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  • A meter bridge experiment is conducted to determine the resistivity of a wire of unknown material. The bridge is balanced using a standard resistor of 15 ohms in one gap and the unknown wire in the

A meter bridge experiment is conducted to determine the resistivity of a wire of unknown material. The bridge is balanced using a standard resistor of 15 ohms in one gap and the unknown wire in the other gap. The length of the standard wire is 1.8 meters, and its resistivity is 8 × 10−7 ohm meter. The bridge is balanced at a jockey position of 0.6 meters from the left end of the standard wire. If the unknown wire has a length of 1.5 meters, determine the resistivity (ρ) of the unknown wire material. Express your answer in terms of fundamental constants and known quantities.

Option: 1

7.2 ×10−5ohm


Option: 2

9×10−5 ohm


Option: 3

35 ×10−5ohm


Option: 4

78 ×10−5ohm


Answers (1)

best_answer

1. The formula for resistance (R) of a wire is given by:R=\rho\left(\frac{l}{A}\right) \text {, where } \rho is the resistivity of the material, l is the length of the wire, and A is the cross-sectional area of the wire.

2. Given the length of the standard wire (lstd) is 1.8 meters, and the jockey position (xstd) is 0.6 meters from the left end, we can find the length of the wire on the other side as lstd, opp = lstd − xstd = 1.8 − 0.6 meters.

3. The bridge is balanced when the ratio of the resistances on the two sides is equal:\frac{R_1}{R_2}=\frac{l_{\mathrm{std}}}{l_{\mathrm{std}, \mathrm{opp}}}

\text { 4. Substituting the values: } \frac{15}{R_2}=\frac{1.8}{1.8-0.6} \text {. }

\text { 5. Solving for } R_2 \text {, we find: } R_2=\frac{15(1.8-0.6)}{1.8} \text { ohms. }6. Substituting the known resistivity of the standard wire (ρstd), length of the standard wire (lstd), and the formula for its resistance:R_{\mathrm{std}}=\rho_{\mathrm{std}}\left(\frac{l_{\mathrm{std}}}{A}\right),

we can solve for the area A:A=\frac{\rho_{\mathrm{std}} l_{\mathrm{std}}}{R_{\mathrm{std}}} .

7. Substituting the given values of ρstd = 8 × 10−7 ohm meter and lstd = 1.8 meters, and the resistance of the standard wire Rstd = 15 ohms: A =\frac{8 \times 10^{-7} \times 1.8}{15} \mathrm{~m}^2

8. Now, we can use the known area A to find the resistivity ρ of the unknown wire material using the resistance formula: \rho\left(\frac{l}{A}\right)=R_2

\text { 9. Solving for } \rho: \rho=\frac{R_2 A}{l} \text {. }

10. Substituting the known values of R2, A, and the length of the unknown wire (l):\rho=\frac{\frac{15(1.8-0.6)}{1.8} \times \frac{8 \times 10^{-7} \times 1.8}{15}}{1.5} \text { ohm meter. }

\text { 11. Simplifying: } \rho=7.2 \times 10^{-7} \mathrm{ohm} \text { meter. }

Therefore, the resistivity of the unknown wire material is 7.2 × 10−7 ohm meter. Therefore, the correct option is 1.

Posted by

Gautam harsolia

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