A0.2 cm diameter wire must carry a 20-a current. if the maximum power dissipation along the wire is 4w/m, what is the minimum allowable conductivity of the wire in ohm-m? (a) 3.18x10 (b) 3.18x10 () 3.18x10 (d) 3.18x10

The conductivity of the wire is .

Explanation:

Given that,

Diameter = 0.2 cm

Current = 20 A

Power = 4 W/m

We need to calculate the conductivity

We know that,

Using  formula of resistance

....(I)

Where,

= resistivity

A = area

l = length

Using formula of power

Put the value of R in equation (I)

Put the all values into the formula

Hence, The conductivity of the wire is .

answer: the correct answer is 1.

explanation: there are various process in which a radioactive nuclei decays:

1) alpha decay: in this process, a heavier nuclei decays into lighter nuclei by releasing alpha particle. the mass number is reduced by 4 units.

2) electron capture: in this decay process, a parent nuclei absorbs an electron and gets converted into a neutron. simply, a proton and an electron combines together to form a neutron. mass number does not change in this process.

3) gamma ray emission: it is a decay process in which an unstable nuclei gives excess energy by a spontaneous electromagnetic process. this decay releases . this process does not change the mass number.

4) positron emission: it is a type of decay process, in which a proton gets converted to neutron and an electron neutrino. this is also known as -decay. in this the mass number remains same.

from the above information, we see that only alpha-decay is producing a change in the mass number of the nucleus.

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg