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  • In a laboratory experiment, a meter bridge setup is used to determine the resistivity of an unknown material of a given wire. The setup consists of a uniform wire of length 1.5 meters, which is pla

In a laboratory experiment, a meter bridge setup is used to determine the resistivity of an unknown material of a given wire. The setup consists of a uniform wire of length 1.5 meters, which is placed horizontally on a non-conductive table. A jockey is used to make electrical contact along the length of the wire. The bridge is balanced using a standard resistor of known resistance 10 ohms. The resistivity of the material is to be calculated.

Given the following additional information:

Length of the wire (l): 1.5 meters

• Length of the wire segment to the left of the jockey (x): 0.75 meters

• Length of the wire segment to the right of the jockey (l − x): 0.75 meters

• Known resistance (R1): 10 ohms

• The bridge is balanced when the null point is achieved, where the jockey is placed at a specific position on the wire. Using the above data and the principles of meter bridge experiment, calculate the resistivity (ρ) of the unknown material of the wire. Express your answer in terms of fundamental constants and known quantities.

Option: 1

4.77ohm


Option: 2

89.63 ohm


Option: 3

56.3 ohm


Option: 4

23.20 ohm


Answers (1)

1.We know that the resistance of a wire is given by the formula: 

R=\rho\left(\frac{l}{A}\right), where ρ is the resistivity of the material, l is the length of the wire, and A is the cross-sectional area of the wire.

2.In the case of the meter bridge experiment, we have a uniform wire of length l = 1.5 meters. The cross-sectional area (A) of the wire is uniform throughout its length.

3. Let’s consider a unit length of the wire (1 meter) for ease of calculation. Therefore, the resistance R of this unit length of wire is given by:R=\rho\left(\frac{l}{A}\right),

4.The bridge is balanced when the ratio of the resistances on the two sides of the jockey is equal to the ratio of the lengths of the wire segments:\frac{R_1}{R_2}=\frac{x}{l-x} \text {. }

5. Substituting the values: \frac{10}{R_2}=\frac{0.75}{0.75}

6. Solving for R2, we find: R2 = 10 ohms.

7. Now we can substitute the value of R2 into the resistance formula for the unit length of the wire: \rho\left(\frac{1}{A}\right)=10 \text {. }

8. Rearranging the formula to solve for ρ, we get: ρ = 10A.

9. To express the cross-sectional area (A) in terms of known quantities, we can use the formula for the cross-sectional area of a circular wire: A = πr2 , where r is the radius of the wire.

10. The radius of the wire is not provided directly, but we can express it in terms of the length of the wire (l) and the unknown resistivity (ρ): r=\sqrt{\frac{\rho l}{\pi}} .

11. Substituting this value of r back into the formula for the cross-sectional area:A=\pi\left(\sqrt{\frac{\rho l}{\pi}}\right)^2=\frac{\rho l}{\pi}

12. Substituting the value of A back into the resistivity formula:\rho=10\left(\frac{\rho l}{\pi}\right) \text {. }

\text { 13. Solving for } \rho: \rho=10\left(\frac{\rho 1.5}{\pi}\right) \text {. }

14. \text { Simplifying: } \rho=\frac{15 \rho}{\pi} \text {. }

\text { 15. Finally, solving for } \rho: \rho=\frac{15}{\pi} \text { ohm meter. }

16. Approximating the value of π as 3.14, we get: ρ ≈ 4.77 ohm meter.

Therefore, the resistivity of the unknown material of the wire is approximately 4.77 ohm meter.

Therefore, the correct option is 1.

Posted by

Sumit Saini

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