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A student wants to determine the resistivity of the material of a given wire using a meter bridge. The known resistance used in the experiment is 10 Ω. The wire is stretched along the length of the meter bridge. When  a balance point is obtained, the length of the wire on the bridge is found to be 60 cm. The length of the bridge wire is 100 cm, and its resistance is negligible. Calculate the resistivity of the material of the given wire.

Option: 1

6r2Ω · m


Option: 2

38r2 Ω · m


Option: 3

87r2 Ω · m


Option: 4

16r2 Ω · m


Answers (1)

best_answer

Given data: Known resistance, R1 = 10 Ω
Length of wire on the bridge, l1 = 60 cm = 0.6 m
Length of the bridge wire, L = 100 cm = 1 m
Let’s assume the resistivity of the given wire material is ρ.
At balance, the ratio of lengths of the wire on the bridge to the length of the bridge wire is equal to the ratio of resistances:

\frac{l_{1}}{L}=\frac{R_{1}}{R}

We want to find ρ, so we need to express R in terms of ρ and other given
quantities.
Resistance of the wire, R = ρ·l1/A

Where A is the cross-sectional area of the wire. Since the wire is uniform, its cross-sectional area remains constant.

Now we can substitute the expression for R into the balance equation:

\frac{l_{1}}{L}=\frac{R_{1} }{R}

\frac{l_{1}}{L}=\frac{R_{1} . A}{\rho .l_{1}}

\rho =\frac{R_{1}.A.l_{1} }{l_{1}/L}

Now we need to find the value of A, the cross-sectional area of the wire.

Let’s assume the wire is cylindrical in shape with radius r. The cross- sectional area of a cylinder is given by:

A = π · r2
We can substitute this into the expression for ρ:

\rho =\frac{ R_{1}.\Pi ^{2}. r^{2}.l_{1} }{l_{1}/L}

Given that L = 1 m, l1 = 0.6 m, and R1 = 10 Ω, we can rearrange the equation to solve for ρ:

\rho =\frac{10 . \Pi . r^{2}. 0.6}{1/1}

ρ = 6πr2 Ω · m

The resistivity of the material of the given wire is given by ρ = 6πr2 Ω·m. Please note that the above calculation assumes that the wire is a uniform cylinder and that the length is stretched along the meter bridge with a balance point. In practice, other factors like temperature variations and non-uniformity of the wire might affect the accuracy of the result.

 

 

 

Posted by

avinash.dongre

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