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  • A meter bridge experiment is conducted to determine the resistivity of a wire of unknown material. The bridge is balanced with a standard resistor of 18 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 with a standard resistor of 18 ohms in one gap and the unknown wire in the other gap. The length of the standard wire is 2.2 meters, and its resistivity is 1.2 × 10−6 ohm meter. The bridge is balanced at a jockey position of 0.5 meters from the left end of the standard wire. If the unknown wire has a length of 1.8 meters, determine the cross-sectional area (A) of the unknown wire. Express your answer in terms of fundamental constants and known quantities.

Option: 1

2 ×10−8 square meters


Option: 2

9×10−8 squaremeters


Option: 3

7.33×10−8 squaremeters


Option: 4

15 ×10−8 squaremeters


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 } \rhois 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 2.2 meters, and the jockey position (xstd) is 0.5 meters from the left end, we can find the length of the wire on the other side as lstd, opp = lstd − xstd = 2.2 − 0.5 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{18}{R_2}=\frac{2.2}{2.2-0.5} \text {. }

\text { 5. Solving for } R_2 \text {, we find: } R_2=\frac{18(2.2-0.5)}{2.2} \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_{\text {std }}=\rho_{\text {std }}\left(\frac{l_{\text {std }}}{A}\right)

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

 

7. Substituting the given values of ρstd = 1.2×10−6 ohm meter and lstd = 2.2 meters, and the resistance of the standard wire Rstd = 18 ohms: A =\frac{1.2 \times 10^{-6} \times 2.2}{18} \mathrm{~m}^2

Therefore, the cross-sectional area of the unknown wire material is\frac{1.32 \times 10^{-6}}{18}\mathrm{m}^2 \text {. which is } \frac{1.32 \times 10^{-6}}{18} \mathrm{~m}^2 \text {. } This seems to be a division operation involving scientific notation.

To simplify the expression, we can perform the division:

\frac{1.32 \times 10^{-6}}{18}=7.33 \times 10^{-8} \mathrm{~m}^2

So, the simplified cross-sectional area of the unknown wire material is 7.33× 10−8 square meters.

Therefore, the correct option is 3.

Posted by

Divya Prakash Singh

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