Question

A current of 3A flows throughs a metal rod of diametes 0.2 mm and lengths 1.5cm when a potential difference of 40 v is applied. The resistivity of material is
(а)28 x 10^-6ohm-m
(b)36 x10^-6ohm-m
(c)16 x 10^-6ohm-m
(d)4 x10^-6ohm-m​

Answer 1

Answer:

OPTION B

Explanation:

R=VI

R=3×40=120Ω

R=βL/2πA                       β=Resistivity

β=2πA/L

find area using equation πr² = (0.0002)²×3.14= 0.000000126= 1.26×$10^{-7}$

β=2πA/L =0.000000792/0.000225=0.00352≈  36 x$10^{-6}$ ohm-m

PLEASE MAKE ME BRAINLIEST

Answer 2

Answer:

The resistivity of material is 28 × $10^{-6}$   Ohm - m.

Explanation:

Given,

I = 3A

d = 0.2 mm   = 0.0002 m

∴ r =  0.0001 m

L = 1.5 cm = 0.015 m

V = 40 v

We known that,

Cross-section Area (A) =  π$r^{2}$

                                      = 3.141 × $(0.0001)^{2}$

                                      = 3.141 × $10^{8}$ m

Resistance (R) = $\frac{V}{I}$

                        = $\frac{40}{3}$

                        = 13.34   ohm

Now,

Resistivity of material ( ρ ) = $\frac{RA}{L}$

                                           = $\frac{13.34 * 3.141 * 10^{-8} }{0.015}$

                                           = $28$ × $10^{-6}$   ohm - m