Question

1. a 1.0 m length of metal wire is connected to a 1.5 v battery, and a current of 8.0 ma flows through it. what is the diameter of the wire? (the resistivity of metal is 2.24×10^-8ω.m)

a. 1.22×10^-5m
b. 22×10^-7m
c. 2.4×10^-5m
d. 60×10^-7m

1. a metal sphere with unknown electrical charge is attached to the ceiling through an insulated cable. a positively charged rod is brought nearby (but does not touch the sphere) and the sphere gets attracted to the rod. the charge on the ball must be

a. negative
b. positive
c. neutral
d. either negative or neutral

1. you are building an electrical circuit and decide to exchange the wire to one with the same length but twice the resistance. how is the resistivity changed?

a. it is reduced by a factor of 2
b. it is reduced by a factor of 4
c. it is increased by a factor of 2
d. it is reduced by a factor of 4

Explanation:

Question 23

Step1: Calculate the resistance

Using Ohm's law \(R=\frac{V}{I}\). Given \(V = 1.5\space V\) and \(I=8.0\times10^{- 3}\space A\), then \(R=\frac{1.5}{8.0\times10^{-3}}=\frac{1500}{8}=187.5\space\Omega\)

Step2: Use the resistance formula \(R=

ho\frac{l}{A}\)
We know \(R = 187.5\space\Omega\), \(
ho=2.24\times 10^{-8}\space\Omega m\), \(l = 1.0\space m\). Rearranging for \(A\) (cross - sectional area), \(A=
ho\frac{l}{R}\). Substituting values: \(A=\frac{2.24\times 10^{-8}\times1}{187.5}=1.1947\times10^{-10}\space m^{2}\)

Step3: Relate area to diameter

Since \(A=\pi r^{2}=\frac{\pi d^{2}}{4}\), then \(d = 2\sqrt{\frac{A}{\pi}}\). Substituting \(A = 1.1947\times10^{-10}\space m^{2}\), \(d=2\sqrt{\frac{1.1947\times 10^{-10}}{\pi}}\approx2.2\times10^{-5}\space m = 22\times10^{-7}\space m\)

A positively charged rod can attract a neutral object (due to charge induction) or a negatively charged object (opposite charges attract).

The resistance formula is \(R=
ho\frac{l}{A}\). If \(l\) (length) and \(A\) (cross - sectional area) are the same (since the wire is just changed in a way that only resistance changes, assume same material geometry in terms of cross - section and length), and \(R_{2} = 2R_{1}\). From \(R=
ho\frac{l}{A}\), if \(l\) and \(A\) are constant, \(R\propto
ho\). So if \(R\) doubles, \(
ho\) doubles.

Answer:

B. \(22\times 10^{-7}m\)

Question 24