The resistance between ends will be `(if (rho L)/(pi R^2) = R_0)`
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The resistance R of a wire is given by the relation R= (rhol)/( pir^(2)) . Percentage error in the measurement of rho, l and r is 1 % , 2% and 3% respectively. Then the percentage error in the measurement of R is

The resistance R of a wire is given by the relation R= (rhol)/( pir^(2)) . Percentage error in the measurement of rho, l and r is 1 % , 2% and 3% respectively. Then the percentage error in the measurement of R is

The resistance R of a wire is given by the relation R= (rhol)/( pir^(2)) . Percentage error in the measurement of rho, l and r is 1 % , 2% and 3% respectively. Then the percentage error in the measurement of R is

The specific resistance rho of a circular wire of radius r , resistance R , and length l is given by rho = pi r^(2) R//l . Given : r = 0.24 +- 0.02 cm , R = 30 +- 1 Omega , and l = 4.80 +- 0.01 cm . The percentage error in rho is nearly

The specific resistance rho of a thin wire of radius r cm, resistance R ohm and length L is given by rho = (pi r^2R)/(L). If L =78+-0.01cm r = 0.26 +- 0.02 and R =32 +- 1 Omega, What is the percentage error in rho?

The specific resistance rho of a circular wire of radius r , resistance R , and length l is given by rho = pi r^(2) R//l . Given : r = 0.24 +- 0.02 cm , R = 30 +- 1 Omega , and l = 4.80 +- 0.01 cm . The percentage error in rho is nearly

If L and R denote inductance and resistance respectively, then the dimension of (L)/(R ) is

Space between tow concentric spheres of radii r_(1) and r_(2) such that r_(1) lt r_(2) is filled with a material of resistivity rho . Find the resistance between inner and outer surface of the material

Space between tow concentric spheres of radii r_(1) and r_(2) such that r_(1) lt r_(2) is filled with a material of resistivity rho . Find the resistance between inner and outer surface of the material