A mixed cost contains a variable element and a fixed element.
Option a
<u>Explanation:</u>
Mixed costs are those costs that has both variable and fixed component. Example: operating cost of a machinery includes fixed costs that cannot be changed with other variable costs like fuel, insurance, depreciation, etc.
It is also named as semi-variable costs. And the formula to calculate mixed cost is as follows,

where,
- y is the "total cost
"
- a is the "fixed cost per period"
- b is the "variable rate per unit of activity"
- x is the "number of units of activity"
Explanation:
It is given that,
Diameter of loop, d = 1.4 cm
Radius of loop, r = 0.7 cm = 0.007 m
Magnetic field, 
(A) Magnetic field of a current loop is given by :

I is the current in the loop


I = 27.85 A
(B) Magnetic field at a distance r from a wire is given by :



r = 0.00222 m

Hence, this is the required solution.
Answer:
We need about 8769 meters of wire to produce a 2.6 kilogauss magnetic field.
Explanation:
Recall the formula for the magnetic field produced by a solenoid of length L. N turns, and running a current I:

So, in our case, where B = 2.6 KG = 0.26 Tesla; I is 3 amperes, and L = 0.57 m, we can find what is the number of turns needed;

Therefore we need about 39312 turns of wire. Considering that each turn must have a length of
, where D is the diameter of the plastic cylindrical tube, then the total length of the wire must be:

We can round it to about 8769 meters.
Answer:
I think the Bulb No. 2 will stop emitting light if the bulb No. 1 burns out.