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

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 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

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

A uniform wire of length l and radius r has a resistance of 100 Omega . It is recast into wire of radius (R )/(2)

Given R_(1) = 5.0 +- 0.2 Omega, and R_(2) = 10.0 +- 0.1 Omega . What is the total resistance in parallel with possible % error?

A wire of length l = 6+- 0.06 cm and radius r = 0.5 +- 0.005 cm and mass m = 0.3 +- 0.003 g . Maximum percentage error in density is :

A wire of length l=6+-0.06cm and radius r=0.5+-0.005cm and m=0.3+-0.003g . Maximum percentage error in density is: