

Solving for A:

\begin{gathered} A=\frac{\rho \cdot 7200.0 \mathrm{~cm}^2}{1500.3 \mathrm{ohm} \cdot \mathrm{cm}} \\ A \approx 3.1912 \mathrm{~cm}^2 \end{gathered}

Finally, we can solve for ρ:

\rho=\frac{3 A}{\mathrm{~cm}^2}

Substituting the calculated value of A:

\rho \approx \frac{3}{\mathrm{~cm}^2} \cdot 3.1912 \mathrm{~cm}^2 \approx 9.5736 \mathrm{ohm} \cdot \mathrm{cm}

Hence, the resistivity (ρ) of the given wire is approximately 9.5736 ohm · cm. Therefore, the correct answer is A.

Posted by

Suraj Bhandari

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