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

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

For finding the specific resistance of a wire, a student took the following readings by using a metre scale, a micrometer and a resistance box. Length of the wire (L) = 100.0 cm Radius (r ) = 0.100 cm and R = 10 Omega What is the maximum possible percentage error in the value of the specific resistance?

The specific resistance of a circular cross - section of a cylindrical wire of radius b and resistance R is given is s=(pi b^(2)R)/(l) , where l is it's length. If b=(24+0.02)cm , R=(130+1)Omega , l = (4.0+0.01)cm , the percentage error in s is nearly.

The radius (r) , length (l) and resistance (x) of a thin wire are (0.2 +- 0.02)cm, (80 +- 0.1)cm , and (30 +- 1) Omega respectively. The percentage error in the specific resistance is

If the length of a cylinder is L = (4.00 +-0.01)cm radius r = (0.250 +- 0.001)cm and mass m = (6.25+- 0.01)g . Calculate the percentage error in determination of density.

The resistance between ends will be (if (rho L)/(pi R^2) = R_0) .

If L is the inductance and R is the resistance, then the unit of - L/R is