#### A student wants to determine the resistivity of the material of a given wire using a meter bridge experiment. The wire is placed on the meter bridge, and various measurements are taken to calculate the resistivity. The known resistors are connected in a ratio of 1 : 9 on one side of the meter bridge. The student balances the bridge by sliding the jockey along the wire. The length of the wire between the jockey and the point of balance is measured to be L = 80 cm. The length of the wire between the jockey and the known resistors is 100 cm, and the known resistance is R = 10 Ω. Calculate the resistivity (ρ) of the material of the given wire.Option: 1 5.81 ×10−7 Ω · mOption: 2 68.9 ×10−7 Ω · mOption: 3 5.45 ×10−7 Ω · mOption: 4 9.81 ×10−7 Ω · m

Given data:
Length of wire between jockey and balance point (L) = 80cm  = 0.80 m
Length of wire between jockey and known resistors = 100 cm = 1.00 m ,Known resistance (R) = 10 Ω
Using the principle of the meter bridge experiment, the ratio of resistances is equal to the ratio of lengths:

$\frac{R}{Unkown resistance} = \frac{Length of wire with known resistance }{Length of wire with unknown resistance }$

Substituting the given values:
$\frac{10}{Unkown resistance} = \frac{1.00 }{0.80 }$

Solving for the unknown resistance:

$Unkown resistance = \frac{10 \times 0.80 }{1.00 } = 8\Omega$

Now, we can use the formula for resistivity:

$\rho = \frac{R\times A}{L}$

Where A is the cross-sectional area of the wire. Let’s assume the wire’s diameter (d) is 0.1 cm (0.001 m). Therefore, radius (r) = 0.0005 m. The cross-sectional area (A) is πr2 :

$A= \Pi \times (0.0005)^{2} = 7.85 \times 10^{-7}m^{2}$
Now, substituting the values into the resistivity formula:

$\rho = \frac{10\times 7.85 \times 10^{-7}}{0.80} = 9.81\times 10^{-7}\Omega m$

Therefore, the resistivity (ρ) of the material of the given wire is $9.81\times 10^{-7}\Omega m$