If the measured value of resistance `R = 1.05 Omega`, wire diameter ` d = 0.60 mm` , and length `l = 75.3 cm`, then find the maximum permissible error in resistivity ,
` rho = ( R ( pi d^(2) // 4)) /( l)`.

If measured value of resistance R = 1.05Omega wire diameter d = 0.60 mm and length l = 75.3 cm If maximum error in resistance measurment is 0.01 Omega and least count of diameter and length measuring device are 0.01 mm and 0.1cm respectively then find max permissible error in resistiviry rho=(R((pid^(2))/(4)))/(l) .

In the experiment of Ohm's law, when potential difference of 10.0 V is applied, current measured is 1.00 A. If length of wire is found to be 10.0cm and diameter of wire 2.50 mm, then find maximum permissible percentage error in resistivity.

In the experiment of Ohm's law, when potential difference of 10.0 V is applied, current measured is 1.00 A. If length of wire is found to be 10.0cm and diameter of wire 2.50 mm, then find maximum permissible percentage error in resistivity.

The relative error in resistivity of a material where resistance = 1.05 + 0.01 W diameter = 0.60 + 0.01 mm length = 75.3 + 0.1 cm is

The relative error in resistivity of a material where resistance = 1.05 + 0.01 W diameter = 0.60 + 0.01 mm length = 75.3 + 0.1 cm is

The relative error in resistivity of a material where resistance = 1.05 + 0.01 W diameter = 0.60 + 0.01 mm length = 75.3 + 0.1 cm is

Resistance of wire having diameter 2 mm and length 100 cm is 0.7 Omega , so resistivity of wire = .....

The resistance of a wire of uniform diameter d and length L is R . The resistance of another wire of the same material but diameter 2d and length 4L will be

A wire has mass m=0.3+-0*003g, radius r=0*5+-0*005mm and length l=6+-0*06cm. The maximum percentage error in the measurement of its density is :